Question

1. For each of the following, explain whether the substances can form a solution or not....

1. For each of the following, explain whether the substances can form a solution or not.

A. NH3 and NaCl CH4

B. KBr C4H10

C. CO2

2. Classify the following compounds as strong electrolytes, weak electrolytes or non-electrolytes based on their dissolution reactions in water.

A. C6H12O6(s)→C6H12O6(aq)

B. Na3PO4(s)→3 Na(aq)+PO43(aq)

C. AgCl(s)↔Ag+(aq)+Cl+(aq)

D. Mg(NO3)2(s)→Mg2+(aq)+2 NO3-(aq)

3. Write the balanced total ionic equation and net ionic equation for the reaction that occurs when and (NH4)2CO3(aq) and MgSO4(aq) are mixed.

4. What is the molarity of a solution in which you dissolve 45.9 g CuSO4(s) in 3.25 L of water?

5. What is the molarity of an acetic acid (C2H4O2) solution in which 95.0 g of acetic acid is dissolved in 150.00 mL of water?

6. Given 300. g of a 27.5% m/m solution of NaCl(aq), how many moles of NaCl do you have?

7. You react 9.00 mL of 0.750 M Hg2(ClO3)2(aq) with an unlimited supply of a 0.100 M solution of NaCl which will produce Hg2Cl2(s) and NaClO3(aq).

A. Write the balanced chemical equation.

B. How many grams of Hg2Cl2(s) could you theoretically produce?

C. Assuming your reaction has a yield of 87.0%, how many liters of 0.100 M NaCl(aq) would you need to add to remove the mercury(I)?

8. You dilute a 15.0 mL of a 12.0 M solution of BaCl2(aq) by adding 75.0 mL of water. What is the molarity of the diluted solution?

9. Calculate the freezing point of a solution containing 48.5 g CaCl2 in 1.00 kg of water. Note, water’s freezing point decreases by 1.86 °C for every 1 mole of particles dissolved in it.

Homework Answers

Answer #1

1. For each of the following, explain whether the substances can form a solution or not.

A. NH3 and NaCl CH4 --> only NH3 and NaCl will; CH4 is gas, will not be soluble in polar compounds

B. KBr C4H10 --> not likely ionic + nonpolar will not form solutions

C. CO2 --> not likely CO2 is not that soluble in water

2. Classify the following compounds as strong electrolytes, weak electrolytes or non-electrolytes based on their dissolution reactions in water.

A. C6H12O6(s)→C6H12O6(aq) not eletrolyte, will not form 100% ions

B. Na3PO4(s)→3 Na(aq)+PO43(aq) --> strong electroltye, 100% ions

C. AgCl(s)↔Ag+(aq)+Cl+(aq) --> weak electrolyte, since this is not likely to occur ( euqilibrium shifted)

D. Mg(NO3)2(s)→Mg2+(aq)+2 NO3-(aq) --> 100% ionizaiton, implies very strong electrolyte.

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