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lianna [129]
3 years ago
9

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
Mathematics
2 answers:
Alona [7]3 years ago
8 0
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
Igoryamba3 years ago
3 0

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\times V=0.8\times 500\times 0.82051

V=\frac{0.8\times 0.082051\times 500}{5.3}=6.19 L

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

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SVETLANKA909090 [29]

Answer:

a. 72

b. 816

c. 570

d. 30

Step-by-step explanation:

Given the graph is a bell - shaped curve. So, we understand that this is a normal distribution and that the bell - shaped curve is a symmetric curve.

Please refer the figure for a better understanding.

a. In a normal distribution, Mean = Median = Mode

Therefore, Median = Mean = 72

b. We have to know that 68% of the values are within the first standard deviation of the mean.

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This implies scores between 63 and 81 constitute 68% of the values, 34% each, since the curve is symmetric.

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c. We have to know that 95% of the values lie between second Standard Deviation of the mean.

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Scores between 54 and 90 totally constitute 95% of the values. So, Scores between 72 and 90 should amount to $ \frac{95}{2} \% $ of the values.

Therefore, Scores between 72 and 90 = $ \frac{95}{2(100)} \times 1200 = \frac{95}{200} \times 1200  $

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That is a total of 570 students scored between 72 and 90.

d. We have to know that 5 % of the values lie on the thirst standard Deviation of the mean.

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Since, we are asked to find scores below 54. It should be 2.5% of the values.

So, Scores below 54 = $ \frac{2.5}{100} \times 1200 $

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That is, 30 students have scored below 54.

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