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Licemer1 [7]
3 years ago
8

For an exothermic reaction at equilibrium, how will increasing the temperature affect Keq?

Chemistry
1 answer:
Sonja [21]3 years ago
8 0

Question:

<em>For an exothermic reaction at equilibrium, how will increasing the temperature affect Keq?</em>

Answer:

<em>The reaction will proceed towards the liquid phase. Heat is on the reactant side of the equation. Lowering temperature will shift equilibrium left, creating more liquid water. A reaction that is exothermic releases heat, while an endothermic reaction absorbs heat.</em>

<em>If you increase the temperature, the position of equilibrium will move in such a way as to reduce the temperature again. It will do that by favouring the reaction which absorbs heat. In the equilibrium, that will be the back reaction because the forward reaction is exothermic.</em>

Hope this helps, have a good day. c;


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you would have to apply both of the low and high pressure together to make something?

Explanation:

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A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 385.mg of oxalic acid H2
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Answer:

The molarity of the sodium hydroxide solution is 0.0692 M

Explanation:

<u>Step 1: </u>Data given

Mass of H2C2O4 = 385 mg = 0.385 grams

volume = 250 mL = 0.250 L

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Molar mass H2C2O4 = 90.03 g/mol

<u>Step 2</u>: The balanced equation

2NaOH + H2C2O4 → Na2C2O4 + 2H2O

<u>Step 3:</u> Calculate moles H2C2O4

Moles H2C2O4 = mass H2C2O4 / molar mass H2C2O4

Moles H2C2O4 = 0.385 grams / 90.03 g/mol

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<u>Step 4: </u>Calculate molarity of H2C2O4

Molarity H2C2O4 = moles / volume

Molarity H2C2O4 = 0.00428 moles / 0.250 L

Molarity H2C2O4 = 0.01712 M

<u>Step 5:</u> Calculate molarity of NaOH

2*Ca*Va = n*Cb*Vb

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⇒ Va = volume of H2C2O4 = 0.250 L

⇒Cb = molarity of NaOH = TO BE DETERMINED

⇒ Vb = volume of NaOH = 0.1237 L

Cb = (2*0.01712*0.250)/0.1237

Cb = 0.0691 M

The molarity of the sodium hydroxide solution is 0.0692 M

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In an experiment designed to determine the concentration of Cu 2 ions in an unknown solution, you need to prepare 100 mL of 0.10
Alexxandr [17]

Answer:

1.6 grams

Explanation:

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We should use 1.6 grams of CuSO₄.

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