Chemistry Unit 4 Worksheet 4 Answers

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Chemistry Unit 4 (C4) - Revision Packs | Teaching Resources

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Chemistry Unit 4 (C4) - Revision Packs | Teaching Resources

PDF Chemistry Unit 4 Worksheet 1 Answers

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Molarity Worksheet # 1

Lesson             Day                  Date                 Topic

1.                                                                     Molarity 1

2.                                                                     Molarity Lab          Molarity 2

3.                                                                     (*4*)

4.                                                                     Titrations lab 1       Homework WS # 4

5.                                                                     Titrations Lab 2      Homework WS # 5

6.                                                                     Dilutions

7.                                                                     Spectrophotometry Lab

8.                                                                     Molarity and Dilutions

9.                                                                     Ion Concentration

10.                                                                   Molarity Unit Review # 1

11.                                                                   Molarity Unit Review # 2

12.                                                                   Chemistry 11 Calculations Practice Test # 1

13.                                                                   Chemistry 11 Calculations Practice Test # 2

         

Molarity Worksheet # 1

1.       15.Eight g of OkCl is dissolved in 225 mL of water. Calculate the molarity.

                                        15.8 g          x        1 mole        

          Molarity      =                                      74.6 g                    =        0.941 M

                                                 

                                                  0.225 L

2.       Calculate the mass of KCl required to organize 250. mL of 0.250 M answer.

          0.250 L        x        0.250 moles x        74.6 g          =        4.Sixty six g         

                                                  1 L               1 mole

3.       Calculate the volume of 0.30 M KCl resolution that contains 6.00 g of KCl.

          6.00 g          x        1 mole         x        1 L               =        0.27 L

                                        74.6 g                   0.30 mol

4.       Calculate the quantity of 0.250 M H2SO4 that incorporates 0.250 g H2SO4.

                   

         

          0.250 g H2SO4      x        1 mole                   x        1 L                         =          0.0102 L

                                                  98.12 g                            0.250 mole

5.       1.50 g of NaCl is dissolved in 100.Zero mL of water. Calculate the concentration.

                             

                                        1.50 g            x        1 mole        

          Molarity      =                                      58.Five g                    =        0.256 M

                                                 

                                                  0.One thousand L

6.       How many moles of NaCl are in 250. mL of a 0.2 hundred M resolution?

          0.250 L      x        0.2 hundred mole     =     0.0500 moles        

                                            1 L                        

7.       How many litres of a zero.200 M OkCl solution comprise 0.250 moles?

          0.250 moles x        1 L                         =        1.25 L

                                       0.2 hundred moles

8.       How many millilitres of 0.2 hundred M H2SO4 are required to fully neutralize 250. mL of 0.250 M NaOH?

         

          H2SO4         +        2NaOH       →      Na2SO4       +        2HOH

          ? mL                     0.250 L                 

0.250 L NaOH  x  0.250 mole  x  1 mole H2SO4    x        1 L     x        1000 mL          =        156 mL

                                        1 L          2 mole NaOH            0.2 hundred mole            1 L

         

9.       Calculate the mass of CuSO4.5H2O required to arrange 100.0 mL of 0.100 M solution.

0.One hundred L        x        0.A hundred mole            x        249.7 g        =        2.50 g

                              1 L                                   1 mole

10.     Calculate the mass of Cu(NO3)2.6H2O required to arrange 100.Zero mL of 0.Two hundred M resolution.

   

              5.Ninety one g

            

11.     Calculate the mass of CoCl3.6H2O required to prepare 500.0 mL of a 0.Two hundred M       resolution.

 

           27.4 g

 

 

 

12.     50.0 g of NaCl is dissolved in 200.Zero mL of water, calculate the molarity.

 

 

        4.27 M

 

 

 

 

13.     25.Zero g of CuSO4.8H2O is dissolved in 25.0 mL of water, calculate the molarity.

 

 

        3.29 M

 

 

 

14.     Calculate the mass of NaCl required to prepare 500.0 mL of a zero.500 M answer.

 

 

       14.6 g

 

15.     Calculate the volume of 0.500 M NaCl solution required to include 0.0500 g of NaCl.

       0.00171 L

 

16.     Calculate the quantity of 0.Two hundred M NaCl resolution required to comprise

          0.653 g of NaCl.

 

 

         0.0558L

 

 

17.     Calculate the mass of NaCl required to organize 256 mL of a 0.35 M solution.

 

 

         5.2 g

 

 

18.     25.2 g of NaCl is dissolved in 365 mL of water, calculate the molarity.

 

 

         1.18 M

        

 

 

19.     56.3 g of CuSO4.8H2O is dissolved in 30. mL of water, calculate the molarity.

 

 

         6.2 M

 

              

Worksheet # 2      Molarity     

 

1.       Calculate the mass of CuSO4.6H2O required to arrange 200.0 mL of 0.300 M solution.

0.200 L        x        0.Three hundred moles x        267.72 g      =        16.1 g         

                              1 L                         1 mole

2.       Calculate the mass of CoCl3.8H2O required to arrange 300.Zero mL of a zero.520 M solution.

0.300 L        x        0.520 moles x        309.Fifty six g      =        48.Three g         

                              1 L                         1 mole

3.       150.Zero g of NaCl is dissolved in 250.Zero mL of water, calculate the molarity.

                              150.0 g        x        1 mole        

Molarity      =                                      58.5 g (3sig figs)                    =        10.Three M

                                                 

                                        0.250 L

         

4.       25.2 g of CuSO4.6H2O is dissolved in 28.0 mL of water, calculate the molarity.

                              25.2 g          x        1 mole        

Molarity      =                                      267.seventy two g      =        3.36 M

                                                 

                                        0.0280 L

5.       Calculate the mass of NaCl required to organize 565.0 mL of a 0.450 M solution.

0.5650 L      x        0.450 moles           x        58.Five g          =        14.Nine g         

                              1 L                                   1 mole

6.       Calculate the volume of 0.250 M NaCl solution required to include

          0.0300 g of NaCl.

0.0300 g NaCl       x        1 mole         x        1 L               =        0.00205 L

                                        58.5 g                    0.250 mole

7.       Calculate the volume of 0.500 M NaCl answer required to contain 0.52 g of NaCl.

0.Fifty two g NaCl x        1 mole                   x        1 L                         =        0.018 L

                              58.5 g                              0.500 mole

8.       Calculate the mass of NaCl required to prepare 360.0 mL of a 0.35 M answer.

         

0.3600 L      x        0.35 moles             x        58.Five g          =        7.4 g 

                              1 L                                   1 mole

9.       55.6 g of NaCl is dissolved in 562 mL of water, calculate the molarity.

                              55.6 g          x        1 mole        

Molarity      =                                      58.5 g                    =        1.Sixty nine M

                                                 

                                        0.562 L

10.     78.9 g of CuSO4.8H2O is dissolved in 500.0 mL of water, calculate the molarity.

                              78.9 g          x        1 mole        

Molarity      =                                      303.seventy six g                =        0.519 M

                                                 

                                        0.5000 L

(*4*) Worksheet # 3

1.       Excess sodium hydroxide resolution is added to 20.0 mL of 0.184 M ZnCl2, calculate the mass of zinc hydroxide that will           precipitate.

          NaOH(aq)      +        ZnCl2(aq)       →      Zn(OH)2(s)    +       2NaCl(aq)

                                        0.0200 L                ? g

0.0200 L ZnCl2  x   0.184 mole   x  1 mole Zn(OH)2  x   99.Forty two g        =          0.366 g Zn(OH)2

                                        1 L               1 mole ZnCl2         1 mole

         

2.       How many millilitres of 1.09 M HCl are required to react with an answer formed by means of dissolving 0.775 g of sodium        carbonate?

          Na2CO3(aq)    +        2HCl(aq)        →      2NaCl(aq)     +     H2O(l)     +     CO2(g)

          0.775g                   ? mL

0.775 g Na2CO3      x      1 mole  x   2 mole HCl        x      1 L     x        One thousand mL          =  13.4 mL

                                        106 g       1 mole Na2CO3         1.09 moles             1 L

3.       Calculate the choice of grams barium carbonate that can be brought about through      including 50.0 mL of 0.424 MBa(NO3)2.

          Ba(NO3)2(aq)  +      K2CrO4(aq)    →       BaCrO4(s)        +     2KNO3(aq)

0.0500 L Ba(NO)2   x     0.424 mole  x    1 mole BaCO3    x     253.Three g   =          5.37 g BaCrO4

                                        1 L                     1 mole Ba(NO)2        1 mole

4.       Determine the collection of millilitres of 0.246 M AgNO3 required to precipitate all of the phosphate ion in a solution containing 2.10 g of sodium phosphate.

          3AgNO3(aq)    +    Na3PO4(aq)     →    Ag3PO4(s)      +     3NaNO3(aq)

          156 mL

5.       How many grams of silver nitrate must be used within the preparation of 150. mL of 0.One hundred twenty five M solution.

          3.19 g

6.       What volume of SO2 is generated by means of the whole reaction of 35.0 mL of 0.924 M Na2SO3?

          Na2SO3(aq)    +        2NaOH(aq)     →      2NaCl(aq)    +          H2O(l) +     SO2(g)

          0.724 L

7.       How many milliliters of 6.2 M NaOH should react to liberate 2.4 L of hydrogen at STP?

          2Al(s)  +        6NaOH(aq)    →        2Na3AlO3(aq)              +         3H2(g)

          35 mL

8.       Calculate the burden of H2C2O4.2H2O required to make 750.Zero mL of a zero.480 M resolution.

          45.4 g

9.       25.4 L of HCl fuel at STP are dissolved in 2.Five L of water to produce an acid solution. What quantity of 0.2 hundred M Ba (OH) 2 will this solution neutralize?

         

          2HCl           +        Ba(OH)2     →      BaCl2          +        2HOH

                   

          25.4 L HCl  x        1 mole         x        1mole Ba (OH) 2  x            1L            =       2.83 L

                                        22.4 L                    2 mole HCl                      0.Two hundred mol

10.     8.25 L at STP of HCL gas is dissolved in 500 ml of water to supply an acid solution. What quantity of 0.2 hundred M Ca    (OH) 2 will this solution neutralize?               

          Ca (OH) 2    +        2HCl           →      CaCl2          +        2H2O

         

8.25 L HCL     x   1 mole    x   1 mole Ca (OH) 2                  x        1 L                  =        0.921L

                              22.4 L          2 mole HCL                              0.2 hundred mol

11.     250 mL of water is added to One hundred mL of 0.0200M H2SO4. What volume of 0.100M KOH will it neutralize?

         

          H2SO4                   +        2KOH         →      2K2SO4                 +          2HOH

0.1 L H2SO4    x   0.0200mol       x    2 mole KOH    x   1 L   x          One thousand mL

                              1L                          1 mole H2SO4       0.A hundred mol    1 L

         

=        40ml

12.     What volume of 0.924 M Na2SO3 is needed for the manufacturing of 350.0 mL of SO2 at STP?

 

          2HCl   +   Na2SO3(aq)       +  2NaOH(aq)          →      2NaCl(aq)    +          H2O(l) +     SO2(g)   

 

 

 

1.69 x 10-2L

 

 

 

 

 

13.     How many milliliters of hydrogen at   STP  can be generated by means of 500.Zero mL 6.2 M NaOH completely reacting with excess Al. 

          2Al(s)  +        6NaOH(aq)    →        2Na3AlO3(aq)            +         3H2(g)

 

 

 

3.5  x 104 mL

 

 

 

 

14.     64.5 L of HCl gasoline at STP are dissolved in water to supply an acid           resolution. What quantity of 0.Two hundred M Ba (OH)2 will this resolution neutralize?

7.20 L

Titration Calculations    Worksheet # 4                                    

1.       In a titration 12.5 mL of 0.Two hundred M NaOH ia required to neutralize 10.0 mL of H2SO4.

          Calculate the focus of the acid ?

          H2SO4               +        2NaOH         →      Na2SO4                   +          2HOH

         

          0.0100 L                     0.0125 L

          ? M                             0.Two hundred M         

          

    Molarity    =    0.0125 L NaOH   x      0.Two hundred moles   x      1 mole H2SO4  

                                                                         1 L                 2 mole NaOH

                                                                                    0.0100 L

                     =        0.125 M                  

         

 

2.       In a titration 22.5 mL of 0.A hundred M HCl ia required to neutralize 20.Zero mL of Ba(OH)2 .

          Calculate the focus of the bottom ?

          2HCl               +        Ba(OH)2         →      BaCl2                  +          2HOH

         

          0.0225 L                     0.0200 L

          0.A hundred M                           ? M         

          

    Molarity    =    0.0225 L HCl   x      0.One hundred moles   x        1 mole Ba(OH)2  

                                                                         1 L               2 mole HCl

                                                                                    0.0200 L

                     =        0.0563 M   

3.       A burette full of 1.Fifty two M nitric acid solution reads 33.10 mL to begin with. After titrating a 25.00 mL pattern of

          barium hydroxide the endpoint was reached and the burette confirmed 46.30 mL. What is the barium hydroxide

          focus?

          46.30  -  33.10   =  13.20 mL  =  0.01320 L

          2HNO3         +          Ba(OH)2         →      Ba(NO3)2                  +          2HOH

          0.01320 L                0.02500 L

          1.52 M                     ? M

                              0.0132L HNO3      x  1.52 mole   x  1 mole Ba(OH)2        

Molarity      =                                                1 L            2 mole HNO3            =          0.402 M

                                                 

                                                               0.02500 L

4.       A burette stuffed with 2.557 M sodium hydroxide answer reads 15.Sixty two mL to start with. After titrating a 25.00 mL pattern of phosphoric acid the endpoint was once reached and the burette now confirmed 39.22 mL. What is the [phosphoric acid]?

          39.22  -  15.62   =  23.60 mL  =  0.02360 L

          3NaOH       +        H3PO4                   →      Na3PO4       +        3H2O

          0 .02360 L                0.02500 L

          2.557 M                ? M

                              0.02360 L NaOH  x  2.557 mole  x   1 mole H3PO4       

Molarity      =                                                1 L             Three mole NaOH                 =        0.8046 M

                                                 

                                                               0.02500 L

5.       A 10.00 mL pattern of 2.120 M sodium hydroxide answer is positioned in a 250.Zero mL Erlenmeyer flask. An indicator known as       bromothymol blue is added to the solution. The answer is blue. Hydrochloric acid was added from a burette until there used to be a inexperienced colour (endpoint were reached). Determine the focus of hydrochloric acid given the next           burette readings:

                                        Burette ultimate                   =        22.04 mL

                                        Burette preliminary                 -         12.08 mL

                                        Difference                       =          9.96 mL              Beware subtraction!!!! One sig fig is lost!!!

          NaOH         +        HCl             →      NaCl  +        H2O

          .01000 L                .00996 L

          2.120 M                ? M

                              0.01000 L NaOH   x   2.A hundred and twenty mole      x      1 mole HCl 

Molarity      =                                                  1 L                      1 mole NaOH    =   2.Thirteen M HCl

                                                 

                                                               0.00996 L

6.       The following data used to be bought all through the titration of 1.0097 M sodium hydroxide with a 25.00 mL aliquot of hydrofluoric acid:   

                                                                      Trial 1                   Trial 2                    Trial 3

                    Burette Final Reading                 34.56 mL                39.Forty two mL                44.20 mL

                    Burette Initial Reading               14.Ninety four mL                19.86 mL                24.66 mL

                    Vol. of NaOH Added                                                                        

         

          Use the above information to resolve the focus of the acid.

          NaOH         +        HF               →      NaF   +        H2O

          0 .01955 L                0 .0250 L

          1.0097 M                ? M

                              0.01955 L NaOH   x   1.0097 mole      x      1 mole HF  

[HF]            =                                                  1 L                      1 mole NaOH          =        0.7896 M

                                                 

                                                               0.0250 L

         

7.       The following data was bought all the way through the titration of 0.0998 M sodium hydroxide with a

10.00 mL aliquot of sulphuric    acid:

                                                            Trial 1                   Trial 2                    Trial 3

                    Burette Final Reading              26.05 mL      48.52 mL      33.Seventy eight mL

                    Burette Initial Reading               2.Forty six mL       34.Ninety four mL      20.22 mL                   

                    Vol. of NaOH added                 

          Use the above information to decide the concentration of the acid.

          6.77  x  10-2 M

8.       The following information was once got all through the titration of 2.0554 M hydrochloric acid with a 25.00 mL aliquot of barium hydroxide:

                                                                      Trial 1                   Trial 2                    Trial 3

                    Burette Final Reading               22.92 mL                25.32 mL                41.30 mL

                    Burette Initial Reading                 0.06 mL                 2.58 mL                 18.Fifty four mL

                    Volume of Acid Added             

          Use the above data to decide the concentration of the barium hydroxide.

          0.9352 M

 

9.       What volume of 2.549 M NaOH is needed to absolutely titrate 50.0 mL of one.285 M HCl resolution ?

 

 

         HCl   +        NaOH                   →

 

         0.0500 L HCl  x   1.285 mol   x   1 mole NaOH      x   1L          =       0.0252 L

                                        1 L                1 mole HCl            2.549 mole

 

 

 

10.     What quantity of 1.146 M KOH is had to totally titrate 20.Eight mL of 0.557 M H2SO4 solution ?

 

         H2SO4         +        2KOH                   →

 

 

         0.0208 L H2SO4   x   0.557 mol   x   2 mole KOH      x     1L      =        0.0202 L

                                            1 L                   1 mole H2SO4         1.146 mole

                            

Titrations Worksheet # 5

1.       Calculate the mass of H2C2O4.2H2O required to prepare 500.0 mL of a nil.200M answer.

          12.6 g                             

2.       Calculate the mass of Cu2SO4.6H2O required to arrange 200.0 mL of a zero.300M solution.

          19.9 g

3.       In a titration 0.Two hundred M NaOH is used to neutralize 10.0 mL of H2SO4. In 3 runs, the following data was once accrued. Calculate the concentration of the acid.

                                                            Volume of 0.20 M NaOH (mL)

          Initial Burette Reading               12.90            15.70            18.50

          Final Burette Reading                 15.70            18.50            21.50

0.028 M

4.       In a titration 0.250 M KOH is used to neutralize 25.Zero mL of H3PO4. In 3 runs, the following data used to be collected. Calculate the concentration of the acid.

                                                            Volume of 0.250 M KOH (mL)

          Initial Burette Reading                         2.90              15.70            28.70

          Final Burette Reading                           15.70            28.70            42.70

                                                                    12.80            13.00           14.00

                   

                                                  Use 12.90 mL      

         

0.0430 M

5.       Calculate the volume of 0.500M H3PO4 required to neutralize 25.Zero mL of

          0.200M NaOH.

          0.00333L

6.       Calculate the volume of 0.50 M NaOH required to neutralize 35.Zero mL of

          0.100M H2C2O4.

                   

          0.014L

7.       In a titration 35.7 mL of 0.250 M H3PO4 is used to neutralize 25.Zero mL of KOH. Calculate the molarity of the base.

          2.08 M

8.       In a titration 35.2 mL of 0.20 M H2C2O4 is used to neutralize 10.0 mL of NaOH.  Calculate the molarity of the base.

          1.4 M

9.                                     2 Al    +                  Three I2               →      2 AlI3

          Initial                     12.0 mol                 15.0 mol                    0

          Change:      

           

End:                       2.Zero mol                  0mol                      10.Zero mol     

10.                                   C        +        2Cl2              →                CCl4

          Initial                     16.0 mol       34.0 mol                            0

          Changes:     

           

End:                       0 mol           2.0 mol                            16.Zero mol               

11.                                   4 Fe     +      Three O2             →                2 Fe2O3

          Initial                     12.Zero mol       8.0 mol                            0

          Change:      

           

End:                       1.3 mol        0 mol                               5.3 mol

12.                         2 NO            +                  O2                         →                2 NO2

                              100. g   x  1 mole             100. g   x  1 mole                        0

                                              30.Zero g                               32.0 g

           

Init:              3.333                               3.125                                         3.333

           

Change:       3.333                               1.667                                         3.333

           

End:             0                                      1.458 mol  x  32.Zero g                  3.333  x  46.Zero g

                                                                                          1 mole                               1 mole

Grams:         0 g                                   46.7 g                                        153 g

                   

13.     Calculate the volume of H2 fuel produced at STP by the response of 300. mL of 0.500 M HCl with extra Zn.

                              Zn    +    2HCl    →   H2    +     ZnCl2

0.Three hundred L         x   0.500 moles   x    1 mole H2      x       22.4 L     =   1.Sixty eight L

                              1 L                 2 mole HCl             1 mole

14.     Calculate the volume of 0.30 M OkayCl resolution that contains 9.00 g of KCl.

         

          0.40 L

Dilutions Worksheet # 6

1.       20.Zero mL of 0.Two hundred M NaOH resolution is diluted to a last quantity of 100.Zero mL, calculate the brand new concentration.

          M1V1           =        M2V2

         

          (20.0)(0.200)         =        M2(100.0)

          M2     =        0.0400 M

         

2.       15.Zero mL of an answer of NaOH is diluted to a final quantity of 250.0 mL and the brand new molarity is 0.0500 M. Calculate the unique molarity of the bottom.

          M1V1           =        M2V2

         

          (15.0) M1    =        (250.0) (0.0500)

          M1     =        0.833 M

         

3.       50.Zero mL of 0.025 M NaOH solution is added to 150.0 mL of water. Calculate the new molarity.

          V2      =        50.Zero mL  +  150.0 mL       =        200.0

                   

M1V1           =        M2V2

         

          (0.025)(50.0)         =        M2 (200.0)

          M2     =        0.0063M

4.       45.Zero mL of an answer of NaOH is diluted by including 250.0 mL of water to provide a brand new molarity of 0.0500 M. Calculate the molarity of the base.

         

          0.328 M

5.       A 0.One hundred twenty five M solution is focused by evaporation to a reduced final quantity of 100.Zero mL and a molarity of 0.A hundred and fifty M. Calculate the original volume.

          120. mL

6.       850.Zero mL of 0.280 M KOH answer is diluted to a last quantity of 1000.0 mL, calculate the new focus.

          0.238 M

7.       95.0 mL of an answer of NaOH is diluted to a last quantity of a hundred thirty five mL and the new molarity is 0.0500 M. Calculate the   unique molarity of the bottom.

         

          0.0711 M

Molarity Review # 7

1.       Convert 250. g AgNO3 to formulation devices after which to atoms of O.

          2.66  x  1024 at O

2.       Convert 5.9 x1025 H2 molecules to grams.

          2.0  x  102 g H2

3.       Calculate the share composition of MgSO4.

          20.2 % Mg           26.7 % S               53.2 % O

4.       A compound is 42.3 % C, 5.94 % H, 32.9 % N, and18.8 % O and has a molecular     mass of 425.25 g/mol. Calculate the empirical and molecular formula.

          C3H5N2O              C15H25N10O5

5.       How many grams O2 are required to devour 56.3 g Al?

4Al      +         3O2       →      2Al2O3

          50.Zero g O2

6.       25.Five mL of 0.One hundred M HCl reacts with excess Zn to produce 25.Three mL of H2 fuel at STP. Calculate the theoretical yield in mL and the share yield of H2 gas.

          Zn      +   2HCl       →    H2     +           ZnCl2.

          0.0255 L    x    0.A hundred mole     x    1 mole H2      x      22.4 L    =   0.0286 L                88.6 %

                                        1 L               2 mole HCl            1 mole

7.       Calculate the energy produced by means of all the response of 150. g H2.

          2H2       +        O2     →         2H2O   +   130. KJ

          4.83  x  103 KJ

8.       84.Zero g of Al reacts with 122 g O2 to produce Al2O3.  How many grams of Al2O3 are produced?  Determine the mass of the reactant in excess and the restricting reactant.

                    4Al               +                  3O2                        →                2Al2O3

         

          84.Zero g  x  1 mole             122 g  x  1 mole

                         27.0 g                                32.0 g

         

I         3.111 mole                                3.8125 mole                                        0

C       3.111 mole                                2.333 mole                                          1.5556 mole

E        0                                                1.4795 mole                                        1.5556 mole

                                                            47.3 g                                                  159 g

9.       15.2 g of Al reacts with 14.3 g O2 to supply Al2O3.  How many grams of Al2O3 are produced?  Determine the mass of the reactant in extra and the proscribing        reactant.

                    4Al               +                  3O2                        →                2Al2O3

         

          15.2 g  x  1 mole             14.Three g  x  1 mole

                         27.0 g                               32.0  g

         

I         0.5630 mole                              0.4469 mole                                        0

C       0.5630 mole                              0.4222 mole                                        0.2815 mole

E        0                                                0.0247 mole                                        0.2815 mole

                                                            0.Seventy nine g                                                   28.7 g

10.     15.Eight g of OkCl is dissolved in 225 mL of water. Calculate the molarity.

          [OkCl]           =        15.8 g          x        1 mole         =        0.941 M

                                                                      74.6 g

                                                  0.225 L

11.     Calculate the mass of OkayCl required to organize 250.Zero mL of 0.250 M solution.

          0.2500L       x        0.250 mole            x        74.6 g          =        4.Sixty six g KCl

                                        1 L                                   1 mole

12.     Calculate the volume of 0.30 M BaCl2 resolution that comprises 6.00 g of OkCl.

          6.00 g          x        1 mole         x        1 L               =        0.27 L

                                        74.6 g                    0.30 mole

13.     Calculate the amount of 0.250 M H3PO4 that contains 0.250 g H2SO4.

          0.250 g        x        1 mole         x        1 L               =        0.0102 L

                                       98.03 g              0.250 mole

14.     1.Five g of BaCl2 is dissolved in 100.0 mL of water. Calculate the focus.

          0.072 M

15.     How many moles of BaCl2 are in 250.0 mL of a 0.Two hundred M resolution?

          0.0500 moles

16.     How many litres of a nil.Two hundred MBaCl2 resolution comprise 0.250 moles?

          1.25 L

17.     Calculate the amount of H2 gasoline produced at STP by way of the response of 400.Zero mL of  0.800 M HCl with extra Zn.

                              Zn    +    2HCl    →   H2    +     ZnCl2

          3.Fifty eight L

18.     Calculate the volume of 0.250 M H3PO4 required to neutralize 25.Five mL of  0.200 M NaOH.

          H3PO4         +        3NaOH       ®      Na3PO4       +        3HOH

          ? L                         0.0255 L

          0.0255 L NaOH    x        0.200 mole     x     1 mole H3PO4       x        1 L                  =        0.00680 L

                                                  1 L                         Three mole NaOH                 0.250 mole

19.     Calculate the quantity of 0.500 M KOH required to neutralize 45.Three mL of  0.320 M H2SO4 .

H2SO4         +        2KOH         ®      K2SO4         +        2HOH

          0.0453 L                ? L

          0.0453 L H2SO4    x        0.320 mole     x     2 mole KOH      x        1 L                  =        0.0580 L

                                                  1 L                         1 mole H2SO4              0.500 mole

20.     Calculate the mass of CoCl3.6H2O required to prepare 500.Zero mL of a zero.Two hundred M resolution.

          27.4 g

Worksheet # 8 Dilutions and Molarity

1.       40.Zero mL of 0.400 M NaOH resolution is diluted to a final quantity of 200.Zero mL,    calculate the brand new focus.

          M1V1           =        M2V2

         

          (0.400)(40.0)         =        M2(200.0)

          M2     =        0.0800 M

         

         

2.       85.Zero mL of a solution of NaOH is diluted to a last quantity of 290.0 mL and the new molarity is 0.0500 M. Calculate the original molarity of the bottom.

                    M1V1           =        M2V2

         

                    M1(85.0)     =        (0.0500)(290.0)

                    M1               =        0.171 M

         

3.       150.Zero mL of 0.025 M NaOH answer is added to 150.0 mL of water. Calculate the new molarity.

          M1V1                     =        M2V2

         

          (0.025)(150.0)       =        M2(300.0)

          M2     =        0.013 M

         

4.       220.0 mL of a solution of NaOH is diluted by way of including 250.Zero mL of water to provide a brand new molarity of 0.0500 M. Calculate the molarity of the base.

                    M1V1           =        M2V2

         

                   

                    M1(220.0)             =        (0.0500)(470.0)

         

                    M1                         =        0.107 M

5.       A zero.350 M answer is focused by evaporation to a discounted final volume of 100.0 mL and a molarity of 0.825 M. Calculate the unique quantity.

         

          V1     =        236 mL

6.       850.0 mL of 0.280 M KOH answer is diluted to a final quantity of 1000.Zero mL, calculate the brand new focus.

          M2     =        0.238 M

                     

7.       28 g of OkCl is dissolved in 225 mL of water, calculate the molarity.

          1.7 M

8.       Calculate the mass of KCl required to arrange 125 mL of 0.450 M solution.

          4.20 g

9.       Calculate the quantity of 0.Forty M OkayCl resolution that accommodates 8.00 g of OkayCl.

          0.27 L

10.     Calculate the volume of 0.Four hundred M H2SO4 required to neutralize 25.Zero mL of 0.200 M NaOH.

          0.00625 L

11.     Calculate the quantity of H2 gas produced at STP via the response of 250.Zero mL of 0.Six hundred M HCl with excess Zn.               Zn    +    2HCl    →   H2    +     ZnCl2

          1.Sixty eight L

12.     8.5 L of HCl fuel at STP is dissolved in 325 mL of water, calculate the molarity of the acid resolution.

          1.2 M

13.     How many moles of NaCl are in 350.Zero mL of a zero.400 M answer?

          0.140M

14.     How many litres of a zero.Three hundred M OkCl resolution comprise 0.350 moles?

          1.17 L

15.     Calculate the mass of 8.25 x one zero five mL of H2 fuel at STP.

          74.4 g

16.     Calculate the number of system gadgets of KCl in 200.0 mL of 0.Three hundred M solution.

          3.61  x  1022 FU

Worksheet # 9     Ion Concentration            

1.       What is the focus of each and every ion in a ten.Five M sodium sulphite solution?

           Na2SO3      →      2 Na+           +        SO32-         

          10.5 M                  21.0M                   10.5M

2.       What is the focus of every ion within the answer formed when 94.Five g of nickel (III) sulphate is dissolved into 850.Zero mL of water?

                                                                                                             

Ni2 (SO4) 3            →      2Ni3+            +        3S042-

0.274 M                          0.548M                 0.822M      

         

Molarity =   94.Five g          x        1 mole                            

                                                            405.7 g                  =        0.274M      

0.8500L

3.       If 3.Seventy eight L of 0.960 M calcium fluoride solution is added to six.36 L of water, what is the resulting focus of every ion?

          M1V1 = M2V2                                                        CaF2          →                Ca2+              +        2F-      

               (0.960) (3.78) = M2 (10.14)                0.358 M                          0.358 M                    0.716 M

          M2= 0.358 M       

4.       What is the focus of every ion in a 5.Fifty five M zinc phosphate resolution?

         

                    Zn3 (PO4) 2           →      3Zn2+          +        2PO43-

                   

                    5.Fifty five M                            16.7                       11.1 M       

5.       What is the concentration of every ion in the resolution shaped when 94.Seventy eight g of iron (III) sulphate

          is dissolved into 550.Zero mL of water?                                                               

                             

                              Fe2 (SO4)3  →      2Fe3+                +        3SO42-

                                                 

                              0.4309 M              0.8619 M              1.293 M

                                                  94.78 g        x        1 mol                                               

          [Fe2 (SO4) 3]          =                                    399.9 g          = 0.4309 M

                                                                0.5500 L                     

6.       If 6.25 L of 0.560 M sodium bromide answer is added to three.Forty five L of water, what's the ensuing

          Concentration of each and every ion?

         

M1V1 = M2V2                                         NaBr           →                    Na  +        +            Br -   

(0.560) (6.25) = M2 (9.70)        0.361 M                              0.361 M                0.361 M

M2= 0.361 M

7.       50.0 mL of 0.200 M Na3PO4 resolution is mixed with 150.0 mL of 0.Four hundred M Na2CO3.

         Calculate all ion concentrations.

                    Na3PO4             →      3 Na+           +        PO43-

                   

    50.0          0.200 M                     0.150 M                0.0500 M

     200.0

                    Na2PO4             →      2 Na+           +        CO32-

                   

    150.0       0.400 M                      0.Six hundred M                   0.Three hundred M

     200.0

[Na+] = 0.600 M + 0.One hundred fifty M = 0.750 M

8.       What is the focus of each and every ion in the resolution formed when 16.Five g of Aluminum sulphate is

          dissolved into 600.Zero mL of water?

                             

                                          16.Five g                  x        1 mol                                 

          [Al2 (SO4) 3] =                                               342.Three g        =   0.0803 M           

                                                            0.600 L       

                    Al2 (SO4) 3   →      2Al3+ +         3SO42-

                    0.0803 M              0.161 M      0.241M

9.       If 1.78 L of 0.420 M barium fluoride answer is added to 2.Fifty six L of water, what is the resulting

          concentration of each and every ion?

          M1V1 = M2V2                                                          BaF2         →         Ba2+             +          2F-                                                                               

               (0.420)(1.78) = M2 (4.34)                   0.172 M                0.172 M                0.345 M

          M2 = 0.172 M

10.     What is the concentration of each ion in a 1.22 M zinc acetate answer?

                    Zn (CH3COO) 2    → Zn 2+       +        2CH3COO-

                    1.22 M                       1.22M              2.44M                  

11.     What is the concentration of each ion within the answer formed when 94.Seventy eight g of cobalt (III) sulphate

          is dissolved into 400.Zero mL of water?

          [Co2(SO4)3]           =        94.Seventy eight g        x        1 mole

                                                                                406.1 g        =        0.5835 M

                                                            0.4000 L

                    Co2(SO4)3             →                2Co3+          +        3SO42-

                   

                    0.5835 M                                  1.167 M                1.750 M

12.     If the chloride concentration in 2.00 L of solution is 0.0900 M, calculate the [Al3+] and the molarity of the AlCl3 answer.

          AlCl3            →      Al3+              +                  3Cl-

          0.0300 M              0.0300 M                        0.0900 M

13.     If the [Ga3+] focus in 2.00 L of answer is 0.0300 M, calculate the [SO42-] and the molarity of the Ga2(SO4)3 solution.

          Ga2(SO4)3  →      2Ga3+          +        3SO42-

          0.0150 M              0.0300 M              0.0450 M

14.     In a titration 12.5 mL of 0.2 hundred M NaOH is had to neutralize 10.Zero mL of H3PO4, calculate the acid focus.

          H3PO4         +        3NaOH       →      Na3PO4       +        3HOH

          0.0100 L                0.0125 L

          ? M                       0.200 M

                    [H3PO4]       =        0.0125 L NaOH    x      0.200 mole    x       1 moles H3PO4                  

                                                                                          1 L                         3 mole NaOH     =        0.1/3 M

                                                                               

                                                                                0.0100 L

         

15.     What volume of 0.2 hundred M H2SO4 is required to neutralize 25.Zero mL of 0.300 M NaOH?

          H2SO4         +        2NaOH       →      Na2SO4       +        2HOH

          ? L                         0.0250 L

          0.2 hundred M                0.300 M

 0.0250 L NaOH    x        0.300 mole    x      1 moles H2SO4      x        1 L             =    0.0188 L

                               1 L                        2 NaOH                           0.200 mole

16.     The [Cl-]    =  0.Six hundred M in 100.Zero mL of a AlClThree answer. How many grams AlCl3 are within the resolution?

          AlCl3            ®      Al3+    +        3Cl-

          0.200 M                                    0.600 M

 0.1000 L     x    0.200 mole   x   133.Five g       =        2.Sixty seven g

                               1 L                1 mole

17.     The [SO42-]   =  0.600 M in 100.Zero mL of a Al2(SO4)Three resolution. How many grams Al2(SO4)Three are in the answer?

                    6.85 g

Worksheet # 10              Molarity Unit Review # 1                   

1.       200.Zero mL of 0.200 M H2SO4 reacts with 250.Zero mL of 0.Forty M NaOH, calculate the concentration of the surplus base.

         

                    H2SO4                   +                  2NaOH                 →      Na2SO4          +        2HOH

          0.2000 L  x  0.2 hundred mole            0.250 L  x  0.40 mole

                              1 L                                             1 L

I                   0.0400 mole                              0.One hundred mole           

C                 0.0400 mole                              0.0800 mole

E                  0                                                0.020                               Note the loss of one sig fig!

                              [NaOH]       =        0.020 mole  =        0.044 M

                                                             0.4500 L                                    Note that the general quantity is 250.0 + 200.Zero mL

2.       100.Zero mL of 0.100 M H2SO4 reacts with 50.Zero mL of 0.20 M NaOH, calculate the concentration of the excess acid.

                    H2SO4                   +                  2NaOH                 →      Na2SO4          +        2HOH

          0.One thousand L  x  0.A hundred mole            0.050 L  x  0.20 mole

                              1 L                                             1 L

I                   0.0100 mole                              0.010 mole           

C                 0.0050 mole                              0.010 mole

E                  0.0050 mole                              0                                     

                              [H2SO4]       =        0.0050 mole          =        0.033 M

                                                             0.1500 L                                   

Note that the overall quantity is 100.0 + 50.Zero mL

3.       500.0 mL of 0.A hundred M H2SO4 reacts with 400.Zero mL of 0.Four hundred M NaOH, calculate the focus of the excess base.

                    H2SO4                   +                  2NaOH                 →      Na2SO4          +        2HOH

          0.5000 L x  0.A hundred mole             0.4000 L x  0.40 mole

                              1 L                                             1 L

I                   0.0500 mole                                0.16 mole           

C                 0.0500 mole                                 0.100 mole

E                  Zero mole                                         0.06 mole                               

Note the loss of sig figs!

                              [NaOH]       =        0.06 mole  =        0.07 M

                                                           

                                                            0.900 L                                     

Note that the overall volume is 500.0 + 400.0 mL

4.       100.0 mL of 0.Two hundred M MgCl2 reacts with 300.0 mL of 0.Four hundred M AlCl3, calculate all ion concentrations.

                    MgCl2             →      Mg2+           +        2Cl-

                   

    100.0       0.Two hundred M                     0.0500 M               0.100 M

     400.0

                    AlCl3             →        Al3+           +          3Cl-

                   

    300.0       0.Four hundred M                      0.300 M                 0.900 M

     400.0

[Cl-] = 0.100 M + 0.900 M = 1.000 M

5.       Change 2.66 moles of H2O to molecules.

          1.60 x 1024 molecules

6.       Change 9.7x1019 atoms Fe to moles.

          1.6 x 10-4 mole     

7.       Convert 88.3 g AgNO3 to system gadgets and then to atoms of O.

88.3 g   x   1 mol       x    6.02 x 1023 FU      x        Three atoms O   =        9.39 x  1023 atoms O

                  169.9 g                        1 mol                 1 FU

8.       Convert 3.Eight x 1025 H2 molecules to grams.

          1.3 x 102 g 

9.       Calculate the empirical method of a compound this is 62.2 % Pb, 8.454 % N, and 28.8 % O. 

          Is this compound ionic or covalent?

          Pb (NO3) 2

10.     A compound is 42.3 % C, 5.94 % H, 32.9 % N, and 18.8 % O and has a molecular mass of 425.25 g/mol. Calculate the empirical and molecular method.

          C15H25N10O5

11.     How many moles of Al2O3 are produced by means of the reaction 200. g Al?

4Al       +      302      →     2Al2O3

          3.70 mole

12.     How many moles Al are required to provide 300. g Al2O3?

                    4Al       +      302      →     2Al2O3

          5.88 mole

13.     100. g Al reacts with excess O2 to produce 150. g Al2O3 in line with

          Calculate the theoretical and percentage yield. 4Al  +  302   →      2 Al2O3.

          79.4 %       

14.     Calculate the power produced via your complete reaction of 150. g H2.

          2H2       +        O2         →      2H2O  +  130. KJ

          4.83 x 103 kJ

15.     How many grams of H2 could be had to produce 260. KJ of power?

          2H2  +  O2        →  2H2O  +  130. KJ

          8.08 g

16.     20. mol H2 reacts with 8.0 mol O2 to produce H2O.  Determine the choice of grams reactant in extra and selection of grams H2O produced.  Identify the proscribing reactant.

          Eight g H2          ,         2.9 x 102 g H2O

17.     How many litres of O2 gasoline are required to supply 100. g Al2O3?

                    4Al       +      302      →     2Al2O3

          32.Nine L

18.     15.Eight g of OkayCl is dissolved in 225 mL of water. Calculate the molarity.

          0.941 M

19.     Calculate the mass of OkayCl required to organize 250.0 mL of 0.250 M resolution.

          4.Sixty six g

20.     Calculate the quantity of 0.30 M OkayCl solution that contains 6.00 g of KCl.

          0.27 L

21.     Calculate the volume of 0.250 M H2SO4 required to neutralize 20.Zero mL of 0.One hundred M NaOH.

          0.00400 L

22.     Calculate the quantity of H2 fuel produced at STP via the response of 150.Zero mL of 0.500 M HCl with extra Zn.

                    Zn    +    2HCl    →   H2    +     ZnCl2

          0.840 L

23.     1.5 L of HCl gasoline at STP is dissolved in 225 mL of water, calculate the molarity of   the acid answer.

          0.30 M

24.     How many moles of NaCl are in 250. mL of a 0.Two hundred M solution?

          0.0500 moles

25.     How many litres of a 0.2 hundred M OkayCl resolution comprise 0.250 moles?

          1.25 L

26.     Calculate the mass of 2.25 x one zero five mL of H2 fuel at STP.

          20.3 g

27.     Calculate the number of components devices of OkCl in 100.0 mL of 0.200 M answer.

          1.20 x 1022 FU

28.     40.6g of OkayBr is dissolved in 500.Zero mL of water, calculate the molarity.

          0.682 M

29.     Calculate the mass of OkayBr required to organize 450.0 mL of 0.350 M answer.

          18.7 g

30.     Calculate the quantity of 0.50 M OkCl solution that comprises 3.00 g of KCl.

          0.080 L

31.     Calculate the quantity of 0.250 M H3PO4 required to neutralize 25.5 mL of

          0.Two hundred M NaOH.

          0.00680 L

32.     In a titration 22.5 mL of 0.2 hundred M H3PO4 is required to neutralize 10.Zero mL of KOH. What is the molarity of the bottom?

          1.35 M

Worksheet # 11    Molarity Unit Review # 2             

 

1.         100.Zero mL of 0.Two hundred M HCl, 200.0 mL of 0.100 M HBr, and one hundred seventy five mL of 0.One hundred M Ba(OH)2.

             Calculate the focus of the excess acid or base.

 

 

 

0.A thousand L  x  0.Two hundred mole   =    0.0200 mole HCl

                               1L

 

0.2000 L  x  0.One hundred mole   =    0.0200 mole HBr

                               1L

 

                                          =    0.0400 mole HX

 

           

0.175L  x  0.100 mole   =    0.0175 mole Ba(OH)2

                               1L

 

 

            2HX              +           Ba(OH)2    

 

I           0.0400 mole                0.0175 mole

 

C        0.0350 mole           0.0175 mole

E         0.0050 mole                0.0000

Total Volume =  100.0 mL  +  200.0 mL  +  one hundred seventy five mL   =   475 mL

 

 

[HX] = 0.0050 mole

             0.475 L  

 

[HX] =  0.011 M 

2.      150.Zero mL of 0.Two hundred M HCl and 250 mL of 0.300 M HNO3 react with extra CaCO3. 

         Calculate the theoretical yield of CO2. Start by means of writing an equation.

 

HCl         +      CaCO3        ®      CO2      +      CaCl2          +    H2O

 

 

0.1500 L HCl    x   0.200 mole    x    1 mole CO2   x   44.0 g     =    0.660 g

                                     1 L                 2 mole HCl        1 mole

 

 

 

 

HNO3        +      CaCO3        ®      CO2      +      Ca(NO3)2              +    H2O

0.2500 L HNO3    x   0.300 mole    x    1 mole CO2   x   44.0 g     =    1.Sixty five g

                                     1 L                    2 mole HCl        1 mole

Total               2.31 g

3.       Calculate the proportion composition of Al2(SO4)3 to 3 vital figures.

2 Al              54.0             %Al =         54.0             x        100%                    =          15.8 %

                                                            342.3

Three S               96.3             %S =           96.3             x        100%                    =          28.1 %

                                                            342.3

12 O            192.0           % O =         192.0           x        100%                    =          56.1%        

                    342.3                               342.3                    

4.      A compound is 42.3 % C, 5.94 % H, 32.9 % N, and and18.8 % O and has a molecular mass of 850.5g/mol. Calculate the empirical and molecular system.

          42.Three g C      x        1 mol           =        3.525 mol    =        3                  C3H5N2O    →      C30H50N20O10

                                        12.Zero g

          5.Ninety four g H      x        1 mol           =        5.881 mol    =        5

                                        1.01 g

          32.9 g N      x        1 mol           =        2.350 mol    =        2       

                                        14.0 g

          18.Eight g O      x        1 mol           =        1.175 mol    =        1

                                        16.0 g

5.       How many grams of 02 are required to eat 56.3 g Al?

          4Al      +         302       →      2Al2O3

          56.3 g Al      x        1 mol           x        3 mol O2      x        32.0 g                    =        50.0 g O2

                                        27.0 g                    4 mol Al                 1 mol

6.       15.8 g of AlCl3 is dissolved in 225 mL of water, calculate the molarity.

          Molarity      =        15.Eight g          x        1 mol

                                                                      133.Five g        =        0.526 M

                                                  0.225 L

7.       Calculate the mass of AlCl3 required to prepare 250.Zero mL of 0.250 M answer.

          0.250 L        x        0.250 mol    x        133.5 g        =        8.34 g

                                              L                      1 mol

8.       Calculate the amount of 0.30 M AlCl3 answer that comprises 6.00 g of AlCl3.

          6.00 g                    x        1 mol           x        1 L               =        0.15 L

                                                  33.5 g                    0.30 mol

9.       Calculate the amount of 0.450 M H2SO4 required to neutralize 25.0 mL of

          0.2 hundred M NaOH.

          H2SO4         +        2NaOH       →      Na2SO4       +        2HOH

            ? L                       0.025 L

          0.025 L NaOH      x        0.Two hundred mol    x        1 mol H2SO4              x           L                  =        0.00556 L

                                                      L                        2 mol NaOH                   0.450 mol

10.     Calculate the volume of H2 gas produced at STP via the response of 350.0 mL of 0.Six hundred M HCl with excess Zn.

                              Zn    +    2HCl    →   H2    +     ZnCl2

          0.350 L HCl          x        0.600 mol    x        1 mol H2      x        22.4 L          =        2.35 L

                                                      L                        2 mol HCl             1 mol

11.     2.9 L of HCl fuel at STP is dissolved in 225 mL of water, calculate the molarity of the acid answer.

          Molarity      =        2.9 L  x        1 mol           =        0.Fifty eight M

                                                            22.4 L                   

                                                  0.225 L

12.     How many moles of NaCl are in 500.0 mL of a zero.Three hundred M solution?

          0.500 L        x        0.300 mol    =        0.A hundred and fifty mol

                                            L

13.     How many litres of a .2300 M OkayCl resolution comprise 0.250 moles?

          0.250 mol    x              L                      =        1.09 L

                                        0.2300 mol

14.     Calculate the mass of 560. mL of CO2 fuel at STP.

          0.560 L        x        1 mol           x           44.0 g                 =        1.10g

                                        22.4 L                       1 mol

15.     Calculate the selection of components units of NaCl in 100.0 mL of 0.200 M solution.

          0.One hundred L        x        0.Two hundred mol    x        6.02 x 1023 FU       =        1.20 x 1022 FU

                                             L                         1 mol

16.     25.Five mL of 0.One hundred M HCl reacts with excess Zn to produce 25.3 mL of H2 gas at STP.

          Calculate the theoretical yield in mL and the share yield of H2 gasoline.

          Zn  +   2HCl   →  H2     + ZnCl2.

          0.0255 L      x        0.A hundred mol    x        1 mol H2      x        22.4 L          =          0.0286 L     

                                           L                       2 mol HCL            1 mol

          % Yield       =        25.3   x        100 %         =        88.6 %

                                        28.6

17.     Calculate the energy produced by all the reaction of 150. g H2.

          2H2+O2             →      2H2O  +        130KJ

          One hundred fifty g H2      x        1mol            x        A hundred thirty KJ        =        4.Eighty three x 103 KJ

                                        2.02 g                              2 mol

18.     84.0 g of Al reacts with 122g O2 to produce Al2O3.  How many grams of Al2O3 are produced?  Determine the mass of the reactant in extra and the restricting reactant.

                    4 Al                        +                            3O2              →                2Al2O3

          84.Zero g          x        1 mol                               122g  x        1mol

                                       27.0 g                                                 32.Zero g         

 

          I         3.111 mol                                            3.813 mol                                  0

          C       3.111 mol                                            2.333 mol                                  1.556

         

          E        0                                                          1.48   mol                                1.556

          Limiting                                                        47.4 g O2 extra                        159 g             

19.     Calculate the percentage composition of Na2SO4.

          2 Na            46.0 g                              % Na                    =        32.4 %

         

          1 S               32.1 g                              % S                       =        22.6 %

          4 O              64.Zero g                              % O                      =        45.0 %

                              142.1

20.     How many litres of O2 gas are required to produce 100. g Al2O3?

                    4Al       +      O2     →     2Al2O3

          100. g Al2O3          x        1 mole         x        Three mole O2   x        22.4 L  =   32.Nine L

                                                  102 g                     2 mole Al2O3                   1 mole

21.     Calculate the molar mass of a fuel that weighs 19.forty three g and has a STP quantity of 9.894 L.

          If the gasoline is an excessively humorous one containing nitrogen and utilized by the dentist, determine the

          molar mass and molecular formulation for the gasoline.

          9.894 L        x        1 mole         =        0.4417 mole

                                        22.4 L

          Molar Mass          =        19.forty three g                            =        44.Zero g/mole           N2O

                                                  0.4417 mole

Write a balanced formulation equation, whole ionic equation, and net ionic equation for each response. There are two no reactions.

22.     Zn(s)  +      2AgNO3(aq)     →       2Ag(s)          +        Zn(NO3)2(aq)

          Zn(s)  +      2Ag+    +        2NO3-     →  2Ag(s)          +        Zn2+             +          2NO3-

          Zn(s)  +      2Ag+         →  2Ag(s)          +        Zn2+            

23.     BaS (aq)        +        2KOH(aq)     →      Ba(OH)2(s)   +        K2S(aq)

          Ba2+    +        S2-      +        2K+    +        2OH-           →      Ba(OH)2(s)   +          2K+     +            S2-

                              Ba2+   +        2OH-           →      Ba(OH)2(s)

24.     2NaCl(aq)     +        F2(g)   →      2NaF(aq)       +        Cl2(g)

          2Na+     +     2Cl-    +        F2(g)   →      2Na+    +      2F-     +        Cl2(g)

                    2Cl-    +        F2(g)   →                2F-     +        Cl2(g)

25.     Sr(OH)2 (aq) +        CuSO4(aq)    →      Cu(OH)2 (s)  +        SrSO4(s)

          Sr2+   +        2OH-          +        Cu2+            +        SO42-           →          Cu(OH)2 (s)  +        SrSO4(s)

          Sr2+   +        2OH-          +        Cu2+            +        SO42-           →          Cu(OH)2 (s)  +        SrSO4(s)

26.     NaCl(aq)        +        Cu(NO3)2(aq)          →     

No reaction, both possible merchandise have prime solubility

27.     NaCl(aq)     +    ZnF2(aq)                    →               

No response, each possible products have high solubility

28.     100.Zero g of an aqueous compound this is 45.49 % Pb, 12.31 % N, and 42.20 % O reacts with every other compound this is 28.16 % N, 8.13 % H, 20.79 % P, and 42.91 % O. If the actual yield of the product containing lead is 60.Zero g, calculate the         percentage yield.

                                                                                          60.Zero g          Actual Yield

3Pb(NO3)4(aq)        +        4(NH4)3PO4(aq)      →      Pb3(PO4)4(s)          +          12NH4NO3(aq)

100.Zero g                                                                    ? g               Theoretical Yield

100.Zero g Pb(NO3)4 x  1 mole     x        1 mole Pb3(PO4)4            x        1001.6 g          =        73.3 g

                                  455.2 g              Three mole 3Pb(NO3)4                    1 mole

          % yield        =        60.0 g                    x        100%                    =          81.8 %

29.     The following information used to be obtained all over the titration of 2.0554 M hydrochloric acid with a 25.00 mL aliquot of barium hydroxide:

                                                                      Trial 1                   Trial 2                    Trial 3

                    Burette Final Reading                 22.ninety two mL                25.32 mL                41.30 mL

                    Burette Initial Reading               0.06 mL                 2.58 mL                 18.54 mL

                    Vol. of Acid Added                   22.86 mL             22.Seventy four mL                22.76 mL                                                  

         

          Use the above knowledge to determine the concentration of the barium hydroxide.

          2HCl           +        Ba(OH)2     →

          0.02275 L              0.02500 L

          2.0554 M              ? M

Molarity  Ba(OH)2         =        0.02275 L    x        2.0554 mole          x        1 mole Ba(OH)2

                                                                                1 L                                   2 mole HCl

                                                                                                    0.02500 L

                                                            =        0.9352 M

30.       20.Zero mL of 0.2 hundred M NaOH solution is diluted to a final quantity of 100.Zero mL, calculate the brand new focus.

M1V1                     =        M2V2

(0.200)(20.0)         =        M2(100)

M2     =        0.0400 M

31.     20.Zero mL of 0.Three hundred M AlCl3 is mixed with 20.Zero mL of 0.300 CaCl2, calculate all ion concentrations.

                    AlCl3             →          Al3+           +        3Cl-

                   

    20.0         0.300 M                     0.150 M                0.450 M

     40.0

                    CaF2              →         Ca2+           +        2Cl-

                   

    20.0         0.Three hundred M                      0.One hundred fifty M                 0.300 M

     40.0

[Cl-] = 0.450 M + 0.Three hundred M = 0.750 M

32.     A burette stuffed with 2.000 M sodium hydroxide answer reads 20.20 mL to begin with. After titrating a

25.00 mL sample of phosphoric acid the endpoint was once reached and the burette now confirmed 40.20 mL. What is the    [phosphoric acid]?

          H3PO4         +        3NaOH       →

          0.02500 L              0.02000 L

          ? M                       2.000 M

          Molarity  H3PO4                  =        0.02000 L    x        2.000 mole  x          1 mole H3PO4

                                                                                                    1 L                         Three mole NaOH

                                                                                                    0.02500 L

                                                            =        0.5333 M

33.     Calculate the quantity of 0.500M KOH required to neutralize 45.Zero mL of 0.320 M H2SO4.

          H2SO4                   +        2KOH         →

          0.0450 L                          ? L

          0.0450 L H2SO4    x        0.320 moles x        2 mole KOH         x        1 L                  =        0.0576 L

                                                  1 L                         1 mole H2SO4                0.500 mole

34.     100.Zero mL of 0.2 hundred M H2SO4 reacts with 150.0 mL of 0.Forty M NaOH, calculate the concentration of the surplus base.

          H2SO4                   +                  2NaOH                           →

         

0.One thousand L  x  0.2 hundred mole                      0.1500 L   x  0.Forty mole

                    1 L                                                       1L

I         0.0200 mole                              0.060 mole

C       0.0200 mole                              0.0400 mole

E        0                                                0.020 mole Note the loss of one sig fig

          [NaOH]       =        0.020 mole            =        0.080 M

                                        0.250 L

35.     200.Zero mL of 0.10 M H2SO4 reacts with 100.0 mL of 0.20 M NaOH, calculate the concentration of the surplus acid.

          H2SO4                   +                  2NaOH       →

0.2000 L  x  0.One hundred mole                      0.1000 L   x  0.20 mole

                    1 L                                                         1L

I         0.0200 mole                              0.020 mole

C       0.010 mole                                0.020 mole

E        0.010    loss of one sig fig         0                 

          [H2SO4]       =        0.010 mole            =        0.033 M

                              0.Three hundred L

36.       250.0 mL of 0.100 M H2SO4 reacts with 100.0 mL of 0.Four hundred M NaOH and

            200.Zero mL of 0.2 hundred M KOH, calculate the concentration of the excess base.

          H2SO4                   +                  2XOH       →

0.2500 L  x  0.One hundred mole                      0.One thousand L   x  0.400 mole        +     0.2000 L   x  0.Two hundred mole

                    1 L                                                         1L                                                            1L

I         0.0250 mole                              0.0800 mole

C       0.0250 mole                               0.0500 mole

E        0                                               0.0300 mole                   

          [NaOH]       =        0.0300 mole                    =        0.0545 M

                              0.550 L

 

37.       If the [Al3+] focus in 3.00 L of answer is .0275 M, calculate the entire ion concentrations and the molarity of a Al2(SO4)3 resolution.

          

 

             Al2(SO4)3             →          2Al3+           +        3SO42-

          

              0.0138 M                          0.0275 M               0.0413 M

          

38.       If the sodium ion focus is 0.450 M in 150.0  mL of a Na3PO4 resolution.  Determine the mass of Na3PO4 within the answer.

          

 

   Na3PO4              →            3Na+           +        PO43-

          

       0.150 M                          0.450 M                 0.One hundred fifty M

       3.69 g

Worksheet # 12   Chemistry 11 Calculations Practice Test # 1        

Pick two formulas that fit each and every classification:

1.   a                   b                  Acid                                 a)  HCl                   e) KOH

2.   c                   d                  Covalent Nonacid            b) CH3COOH          f) NH4Cl

3.   h                   f                   Salt                                  c)  CH4                   g) Ba(OH)2

4.   e                   g                  Base                                 d)  HOH                 h)  AgNO3

5.   Calculate the molarity of the solution shaped when 2 hundred g of NaCl is dissolved in

      A hundred mL of H2O.

Molarity = 200g   x   1 mole

                                   58.5g           =     34.2 M

                        0.One hundred L

                     

6.       How many grams of AgCl are required to prepare One hundred fifty mL of 0.2 hundred M resolution?

                0.150L   x   0.Two hundred mole   x    143.4 g       =    4.30 g

                                               1 L              1 mole

7.       How many litres of 0.200 M AgCl are needed to offer 50 g of AgCl?

                50g   x      1 mole   x           1 L       =    1.7 L

                                       143.4g         0.Two hundred mole

8.       A hundred g of AlClThree is dissolved in 200 mL H2O, calculate [Al3+] and [Cl-].

               100 g   x 1 mole

Molarity =                     133.5 g           =     3.745 M     AlCl3    →    Al3+       +       3Cl-

                              0.200 L    

                                                                     3.745 M                     3.75 M       11.2M

9.       In 3 runs of a titration 36.9, 34.4 and 34.Three mL of 0.Two hundred M NaOH was required to neutralize a 25.0 mL sample of H2CO3. Calculate the molarity of the acid.

                                               H2SO4              +             2NaOH    →         Na2SO4      +              2HOH

                                               0.0250 L                          0.3435 L

                                               ? M                                 0.Two hundred M

                              [ H2SO4]                =        0.03435 L   x   0.Two hundred mole   x  1 mole H2SO4       =    0.137 M

                                                                                                 1 L               2 mole  NaOH

                                                                                                           0.0250 L

                                                                  

=        0.137 M

10.    Calculate the [NaOH] because of excess NaOH in the new solution produced by way of blending 100. mL

0.Two hundred M HCl and 100. mL 0.300 M NaOH.

                                     HCL            +                  NaOH         →      NaCl      +                  HOH

0.100L   x   0.200 mole  =   0.0200 mol         0.100L x 0.Three hundred mol  =   .030 mole

                     1 L                                                         1 L

I                                    0.0200 mole                    0.0300 mole

C                                  0.0200 mole                    0.0200 mole

E                                   0  mole                            0.0100 mole

Total Volume = 2 hundred mL = 0.200 L           Molarity     =         0.0100 mole           =  0.0500 M

                                                                                                           0.Two hundred L

11.    A empty beaker has a  mass of 29.86 g. The similar beaker is full of 0.250 L with a solution of CaCl2 and weighs 87.26 g. The answer is evaporated to dryness and the mass of the beaker and forged is 62.31 g. Calculate the molarity of the solution.  

Mass of CaCl2 = 62.31 – 29.86 = 32.45g

Molarity = 32.45g   x   1 mole

                                     111.1g           =     1.17 M

                            .250 L

12.     Complete the reaction equations.

      i) Formula Equation/Chemical Equation

2AgNO3 (aq)  +  Na2SO4 (aq)   →      Ag2SO4(s)       +      2NaNO3(aq)

      ii) Total Ionic Equation

2Ag+(aq)      +     2NO3-(aq)    +    2Na+(aq)     +  SO42-(aq)   →     Ag2SO4(s)       +      2Na+(aq)       +      2NO3-(aq)

      iii) Net Ionic Equation

2Ag+(aq)     +  SO42-        →     Ag2SO4(s)      

                          

13. Complete the components equation:

2H3PO4(aq)    +    3Sr(OH)2(aq)     →    Sr3(PO4)2(s) +        6HOH(l)

Complete the entire ionic equation:

6H+(aq)      +     2PO43-     +    3Sr2+(aq)   +    6OH-  →   Sr3(PO4)2(s)         +        6HOH(l)

Complete the online ionic equation:

6H+(aq)      +     2PO43-     +    3Sr2+(aq)   +    6OH-      →        Sr3(PO4)2(s)   +        6HOH(l)

14. Complete the formulation equation:

Fe3(PO4)2(aq)    +    3Zn(s)     →           3Fe(s)     +     Zn3(PO4)2(s)

Complete the complete ionic equation:

3Fe2+(aq)     +    2PO43-    +    3Zn(s)     →        3Fe(s)     +     Zn3(PO4)2(s)

Complete the online ionic equation:

3Fe2+(aq)     +    2PO43-    +    3Zn(s)     →        3Fe(s)     +     Zn3(PO4)2(s)

Worksheet # 13              Chemistry Calculations Practice Test # 2

1.       Calculate the number of formulation gadgets in 250. g  CaCl2.

          1.35 x 1024   FU

2.       Calculate the mass of 2.35 x 1020  molecules of CO2.

          0.0172 g

3.       Calculate the STP volume of 10.Zero g of CO2 gasoline.

          5.09 L

4.       Calculate the choice of grams CaCl2 in 350. mL of a zero.250M resolution.

          9.72g

5.       Calculate the quantity of 0.250 M NaCl solution that may contain 0.17 g NaCl.

          0.012L        

6.       1.26 g of AlCl3are dissolved in 160.Zero ml of water. Calculate the molarity of the answer.

         

          0.0590M

                   

7.       12.5 ml of CO2 gasoline at STP are dissolved in 250.0 ml of water. Calculate the molarity of the solution.

          0.00223M

8.       10.0 g of Al2(SO4)Three is dissolved in one hundred fifty five ml of water. Calculate the two ion concentrations.

          0.377 M

          0.565 M

9.       200.0 ml of 0.200M H3PO4 reacts with 200.Zero ml of 0.300M KOH. Calculate the molarity of the excess acid within the new answer shaped.

          0.0500M

10.     Sixteen g of Ca react with water. Calculate the amount of H2 gasoline produced at STP.     

          Ca   +   2H2O   ®   H2   +   Ca(OH)2  

          8.9L  

         

11.     In a titration 0.2 hundred M NaOH is used to neutralize 10.0 mL of H2SO4. In three runs the next information was amassed. Calculate the focus of the acid.      

          Volume of 0.2 hundred M NaOH         25.3 mL       25.8 mL       25.6 mL

          0.256 M

12.     60.0 g of Al react with 60.Zero g of O2. Calculate the amount of excess reactant.   

          4Al      +       3O2   →      2Al2O3

          6.66 g O2

13.     Calculate the proportion composition of the elements in Ga2 (SO4)3 to 3 significant digits.

                              32.6%         ,           22.5%       ,          44.9%

14.     What volume of 0.300 M resolution should be diluted to a final volume of 1200.Zero mL and have a molarity of 0.2500M.

          1.00L

15.     Calculate the selection of grams NaCl produced through your complete reaction of

          520 g Cl2.               2Na              +        Cl2    →    2NaCl  

          857g

16.     If the true yield of NaCl within the closing query used to be 200. g, calculate the proportion yield of NaCl.

          23.3%        

17.     200.0 ml 0.2 hundred M HCl reacts with 400.Zero ml  0.150M NaOH. Calculate the molarity of excess base.  

          HCl   +         NaOH   →    NaCl    +      H2O

          0.0333M

18.     100.0 mL of 0.250 M HCl answer is diluted by adding 250.Zero mL of water, calculate the new concentration.

          0.0714M

19.     65.Five mL of 0.Three hundred M is diluted to a new molarity of 0.0600 M, how much water was added?

          262mL

20.     56.0 mL of 0.A hundred M HCl reacts with 0.250 M Ba (OH) 2, calculate the volume of base required to totally neutralize the acid.

          0.0112L

21.     Write the components, complete, and net ionic equation for each and every.

          H3PO4 (aq) and NaOH (aq).

          H3PO4(aq) + 3NaOH (aq) → Na3PO4(aq) + 3HOH (l)

         

          3H+(aq) + PO4-3(aq) + 3Na+(aq) + 3OH-(aq) → 3Na+(aq) + PO4-3(aq) + 3HOH (l)

         

          H+(aq) + OH-(aq) → HOH (l)

22.     Write the formula, whole, and web ionic equation for each.

          Na3PO4 (aq) and Ca (NO3) 2(aq).

          2Na3PO4(aq) + 3Ca(NO3) 2(aq)→ Ca3 (PO4) 2(s) + 6NaNO3(aq)

          Na+(aq) + 2PO43-(aq) +3Ca2+(aq) + 6NO3-(aq) → Ca3 (PO4) 2(s) + 6Na+(aq)+ 6NO3-(aq)

          3Ca2+(aq) + 2PO43-(aq) → Ca3 (PO4) 2(s)

23.     Write the formula, whole, and web ionic equation for each.

          Cu (NO3) 2(aq) and Ag(s).

          Ag (s) + Cu (NO3) 2(aq)→ No Reaction

24.     A empty beaker has a mass of 25.86 g. The identical beaker is filled with 0.250 L with a solution of Cl2 and weighs 87.26 g. The resolution is evaporated to dryness and the mass of the beaker and solid is 36.31 g. Calculate the molarity of the answer.

          0.376 M

25.     125.0 g of an aqueous compound this is 3.091 % H, 31.62  % P, and 65.29 % O reacts with another compound this is 80.14 % Ba, 18.68 % O, and 1.179 % H. If the real yield of the forged product is 350. g, calculate the proportion yield of the cast.

          91.2%

26.        0.0250 M

27.        0.0091 M

28.       [Ca2+]   =    0.0714 M         [Al3+]  =  0.193 M          [Cl-]  =    0.722M

  29        [Na+]   =    0.333 M         [PO43-]  =  0.0667 M          [SO42-]  =   0.0667 M

  30.        0.0827 M

31.      150.0 mL of 0.200 M HCl and 250.0 mL of 0.Three hundred M HNO3 react with extra CaCO3.

           Calculate the theoretical yield of CO2. Start by way of writing an equation.

         

          2.31 g

Unit 8 Worksheet 4

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