K2C2O4is reacted with KMnO4 according to the following chemical equation.K2C2O4 (aq) + 2 KMnO4 (aq) + 8 H2O → 7 CO2 (g) + 2 Mn(OH)2 (s) + 2 KOH (aq)How many grams of K2C2O4is needed to react with 50.0 g KMnO4?How many grams of CO2is produced? A freezer containing 2.0 kg of ice had a temperature of -5.0 oC initially. Suddenly, a power interruption occurred. Right before the power returned, only 250 g ice was left. Calculate the heat flow of the water during the blackout. The industrial method currently used to produce butanol is the hydration of butene.C4H8(g) + H2O(g) → C4H10O(g)The enthalpy values for the following reactions are:4C(s) + 4H2(g) → C4H8(g) ΔH = – 7·1 kJ mol–14C(s) + 5H2(g) + ½O2(g) → C4H10O(g) ΔH = – 292·8 kJ mol–12H2(g) + O2(g) → 2H2O(g) ΔH = – 483·6 kJ mol–1Using the data above, calculate the enthalpy change, in kJ mol–1, to produce butanol by hydration of butene.

Given, K2C2O4is reacted with KMnO4 according to the following chemical equation : K2C2O4 (aq) + 2…

Answered: K2C2O4is reacted with KMnO4 according…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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K₂C₂O₄is reacted with KMnO₄ according to the following chemical equation.

K₂C₂O₄ (aq) + 2 KMnO₄ (aq) + 8 H₂O → 7 CO₂ (g) + 2 Mn(OH)₂ (s) + 2 KOH (aq)

How many grams of K₂C₂O₄is needed to react with 50.0 g KMnO₄?
How many grams of CO₂is produced?

A freezer containing 2.0 kg of ice had a temperature of -5.0 ^oC initially. Suddenly, a power interruption occurred. Right before the power returned, only 250 g ice was left. Calculate the heat flow of the water during the blackout.

The industrial method currently used to produce butanol is the hydration of butene.

C₄H₈(g) + H₂O(g) → C₄H₁₀O(g)

The enthalpy values for the following reactions are:

4C(s) + 4H₂(g) → C₄H₈(g) ΔH = – 7·1 kJ mol–1

4C(s) + 5H₂(g) + ½O₂(g) → C₄H₁₀O(g) ΔH = – 292·8 kJ mol–1

2H₂(g) + O₂(g) → 2H₂O(g) ΔH = – 483·6 kJ mol–1

Using the data above, calculate the enthalpy change, in kJ mol–1, to produce butanol by hydration of butene.

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