Oxidation numbers play an important role in the systematic nomenclature of chemical compounds. By definition, the oxidation number of an atom is the charge that atom would have if the compound was composed of ions. The oxidation number of simple ions is equal to the charge on the ion. The oxidation number of sodium in the Na+<span> ion is +1.</span>
Answer:
50 mL; 7
Explanation:
By looking at the graph, the boundary point where the solution turns from acidic to basic appears at approximately 50 mL.
Because this is a titration between a strong acid and strong base, the pH of the equivalence point is always 7 (aka neutral).
to solve the amount of sulfuric acid to maximize the profit
we need to establish the profit
let S(x) = 100x the selling price
P = S(x) – C(x)
P = 100x – ( 100,000 + 50x + 0.0025x^2)
P = 50x – 100000 – 0.0025x^2
Getting the first derivative and equating it to 0
0 = 50 – 0.005x
X = 10000 units of sulfuric acid
But the capacity is only 7000 units
<span>So 7000 units should be manufactured</span>
(2) U-238 is the radioisotope used in dating geological formations. Uranium-238 has a half-life of 4.46 billion years, which makes it suitable to date such old formations.
The other options do not have such long durabilities and half-lives.
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.
