For the given reaction, according to the Law of Conservation of Energy, the energy required to decompose Hcl and produce are equal.
Answer: Option C
<u>Explanation:</u>
According to law of conservation's of energy, energy can only be transferred from reactants to product side. So in this process, it is stated that 185 kJ of energy will be needed to decompose it. So that 185 kJ of energy will be getting transferred to produce the creation of hydrogen and chloride in the product side.
So if we see from the reactants side, the energy of 185 kJ is required for decomposition of hydrogen chloride. Similarly, if we see from the product side, the 185 kJ utilized for decomposition is transferred as energy required to create hydrogen and chlorine atoms. This statement will be in accordance with the law of conservation's of energy.
The volume of HCl required is 23.89 mL
Calculation of volume:
The reaction:
As HCl and NaOH react in 1 : 1 ratio.
Volume of NaOH= 718 mL
Concentration= 0.183M
Volume of HCl= ?
Concentration= 5.50M
Using the dilution formula:
Therefore,
Volume of HCl required will be 23.89 mL.
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<u>Answer:</u> S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction
<u>Explanation:</u>
The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when oxidation number of a species increases.
A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when oxidation number of a species decreases.
For the given chemical reaction:
<u>On the reactant side:</u>
Oxidation number of H = +1
Oxidation number of N = +5
Oxidation number of O = -2
Oxidation number of S = 0
<u>On the product side:</u>
Oxidation number of H = +1
Oxidation number of N = +2
Oxidation number of O = -2
Oxidation number of S = +6
As the oxidation number of S is increasing from 0 to +6. Thus, it is getting oxidized. Similarly, the oxidation number of N is decreasing from +5 to +2. Thus, it is getting reduced.
The oxidation numbers of O and H remain the same on both sides of the reaction. Thus, they are neither getting oxidized or reduced.
Hence, S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction