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Bezzdna [24]
3 years ago
11

Which process is an oxidation?

Chemistry
1 answer:
Zarrin [17]3 years ago
4 0
<span>The problem has to do with oxidation states of the matter. The oxidation state of oxygen will always be -2 with the exception of peroxides which will have a state of -1. The overall balanced state of chemical compounds will be 0, so the oxidation state of Mn in MnO2 will be +4. The oxidation state of MnO4- will then be +7 to balance out to the negative one charge. The state change from +4 to +7 is 3, thus three electrons have to be lost in order for this to happen; a loss of a charge of -3 results in an increase of charge of 3. Oxidation is always the process of 'losing' electrons.

</span><span>E] MnO2(s) MnO4-(aq</span>
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Answer:

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Explanation:

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4 years ago
9. Calculate the standard enthalpy of the 3rd reaction using the given data: A,H° = +52.96 kJ/mol AFH= -483.64 kJ/mol H2s)+I202
Mice21 [21]

Answer:

-586.56 kJ/mol is the standard enthalpy of the 3rd reaction.

Explanation:

H_2(g) +I_2(s) \rightarrow 2 HI(g) ,\Delta H^{o}_{1}= +52.96 kJ/mol...[1]

2 H_2(g) + O_2(g)\rightarrow 2 H_2O(g),\Delta H^{o}_{2}=-483.64 kJ/mol...[2]

4 HI(g)+O_2(g)\rightarrow 2 I_2(s)+2 H_2O(g) ,\Delta H^{o}_{3} =?..[3]

The unknown standard enthalpy of third reaction can be calculated by using Hess's law:

The law states that 'the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps'.

[2] - 2 × [1] = [3]

O_2+4HI\rightarrow 2H_2O(g)+2I_2(s)

\Delta H^{o}_{3}=\Delta H^{o}_{2}-2\times \Delta H^{o}_{1}

=-483.64 kJ/mol - 2\times (52.96 kJ/mol)=-586.56 kJ/mol

The standard enthalpy of the 3rd reaction is -586.56 kJ/mol.The negative sign indicates that energy is released during this reaction.

6 0
3 years ago
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Papessa [141]

Answer:

the molar mass

Explanation:

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3 years ago
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alekssr [168]

Answer:

B

Explanation:

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4 0
3 years ago
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