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n200080 [17]
2 years ago
6

Write a half-reaction for the oxidation of the manganese in MnCO3(s) to MnO2(s) in neutral groundwater where the carbonate-conta

ining species in the product is HCO3–(aq). Add H2O and H+ to balance the H and O atoms in the equation. Do not add electrons; you may leave the half-reaction unbalanced with respect to charge.
Chemistry
1 answer:
Leokris [45]2 years ago
3 0

Answer:MnCO3+2H2O----->MnO2+ HCO3-+2e-+3H+

Explanation:The equation to be balanced is

MnCO3 ------> MnO2+HCO3-

The oxidation number of Mn changes from +2 in MnCO3 to +4 in MnO2

Therefore two electrons must be added to the right as shown below:

MnCO3 -------> MnO2+ HCO3-+ 2e-Now,there is one negative charge HCO3- and 1 negative charge on the two electrons making a total of -3 charges on the right. There is zero charge on the left.

To balance the equation,add3H+on the right,to cancel out the charges.

MnCO3 --------> MnO2+HCO3-+2e-+3H+

Adding H2O to balance Hydrogen and Oxygen atoms:

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

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Answer:

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Explanation:

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3 0
2 years ago
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Consider a beaker of water sitting on the pan of an electronic scale that has been tared. A metal weight hanging from a string i
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Answer:

See explanation below for answers

Explanation:

We know that the balance is tared, so the innitial weight would be zero. Now, let's answer this by parts.

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In this case all we need to do is to substract the 0.70 with the 0.13 g. so:

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b) Volume of water.

In this case, we have the density of water, so we use the formula for density and solve for volume:

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2 years ago
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tensa zangetsu [6.8K]
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