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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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I think it should be Ca and CI

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The molar mass of vitamin A (C20H30O).
vivado [14]
Molar mass of C:  12.011 g/mol
The equation says C20, which means there are 20 carbon atoms in each molecule of Vitamin A.  So, we multiply 12.011 by 20 to get 240.22 g/mol carbon.

Molar mass of H:  1.0079 g/mol
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Then, just add it up:
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4 0
3 years ago
PLEASE HELP!!!!!!!
dsp73

Answer: 1. 2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu

2. 3 moles of CuCl_2 : 2 moles of Al

3. 0.33 moles of CuCl_2 : 0.92 moles of Al

4. CuCl_2 is the limiting reagent and Al is the excess reagent.

5. Theoretical yield of AlCl_3 is 29.3 g

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Al=\frac{25.0g}{27g/mol}=0.92moles

\text{Moles of} CuCl_2=\frac{45.0g}{134g/mol}=0.33moles

The balanced chemical equation is:

2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu  

According to stoichiometry :

3 moles of CuCl_2 require = 2 moles of Al

Thus 0.33 moles of CuCl_2 will require=\frac{2}{3}\times 0.33=0.22moles  of Al

Thus CuCl_2 is the limiting reagent as it limits the formation of product and Al is the excess reagent.

As 3 moles of CuCl_2 give = 2 moles of AlCl_3

Thus 0.33 moles of CuCl_2 give =\frac{2}{3}\times 0.33=0.22moles  of AlCl_3

Theoretical yield of AlCl_3=moles\times {\text {Molar mass}}=0.22moles\times 133.34g/mol=29.3

Thus 29.3 g of aluminium chloride is formed.

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If it take 20 second for 200cm3 of H25 to diffuse through a porovs pot,hours,long will it take 400cm3 of NH3 of diffuse under th
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