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MaRussiya [10]
3 years ago
15

How many grams of oxygen (O) are present in 0.0207 moles of Ca(HCO 3) 2

Chemistry
1 answer:
Zolol [24]3 years ago
5 0

Answer:

1.99grams

Explanation:

- First, we need to calculate the molar mass of the compound: Ca(HCO3)2

Ca = 40g/mol, H = 1g/mol, C = 12g/mol, O = 16g/mol

Hence, Ca(HCO3)2

= 40 + {1 + 12 + 16(3)}2

= 40 + {13 + 48}2

= 40 + {61}2

= 40 + 122

= 162g/mol

Molar mass of Ca(HCO3)2 = 162g/mol

- Next, we calculate the mass of oxygen in one mole of the compound, Ca(HCO3)2.

Oxygen = {16(3)}2

= 48 × 2

= 96g of Oxygen

- Next, we calculate the percentage composition of oxygen by mass by dividing the mass of oxygen in the compound by the molar mass of the compound i.e.

% composition of O = 96/162 × 100

= 0.5926 × 100

= 59.26%.

- The number of moles of the compound, Ca(HCO3)2, must be converted to mass by using the formula; mole = mass/molar mass

0.0207 = mass/162

Mass = 162 × 0.0207

Mass = 3.353grams

However, in every gram of Ca(HCO3)2, there is 0.5926 g of oxygen

Hence, in 3.353grams of Ca(HCO3)2, there will be;

0.5926 × 3.353

= 1.986

= 1.99grams.

Therefore, there is 1.99grams of Oxygen in 0.0207 moles (3.353g) of Ca(HCO3)2.

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Answer:

Detail is given below

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As the size of atom increases the ionization energy from top to bottom also  decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus.

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