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SashulF [63]
3 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]3 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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A 5.024 mg sample of an unknown organic molecule containing carbon, hydrogen, and nitrogen only was burned and yielded 13.90 mg
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Answer:

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Explanation:

Mass of the unknown compound = 5.024 mg

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Mass of H2O = 6.048 mg

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For carbon, C:

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C = 12/44 x 13.90 = 3.791 mg

For hydrogen, H:

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H = 2/18 x 6.048 = 0.672 mg

For nitrogen, N:

Mass N = mass of unknown – (mass of C + mass of H)

Mass of N = 5.024 – (3.791 + 0.672)

Mass of N = 0.561 mg

Now, we can obtain the empirical formula for the compound as follow:

C = 3.791 mg

H = 0.672 mg

N = 0.561 mg

Divide each by their molar mass

C = 3.791 / 12 = 0.316

H = 0.672 / 1 = 0.672

N = 0.561 / 14 = 0.040

Divide by the smallest

C = 0.316 / 0.04 = 8

H = 0.672 / 0.04 = 17

N = 0.040 / 0.04 = 1

Therefore, the empirical formula for the compound is C8H17N

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