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vichka [17]
3 years ago
13

To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 2.2-L bulb, then filled

it with the gas at 2.00 atm and 27.0 ∘C and weighed it again. The difference in mass was 5.1 g.
Identify the gas. Express your answer as a chemical formula.
Chemistry
1 answer:
Ber [7]3 years ago
7 0

Answer:

N2

Explanation:

We use the ideal gas equation to calculate the number of moles of the diatomic gas. Then from the number of moles we can get

Given:

P = 2atm

1atm = 101,325pa

2atm = 202,650pa

T = 27 degrees Celsius = 27 + 273.15 = 300.15K

V = 2.2L

R = molar gas constant = 8314.46 L.Pa/molK

PV = nRT

Rearranging n = PV/RT

Substituting these values will yield:

n = (202,650 * 2.2)/(8314.46* 300.15)

n = 0.18 moles

To get the molar mass, we simply divide the mass by the number of moles.

5.1/0.18 = 28.5g/mol

This is the closest to the molar mass of diatomic nitrogen N2.

Hence, the gas is nitrogen gas

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A gas containing 80% CH4 and 20% He is sent through a quart/ diffusion tube (see Figure P8.8) to recover the helium, Twenty perc
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For a gas containing 80% CH4 and 20% He is sent through a quart diffusion tube, the composition is mathematically given as

%He=12.5%

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<h3>What is the composition of the waste gas if 100 kg moles of gas are processed per minute?</h3>

Generally, the equation for the Material balance  is mathematically given as

F=R+W

Therefore

100=0.20*1000+W

W=80kmol/min

In conclusion, waste gas compose

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Hence

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brainly.com/question/16416932

3 0
2 years ago
The half life of oxygen is 2 minutes. What fraction of a sample of 0.15 will remain after 5 half lives?​
Natasha_Volkova [10]

Answer:

3.13%.

Explanation:

The following data were obtained from the question:

Original amount (N₀) = 0.15

Half life (t½) = 2 mins

Number of half-life (n) = 5

Fraction of sample remaining =.?

Next, we shall determine the amount remaining (N) after 5 half-life. This can be obtained as follow:

Amount remaining (N) = 1/2ⁿ × original amount (N₀)

NOTE: n is the number of half-life.

N = 1/2ⁿ × N₀

N = 1/2⁵ × 0.15

N = 1/32 × 0.15

N = 0.15/32

N = 4.69×10¯³

Therefore, 4.69×10¯³ is remaining after 5 half-life.

Finally, we shall the fraction of the sample remaining after 5 half-life as follow:

Original amount (N₀) = 0.15

Amount remaining (N) = 4.69×10¯³

Fraction remaining = N/N₀ × 100

Fraction remaining = 4.69×10¯³/0.15 × 100

Fraction remaining = 3.13%

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The answer is c

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