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Sedaia [141]
3 years ago
8

Only 5 questions! All you have to do is check my answers...1. A flask contains four gases: CH4, O2, C2H5, and N2. When the stopp

er is removed, which gas will diffuse the fastest?
CH4 <---
O2
C2H5
N2

2. Which gas effuses 2.39 times slower than nitrogen gas?

O2
Cl2
Br2 <---
I2

3. What is the molecular mass of a gas that effuses three times faster than radon?

16 g/mol
25 g/mol <---
50 g/mol
67 g/mol,
Chemistry
1 answer:
jarptica [38.1K]3 years ago
3 0

1. CH4
2. Br2
3. 25 g/mol

Graham's law indicates that the rate of diffusion is inversely proportional to the square root of its molecular weight. So the lighter the gas, the more rapid the diffusion. With that in mind, let's look at your questions:

1. A flask contains four gases: CH4, O2, C2H5, and N2. When the stopper is removed, which gas will diffuse the fastest?The answer here will be the lightest of the 4 gasses. So let's look at the molar mass of each gas:
CH4 = 16.04 g/mol
O2 = 32.00 g/mol
C2H5 = 29.06
N2 = 28.01
And CH4 is the lightest and therefore the fastest.

2. Which gas effuses 2.39 times slower than nitrogen gas? O2 Cl2 Br2 <--- I2
Let's start with the square root of 28.01 which is 5.292447449.
Now multiply by 2.39 and square the result to get the molar mass of the
desired gas which is (5.29*2.39)^2 = 12.64^2 = 159.77Now let's look at the molar masses.
O2 = 32
Cl2 = 70.91
Br2 = 159.81
I2 = 253.81
And the match goes to Br2

3. What is the molecular mass of a gas that effuses three times faster than radon? 16 g/mol 25 g/mol <--- 50 g/mol 67 g/mol,The molar mass of radon is 222 g/mol. So let's get the square root, divide by
3, then square the result. So(sqrt(222)/3)^2 = (14.89966443/3)^2 = 4.966554809^2 = 24.66666667 g/mol.And of the available choices, 25 g/mol is the best match.
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Determine the value of the equilibrium constant, Kgoal, for the reaction C(s)+12O2(g)+H2(g)⇌12CH3OH(g)+12CO(g), Kgoal=? by makin
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1.71x10²⁷

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If we sum 1/2 of (3) + 1/2 of (1):

1/2 (3.) C(s) + 1/2O₂(g) ⇌ CO(g), K₃ = √2.10×10⁴⁷  = 4.58x10²³

1/2 (1)   1/2CO₂(g) + 3/2H₂(g) ⇌ 1/2CH₃OH(g) + 1/2H₂O(g), K₁ = √1.40×10² = 11.8

C(s) + 1/2O₂(g) +<u> 1/2CO₂(g) </u>+<u> 3/2H₂(g</u>) ⇌ 1/2CH₃OH(g) + <u>1/2H₂O(g)</u> + <u>CO(g)</u>

K' = 4.58x10²³ * 11.8 = 5.42x10²⁴

+1/2 (2):

<u>1/2 CO(g)</u> +<u> 1/2H₂O(g)</u> ⇌<u> 1/2CO₂(g)</u> + <u>1/2H₂</u> (g), K = √1.00×10⁵ = 316.2

C(s) + 1/2O₂(g) + H₂(g) ⇌ 1/2 CHO₃H(g) + 1/2CO(g)

K'' = 5.42x10²⁴* 316.2 =

<h3>1.71x10²⁷</h3>

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