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alexgriva [62]
3 years ago
13

The density of nitric oxide (NO) gas at 0.866 atm and 46.2 degrees celcius is ____ g/L

Chemistry
1 answer:
Artyom0805 [142]3 years ago
8 0

Answer:

 d = 0.992 g/L

Explanation:

Data Given:

Pressure of nitric oxide (NO) = 0.866 atm

Temperature of a gas = 46.2° C

Convert the temperature to kelvin = 46.2° C + 273  

temperature in kelvin = 319.2 K

density of nitric oxide (NO) = ?

Solution:

Density of a gas can be calculated by

                       d = PM /RT

Where

d = density

P = Pressure

M = molar mass of gas

R = ideal gas constant  = 0.0821 L atm mol⁻¹ K⁻¹

T = temperature

So,

Molar mass of NO = 30 g/mol

Put values in the formula:

                        d = PM /RT

           d = 0.866 atm × 30 g/mol / 0.0821 L atm mol⁻¹ K⁻¹ × 319.2 K

           d = 25.98 atm. g/mol / 26.2 L atm mol⁻¹

           d = 0.992 g/L

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A solution is made by mixing 55.g of thiophene C4H4S and 65.g of acetyl bromide CH3COBr.
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Answer:

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Explanation:

Mole fraction is moles of solute / Total moles

Total moles are the sum of moles of solute + moles of solvent.

Let's find out the moles of our solute and our solvent.

Mass of solute: 55g

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Mol = Mass / molar mass

55 g / 84.06 g/mol = 0.654 moles of C₄H₄S

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A sample of nitrogen gas is at a temperature of 50 c and a pressure of 2 atm. If the volume of the sample remains constant and t
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Answer:

The new temperature of the nitrogen gas is 516.8 K or 243.8 C.

Explanation:

Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.

Gay-Lussac's law can be expressed mathematically as follows:

\frac{P}{T} =k

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You want to study two different states, an initial state and a final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. By varying the temperature to a new value T2, then the pressure will change to P2, and the following will be fulfilled:

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In this case:

  • P1= 2 atm
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Replacing:

\frac{2 atm}{323 K} =\frac{3.2 atm}{T2}

Solving:

T2*\frac{2 atm}{323 K} =3.2 atm

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T2= 516.8 K= 243.8 C

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