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
P = 11.133 atm (purple)
T = -236.733 °C(yellow)
n = 0.174 mol(red)
<h3>Further explanation </h3>
Some of the laws regarding gas, can apply to ideal gas (volume expansion does not occur when the gas is heated),:
- Boyle's law at constant T, P = 1 / V
- Charles's law, at constant P, V = T
- Avogadro's law, at constant P and T, V = n
So that the three laws can be combined into a single gas equation, the ideal gas equation
In general, the gas equation can be written
where
P = pressure, atm
V = volume, liter
n = number of moles
R = gas constant = 0.08206 L.atm / mol K
T = temperature, Kelvin
To choose the formula used, we refer to the data provided
Because the data provided are temperature, pressure, volume and moles, than we use the formula PV = nRT
T= 10 +273.15 = 373.15 K
V=5.5 L
n=2 mol
V=8.3 L
P=1.8 atm
n=5 mol
T = 12 + 273.15 = 285.15 K
V=3.4 L
P=1.2 atm
MnCl2(aq) is an ionic compound which will have the releasing of 2 Cl⁻ ions ions in water for every molecule of MnCl2 that dissolves.
MnCl2(s) --> Mn+(aq) + 2 Cl⁻(aq)
[Cl⁻] = 0.92 mol MnCl2/1L × 2 mol Cl⁻ / 1 mol MnCl2 = 1.8 M
The answer to this question is [Cl⁻] = 1.8 M
Hi, your answer is correct.