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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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2MnCO_3(s)+O_2(g)\rightarrow 2MnO_2(s)+2CO_2(g)

In the second step, manganese (IV) oxide and aluminum react to form manganese and aluminum oxide:

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Explanation :

The given two chemical reactions are:

(1) 2MnCO_3(s)+O_2(g)\rightarrow 2MnO_2(s)+2CO_2(g)

(2) 3MnO_2(s)+4Al(s)\rightarrow 3Mn(s)+2Al_2O_3(s)

First we are multiplying reaction 1 by 3, and reaction 2 by 2, we get:

(1)

(2) 6MnO_2(s)+8Al(s)\rightarrow 6Mn(s)+4Al_2O_3(s)

Now we are adding both the reactions, we get the overall chemical reaction.

6MnCO_3(s)+3O_2(g)+6MnO_2(s)+8Al(s)\rightarrow 6MnO_2(s)+6CO_2(g)+6Mn(s)+4Al_2O_3(s)

The  MnO_2 is common on both side, by cancelling it, we get:

The net chemical equation for the production of manganese is:

6MnCO_3(s)+3O_2(g)+8Al(s)\rightarrow 6CO_2(g)+6Mn(s)+4Al_2O_3(s)

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