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Blizzard [7]
3 years ago
5

Which statement best explains why calcium has a larger atomic radius than magnesium? A. Calcium has a stronger nuclear charge th

an magnesium. B. Calcium has electrons in more energy levels than magnesium. C. Calcium has more neutrons than magnesium. D. Calcium’s third energy level contains more electrons than magnesium
Chemistry
2 answers:
ahrayia [7]3 years ago
7 0

Answer:

B. calcium has electrons in more energy levels than magnesium

Explanation:

atoms are made of three types of subatomic particles - electrons, protons and neutrons.

Neutrons and protons reside in the nucleus of the atom and electrons reside in energy shells

electronic configuration for both Ca and Mg are as follows

Mg - 2,8,2

Ca - 2,8,8,2

outermost energy shell of Mg with electrons is the third energy level

whereas outermost energy shell of Ca with electrons is the fourth energy shell

therefore Ca has a larger atomic radius than Mg as it has one more energy shell than Mg in which electrons reside

Stella [2.4K]3 years ago
7 0

Answer: Option (B) is the correct answer.

Explanation:

Atomic number of calcium is 20 so its electronic distribution is 2, 8, 8, 2, that is, number of shells occupied will be K, L, M and N.

Whereas atomic number of magnesium is 12 so its electronic distribution is 2, 8, 2, that is, number of shells occupied will be K, L and M.

Therefore due to increase in number of shells there will be increase in atomic size of the atom.

Hence, we can conclude that the statement calcium has electrons in more energy levels than magnesium best explains why calcium has a larger atomic radius than magnesium.

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a) Percentage by mass of carbon: 18.3%

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b)  Percentage by mass of chlorine: 80.37%

c) Molecular formula: C_{2} H Cl_{3}

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0.089g H_{2}O*\frac{2g H}{18g H_{2}O }  = 0.010g H

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Let's calculate the percentage by mass of chlorine:

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0.77g H*\frac{1mol H}{1g H} = 0.77mol H

80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl

Dividing each of the quantities above by the smallest (0.77mol), the  subscripts in a tentative formula would be

C=\frac{1.52}{0.77} = 1.97 ≈ 2

H = \frac{0.77}{0.77} = 1

Cl =\frac{2.27}{0.77}=2.94≈3

The empirical formula for the compound is:

C_{2} H Cl_{3}

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

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