Half life of a radio-active material is refereed to the time required for reducing its activity to 50 %.
Now, when I-131 passes 1st half life, it's activity will reduce to half i.e. 50%
When, I-131 passes 2nd half life, activity of I-131 will be reduced to 25%
On passing, 3rd half-life, percent activity of the I-131 <span>remaining will be 12.5 %.</span>
mol = 25/12
= 2.083 mol
1 mol = 6.02 × 10^23 atoms
2.083 mol = X
X = 2.083/1 × 6.02 × 10^23
= 1.254 × 10^24 atoms
Hydrogen - 7.44%
Carbon - (100-7.44)% = 92.56%
Lets take 100 g of benzene, then we have
Hydrogen - 7.44 g
Carbon - 92.56 g
n - number of moles
n(H) = 7.44g *1 mol/1.0g = 7.44 mol
n(C) = 92.56 g* 1mol/12.0 g ≈ 7.713 mol
n(C) : n(H) = 7.713 mol : 7.44 mol = 1:1
Empirical formula is CH.
M(CH) = (12.0+ 1.00) g/mol = 13.0 g/mol
M (benzene) = 78.1 g/mol
M (benzene)/M(CH)= 78.1 g/mol/13.0 g/mol = 6
So, molecular formula of benzene is C6H6.
Answer:
If the volume of a gas is decreased by changing the shape of the container, the number of collisions per area between the molecules and the container walls will increase. This corresponds to an increase in the gas pressure, which is the force exerted per area.
