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3 years ago

Determine the number of atoms of nickel (Ni) in a 5.05 grams solid nickel coin

Chemistry

1 answer:

Rzqust [24]3 years ago

6 0

Answer:

0.517× 10²³ atoms

Explanation:

Given data:

Mass of nickel = 5.05 g

Number of atoms of nickel = ?

Solution:

Formula:

Number of moles = mass/molar mass

Molar mass of nickel = 58.7 g/mol

by putting values,

Number of moles = 5.05 g/ 58.7 g/mol

Number of moles =0.086 mol

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms

0.086 mol × 6.022 × 10²³ atoms / 1mol

0.517× 10²³ atoms

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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

SCORPION-xisa [38]

Explanation:

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.