<span>Important information to solve the exercise :
Substance ΔHf (kJ/mol):
HCl(g)= −92.0 </span><span>kJ/mol
Al(OH)3(s)= −1277.0 </span><span><span>kJ/mol
</span> H2O(l)= −285.8 </span><span>kJ/mol
AlCl3(s) =−705.6 </span><span>kJ/mol
</span><span>Al(OH)3(s)+3HCl(g)→AlCl3(s)+3H2O(l)
reactants products
products- reactants:</span><span>
(−705.6) + (3 x −285.8) - ( −1277.0 ) - (3 x −92.0 ) = - 10.0 </span>kJ per mole at 25°C
<span>
</span>
Answer:
178.55
Explanation:
176
×
0.05+
177
×
0.19
+
178
×
0.27
+
179
×
0.14
+
180
×
0.35
=
178.55
Atom is the smallest indivisible particle of matter.
P = 2.30 atm
Volume in liter = 2.70 mL / 1000 => 0.0027 L
Temperature in K = 30.0 + 273 => 303 K
R = 0.082 atm
molar mass O2 = 31.9988 g/mol
number of moles O2 :
P * V = n * R* T
2.30 * 0.0027 = n * 0.082 * 303
0.00621 = n * 24.846
n = 0.00621 / 24.846
n = 0.0002499 moles of O2
Mass of O2:
n = m / mm
0.0002499 = m / 31.9988
m = 0.0002499 * 31.9988
m = 0.008 g
Answer:- C. H
Explanations:- Reduction is gain of electron. In other words we could say that decrease in oxidation number is reduction.
As per the rules, oxidation number of hydrogen in its compounds is +1(except metal hydrides) and the oxidation number of oxygen in its compounds is -2.
The oxidation number in elemental form is zero.
In 
, the oxidation number of H is +1 and oxidation number of O is -2. Oxidation number of Cl in 
is -1. On product side, the oxidation number of hydrogen in 
is zero and in 
the oxidation number of H is +1 and that of O is -2. Oxidation number of Cl in 
is 0.
From above data, Oxidation number of O is -2 on both sides so it is not reduced.
Oxidation number of Cl is changing from -1 to 0 which is oxidation.
Oxidation number of H is changing from +1 to 0 which is reduction.
So, the right choice is C.H
