(P1)(V1)=(P2)(V2)
(1.50)(5.00)=(1240/760)(V2)
(7.5)/(1240/760)=V2
V2=4.596774194 L
It is fact that
6.023
×
10
23
formula units of barium nitrate have a mass of
16.6*10^23
⋅
g
. This is what we specify when we say molar mass. And thus the mass of
5.30
×
10
22
formula units of barium nitrate is the quotient multiplied by the molar mass:
5.30
×
10
22
6.023
×
10
23
m
o
l
×
16.6*10^23
⋅
g
⋅
m
o
l
−
1
A compound is analyzed and found to contain 22.10%Al, 25.40%P, and 52.50%O
Let the total mass of compound = 100g
The mass of Aluminum = 22.10 g
Moles of Al = mass / atomic mass = 22.10 /23= 0.96
The mass of P = 25.40 g
Moles of P = 25.40 / 31 = 0.819
The mass of oxygen = 52.50 g
Moles of oxygen = 52.50 / 16= 3.28
The mole ratio of the two elements will be
Al = 0.96/0.819 = 1.17
P = 0.819/0.819 = 1
O = 3.28 / 0.819 = 4
The whole number ratio will be 1 : 1 : 4
So formula will be AlPO4
