Answer:
5 moles of H2
Explanation:
2 mol HCl * x mol H2 = 10 mol HCl
x=5
Check work:
5 mol H2 * 2 mol HCl/1 mol H2 = 10 mol HCl
Molar mass O2 = 31.99 g/mol
Molar mass CO2 = 44.01 g/mol
Moles ratio:
<span>C3H8 + 5 O2 = 3 CO2 + 4 H2O
</span>
5 x 44.01 g O2 ---------------- 3 x 44.01 g CO2
( mass of O2) ------------------ 37.15 g CO2
mass of O2 = 37.15 x 5 x 44.01/ 3 x 44.01
mass of O2 = 8174.8575 / 132.03
mass of O2 = 61.916 g
Therefore:
1 mole O2 ----------------- 31.99 g
moles O2 -------------------- 61.916
moles O2 = 61.916 x 1 / 31.99
moles = 61.916 / 31.99 => 1.935 moles of O2
Shape and size has no effect on an objects density
Answer:
3
Explanation:
The first election shell can only hold 2 electrons, but the next one can hold up to 8
Answer: 58.305torr
Explanation:
We shall solve this using Raoult law.
Mathematical expression is;
Ptotal=XaPa + XbPb
Where Xa, Xb and Pa Pb are the mole fraction and vapor pressure of each component. we shall now solve this by steps
STEP1: Find the mole of benzene and Toluene;
Mole= mass÷molar mass
For toluene:
Mass= 398g
Molar mass= 92.1g/mol
Mole= 398÷92.1 =4.32mol
For benzene:
Mass=734g
Molar mass= 78.1g/mol
Mole= 734÷78.1 = 9.398mol
STEP2: Find the mole fraction of each component:
Let Xb be mole fraction for benzene
Let Xt be mole fraction for toluene.
Therefore; nB+nT = X
nB and nT are the mole of component B and T
X= 4.32+9.398= 13.718
Mole fraction for Toluene;
Xt= nT÷X
4.32÷13.718= 0.315
Mole fraction for benzene;
Xb= nB÷X
9.398÷13.718= 0.685
STEP3: Find the total total pressure;
We will use Rauolt law.
Ptotal= XbPb + XtPt
Pt = 22torr
Pb = 75torr
Therefore Ptotal is;
(0.315×22) + (0.685×75) = 6.93+51.375= 58.305
Therefore the total pressure is 58.305torr