The answer is D. Use the equation PV=nRT
P=(.567mol)(.0821)(300K)/4.5L
Mass of KCl= 1.08 g
<h3>Further explanation</h3>
Given
1 g of K₂CO₃
Required
Mass of KCl
Solution
Reaction
K₂CO₃ +2HCl ⇒ 2KCl +H₂O + CO₂
mol of K₂CO₃(MW=138 g/mol) :
= 1 g : 138 g/mol
= 0.00725
From the equation, mol ratio K₂CO₃ : KCl = 1 : 2, so mol KCl :
= 2/1 x mol K₂CO₃
= 2/1 x 0.00725
= 0.0145
Mass of KCl(MW=74.5 g/mol) :
= mol x MW
= 0.0145 x 74.5
= 1.08 g
The freezing point of water is 0° C. The Celsius (centigrade) scale is based off of water, with the freezing point at 0° and the boiling point at 100°. (Google will probably tell you this in 10 seconds, then you wouldn't have had to wait 7 hours!)
Answer:
D is the answer may be. I'm not sure.
