Explanation:
(a) potassium oxide with water
According to reaction,1 mole of potassium oxide reacts with 1 mole of water to give 1 mole of potassium hydroxide.
(b) diphosphorus trioxide with water
According to reaction,1 mole of diphosphorus trioxide reacts with 2 moles of water to give 2 moles of phosphorus acid.
(c) chromium(III) oxide with dilute hydrochloric acid,
According to reaction,1 mole of chromium(III) oxide reacts with 6 moles of hydrochloric acid to give 2 moles of chromium(III) chloride and 3 moles of water.
(d) selenium dioxide with aqueous potassium hydroxide
According to reaction,1 mole of selenium dioxide reacts with 2 moles of potassium hydroxide to give 1 mole of potassium selenite and 1 mole of water.
Answer:
b) 3.10
Explanation:
HF ⇄ H
+ + F
Using Henderson-Hasselbalch Equation:
pH = pKa + log [A-]/[HA].
Where;
pKa = Dissociation constant = -log Ka
Hence, pKa of HF = -log 7.2 x 10^-4 = 3.14266
[A-] = concentration of conjugate base after dissociation = moles of base/total volume
= 0.15 x 0.3/0.8
= 0.05625 M
[HA] = concentration of the acid = moles of acid/total volume
= 0.10 x 0.5/0.8
= 0.0625 M
Note: <em>Total volume = 500 + 300 = 800 mL = 0.8 dm3</em>
pH = 3.14266 + log [0.05625/0.0625]
= 3.14267 + (-0.04575749056)
= 3.09691250944
<em>From all the available options below:</em>
<em>a) 2.97
</em>
<em>b) 3.10
</em>
<em>c) 3.19
</em>
<em>d) 3.22
</em>
<em>e) 3.32</em>
The correct option is b.
Answer:
-0.050 kJ/mol.K
Explanation:
- A certain reaction is thermodynamically favored at temperatures below 400. K, that is, ΔG° < 0 below 400. K
- The reaction is not favored at temperatures above 400. K, that is. ΔG° > 0 above 400. K
All in all, ΔG° = 0 at 400. K.
We can find ΔS° using the following expression.
ΔG° = ΔH° - T.ΔS°
0 = -20 kJ/mol - 400. K .ΔS°
ΔS° = -0.050 kJ/mol.K
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