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avanturin [10]
3 years ago
11

The volume of air in a person’s lungs is 615 mL at a pressure of 760. mmHg. Inhalation occurs as the pressure in the lungs drops

to 752 mmHg with no change in temperature and amount of gas. To what volume, in milliliters, did the langs expand?
Chemistry
1 answer:
chubhunter [2.5K]3 years ago
8 0

<u>Answer:</u> The final volume of lungs is 621.5 mL

<u>Explanation:</u>

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.

The equation given by this law is:

P_1V_1=P_2V_2

where,

P_1\text{ and }V_1 are initial pressure and volume.

P_2\text{ and }V_2 are final pressure and volume.

We are given:

P_1=760mmHg\\V_1=615mL\\P_2=752mmHg\\V_2=?mL

Putting values in above equation, we get:

760mmHg\times 615mL=752mmHg\times V_2\\\\V_2=\frac{760\times 615}{752}=621.5mL

Hence, the final volume of lungs is 621.5 mL

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2 years ago
How many moles of NaCl are produced if 239.7 grams of Na2S reacts with plenty of AlCl3?
lana66690 [7]

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6.142 moles of NaCl

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We'll begin by writing the balanced equation for the reaction. This is given below:

2AlCl3 + 3Na2S —> Al2S3 + 6NaCl

Next, we determine the number of mole in 239.7 g of Na2S. This is illustrated below:

Mass mass of Na2S = 78.048g/mol

Mass of Na2S = 239.7g

Number of mole Na2S =..?

Mole = Mass /Molar Mass

Number of mole Na2S = 239.7/78.048 = 3.071 moles

Finally, we can obtain the number of mole of NaCl produced from the reaction as follow:

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3 moles of Na2S reacted to produce 6 moles of NaCl.

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Substitute the expression for K:

\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}

Now we may use the values given to solve:

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7 0
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