Achemist obtains 500.0 ml of a solution containing an unknown concentration of calcium iodide, cai2. he pipets 15 ml of this solution into a 100 ml volumetric flask and dilutes to the mark. he then pipettes 10 ml of this diluted solution into a 25 ml volumetric flask and dilutes to the mark. he analyzes some of the solution from the final volumetric flask and finds that the iodide ion concentration is 0.41 m. (note: in solution, calcium iodide breaks apart into one ca2+ ion for every two i- ions, so a solution that is 1.0 m in cai2 is 2.0 m in determine the molar concentration of calcium iodide in the original solution. latex: cai _{2} (s)→ ca^{ 2+}(aq)+2 i^{-}(aq)

3.41735M

Explanation:

Step1: Balance the equation

CaI2 (s) → Ca2+ (aq ) + 2I- (aq)

We have 15 mL of the original sample and pipet it in a 100mL solution.

100/15 = 6.67 ⇒ this is a 1:6.67 dilution

We take 10mL of this dilution into a 25 mL flask ⇒this is a 1:2.5 dilution

6.67 * 2.5 = 16.675

⇒the total dilution is 1:16.675  

Let's imagine the conc of CaI2 [CaI2] = 1M

⇒ 15mL of this 1M conc in a 100mL = 0.15M

⇒10mL of this 0.15 M in 25mL = 0.06M

1M original conc/ O.O6M final concentration = 16.67

⇒this means the dilution is 16.67 (1:16.67)

After analyzes we found that the iodide ion concentration = 0.41 M

⇒original conc / 0.41 M  = 16.67

original conc I- = 16.67 * 0.41M = 6.8347 M

Since we have one Ca2+ ion for every 2 I- ions, we have to divide this concentration by 2 to find  the [Ca2+]

6.8347 / 2 = 3.41735 M  =[Ca2+] = [CaI2]

The concentration calcium iodide = 3.41735M

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.