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SashulF [63]
2 years ago
7

The mineral rhodochrosite [manganese(II) carbonate, MnCO3] is a commercially important source of manganese. Write a half-reactio

n for the oxidation of the manganese in MnCO3 to MnO2 in neutral groundwater where the principal carbonate species is HCO3–. Add H2O, H+, and electrons as needed to balance the half-reaction.
Chemistry
1 answer:
aleksklad [387]2 years ago
5 0

Answer:

MnCO3 + 2H2O ⇄ MnO2  + HCO3-  + 2e- +3H+

Explanation:

<u>The</u> unbalanced equation

MnCO3 ⇄ MnO2  + HCO3-

In MnCO3, the oxidation number of Mn is +2

In Mno2, the oxidation number of Mn is +4

The change from +2 to +4 requires an addition of  2 electrons (to the right side).

MnCO3 ⇄ MnO2  + HCO3-  + 2e-

The total charge now is -3 on the right side. To balance this we add 3 hydrogen atoms on the right side.

MnCO3 ⇄ MnO2  + HCO3-  + 2e- +3H+

On the right side we have 4 hydrogen atoms in total. On the left side we have 0 hydrogen atoms. So to balance, we have to add 2H2O on the left side

MnCO3 + 2H2O ⇄ MnO2  + HCO3-  + 2e- +3H+

Now the reaction is balanced.

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The molar heat of fusion of gold is 12.550 kJ mol–1. At its melting point, how much mass of melted gold must solidify to release
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The mass of melted gold to release the energy would be  3, 688. 8 Kg

<h3>How to determine the mass</h3>

The formula for quantity of energy is given thus;

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To find the number of moles, we have

235.0 = n × 12.550

number of moles = \frac{235}{12. 550} = 18. 725 moles

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Thus, the mass of melted gold to release the energy would be  3, 688. 8 Kg

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