Total mass of CaCO3 = 40 amu of Ca + 12amu of C + 16×3 amu of oxygen = 100amu of CaCO3
i.e 100 tonnes of CaCO3 .
mass of CO2 = 12amu of C + 2× 16amu of O = 44 amu of CO2
mass % of CO2 in CaCO3 = (44/100)×100 =44%
i.e
44% of 100 tonnes is CO2.
=44 tonnes of CO2.
therefore, 44% of CO2 is present in CaCO3.
Answer: The empirical formula of the compound is 
Explanation:
Empirical formula is defined formula which is simplest integer ratio of number of atoms of different elements present in the compound.
Percentage of iron in a compound = 36.76 %
Percentage of sulfur in a compound = 21.11 %
Percentage of oxygen in a compound = 42.13 %
Consider in 100 g of the compound:
Mass of iron in 100 g of compound = 36.76 g
Mass of iron in 100 g of compound = 21.11 g
Mass of iron in 100 g of compound = 42.13 g
Now calculate the number of moles each element:
Moles of iron=
Moles of sulfur=
Moles of oxygen=
Divide the moles of each element by the smallest number of moles to calculated the ratio of the elements to each other
For Iron element = 
For sulfur element =
For oxygen element = 
So, the empirical formula of the compound is 
Answer:

Explanation:
Charge on 1 electron =
The expression for charge is:-

Given that:- Charge =


Total number of electrons, n = 
No, they do not.
Carbon dioxide has a linear geometry because the lone pair and bond pair repulsion cancels out; however, water has a bent structure because only the oxygen atom possesses a lone pair which brings the bonding electron pairs closer.