Chemistry: O2(g) effuses at a rate that is times that of br2(g) under the same conditions.

O2(g) effuses at a rate that is times that of br2(g)

Ответ:

dianactorres

15.03.2022

Chemistry

Oxygen gas effuses at a rate that is 2.236 times that of bromine gas under the same conditions.

Explanation:

Graham's Law:

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

We are given:

Molar mass of oxygen gas= 32 g/mol

Molar mass of bromine gas = 160 g/mol

By taking their ratio, we get:

$\frac{r_{O_2}}{r_{Br_2}}=\sqrt{\frac{160 g/mol}{32 g/mol}}$

$\frac{r_{O_2}}{r_{Br_2}}=\sqrt{5}=2.236$

$r_{O_2}=2.236\times r_{Br_2}$

Oxygen gas effuses at a rate that is 2.236 times that of bromine gas under the same conditions.

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Ответ:

goldenarrow

17.07.2021

Chemistry

5 gallons

Step-by-step explanation:

$\dfrac{5/12\ gallons}{1/24\ minutes}=\dfrac{5}{12}\times \dfrac{24}{1}=\dfrac{10\ gallons}{1\ minute}\\\\\\\dfrac{10\ gallons}{1\ minute}=\dfrac{x}{1/2\ minute}\\\\\\5\ gallons = x$

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O2(g) effuses at a rate that is times that of br2(g) under the same conditions.

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