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3 years ago

What is the mass of a sample of sulfur with a volume of 5.0 cm3

Chemistry

1 answer:

qwelly [4]3 years ago

8 0

Answer:

0.00714 g

Explanation:

We know that the volume of one mole of a substance is 22400 cm^3

Thus;

1 mole of sulphur occupies 22400 cm^3

x moles of sulphur occupies 5.0 cm3

x = 5.0 cm3 * 1 mole/22400 cm^3

x = 2.23 * 10^-4 moles

Molar mass of sulphur = 32 g/mol

Mass of sulphur = 2.23 * 10^-4 moles * 32 g/mol

Mass of sulphur = 0.00714 g

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omeli [17]

Answer : The rate law for formation of NOBr based on this mechanism is, $\frac{k_1\times k_2}{k_1^-}[Br_2][NO]^2$

Explanation :

The overall reaction is:

$Br_2(g)+2NO(g)\rightleftharpoons 2NOBr(g)$

Rate law = $k[Br_2][NO]^2$

The first step of the overall reaction is:

$NO(g)+Br_2(g)\overset{k_1}{\rightarrow} NOBr_2(g)$

$NO(g)+Br_2(g)\overset{k_1^-}{\leftarrow} NOBr_2(g)$

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Rate law 2 = $k_1^-[NOBr_2]$

The second step of the overall reaction is:

$NOBr_2(g)+NO(g)\overset{k_2}{\rightarrow} 2NOBr$

Rate law 3 = $k_2[NOBr_2][NO]$

Now rate law of overall reaction can be obtained as follows.

We are multiplying rate law 1 and rate law 3 and dividing by rate law 2, we get:

Rate law = $\frac{[k_1[Br_2][NO]]\times [k_2[NOBr_2][NO]]}{[k_1^-[NOBr_2]]}$

Rate law = $\frac{k_1\times k_2}{k_1^-}[Br_2][NO]^2$

Rate law = $k[Br_2][NO]^2$

The rate law for formation of NOBr based on this mechanism is, $\frac{k_1\times k_2}{k_1^-}[Br_2][NO]^2$

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jolli1 [7]

Answer:

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Explanation:

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