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Lady_Fox [76]
3 years ago
12

A gas in a sealed container has a pressure of 125 atm at a temperature of 303 K. If the pressure in the container is increased t

o 200 atm, what is the new temperature if the volume remains constant?
(Show work pls :)!)

Chemistry
1 answer:
victus00 [196]3 years ago
6 0

Answer:

T_2=484.8K

Explanation:

Hello there!

In this case, according to the the variable temperature and pressure and constant volume, it turns out possible for us to calculate the new temperature via the Gay-Lussac's law as shown below:

\frac{T_2}{P_2} =\frac{T_1}{P_1}

Thus, by solving for the final temperature, T2, we obtain:

T_2 =\frac{T_1P_2}{P_1}

So we plug in the given data to obtain:

T_2 =\frac{303K*200atm}{125atm}\\\\T_2=484.8K

Best regards!

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Given: 2H2O2 → 2H2O + O2 structure of H2O2: H–O–O–H Bond Bond Energy (kJ/mol) O–H 459 O=O 494 O–O 142 Based on the given bond en
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Answer:

B. - 210 kJ

Explanation:

<em>∵ ΔHrxn = ∑(bond energies)products - ∑(bond energies)reactants.</em>

  • The bond formation in the products releases energy (exothermic).
  • The bond breaking in the reactants requires energy (endothermic).

The products:

  • H₂O contains 2 O-H (- 459 kJ/mol) bonds.
  • O₂ contain 1 O=O (- 494 kJ/mol) bond.

The reactants:

  • H₂O₂ contain 2 O–H (459 kJ/mol) bonds and 1 O–O (142 kJ/mol) bond.

∵ ΔHrxn = ∑(bond energies)products - ∑(bond energies)reactants.

<em>∴ ΔHrxn = [2 (2 x (O–H bond energy) + (1 x (O=O bond energy)] - 2 [(2 x (O–H bond energy) + (1 x (O–O bond energy)] </em>= [2 (2 x - 459 kJ/mol) + (1 x - 494 kJ/mol)] - 2 [(2 x 459 kJ/mol) + (1 x 142 kJ/mol)] = (- 2330 kJ) + (2120 kJ) = <em>- 210 kJ.</em>

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