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kap26 [50]
2 years ago
8

Which reactions are oxidation-reduction reactions? check all that apply. fes(s) 2hcl(aq) → h2s(g) fecl2(g) agno3(aq) nacl(aq)

→ agcl(s) nano3 2c3h6(g) 9o2(g) → 6co2(g) 6h2o(g) fe2o3(s) 3co(g) → 2fe(l) 3co2(g)
Chemistry
1 answer:
Dafna1 [17]2 years ago
7 0

oxidation-reduction reactions are -

  1. 2C3H6(g) + 9O2(g) → 6CO2(g) + 6H2O(g)
  2. Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

For reaction,

  1. 2C3H6(g) + 9O2(g) → 6CO2(g) + 6H2O(g)

<u>On reactant side</u>:

Oxidation state of Carbon = +2

Oxidation state of Oxygen = 0

<u>On product side:</u>

Oxidation state of Carbon = +4

Oxidation state of Oxygen = -2

Here, carbon's oxidation state is rising from +2 to +4. As a result, it is oxidizing and the oxygen's oxidation state is decreasing from 0 to -2. As a result, it is decreasing.

For reaction,

                Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

<u>When reacting:</u>

Iron's oxidation state is +3.

Carbon's oxidation state is +2.

<u>On product side:</u>

Iron's oxidation state is zero.

Carbon's oxidation state is +4.

Here, carbon's oxidation state is rising from +2 to +4. As a result, it is being oxidized and the iron's oxidation state is changing from +3 to 0. As a result, it is decreasing.

To learn more about oxidation-reduction from given link

brainly.com/question/5794822

#SPJ4

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HELP!!!! A glass flask containing 100 milliliters of water is sealed with a rubber stopper and placed on a burner. After a while
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Answer: A closed system, because energy can enter or leave the container, but the water molecules cannot

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3 years ago
Given the following data:
bagirrra123 [75]

176.0 \; \text{kJ} \cdot \text{mol}^{-1}

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its \Delta H can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with \Delta H given be denoted as (1), (2), (3), and the last equation (4). Let a, b, and c be letters such that a \times (1) + b \times (2) + c \times (3) = (4). This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, 3 + (-1) = 2 shall resemble the number of \text{H}_2 left on the product side when the second equation is directly added to the third. Similarly

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Thus

a = -1/2\\b = 1/2\\c = -1/2 and

-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)

Verify this conclusion against a fourth species involved- \text{N}_2 \; (g) for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

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Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

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Answer:

The temperature is the same overtime.

Explanation:

Since the line on the graph is straight the temperature will be constant.

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