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

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.

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.