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stepan [7]
3 years ago
7

a sample of gas with a volume of 30.0 ml at 25.0 °c is heated to 50.0 degrees celsius. what is the new volume of the gas?

Chemistry
1 answer:
mart [117]3 years ago
4 0

Answer : The new volume of the gas is, 32.5 mL

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number o moles of gas.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=30.0mL\\T_1=25.0^oC=(25.0+273)K=298K\\V_2=?\\T_2=50.0^oC=(50.0+273)K=323K

Putting values in above equation, we get:

\frac{30.0mL}{298K}=\frac{V_2}{323K}\\\\V_2=32.5mL

Therefore, the new volume of the gas is, 32.5 mL

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Sample means for solutions 1 and 2 are 19.27 and 10.32 respectively

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The total for answer 1 is given by:

9.7+10.5+9.4+10.6+9.3+10.7+9.6+10.4+10.2+10.5 = 192.7

The sample size is 10 and provides us with

192.7/10 = 19.27

For solution 2, the sum is given by:

10.6+10.3+10.3+10.2+10.0+10.7+10.3+10.4+10.1+10.3 = 103.2

The sample size is 10, this gives us

103.2/10 = 10.32

The total for answer 2 is given by:

10.6+10.3+10.3+10.2+10.0+10.7+10.3+10.4+10.1+10.3 = 103.2

The sample size is 10 and provides us with

103.2/10 = 10.32

Learn more about semiconductor manufacturing here brainly.com/question/22779437

#SPJ4.

In semiconductor manufacturing, wet chemical etching is often used to remove silicon from the backs of wafers prior to metalization. The etch rate is an important characteristic in this process and is known to follow a normal distribution. Two different etching solutions have been compared, using two random samples of 10 wafers for each solution. Assume the variances are equal. The etch rates are as follows (in mils per minute): Solution 1 Solution 2 9.7 10.6 10.5 10.3 9.4 10.3 10.6 10.2 9.3 10.0 10.7 10.7 9.6 10.3 10.4 10.4 10.2 10.1 10.5 10.3 Calculate sample means of solution 1 and solution 2

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