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MA_775_DIABLO [31]
3 years ago
5

What is the volume, in liters, of 1.40 mol of oxygen gas at 20.0°C and 0.974 atm?

Chemistry
1 answer:
topjm [15]3 years ago
4 0

Answer:

V = 34.55 L

Explanation:

Given that,

No of moles, n = 1.4

Temperature, T = 20°C = 20 + 273 = 293 K

Pressure, P = 0.974 atm

We need to find the volume of the gas. It can be calculated using Ideal gas equation which is :

PV=nRT

R is gas constant, R=0.08206\ L-atm/mol-K

Finding for V,

V=\dfrac{nRT}{P}\\\\V=\dfrac{1.4\times 0.08206\times 293}{0.974 }\\\\V=34.55\ L

So, the volume of the gas is 34.55 L.

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What is the maximum number of grams of N-acetyl-p-toluidine can be prepared from 70. milliliters of 0.167 M p-toluidine hydrochl
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<u>Answer:</u> The maximum amount of N-acetyl-p-toluidine that can be prepared is 1.7 grams.

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0.167M=\frac{\text{Moles of p-toluidine hydrochloride}\times 1000}{70}\\\\\text{Moles of p-toluidine hydrochloride}=\frac{0.167\times 70}{1000}=0.0117mol

The chemical equation for the reaction of p-toluidine hydrochloride and acetic anhydride follows:

\text{p-toluidine hydrochloride}+\text{Acetic anhydride}\rightarrow \text{N-acetyl-p-toluidine}

By Stoichiometry of the reaction:

1 mole of p-toluidine hydrochloride produces 1 mole of N-acetyl-p-toluidine

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

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Moles of N-acetyl-p-toluidine = 0.0117 moles

Putting values in equation 1, we get:

0.0117mol=\frac{\text{Mass of N-acetyl-p-toluidine}}{149.2g/mol}\\\\\text{Mass of N-acetyl-p-toluidine}=(0.0117g/mol\times 149.2)=1.7g

Hence, the maximum amount of N-acetyl-p-toluidine that can be prepared is 1.7 grams.

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