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Ostrovityanka [42]
3 years ago
11

Which statement best explains why different gases effuse at different rates?

Chemistry
2 answers:
arlik [135]3 years ago
6 0
Your answer is A. <span>At a given pressure, gas particles with smaller masses move faster.

Hope this helps.

</span>
matrenka [14]3 years ago
6 0

Answer: Different gases effuse at different rates because at a given temperature, gas particles with smaller masses move faster.

Explanation: The motion of particles in a gas depends on the kinetic energy of the gas particles which further depend on the temperature of the gas. Thus, the rate of gas given by Graham's law can only be stated when the two gases are at fixed temperature.

Graham's Law: This law states that the rate of effusion of gas is inversely proportional to the square root of its mass.

Mathematically,

\frac{(Rate)_1}{(Rate)_2}=\sqrt{\frac{M_2}{M_1}}

From the above relation, the gas particle with smaller mass will effuse faster at a given temperature.


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It tells us more about how how bigger or smaller the concentration of products is to that of the reactants when a reaction attains equilibrium. From the given data, as the color of the reactant mixture (Br2 is reddish-brown, and H2 is colourless) fades, more of the colorless product (HBr is colorless) is being formed as the reaction approaches equilibrium. This indicates yhat the concentration of products becomes relatively higher than that of the reactants as the reaction progresses towards equilibrium, the equilibrium constant K, must be greater than 1 therefore.

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Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an at
valina [46]

Answer:

Percentage abundance of 121 Sb is = 57.2 %

Percentage abundance of 123 Sb is = 42.8 %

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 121 Sb :

% = x %

Mass = 120.9038 u

For second isotope, 123 Sb:

% = 100  - x  

Mass = 122.9042 u

Given, Average Mass = 121.7601 u

Thus,  

121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042

120.9038x+122.9042\left(100-x\right)=12176.01

Solving for x, we get that:

x = 57.2 %

<u>Thus, percentage abundance of 121 Sb is = 57.2 % </u>

<u>percentage abundance of 123 Sb is = 100 - 57.2 %  = 42.8 %</u>

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3 years ago
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