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Luda [366]
3 years ago
7

If it takes three "breaths" to blow up a balloon to 1.2 LL, and each breath supplies the balloon with 0.060 moles of exhaled air

, how many moles of air are in a 3.0 LL balloon?
Chemistry
1 answer:
lara31 [8.8K]3 years ago
4 0

Answer:

0.45 moles of air are in a 3.0 L balloon

Explanation:

The given problem can be solved by using unitary method.

Firstly, we have to find out how many breaths are equivalent to 3.0 L. Secondly, we have to find out how many moles of air are supplied corresponding to the calculated number of breaths.

1.2 L = 3 breaths

So, 3.0 L = \frac{3}{1.2}\times 3.0 breaths = 7.5 breaths

1 breath supplies 0.060 moles of air

So, 7.5 breaths supplies (7.5\times 0.060) moles of air or 0.45 moles of air.

Hence, 0.45 moles of air are in a 3.0 L balloon.

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Answer:

5.54 m³.

Explanation:

  • We should use the ideal gas law: <em>PV = nRT,</em>

where, P is the pressure of the gas in atm,

V is the volume of the gas in L,

n is the number of moles in mole,

R is the general gas constant (R = 0.082 L.atm/mol),

T is the temperature of the gas in K.

<em>We have two different cases at different (P, V and T) while the number of moles of He and R is constants.</em>

<em>∴ (P₁V₁) / (T₁) = (P₂V₂) / (T₂).</em>

We can use P in KPa and V in m³ that the conversion factor can be canceled by division, but we should use T in K because its conversion factor is additive value.

P₁ = 101.0 kPa, V₁ = <em>??? m³</em>, and T₁ = (20 °C + 273) = 293.0 K.

P₂ = 11.7 kPa, V₂ = 35.4 m³, and T₂ = (-56 °C + 273) = 217 K.

∴ The initial volume (V₁) = (P₂V₂T₁) / (P₁T₂) = (11.7 kPa)(35.4 m³)(293.0 K) / (101.0 kPa)(217.0 K) = (121354.74) / (21917) = 5.537 m³ ≅ 5.54 m³.

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