Answer:
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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
Answer:
= 13.0 moles O2
Explanation:
1] Given the equation: 2C8H18 + 25 O2 ----> 16CO2 + 18H2O
a. How many moles of oxygen gas are required to make 8.33 moles of carbon dioxide?
8.33 moles CO2 X
25mol O2
16mol CO2
= 13.0 moles O2
Answer:
1. an educated guess
2. data
3. what changes in experiment
4. what stays the same in both groups
5. the group where nothing changes, normal
6. group with independent variable, what's being tested
OK in the case of hydrazine 14 grams of nitrogen combine with 2 gram of hydrogen and with ammonia 14 grams combine with 3 grams of hydrogen.
Ratio 2:3