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4 years ago

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What is the boiling point of a solution prepared by dissolving 70.0 g of naphthalene in 220.0 g of benzene. The Kb for benzene=

2.53 C/m. The boiling point of pure benzene is 80.1 degrees Celsius.

2 answers:

leonid [27]4 years ago

7 0

YEEEEEEEEEEEEEEET SAVEAGWES

Marta_Voda [28]4 years ago

5 0

Answer:

2.8 C/m

Explanation:

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What is the general formula for a secondary amine? rnh2 h2nrnh2 rn2h r2nh

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The general fomula of the amine group ought to be R2NH because, the two hydrogen atoms of ammonia have been substituted by alkyl groups.

<h3>What is a secondary amine?</h3>

A secondary amine is one in which two alkyl groups are subtituted on the nitrohen atom of ammonia.

This implies that the general fomula of the amine group ought to be R2NH. This is because, the two hydrogen atoms of ammonia have been substituted by alkyl groups.

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3 years ago

The diagram below shows the distance traveled by two light waves in 1 second.

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3 years ago

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We looked at the acrylonitrile molecule, c3h3n, in the previous exercise. Which of the following statements about the molecule i

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The true statemments that reflect the nature of the compound are statements (i) and (ii).

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2 years ago

On combustion, 1.0 L of a gaseous compound of hydrogen, carbon, and sulfur gives 2.0 L of CO2, 3.0 L of H2O vapor, and 1.0 L of

Murrr4er [49]

Answer:

The empirical formula of the organic compound is = $C_2H_6S_1$

Explanation:

At STP, 1 mole of gas occupies 22.4 L of volume.

Moles of $CO_2$ gas at STP occupying 2.0 L = n

$n\times 22.4L=2.0L$

$n=\frac{2.0 L}{22.4 L}=0.08929 mol$

Moles of carbon in 0.08920 mol = 1 × 0.08920 mol = 0.08920 mol

Moles of $H_2O$ gas at STP occupying 3.0 L = n'

$n'\times 22.4L=3.0L$

$n'=\frac{3.0 L}{22.4 L}=0.1339 mol$

Moles of hydrogen in 0.1339 moles of water vapor = 2 × 0.1339 mol = 0.2678 mol

Moles of $SO_2$ gas at STP occupying 1.0 L = n''

$n''\times 22.4L=1.0L$

$n''=\frac{1.0 L}{22.4 L}=0.04464 mol$

Moles of sulfur in 0.04464 mol = 1 × 0.04464 mol = 0.04464 mol

Moles of carbon , hydrogen and sulfur constituent of that organic compound .

Moles of carbon in 0.08920 mol = 1 × 0.08920 mol = 0.08920 mol

Moles of hydrogen in 0.1339 moles of water vapor = 2 × 0.1339 mol = 0.2678 mol

Moles of sulfur in 0.04464 mol = 1 × 0.04464 mol = 0.04464 mol

For empirical; formula divide the least number of moles from all the moles of elements.

carbon = $\frac{0.08920 mol}{0.04464 mol}=2$

Hydrogen = $\frac{0.2678 mol}{0.04464 mol}=6$

Sulfur = $\frac{0.04464 mol}{0.04464 mol}=1$

The empirical formula of the organic compound is = $C_2H_6S_1$

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3 years ago