A) solid
b)liquid
c)liquid
d)gas
The given question is incomplete. The complete question is:
When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.
Answer: The vant hoff factor for sodium chloride in X is 1.9
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
= freezing point constant
i = vant hoff factor = 1 ( for non electrolyte)
m= molality =

Now Depression in freezing point for sodium chloride is given by:
= Depression in freezing point
= freezing point constant
m= molality =


Thus vant hoff factor for sodium chloride in X is 1.9
Explanation:
cant answer without context
Answer:
31,380 Joules
Explanation:
Given Data:
Mass = m = 100 g
Temperature 1 = = 25 °C
Temperature 2 = = 100 °C
Specific Heat Constant = c = 4.184
Change in Temp. = ΔT = 100 - 25 = 75 °C
Required:
Heat = Q = ?
Formula:
Q = mcΔT
Solution:
Q = (100)(4.184)(75)
Q = 31, 380 Joules
Hope this helped!
~AH1807