What is the volume of the container needed to store 0.8 moles of argon gas at 5.3 atm and 227°C?. . (Given: R = 0.08205 l · atm/mol · K). . 2.81 liters. . 4.39 liters. . 6.19 liters. . 9.67 liters

Answers

Answer 1

Answer: T = 227 ° C + 273 = 500 K
V = R n T/p = (0.08205 l atm/mol K * 0.8 mol* 500 K) / 5.3 atm = 6.19 lit
Answer: C ) 6.19 liters

Answer 2

Answer:

Answer:

6.19 L

Step-by-step explanation:

We are given that

Number of moles of argon gas=0.8 moles

Pressure =P=5.3 atm

Temperature =T=227 degree Celsius= $227+273=500k$

We have to find the volume of the container needed to store the argon gas.

We know that

$PV=nRT$

Where R=0.082051 L atm/ mol k

Substitute the values in the given formula

$5.3* V=0.8* 500* 0.82051$

$V=(0.8* 0.082051* 500)/(5.3)=6.19 L$

Hence, the volume of container needed to store the gas=6.19 L

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Answers

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Answers

Answer:

MY=80

Step-by-step explanation:

MX=MY ( since M is the midpoint)

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Answers

Answer:

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Step-by-step explanation:

Answer:

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Welcome

LC)Variable c is 9 more than variable w. Variable c is also 3 less than variable w. Which pair of equations best models the relationship between c and w? (5 points) c = w – 9
c = 4w + 3
c = w + 9
c = w – 3
w = c + 9
w = c – 3
w = c – 9
w = 4c + 3