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MakcuM [25]
2 years ago
6

Identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent for the following equation

Chemistry
1 answer:
bazaltina [42]2 years ago
4 0

The element that gains electrons, becomes reduced.

While the one which loses electrons, becomes oxidized.

In this equation,

CH₃OH + Cr₂O₇²⁻---- --> CH₂O + Cr³⁺.

By balancing the equation, we will get:

3CH₃OH + Cr₂O₇²⁻ + 8H⁺ --> 3CH₂O + 2Cr³⁺ + 7H₂O

Here the oxidation state of Cr changes from +6 to +3 that is it is being reduced thus serving as a oxidizing agent while other element retain their charges.

Here Cr₂O₇²⁻ is reduced while CH₃OH is oxidized.

So Cr₂O₇²⁻ serves as a oxidizing agent, while CH₃OH serves as reducing agent .

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Determine the value of the equilibrium constant, Kgoal, for the reaction C(s)+12O2(g)+H2(g)⇌12CH3OH(g)+12CO(g), Kgoal=? by makin
Licemer1 [7]

Answer:

1.71x10²⁷

Explanation:

If we sum 1/2 of (3) + 1/2 of (1):

1/2 (3.) C(s) + 1/2O₂(g) ⇌ CO(g), K₃ = √2.10×10⁴⁷  = 4.58x10²³

1/2 (1)   1/2CO₂(g) + 3/2H₂(g) ⇌ 1/2CH₃OH(g) + 1/2H₂O(g), K₁ = √1.40×10² = 11.8

C(s) + 1/2O₂(g) +<u> 1/2CO₂(g) </u>+<u> 3/2H₂(g</u>) ⇌ 1/2CH₃OH(g) + <u>1/2H₂O(g)</u> + <u>CO(g)</u>

K' = 4.58x10²³ * 11.8 = 5.42x10²⁴

+1/2 (2):

<u>1/2 CO(g)</u> +<u> 1/2H₂O(g)</u> ⇌<u> 1/2CO₂(g)</u> + <u>1/2H₂</u> (g), K = √1.00×10⁵ = 316.2

C(s) + 1/2O₂(g) + H₂(g) ⇌ 1/2 CHO₃H(g) + 1/2CO(g)

K'' = 5.42x10²⁴* 316.2 =

<h3>1.71x10²⁷</h3>

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3 years ago
As an athlete exercises, sweat is produced and evaporated to help maintain a proper body temperature. On average, an athlete los
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<u>Answer:</u> For the given amount of sweat lost, the amount of energy required will be 692,899 Joules.

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Hence, for the given amount of sweat lost, the amount of energy required will be 692,899 Joules.

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