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kari74 [83]
3 years ago
5

Chemical formula of ammonium oxide

Chemistry
1 answer:
11111nata11111 [884]3 years ago
8 0

The answer is (NH4)2O

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As you move from left to right, the nucleus gains protons. This increases the positive charge of the nucleus and its attractive force on the electrons. At the same time, electrons are added to the atoms as you move from left to right across a period.

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For each of the following balanced oxidation-reduction reactions, (i) identify the oxidation numbers for all the elements in the
AlladinOne [14]

Answer:

<h3>1. 10 e⁻</h3>

Oxidation numbers

I₂O₅(s): I (5+); O(2-)

CO(g): C(2+); O(2-)

I₂(s): I(0)

CO₂(g): C(4+); O(2-)

<h3>2. 4 e⁻</h3>

Oxidation numbers

Hg²⁺(aq): Hg(2+)

N₂H₄(aq): N(2-); H(1+)

Hg(l): Hg(0)

N₂(g): N(0)

H⁺(aq): H(1+)

<h3>3. 6 e⁻</h3>

Oxidation numbers

H₂S(aq): H(1+); S(2-)

H⁺(aq): H(1+)

NO₃⁻(aq): N(5+); O(2-)

S(s): S(0)

NO(g): N(2+); O(2-)

H₂O(l): H(1+); O(2-)

Explanation:

In order to state the total number of electrons transferred we have to identify both half-reactions for each redox reaction.

1.  I₂O₅(s) + 5 CO(g) → I₂(s) + 5 CO₂(g)

Oxidation: 10 e⁻ + 10 H⁺(aq) + I₂O₅(s) → I₂(s) + 5 H₂O(l)

Reduction: 5 H₂O(l) + 5 CO(g) → 5 CO₂(g) + 10 H⁺(aq) + 10 e⁻

2. 2 Hg²⁺(aq) + N₂H₄(aq) → 2 Hg(l) + N₂(g) + 4 H⁺(aq)

Oxidation: N₂H₄(aq) → N₂(g) + 4 H⁺(aq) + 4 e⁻

Reduction: 2 Hg²⁺(aq) + 4 e⁻ → 2 Hg(l)

3. 3 H₂S(aq) + 2H⁺(aq) + 2 NO₃⁻(aq) → 3 S(s) + 2 NO(g) + 4H₂O(l)

Oxidation: 3 H₂S(aq) → 3 S(s) + 6 H⁺(aq) + 6 e⁻

Reduction: 8 H⁺(aq) + 2 NO₃⁻(aq) + 6 e⁻ → 2 NO(g) + 4 H₂O

5 0
3 years ago
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