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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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if you can’t already tell by the amount of questions i do not understand chemistry lol, giving brainliest
Vladimir [108]

Answer:

Ok, so the process here is to convert the mass of H2 (hydrogen gas) to moles by dividing the mass by the molar mass of H2. Once you have the moles then you have to multiply by the STP (standard temperature and pressure) molar volume which should be 22.4.

Molar mass of H2 = (1.01)x2 = 2.02g/mol

19.3/2.02 = 9.55 moles

Now just multiply the moles by the molar volume

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3 years ago
How many liters are in a .00813M solution that contains 1.55 g of KBr?
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Answer:

                      1.602 L (or) 1602 mL

Explanation:

             Molarity is the amount of solute dissolved per unit volume of solution. It is expressed as,

                         Molarity  =  Moles / Volume of Solution    ----- (1)

Rearranging above equation for volume,

                         Volume of solution  =  Moles / Molarity    -------(2)

Data Given;

                  Molarity  =  0.00813 mol.L⁻¹

                  Mass  =  1.55 g

First calculate Moles for given mass as,

                   Moles  =  Mass / M.mass

                   Moles  =  1.55 g / 119.002 g.mol⁻¹

                   Moles  =  0.0130 mol

Now, putting value of Moles and Molarity in eq. 2,

                         Volume of solution  =  0.0130 mol / 0.00813 mol.L⁻¹

                         Volume of solution  = 1.60 L

or,

                         Volume of solution  =  1602 mL

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