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3 years ago

Name four products of incomplete combustion

Chemistry

1 answer:

aalyn [17]3 years ago

5 0

Answer: The products are carbon monoxide, water and carbon.

Explanation:

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An electrochemical cell has the following standard cell notation: Al(s) | Al3+(aq) || Mg2+(aq) | Mg(s)

Soloha48 [4]

Answer:

a. Al(s) ⇄ Al⁺³(aq) + 3e⁻ (oxidation)

Mg²⁺(aq) + 2e⁻ ⇄ Mg(s) (reduction)

b. ΔE° = + 0.715 V

c. It's an electrolytic cell, because it's a nonspontaneous reaction.

d. 2Al(s) + 3Mg²⁺(aq) ⇄ 2Al⁺³(aq) + 3Mg(s)

Explanation:

a. By the notation given, first is represented the oxidation reaction and then the reduction reaction, so they are:

Al(s) ⇄ Al⁺³(aq) + 3e⁻ (oxidation)

Mg²⁺(aq) + 2e⁻ ⇄ Mg(s) (reduction)

b. The standard potential of the cell (ΔE°) is the reduction potential of the oxidation less the reduction potential of the reduction. The reduction potentials are:

Al(s) = -1.66 V

Mg(s) = -2.375 V

ΔE° = -1.66 - (-2.375)

ΔE° = + 0.715 V

c. It's an electrolytic cell.

A galvanic cell is spontaneous, so the cathode (reduction) has a higher E° than the cathode (oxidation). In this case, the oxidation reaction has a higher E°, so the reaction is nonspontaneous and it's necessary an external force to it happen, so it's an electrolytic cell.

d. 2Al(s) + 3Mg²⁺(aq) ⇄ 2Al⁺³(aq) + 3Mg(s)

The number of electrons must be the same, so the oxidation reaction is multiplied by 2, and the reduction reaction by 3.

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3 years ago

Kp for the following reaction is 0.16 at 25 degree C. 2 NOBr(g) 2 NO(g) Br_2(g) The enthalpy change for the reaction at standard

finlep [7]

Answer:

Explanation:

Given that:

$2 NOBr_{(g)} \iff 2 NO_{(g)} + Br_{2(g)}$

From above:

$K_p = 0.16 = \dfrac{(P_{NO})^2 (P_{Br})}{(P_{NOBr})^2}$

To predict the effect of the addition of Br₂(g);

The addition of Br₂(g) will favor the equilibrium to shift to the left i.e. formation of NOBr

The removal of some NOBr will cause the equilibrium position to shift to the left side. This is because concentration on the left side is decreased and the concentration on the right side will be increased. Thus, the equilibrium will shift towards where the concentration is reduced which is the left side.

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3 years ago