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kifflom [539]
3 years ago
9

An adult human lung has a volume of 2.9 liters. If its volume decreases to 1.2 liters and it contains 0.049 mole of air after it

exhales, how many moles of air does it contain when full?
Chemistry
1 answer:
Lera25 [3.4K]3 years ago
6 0

Answer:

In full volume it contain 0.12 moles.

Explanation:

Given data:

Total volume= Vt = 2.9 L

Decreased volume= Vd = 1.2 L

Number of moles of air present in decreased volume= n = 0.049 mol

Number of moles of air in total volume= n = ?

Solution:

Formula:

Vt/ Vd = n (in total volume) /n ( decreased volume)

2.9 L / 1.2 L = X / 0.049 mol

2.42 = X / 0.049 mol

X = 2.42 × 0.049

X = 0.12 mol

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Read 2 more answers
A 25.0 L tank of nitrogen gas is at 25 oC and 2.05 atm . If the temperature stays at 25 oC and the volume is decreased to 14.5 L
NeTakaya

Answer:

\boxed {\boxed {\sf P_2 \approx 3.53 \ atm}}

Explanation:

In this problem, the temperature stays constant. The volume and pressure change, so we use Boyle's Law. This states that the pressure of a gas is inversely proportional to the volume. The formula is:

P_1V_1=P_2V_2

Now we can substitute any known values into the formula.

Originally, the gas has a volume of 25.0 liters and a pressure of 2.05 atmospheres.

25.0 \ L * 2.05 \ atm = P_2V_2

The volume is decreased to 14.5 liters, but the pressure is unknown.

25.0 \ L * 2.05 \ atm = P_2 * 14.5 \ L

Since we are solving for the new pressure, or P₂, we must isolate the variable. It is being multiplied by 14.5 liters and the inverse of multiplication is division. Divide both sides by 14.5 L .

\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}=\frac{P_2 *14.5 \ L}{14.5 \ L}

\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}= P_2

The units of liters cancel.

\frac {25.0  * 2.05 \ atm }{14.5 }=P_2

\frac {50.25\  atm }{14.5 }=P_2

3.53448276 \ atm = P_2

The original values of volume and pressure have 3 significant figures, so our answer must have the same.

For the number we found, that is the hundredth place.

  • 3.53448276

The 4 in the thousandth place (in bold above) tells us to leave the 3 in the hundredth place.

3.53 \ atm \approx P_2

The new pressure is approximately <u>3.53 atmospheres.</u>

8 0
3 years ago
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