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
Answer:
6s
Explanation:
Barium is in group 2 of the s block and is in period 6.
Answer : The final concentration of
is, 2.9 M
Explanation :
Expression for rate law for first order kinetics is given by:

where,
k = rate constant = 
t = time passed by the sample = 3.5 min
a = initial concentration of the reactant = 3.0 M
a - x = concentration left after decay process = ?
Now put all the given values in above equation, we get


Thus, the final concentration of
is, 2.9 M
The number of moles of aluminium that are needed to react completely with 13.2 moles of FeO is 8.8 moles
calculation
2Al + 3FeO → 3aFe +Al2O3
by use of of mole ratio of Al: FeO from equation above = 2:3 the moles of Al is therefore
= 13.2 x 2/3=8.8 moles of Al
Answer:
D. 2:5
Explanation:
It has 5 valency electrons
