Answer:
D
Explanation:
Meteorologists use this symbol to show that the front is a stationary front.
Answer:
A) [H3PO4] will increase, [KH2PO4] will decrease, and pH will slightly decrease.
Explanation:
A buffer is a solution which resists changes to its pH when a small amount of acid or base is added to it.
Buffers consist of a weak acid (HA) and its conjugate base (A–) or a weak base and its conjugate acid. Weak acids and bases do not completely dissociate in water, and instead exist in solution as an equilibrium of dissociated and undissociated species. When a small quantity of a strong acid is added to a buffer solution, the conjugate base, A-, reacts with the hydrogen ions from the added acid to form the weak acid and a salt thereby removing the extra hydrogen ions from the solution and keeping the pH of the solution fairly constant. On the other hand, if a small quantity of a strong base is added to the buffer solution, the weak acid dissociates further to release hydrogen ions which then react with the hydroxide ions of the added base to form water and the conjugate base.
For example, if a small amount of strong acid is added to a buffer solution that is 0.700 M H3PO4 and 0.700 M KH2PO4, the following reaction is obtained:
KH₂PO₄ + H+ ----> K+ + H₃PO₄
Therefore, [H₃PO₄] will increase, [KH₂PO₄] will decrease, and pH will slightly decrease.:
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.