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Ilya [14]
3 years ago
6

A gas in a sealed container had its volume increased from 12.1 liters to 21.1 liters.

Chemistry
1 answer:
ArbitrLikvidat [17]3 years ago
8 0

Answer:

The answer to your question is   P2 = 170.9 torr

Explanation:

Data

Volume 1 = 12.1 l                                Volume 2 = 21.1 l

Temperature 1 = 241 °K                      Temperature 2 = 298°K

Pressure 1 = 546 torr                           Pressure 2 = ?

Process

To solve this problem use the combined gas law.

                P1V1/T1 = P2V2/T2

-Solve for P2

                P2 = T1V1T2 / T1V2

-Substitution

                P2 = (241 x 12.1 x 298) / (241 x 21.1)

-Simplification

                P2 = 868997.8 / 5085.1

-Result

                P2 = 170.9 torr

       

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Answer:

Q = 28.9 kJ

Explanation:

Given that,

Mass of Aluminium, m = 460 g

Initial temperature, T_i=15^{\circ} C

Final temperature, T_f=85^{\circ}

We know that the specific heat of Aluminium is 0.9 J/g°C. The heat required to raise the temperature is given by :

Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\Q=460\ g\times 0.9\ J/g^{\circ} C\times (85-15)^{\circ} C\\\\Q=28980\ J\\\\\text{or}\\\\Q=28.9\ kJ

So, 28.9 kJ of heat is required to raise the temperature.

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The standard reduction potentials of lithium metal and chlorine gas are as follows:Reaction Reduction potential(V)Li+(aq)+e−→Li(
meriva

Answer:

A) E° = 4.40 V

B) ΔG° = -8.49 × 10⁵ J

Explanation:

Let's consider the following redox reaction.

2 Li(s) +Cl₂(g) → 2 Li⁺(aq) + 2 Cl⁻(aq)

We can write the corresponding half-reactions.

Cathode (reduction): Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq)      E°red = 1.36 V

Anode (oxidation):  2 Li(s) → 2 Li⁺(aq) + 2 e⁻         E°red = -3.04

<em>A) Calculate the cell potential of this reaction under standard reaction conditions.</em>

The standard cell potential (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.

E° = E°red, cat - E°red, an = 1.36 V - (-3.04 V) 4.40 V

<em>B) Calculate the free energy ΔG° of the reaction.</em>

We can calculate Gibbs free energy (ΔG°) using the following expression.

ΔG° = -n.F.E°

where,

n are the moles of electrons transferred

F is Faraday's constant

ΔG° = - 2 mol × (96468 J/V.mol) × 4.40 V = -8.49 × 10⁵ J

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