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zavuch27 [327]
3 years ago
15

How many moles of a gas sample are in a 20.0 L container at 373 K and 203 kPa? The gas constant is 8.31 L-kPa/mol-K. 0.33 moles

3.05 moles 1.30 moles 0.65 moles
Chemistry
2 answers:
Gennadij [26K]3 years ago
8 0
We need to use the ideal gas formula--> PV= nRT 
P, pressure= 203 kPa
V, volume= 20.0 Liters
n, moles= ?
R, constant= <span>8.31 L-kPa/mol-K
T, temperature= 373 K

now, lets plug these values into the formula

(203)(20.0)= (n)(8.31)(373)

n= 1.31 moles</span>
Degger [83]3 years ago
3 0

<u>Answer:</u> The number of moles of the gas is 1.31 mol.

<u>Explanation:</u>

To calculate the number of moles of gas, we use the equation given by ideal gas, which is:

PV=nRT

where,

P = pressure of the gas = 203 kPa

V = volume of the gas = 20 L

n = Number of moles of gas = ? mol

R = Gas constant = 8.31\text{L kPa }mol^{-1}K^{-1}

T = temperature of the gas = 373 K

Putting values in above equation, we get:

203kPa\times 20L=n\times 8.31\text{L kPa }\times 373K\\\\n=1.31mol

Hence, the number of moles of the gas is 1.31 mol.

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Planck's constant = 6.626×10^-34 Js

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2 years ago
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3 0
2 years ago
Hello, everyone!
marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the <em>law of ideal gases,</em>  

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x \frac{1 g}{1000 mg} x \frac{1 mol}{28.01 g}

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V =  \frac{0.00125 mol x 0.082057 \frac{atm L}{mol K}  x 243 K}{0.92 atm}

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x \frac{1000 cm^{3} }{1 L} = 27.09 cm³

<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>

c = 27 cm³ / m³ = 27 ppm

<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 \frac{cm^{3} }{m^{3} } x \frac{1 m^{3} }{1 000 000 cm^{3} } x 100%

c = 0.003 %

So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.

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Answer:

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