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2 years ago

9

A student has a balloon with a volume of 2.5 liters that contains 4.0 moles of air. The ballon has a small leak, allowing one mo

le to escape, leaving just 3.0 moles of air inside the balloon. What is the balloon's new volume after the air has escaped?

1 answer:

Over [174]2 years ago

6 0

Answer:

1.9 L

Explanation:

Step 1: Given data

Initial number of moles of air (n₁): 4.0 mol

Initial volume of the balloon (V₁): 2.5 L

Final number of moles of air (n₂): 3.0 mol

Final volume of the balloon (V₂): ?

Step 2: Calculate the final volume of the balloon

According to Avogadro's law, the volume of an ideal gas is directly proportional to the number of moles. We can calculate the final volume of the balloon using the following expression.

V₁ / n₁ = V₂ / n₂

V₂ = V₁ × n₂ / n₁

V₂ = 2.5 L × 3.0 mol / 4.0 mol

V₂ = 1.9 L

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The given cell is:

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<u>Reduction half reaction:</u> $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

<u>Net reaction:</u> $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

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Putting values in above equation, we get:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

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$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

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nadya68 [22]

Answer:

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