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Fed [463]
3 years ago
7

Ethylene glycol (C2H6O2) is used as an antifreeze in cars. If 250 g of ethylene glycol is added to 3.00 kg of water, what is the

molality? Calculate how much the freezing point of water will be lowered. The freezing-point depression constant for water is Kf = –1.86°C/m. Show your work.
Chemistry
1 answer:
Zepler [3.9K]3 years ago
3 0

Answer:

2,909 M

Explanation:

molair mass is of.ethylene is 26,04 g/mol

first you need to calculate how much mL 3 kg is. You can do this by using the density of ethylene: 1,1 g/mL.

3000 g x 1.1 = 3300 mL = 3,3 L

Next you need to calculate the amount of moles:

250 g / 26,04 g/mol = 9,60 mol

Now you can calculate the molarity:

9,6/3.3 = 2,909 M

I don't know the answer for the second question. I'm sorry.

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Answer:

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Explanation:

At constant pressure and temperature, the mole ratio of the gases is equal to their volume ratio (a consequence of Avogadro's law).

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Then, you must write the balanced chemical equations for the complete combustion of the four hydrocarbons in the list of choices, and conclude which has such mole ratio (1 mol hydrocarbon : 3 mol oxygen).

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<u>a. C₂H₄:</u>

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The following analysis just shows that the other options are not right.

<u>b. C₂H₂:</u>

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The mole ratio for this reaction is 2 mol C₂H₂ :5 mol O₂.

<u>с. С₃Н₈</u>

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The mole ratio is 1 mol C₃H₈ : 5 mol O₂

<u>d. C₂H₆</u>

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The mole ratio is 2 mol C₂H₆ : 7 mol O₂

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