On wow, forgot this could be on here. Glad to see this before anything important.kenniky wrote:15g of CaCl2 is added to 100 mL of water. What is the resultant boiling point of the water? (assume standard conditions)
100.7 °C?
On wow, forgot this could be on here. Glad to see this before anything important.kenniky wrote:15g of CaCl2 is added to 100 mL of water. What is the resultant boiling point of the water? (assume standard conditions)
100.7 °C?
Incorrect, check your equation again (I'm pretty sure I know where you went wrong)Avogadro wrote:On wow, forgot this could be on here. Glad to see this before anything important.kenniky wrote:15g of CaCl2 is added to 100 mL of water. What is the resultant boiling point of the water? (assume standard conditions)
100.7 °C?
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correct, your turnPiggy wrote:kenniky wrote:15g of CaCl2 is added to 100 mL of water. What is the resultant boiling point of the water? (assume standard conditions)102.1 °C?
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Piggy wrote:An apparatus contains Br2(g) in a solution. What happens to the vapor pressure when the piston lowers and decreases the volume of Br2(g) above the solution by half? (Assume no temperature change).
I think it would double?
Incorrect.Avogadro wrote:Piggy wrote:An apparatus contains Br2(g) in a solution. What happens to the vapor pressure when the piston lowers and decreases the volume of Br2(g) above the solution by half? (Assume no temperature change).I think it would double?
Actually, I think I figured out my issue.Piggy wrote:Incorrect.Avogadro wrote:Piggy wrote:An apparatus contains Br2(g) in a solution. What happens to the vapor pressure when the piston lowers and decreases the volume of Br2(g) above the solution by half? (Assume no temperature change).I think it would double?
Should I give the answer now or wait?
If I'm correct, I didn't pay proper attention to the fact that it's vapor pressure as opposed to "normal pressure"- I believe that the vapor pressure would remain the same, as some of the gas would condense into liquid Br2.
Correct! Your turn.Avogadro wrote:Actually, I think I figured out my issue.Piggy wrote:Incorrect.Avogadro wrote:I think it would double?
Should I give the answer now or wait?
If I'm correct, I didn't pay proper attention to the fact that it's vapor pressure as opposed to "normal pressure"- I believe that the vapor pressure would remain the same, as some of the gas would condense into liquid Br2.
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