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beks73 [17]
2 years ago
13

A gas has a volume of 27. 5 L at 302 k and 1. 40 atm. How many moles are in the sample of gas?

Chemistry
1 answer:
e-lub [12.9K]2 years ago
8 0

There are  1.554 moles are in the sample of gas when a  gas has a volume of 27. 5 L at 302 K and 1. 40 atm.

Calculation,

The given question is solved by the using of ideal gas equation which is shown below.

PV = nRT                               .... ( i )

where , P is he pressure of the gas = 1. 40 atm

V is the volume occupied by the gas = 27.5 L

R is the universal gas constant = 0.082 L atm/K mol

T is the temperature =  302 K

n is he number of moles of the gas = ?

By putting the value of pressure  P , volume V , Universal gas constant and temperature T in the equation ( i ) we get the number of moles.

1. 40 atm × 27.5 L = n × 0.082 L atm/K mol×  302 K

n = 1. 40 atm × 27.5 L/0.082 L atm/K mol×  302 K = 1.554 mole

There are  1.554 moles are in the sample of gas

To learn more about moles

brainly.com/question/26416088

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The new pH is 7.69.

According to Hendersen Hasselbach equation;

The Henderson Hasselbalch equation is an approximate equation that shows the relationship between the pH or pOH of a solution and the pKa or pKb and the ratio of the concentrations of the dissociated chemical species. To calculate the pH of the buffer solution made by mixing salt and weak acid/base. It is used to calculate the pKa value. Prepare buffer solution of needed pH.

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Here, 100 mL  of  0.10 m TRIS buffer pH 8.3

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∴  8.3 = 8.3 + log \frac{[0.005]}{[0.005]}

<em>    </em>inverse log 0 = \frac{[B]}{[A]}

   \frac{[B]}{[A]} = 1

Given; 3.0 ml of 1.0 m hcl.

           pka = 8.3

           0.003 mol of HCL.

pH = 8.3 + log \frac{[0.005-0.003]}{[0.005+0.003]}\\pH = 8.3 + log \frac{[0.002]}{[0.008]}\\\\pH = 8.3 + log {0.25}\\\\pH = 8.3 + (-0.62)\\pH = 7.69

Therefore, the new pH is 7.69.

Learn more about pH here:

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