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Thepotemich [5.8K]
3 years ago
8

Help me please I don’t know this

Chemistry
1 answer:
jekas [21]3 years ago
3 0
I think it’s 62 grams
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Calculate the freezing point of a 12.25 m 12.25m aqueous solution of propanol. Freezing point constants can be found in the list
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Answer:

∆t(f) = 49.755C

Explanation:

Freezing point is defined as the point in which a liquid changes from liquid to solid state. Therefore, to calculate the freezing point of aqueous solution of propanol

∆t = k × I × m

Where k is cryscopic constant =

I = vant Hoff factor = 2

M = molar concentration = 12.25m

∆t (f) = 1.9 × 12.25 × 2 = 49.755C

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C(S)+O2(g)--&gt;CO2(g)
soldi70 [24.7K]

<u>Answer:</u> The correct answer is 1.18 g.

<u>Explanation:</u>

We are given a chemical equation:

C(S)+O2(g)\rightarrow CO_2(g)

We know that at STP conditions:

22.4L of volume is occupied by 1 mole of a gas.

So, 2.21L of carbon dioxide is occupied by = \frac{1}{22.4L}\times 2.21L=0.0986mol of carbon dioxide gas.

By Stoichiometry of the above reaction:

1 mole of carbon dioxide gas is produced by 1 mole of carbon

So, 0.0986 moles of carbon dioxide is produced by = \frac{1}{1}\times 0.0986=0.0986mol of carbon.

Now, to calculate the mass of carbon, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of carbon = 0.0986 mol

Molar mass of carbon = 12 g/mol

Putting values in above equation, we get:

0.0986mol=\frac{\text{Mass of carbon}}{12g/mol}\\\\\text{Mass of carbon}=1.18g

Hence, the correct answer is 1.18 g.

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