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Anvisha [2.4K]
3 years ago
12

Write the balanced half-equation describing the oxidation of mercury to hgo in a basic aqueous solution. Please include the stat

es of matter for each compound.
Chemistry
1 answer:
Cloud [144]3 years ago
8 0

Answer:

Hg^0+2OH^-\rightarrow Hg^{2+}O+H_2O+2e^-

Explanation:

Hello,

In this case, mercury (II) oxide (HgO) is obtained via the reaction:

Hg(l)+O_2\rightarrow HgO

Nonetheless, since it is a reaction carried out in basic solution, mercury's half-reaction only, must be:

Hg^0+2OH^-\rightarrow Hg^{2+}O+H_2O+2e^-

Thus, it is seen that OH ionis should be added due to the basic aqueous solution considering that 2 electrons are transferred from 0 to 2 in mercury.

Best regards.

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4. Magnesium and oxygen undergo a chemical reaction to
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Answer:

About 16.1 grams of oxygen gas.

Explanation:

The reaction between magnesium and oxygen can be described by the equation:
\displaystyle 2\text{Mg} + \text{O$_2$} \longrightarrow 2\text{MgO}

24.4 grams of Mg reacted with O₂ to produce 40.5 grams of MgO. We want to determine the mass of O₂ in the chemical change.

Compute using stoichiometry. From the equation, we know that two moles of MgO is produced from every one mole of O₂. Therefore, we can:

  1. Convert grams of MgO to moles of MgO.
  2. Moles of MgO to moles of O₂
  3. And moles of O₂ to grams of O₂.

The molecular weights of MgO and O₂ are 40.31 g/mol and 32.00 g/mol, respectively.

Dimensional analysis:

\displaystyle 40.5\text{ g MgO} \cdot \frac{1\text{ mol MgO}}{40.31\text{ g MgO}} \cdot \frac{1\text{ mol O$_2$}}{2\text{ mol MgO}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}

In conclusion, about 16.1 grams of oxygen gas was reacted.

You will obtain the same result if you compute with the 24.4 grams of Mg instead:

\displaystyle 24.4\text{ g Mg}\cdot \frac{1\text{ mol Mg}}{24.31\text{ g Mg}} \cdot \frac{1\text{ mol O$_2$}}{1\text{ mol Mg}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}

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Consider the image of a fossil below.
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