The given reaction is a combustion reaction, since a hydrocarbon is burning in presence of oxygen
This
can be solved using Dalton's Law of Partial pressures. This law states that the
total pressure exerted by a gas mixture is equal to the sum of the partial
pressure of each gas in the mixture as if it exist alone in a container. In
order to solve, we need the partial pressures of the gases given. Calculations
are as follows:<span>
P = P1 + P2 + P3
<span>P = (0.08206 atm.L/mol.K)( 298.15
K)/1.50) x (0.158 mole + 0.09 mol + 0.044 mol) = <span>4.76
atm</span></span></span>
Answer:
(a) Benzene = 0.26; toluene = 0.74
(b) Benzene = 0.55
Explanation:
1. Calculate the composition of the solution
For convenience, let’s call benzene Component 1 and toluene Component 2.
According to Raoult’s Law,
where
p₁ and p₂ are the vapour pressures of the components above the solution
χ₁ and χ₂ are the mole fractions of the components
p₁° and p₂° are the vapour pressures of the pure components.
Note that
χ₁ + χ₂ = 1
So,
χ₁ = 0.26 and χ₂ = 0.74
2. Calculate the mole fraction of benzene in the vapour
In the liquid,
p₁ = χ₁p₁° = 0.26 × 75 mm = 20 mm
∴ In the vapour
Note that the vapour composition diagram below has toluene along the horizontal axis. The purple line is the vapour pressure curve for the vapour. Since χ₂ has dropped to 0.45, χ₁ has increased to 0.55.
<span> The nuclear reactor powers plants to inhale heats that are needed to create/produce steam- (a gas that comes from liquid at its boiling point) The process is called fission. Fission entails the splitting on atoms in the nuclear reactor.
</span>
Answer:
The volume when the conditions were altered is 0.5109 L or 510.9 mL
Explanation:
Using the general gas equation,
P1 V1 / T1 = P2 V2 / T2
where;
P1 = 756 mmHg
V1 = 475 ml = 0.475 L
T1 = 23.5°C = 23.5 + 273K = 275.5 K
P2 = 722 mm Hg
T2 = 10°C = 10 + 273 K = 283 K
V2 = ?
Rearranging to make V2 the subject of the formula, we obtain:
V2 = P1 V1 T2 / P2 T1
V2 = 756 * 0.475 * 283 / 722 * 275.5
V2 = 101, 625.3 / 198911
V2 = 0.5109 L or 510.9 mL
