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RideAnS [48]
3 years ago
10

Of the following gases which will have the lowest rate of effusion at a given temperature?A. SO3B. CH4C. NH3D. HBr E. HCl

Chemistry
1 answer:
egoroff_w [7]3 years ago
6 0

Answer: HBr has the lowest rate of effusion at a given temperature.

Explanation: The effusion rate usually increases with increase in temperature because the kinetic energy of the gaseous molecules increases. But it was not true for gases having heavier mass. This was explained by Graham's Law.

Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molecular weight.

(\text{Rate of effusion})_A\propto \frac{1}{\sqrt{\text{Mol. mass}_A}}

We are given different gases with different Molecular masses. The gas having larger Molecular mass will have the lowest rate of effusion.

Mol. Mass of SO_3 = 80 g/mol

Mol. Mass of CH_4 = 16 g/mol

Mol. Mass of NH_3 = 17 g/mol

Mol. Mass of HBr = 81 g/mol

Mol. Mass of HCl = 36 g/mol

As, Mol. mass of HBr is the highest, so its rate of effusion will be the lowest.

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If a 0.320 mM solution of MnO41- has an absorbance of 0.480 at 525 nm in a 1.000 cm cell. What is the concentration of a MnO41-
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Hence the concentration of a MnO41- solution that has absorbance of 0.490 in the same cell at that wavelength is 0.3266.

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A2 / A1 = C2/ C1\\\\0.49/ 0.48 = C2 / 0.32\\\\C2 = 0.3266

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3 years ago
Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =
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<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{c}

For a general chemical reaction:

aA+bB\rightleftharpoons cC+dD

The expression for K_{eq} is written as:

K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)

The expression of K_c for above equation is:

K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}

We are given:

[H_2S]_{eq}=0.671M

[O_2]_{eq}=0.587M

K_c=1.35

Putting values in above expression, we get:

1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}

[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M

Hence, the equilibrium concentration of water is 0.597 M

8 0
3 years ago
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