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
False, the radiation from food and medical procedures have very little consequences. The radiation from a nuclear power plant can kill people which makes it a very large problem if not contained properly.
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Wurtz reaction is a special type of organic reaction involving the synthesis of aliphatic hydrocarbons from two molecules of an alkyl halide and two atoms of sodium in the presence of dry ether solution
Please bear in mind that wurtz reaction fails whenever tertiary alkyl halides are used.
An example of Wurtz reaction is given below:
2R – X + 2Na → R–R + 2Na + X−
<h3>What are organic compounds?</h3>
Organic compounds can simply be defined as those classes of organic molecules which contain carbon atoms covalently bonded to hydrogen atoms (C-H bonds).
Below are some few general characteristics of organic compounds:
- All organic compounds contain carbon.
- Most of them are flammable.
- They are all soluble in non-polar solvents
- Most organic compounds / substances are covalently bonded molecules
Some classes of organic compounds are:
So therefore, Wurtz reaction is a special type of organic reaction involving the synthesis of aliphatic hydrocarbons from two molecules of an alkyl halide and two atoms of sodium in the presence of dry ether solution
Learn more about organic compounds:
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Answer:
We need 78.9 mL of the 19.0 M NaOH solution
Explanation:
Step 1: Data given
Molarity of the original NaOH solution = 19.0 M
Molarity of the NaOH solution we want to prepare = 3.0 M
Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
Step 2: Calculate volume of the 19.0 M NaOH solution needed
C1*V1 = C2*V2
⇒with C1 = the concentration of the original NaOH solution = 19.0 M
⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED
⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M
⇒with V2 = the volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
19.0 M * V2 = 3.0 M * 0.500 L
V2 = (3.0 M * 0.500L) / 19.0 M
V2 = 0.0789 L
We need 0.0789 L
This is 0.0789 * 10^3 mL = 78.9 mL
We need 78.9 mL of the 19.0 M NaOH solution
Answer:
ΔH rxn = -1010 kJ/molC₂H₂
Explanation:
To obtain the enthalpy change for a reaction from bond energies what we do is to make an inventory of the bonds broken and formed for the balanced chemical reaction:
C₂H₂ + 5/2O₂ ⇒ 2CO₂ + H₂O
Bond Broken Bonds Formed
2 C-H + 1 C≡C + 5/2 O=O 4C=O + 2 H-O
Enthalpy bonds broken:
2 mol (456 kJ/mol)+ 1 mol (962 kJ/mol) + 5/2 mol (499 kJ/mol) = 3121.5 kJ
Enthalpy bond formed:
4 mol (802 kJ/mol) + 2 mol (462 kJ/mol) = 4132.0 kJ
ΔH rxn = H broken - H formed = 3121.5 kJ - 4132.0 kJ = - 1010 kJ (per mol C₂H₂ )
