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2 years ago

Question 25———-A, YA are two isotopes of element A.

Chemistry

1 answer:

lakkis [162]2 years ago

6 0

Answer:

Option C. 1

Explanation:

Step 1:

Determination of the Neutron of both isotopes. This is illustrated below.

For isotope y xA:

Mass number = y

Atomic number = x

Neutron =..?

Atomic number = proton number = x

Mass number = Proton + Neutron

y = x + Neutron

Rearrange

Neutron = y – x

For isotope (y + 1) xA:

Mass number = y + 1

Atomic number = x

Neutron =.?

Atomic number = proton number = x

Mass number = Proton + Neutron

y + 1 = x + Neutron

Rearrange

Neutron = y + 1 – x

Step 2:

Determination of the difference between the neutron number of both isotopes. This is illustrated below:

For isotope y xA:

Neutron number = y – x

For isotope (y + 1) xA:

Neutron number = y + 1 – x

Difference in neutron number

=> (y + 1 – x) – (y – x)

=> y + 1 – x – y + x

Rearrange

=> y – y + 1 – x + x

=> 1

Therefore, the difference in the neutron number of both isotopes is 1

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Answer:

Explanation:

The cell reaction properly written is shown below:

Cu|Cu²⁺ $_{aq}$ || Ag⁺ $_{aq}$ | Ag

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Oxidation half:

Cu $_{s}$ ⇄ Cu²⁺ $_{aq}$ + 2e⁻

At the anode, oxidation occurs.

Reduction half:

Ag⁺ $_{aq}$ + 2e⁻ ⇄ Ag $_{s}$

At the cathode, reduction occurs.

To derive the overall reaction, we must balance the atoms and charges:

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we multiply the second reaction by 2 to balance up:

2Ag⁺ $_{aq}$ + 2e⁻ ⇄ 2Ag $_{s}$

The net reaction equation:

Cu $_{s}$ + 2Ag⁺ $_{aq}$ + 2e⁻⇄ Cu²⁺ $_{aq}$ + 2e⁻ + 2Ag $_{s}$

We then cancel out the electrons from both sides since they appear on both the reactant and product side:

Cu $_{s}$ + 2Ag⁺ $_{aq}$ ⇄ Cu²⁺ $_{aq}$ + 2Ag $_{s}$

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Answer:

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