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zubka84 [21]
3 years ago
13

A sample of Kr gas is observed to effuse through a pourous barrier in 8.15 minutes. Under the same conditions, the same number o

f moles of an unknown gas requires 4.53 minutes to effuse through the same barrier. The molar mass of the unknown gas is ____________ g/mol.
Chemistry
1 answer:
Ksju [112]3 years ago
7 0

Answer:

25.88 g/mol

Explanation:

Graham's law is a famous law which states that the diffusion rate or the effusion rate of any gas varies inversely to the square root of the molecular weight the gas.

So from Graham's law, we have,

$\frac{\text{time}}{M^{1/2}}=\text{constant}$

Using the sample of Kr gas having M = 83.8

$\frac{8.15}{(83.8)^{0.5}}= \frac{4.53}{M^{0.5}}$

$M^{0.5}= 5.088$

M = 25.88 g/mol

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The chemical reaction would be written as:

2HgO = 2Hg + O2

We use this reaction and the amount of the reactant to calculate for the moles of oxygen produced. THen, we use avogadro's number to convert it to molecules. We do as follows:

12.5 g HgO (1 mol / 216.59 g) (1 mol O2 / 2 mol HgO) ( 6.022x10^23 molecules O2 / 1 mol O2 ) = 1.74x10^22 molecules O2 produced
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3 years ago
A car can accelerate from a standstill to 100 km/h [E] in 9.60s. calculate the average acceleration
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Answer:

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Explanation:

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Final velocity = 100 km/h

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1 km = 1000 m

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1km/hr=\frac{5}{18}

100km/hr=\frac{5}{18}\times 100

v=\frac{5}{18}\times 100

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u = 0 m/s

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a=\frac{27.78-0}{9.6}

a=\frac{27.78}{9.6}

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