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nikklg [1K]
3 years ago
14

If you have 177 grams of C4H80, how many moles do you have?

Chemistry
1 answer:
quester [9]3 years ago
4 0
I think it’s 2.45, not 100% sure
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Oxidation number ofAl2S12O7
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-2? if wrong im sorry please tell me what I did that was incorrect ty  Explanation:

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An unknown gas Q requires 2.67 times as long to effuse under the same conditions as the same amount of nitrogen gas. What is the
Misha Larkins [42]

Answer:

The correct answer is 199.66 grams per mole.

Explanation:

Based on law of effusion given by Graham, a gas rate of effusion is contrariwise proportionate to the square root of molecular mass, that is, rate of effusion of gas is inversely proportional to the square root of mass. Therefore,  

R1/R2 = √ M2/√ M1

Here rate is the rate of effusion of the gas expressed in terms of number of mole per uni time or volume, and M is the molecular mass of the gas.  

Rate Q/Rate N2 = √M of N2/ √M of Q

The molecular mass of N2 or nitrogen gas is 28 grams per mole and M of Q is molecular mass of Q and based on the question Q needs 2.67 times more to effuse in comparison to nitrogen gas, therefore, rate of Q = rate of N2/2.67

Now putting the values we get,  

rate of N2/2.67/rate of N2 = √28/ √M of Q

√M of Q = √ 28 × 2.67

M of Q = (√ 28 × 2.67)²

M of Q = 199.66 grams per mole

3 0
3 years ago
What is the preasure in atmospheres of 20 mol of nitrogen gas in 36.2 L cylinder at 25 degrees C?
Savatey [412]

Answer:

P = 13.5 atm

Explanation:

Given that

No. of moles, n = 20 moles

Volume of nitrogen gas = 36.2 L

Temperature = 25°C = 298 K

We need to find the pressure of the gas. Using the ideal gas equation

PV = nRT

Where

R is gas constant, R=0.082057\ L-atm/K-mol

So,

P=\dfrac{nRT}{V}\\\\P=\dfrac{20\times 0.082057\times 298}{36.2 }\\\\P=13.5\ atm

so, the pressure of the gas is equal to 13.5 atm.

7 0
3 years ago
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