Question

1. enter the net ionic equation, including phases, for the reaction of AgNO3(aq) and KCI(aq).

2. 76.0 mL of a 1.50 M solution is diluted to a total volume of 288 mL. A 144-mL portion of that solution is diluted by adding 127 mL of water. what is the final concentration? assume the volumes are additive.

3. what is the concentration of a solution made by diluting 35 mL of 6.0 M HCI to a final volume of 750 mL?

4. calculate the molarity of the following solutions.

a) 0.850 mol of Na2S in 1.85 L of solution.

b) 22.3 g of MgS in 981 mL of solution.

5. complete this equation for the dissociation of Na2CO3(aq). omit water from the equation because it is understood to be present.

Na2CO3(aq) --------->

Answer 1

1) for the reaction of AgNO3(aq) and KCI(aq) :

AgNO3 (aq) + KCl (aq) --> AgCl (s) + KNO3 (aq)

okay so first you would want to balance that double replacement reaction but since its already balanced
You split up the aqueous solutions but since AgCl is the precipitate you keep it the same:

Ag⁺(aq) + NO3 ^-(aq) + K⁺ (aq) + Cl^-(aq) --> AgCl (s) + K⁺(aq) + NO3^-(aq)

now you cancel out the ions that are on both sides giving you:

Ag⁺ (aq) + Cl^- (aq) ---> AgCl (s)

2) 76.0 mL of a 1.50 M solution is diluted to a total volume of 288 mL. A 144-mL portion of that solution is diluted by adding 127 mL of water. what is the final concentration? assume the volumes are additive.

(76.0mL)(1.5M) = (288mL)(xM)

x = 144/288 = 0.395 M

concentration of a solution is 0.395 M

A 144-mL portion of that solution is diluted by adding 127 mL of water,then total volume is 271 ml

(144 mL)(0.395 M) = (271 mL)(xM)

x = 56.88/271 = 0.20 M

Final concentration of a solution is 0.20 M

3) what is the concentration of a solution made by diluting 35 mL of 6.0 M HCI to a final volume of 750 mL?

(35mL)(6M) = (750mL)(xM)

x = 210/750 = 0.28M

concentration of a solution is 0.28 M

4)

A) the molarity of a solution of 0.850 moles in 1.85 L of solution.

moles Na₂S = 0.850 mole

molarity =0.850 mole / 1.85 L = 0.45 M

B)the molarity of a solution of 22.3 g of MgS in 981 mL of solution.

Solution: 981 mL = 0.981 L                         Molar Mass of MgS = 56.38 g/mole

moles MgS = 22.3 gm / 56.38 g/mole = 0.395 mole

molarity =0.395 mole / 0.981 L = 0.40 M

5) Complete this equation for the dissociation of Na2Co3:

Na2CO3 → 2 Na⁺ + CO3^2-