Answer:
If the volume of the container is decreased by a factor of 2 the pressure is is increased by the same factor to 1664 torr.
Explanation:
Here we have Boyle's law which states that, at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure
V ∝ 1/P or V₁·P₁ = V₂·P₂
Where:
V₁ = Initial volume
V₂ = Final volume = V₁/2
P₁ = Initial pressure = 832 torr
P₂ = Final pressure = Required
From V₁·P₁ = V₂·P₂ we have,
P₂ = V₁·P₁/V₂ = V₁·P₁/(V₁/2)
P₂ = 2·V₁·P₁/V₁ = 2·P₁ = 2× 832 torr = 1664 torr
Answer:
Fossil B
Explanation:
Fossil B is older because it's later is at the bottom
Answer:
(a) ΔSº = 216.10 J/K
(b) ΔSº = - 56.4 J/K
(c) ΔSº = 273.8 J/K
Explanation:
We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.
First we need to find in an appropiate reference table the standard molar entropies entropies, and then do the calculations.
(a) C2H5OH(l) + 3 O2(g) ⇒ 2 CO2(g) + 3 H2O(g)
Sº 159.9 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 2(213.8) + 3(188.8) ] - [ 159.9 + 3(205.) ] J/K
ΔSº = 216.10 J/K
(b) CS2(l) + 3 O2(g) ⇒ CO2(g) + 2 SO2(g)
Sº 151.0 205.2 213.8 248.2
(J/Kmol)
ΔSº = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K
(c) 2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)
Sº 173.3 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K
Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4 total mol gas reactants to 3, so the entropy change will be negative.
Note we need to multiply the entropies of each substance by its coefficient in the balanced chemical equation.
Radiation or heat by the sun
