1) and 2) 3) and 4) Note: 0.400g NaCl Conductivity at 12,000 uS/cm Based on your
ID: 849847 • Letter: 1
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1) and 2)
3) and 4)
Note: 0.400g NaCl Conductivity at 12,000 uS/cm
Based on your current understanding, draw molecular-level views of what happens when the following solids are added (separately) to water: Calculate the mass of CaC 2 H20 that would be needed to prepare a 50.00-mL aqueous solution that would yield the same conductivity as the NaCI(aq) solution Calculate the mass of NaN03 that would be needed to prepare a 50.00-mL aqueous solution that would yield the same conductivity as the NaCI (aq) solution. Calculate the mass of NH4CI that would be needed to prepare a 50.00-mL aqueous solution that would yield the same conductivity as the NaCI (aq) solution.Explanation / Answer
Answer 1)
a) CaCl2 is soluble in water and dissociates to form Ca2+ and Cl- ions. And no, nothing happens to water, because neither Ca2+ nor Cl- ions hydrolyze (react with water).
That's because Ca is the cation of a strong base, and Cl is the anion of a strong acid.
The only thing that water will do is form a hydration sphere around each of the two ions in solution. But, that isn't a chemical change in the water.
b) sodium Nitrate (NaNO3) + water (H2O) ---> NaOH (Sodium Hydroxide) and (HNO3)
I'm pretty sure that NaOH stays in ion form (Na+ and OH-) because Na is an ion that forms strong bases, and strong bases typically dissociate very close to 100%. I am close to positive that HNO3 (Nitric Acid) would stay together.
c) It is a good solute because highly soluble in water but its aqueous solution is slightly acidic in nature due to formation of weak base Ammonium hydroxide and strong acid Hydrochloric acid after the hydrolysis of salt.
Answer for 2)
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