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2 years ago

What does the number 7 mean in the following example-73Li?

Chemistry

1 answer:

Pavlova-9 [17]2 years ago

4 0

Answer:

7 is the mass number of li and 3 is the atomic number

a 3

b 7

c 3

d 4

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A particular voltaic cell operates with the reaction: Zn(s)+Cl2(g)--->Zn^2+(aq)+2Cl^-(aq) giving a cell potential of .853 V.

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The amount of energy liberated will be 49.38 J.

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The amount of energy liberated (gibbs free energy) can be calculated using the following equation:

ΔG° = -nFε

n: amount of moles of electrons transfered

F: Faraday's constant

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20.0 g of Zn is equal to 0.30 mol.

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Rank the members of each set of compounds according to the ionic character of their bonds. Most ionic bonds?a) PCl3 PBr3 PF3 Mos

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More is the electronegativity difference between the combining atoms more polar is the compound. Hence, more ionic it will be in nature.

(a) Electronegativity value of P = 2.19

Electronegativity value of Cl = 3.16

Electronegativity value of Br = 2.96

Electronegativity value of F = 3.98

Electronegativity difference of a P-Cl bond = 3.16 - 2.19 = 0.97

Electronegativity difference of a P-Br bond = 2.96 - 2.19 = 0.77

Electronegativity difference of a P-F bond = 3.98 - 2.19 = 1.79

Since, a P-F bond has the highest electronegativity difference. Therefore, $PF_{3}$ is the most ionic compound and $PBr_{3}$ is the least ionic compound.

(b) Electronegativity value of B = 2.04

Electronegativity value of N = 3.04

Electronegativity value of C = 2.55

Electronegativity value of F = 3.98

Electronegativity difference of a B-F bond = 3.98 - 2.04 = 1.94

Electronegativity difference of a N-F bond = 3.04 - 3.98 = 0.94

Electronegativity difference of a C-F bond = 3.98 - 2.55 = 1.43

Since, a B-F bond has the highest electronegativity difference. Therefore, $BF_{3}$ is the most ionic compound and $NF_{3}$ is the least ionic compound.

(c) Electronegativity value of Se = 2.55

Electronegativity value of Te = 2.1

Electronegativity value of Br = 2.96

Electronegativity value of F = 3.98

Electronegativity difference of a Se-F bond = 3.98 - 2.55 = 1.43

Electronegativity difference of a Te-F bond = 3.98 - 2.1 = 1.88

Electronegativity difference of a Br-F bond = 3.98 - 2.19 = 1.02

Since, a Te-F bond has the highest electronegativity difference. Therefore, $TeF_{4}$ is the most ionic compound and $BrF_{3}$ is the least ionic compound.

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