Empirical formula is the simplest ratio of components making up a compound.
The percentage composition of each element has been given
therefore the mass present of each element in 100 g of compound is
B N H
mass 40.28 g 52.20 g 7.53 g
number of moles
40.28 g / 11 g/mol 52.20 g / 14 g/mol 7.53 g / 1 g/mol
= 3.662 mol = 3.729 mol = 7.53 mol
divide the number of moles by the least number of moles, that is 3.662
3.662 / 3.662 3.729 / 3.662 7.53 / 3.662
= 1.000 = 1.018 = 2.056
the ratio of the elements after rounding off to the nearest whole number is
B : N : H = 1 : 1 : 2
therefore empirical formula for the compound is B₁N₁H₂
that can be written as BNH₂
Answer:
a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) 0.0035 mole
c) 0.166 M
Explanation:
Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.
The equation of the reaction is expressed as:

1 mole 3 mole
The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

10 ml 17.50 ml
(x) M 0.200 M
Molarity = 
= 0.0035 mole
c) What was the molar concentration of phosphoric acid in the original stock solution?
By stoichiometry, converting moles of NaOH to H₃PO₄; we have
= 
= 0.00166 mole of H₃PO₄
Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:
Molar Concentration = 
Molar Concentration = 
Molar Concentration = 0.166 M
∴ the molar concentration of phosphoric acid in the original stock solution = 0.166 M
Answer:
single orbital, s sub level.
three orbitals, p sub level
5 orbitals , d sub level
seven orbital , f sub level