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zubka84 [21]
2 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]2 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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Can someone how to do these problems!!
Novosadov [1.4K]
You can use the follwing equations.
pH=14-pOH
pOH=-log[OH⁻]

a)  pOH=-log10⁻¹²
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You can also use these equations:
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pH=-log[H⁺]

b) [H⁺]=(1.01×10⁻¹⁴)/10⁻²M
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You can use either method.  It does not really matter.
I hope this helps.  Let me know if anything is unclear and when you do the calculations for c you should get pH=7.
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