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Mrac [35]
2 years ago
5

An unknown gas effuses 2. 165 times faster than xe. what is the molar mass of the gas? show the set up and answer with unit

Chemistry
1 answer:
enot [183]2 years ago
8 0

27.94 g/mole is the molar mass of the unknown gas.

Use Graham's Law, which indicates that the square root of a gas's molar mass determines its rate of effusion.

The following equation compares two distinct gases, (a) and (b), and expresses this:

sqrt(M(b)/M) = R(a)/R(b) (a)

R is the effusion rate.

square root = sqrt

the molar mass, M

In response to your query, if gas (a) is the unknown gas and Xe(M = 131 g/mol) is the second gas, we have:

sqrt(M(Xe)/M(a) = 2.165, where R(a)/R(Xe)

Square both sides, and you get:

(2.165)² = 131 g/mol/ M (a)

M(a)(2.165)² = 131

M(a) is equal to 131/(2.165)² = 27.94 g/mol.

Having a molar mass of 27.94 g/mol, the unknown gas.

Learn more about Graham's law of effusion here brainly.com/question/22359712

#SPJ4

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The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol. How much faster is the decomposition at 625°C th
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Answer:

The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_2 = rate of reaction at T_2

K_1 = rate of reaction at T_1

Ea = activation energy of the reaction

R = gas constant = 8.314 J/K mol

E_a=300 kJ/mol=300,000 J/mol

T_2=625^oC=898.15 K,T_1=525^oC=798.15 K

\log (\frac{K_2}{K_1})=\frac{300,000 J/mol}{2.303\times 8.314 J/K mol}[\frac{1}{798.15 K}-\frac{1}{898.15 K}]

\log (\frac{K_2}{K_1})=2.185666

K_2=153.344\times K_1

The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.

4 0
3 years ago
If 61.5 moles of an ideal gas occupies 97.5 liters at 473 K, what is the pressure of the gas, in mmHg?
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2. How many grams of water can be heated from 20.0 oC to 75oC using 12500.0 Joules?
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Answer;

  = 0.054 kg or 54 g

Explanation;

Using the equation; Q = mcΔT where Q is the quantity of heat transferred, m is the mass, c is specific heat of the substance, ΔT is delta T, the change in temperature.  

ΔT = 75 - 20 = 55 C.  

Solve the equation for m  

m = Q/ cΔT

Mass = 12500 / (55 × 4200)

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