The grams of oxygen that are required to produce 1 mole of H₂O is 16 g ( answer B)
<u><em> calculation</em></u>
2 CH₄ + 2NH₃ +3 O₂ → 2HCN + 6H₂O
step 1: use the mole ratio to find moles of O₂
from equation above the mole ratio of O₂: H₂O is 3:6 therefore the moles of O₂ = 1 mole x3/6 =0.5 moles
step 2: find mass of O₂
mass= moles x molar mass
from periodic table the molar mass of O₂ = 16 x2= 32 g/mol
mass O₂ = 0.5 moles x 32 g/mol = 16 g (answer B)
Answer:
∆H > 0
∆Srxn <0
∆G >0
∆Suniverse <0
Explanation:
We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.
Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.
The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.
Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.
The answer is B) fills all the space in its container
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.
Answer:
Explanation:
There's 2 steps: first find the volume then the mass
volume=lwh
convert 20 cm to mm: 20x10=200mm
volume=(170)(200)(0.99)
volume=33,660 mm
since the density is in cm cubed, convert 33,660 mm to cm
33,660/10=3366
density=mass/volume, rearrange to get mass=density x volume
mass=19.3 x 3366
mass=64963.8
