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dybincka [34]
4 years ago
7

A gas‑filled weather balloon has a volume of 50.0 L at ground level, where the pressure is 759 mmHg and the temperature is 22.3

∘ C. After being released, the balloon rises to an altitude where the temperature is − 9.00 ∘ C and the pressure is 0.0701 atm. What is the weather balloon's volume at the higher altitude?
Chemistry
1 answer:
OlgaM077 [116]4 years ago
7 0

<u>Answer:</u> The volume of balloon at higher altitude is 631.3 L

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law. The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1,V_1\text{ and }T_1 are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2 are the final pressure, volume and temperature of the gas

We are given:

<u>Conversion factor:</u>  1 atm = 760 mmHg

P_1=759mmHg=0.99atm\\V_1=50.0L\\T_1=22.3^oC=[273+22.3]K=295.3K\\P_2=0.0701atm\\V_2=?L\\T_2=-9.00^oC=[273-9]=264K

Putting values in above equation, we get:

\frac{0.99atm\times 50.0L}{295.3K}=\frac{0.0701atm\times V_2}{264K}\\\\V_2=\frac{0.99\times 50.0\times 264}{0.0701\times 295.3}=631.3L

Hence, the volume of balloon at higher altitude is 631.3 L

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1. Write the <em>chemical equation</em> for the reaction.

HNO₃ + KOH ⟶ KNO₃ + H₂O

===============

2. Calculate the <em>moles of HNO₃</em>

c = n/V                                               Multiply each side by V and transpose

n = Vc

V = 0.027 86 L

c = 0.1744 mol·L⁻¹                             Calculate the moles of HNO₃

Moles of HNO₃ = 0.027 86 × 0.1744

Moles of HNO₃ = 4.859 × 10⁻³ mol HNO₃

===============

3. Calculate the <em>moles of KOH </em>

1 mol KOH ≡ 1 mol HNO₃                 Calculate the moles of KOH

Moles of KOH = 4.859 × 10⁻³× 1/1

Moles of KOH = 4.859 × 10⁻³ mol KOH

===============

4. Calculate the <em>molar concentration</em> of the KOH

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