Lets take 100 g of this compound,
so it is going to be 2.00 g H, 32.7 g S and 65.3 g O.
2.00 g H *1 mol H/1.01 g H ≈ 1.98 mol H
32.7 g S *1 mol S/ 32.1 g S ≈ 1.02 mol S
65.3 g O * 1 mol O/16.0 g O ≈ 4.08 mol O
1.98 mol H : 1.02 mol S : 4.08 mol O = 2 mol H : 1 mol S : 4 mol O
Empirical formula
H2SO4
Atomic mass is equal to the total number of electrons neutrons and protons
The volume : 8,526 quarts
<h3>Further explanation</h3>
Given
The density of whole milk = 1.04 g/ml
mass = 18.5 pounds
Required
The volume
Solution
Conversion of mass
1 pound = 453,592 g
18.5 pounds = 8391,45 g
Density formula:
.
Input the value :
V = m : ρ
V = 8391,45 g : 1.04 g/ml
V = 8068.7 ml
1 ml = 0,00105669 quarts
8068.7 ml =8,526 quarts
0.075 L * 1.0 M = 0.075 mol HCl
H2 is half of HCl by the coefficients divide 0.075 by 2 and get 0.0375 mol H2 gas
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