How many grams of sulfuric acid are produced when 68.1 g of lead (ii) sulfate react?

The chemical reaction to produce sulfuric acid fro lead (ii) sulfate is:
PbSO4 + SO2 + 2H2O --> Pb + 2H2SO4

first convert the mass of PbSO4 to moles
moles PbSO4 = 68.1 g ( 1 mol / 303.26 g )
moles PbSO4 = 0.2246 mol PbSO4

mass H2SO4 = 0.2246 mol PbSO4 ( 2 mol H2SO4 / 1 mol PbSO4) ( 98 g / mol)

mas H2SO4 = 44 g H2SO4

There is one s orbital, and there are three p orbitals, five d orbitals, and seven f orbitals. thus, the answer is b. do not confuse the number of orbitals in a subshell with the number of electrons the subshell can hold. each orbital can hold two electrons, so the capacity of an nf subshell is 14 electrons.

True is ur answer : )