TRUE you are correct but the earth is also made up of papyrus, an element.
Answer:
Explanation:
Here we have to use stoichiometry.
First of all, we have to calculate the mass of 100% of yield:
1.7 g ------- 98%
X -------- 100%
X = 1.73 g (approximately)
Second, we have to calculate the mass of N2 that is necessary to react to produce the mass of 1.73g of NH3. To do that, we have to use the Molar mass of N2 and NH3 and don't forget the stoichiometric relationship between them.
Molar Mass N2 : 14x2 = 28 g/mol
Molar Mass NH3: 14 + 3 = 17 g/mol
28g (N2) ------- 17x2 (NH3)
X ------------ 1.73 g
X = 1.42 g (approximately)
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.
Explanation:
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Answer: A. 11.15
Explanation:
The pH of a solution is given by the formular; pH = -log [H+].
pH = - log [ 7.0 x 10 ∧-12]
= 12 - log 7
= 12 - 0.845
= 11.15
