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Rasek [7]
3 years ago
14

What element is being oxidized in the following redox reaction?

Chemistry
1 answer:
DIA [1.3K]3 years ago
8 0

Answer:

The element that has been oxidized is the N

Explanation:

Zn²⁺(aq) + NH₄⁺(aq) → Zn(s) + NO₃⁻(aq)

See all the oxidation states:

Zn²⁺  → acts with +2

In ammonia, H acts with +1 and N with -3

Zn(s), acts with 0. In all the elements in ground state, the oxidation state is 0.

Zn changed from 2+ to 0. The oxidation number, has decreased.

This element has been reduced.

NO₃⁻ (aq) it's a ion, from nitric acid.

N acts with +5

O acts with -2

The global charge is -1

The N, has increased the oxidation state, so this element is the one oxidized.

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If a certain gas occupies a volume of 19 L when the applied pressure is 9.5 atm, find the pressure when the gas occupies a volum
kakasveta [241]

Answer: 37.6 atm

Explanation:

Given that,

Initial volume of gas (V1) = 19L

Initial pressure of gas (P1) = 9.5 atm

Final volume of gas (V2) = 4.8L

Final pressure of gas (P2) = ?

Since pressure and volume are given while temperature remains the same, apply the formula for Boyle's law

P1V1 = P2V2

9.5 atm x 19L = P2 x 4.8L

180.5 atm•L = 4.8L•P2

Divide both sides by 4.8L

180.5 atm•L/4.8L = 4.8L•P2/4.8L

37.6 atm = P2

Thus, the final pressure is 37.6 atmospheres.

3 0
3 years ago
Rhodium crystallizes in a face-centered cubic unit cell. The radius of a rhodium atom is 135 pm. Determine the density of rhodiu
Deffense [45]

Answer:

Density of unit cell ( rhodium) = 12.279 g/cm³

Explanation:

Given that:

The radius (r) of a rhodium atom = 135 pm

The atomic mass of rhodium = 102.90 amu

For a face-centered cubic unit cell,

r = \dfrac{a}{2\sqrt{2}}

where;

a = edge length.

Making "a" the subject of the formula:

a = 2 \sqrt{2} \times r

a = 2 \times 1.414 \times 135 \ pm

a = 381.8 pm

to cm, we get:

a = 381.8 × 10⁻¹⁰ cm

However, recall that:

density \ of \ unit \ cell = \dfrac{mass \ of \ unit \ cell}{volume \ of \unit \ cell}

where;

mass of unit cell = mass of atom × numbers of atoms per unit cell

Also;

mass\  of\ atom =\dfrac{ atomic \ mass}{Avogadro  \  number}

mass\  of\ atom =\dfrac{ 102.9}{6.023 \times 10^{23}}

Recall also that number of atoms in a unit cell for a  face-centered cubic = 4

So;

mass \ of \ unit \ cell= \dfrac{102.90}{6.023 \times 10^{23}}\times 4

mass of unit cell = 6.83380375 × 10⁻²² g

Density  \ of  \ unit \  cell = \dfrac{6.83380375 \times 10^{-22}}{(381.8\times 10^{-10})^3}

Density of unit cell ( rhodium) = 12.279 g/cm³

3 0
3 years ago
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saul85 [17]

Answer:

Magnesium Bromide

Explanation:

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