15.63 mol. You need 15.63 mol HgO to produce 250.0 g O_2.
<em>Step 1</em>. Convert <em>grams of O_2 to moles of O_2</em>
Moles of O_2 = 250.0 g O_2 × (1 mol O_2/32.00 g O_2) = 7.8125 mol O_2
<em>Step 2</em>. Use the molar ratio of HgO:O_2 to convert <em>moles of O_2 to moles of HgO
</em>
Moles of HgO = 0.8885 mol O_2 × (2 mol HgO/1 mol O_2) = <em>15.63 mol HgO</em>
I think it is not possible to make cd without plastic . because except plastic no other material is so durable for making cd.
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:
I believe this is a K-12 test question. If the answers below are what you have on your test . . .
- Precise
- Accurate
- Identical
- None of the above
Then the answer is <u>precise</u>.