The oxidation number of elements in equation below are,
4NH₃ + 3Ca(ClO)₂ → 2N₂ + 6H₂O + 3CaCl₂
O.N of N in NH₃ = -3
O.N of Ca in Ca(ClO)₂ and CaCl₂ = +2
O.N of N in N₂ = 0
O.N of Cl in Ca(ClO)₂ = +1
O.N of Cl in CaCl₂ = -1
Oxidation:
Oxidation number of Nitrogen is increasing from -3 (NH₃) to 0 (N₂).
Reduction:
Oxidation number of Cl is decreasing from +1 [Ca(ClO)₂] to -1 (CaCl₂).
Result:
<span>N is oxidized and Cl is reduced.</span>
Answer:
The molality of isoborneol in camphor is 0.53 mol/kg.
Explanation:
Melting point of pure camphor= T =179°C
Melting point of sample = 
= 165°C
Depression in freezing point = 
Depression in freezing point is also given by formula:
= The freezing point depression constant
m = molality of the sample
i = van't Hoff factor
We have: = 40°C kg/mol
i = 1 ( organic compounds)
The molality of isoborneol in camphor is 0.53 mol/kg.
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Explanation:
To begin calculating, there is one thing you need to remember :
Then we have
As you know decomposition of 2moles now has prodused <span>196kj
So, </span><span>q is made due </span>
I'm sure it will help.
Answer:
Atomic mass of nitrogen = 14.0067 amu.
Explanation:
Isotopes can be defined as two or more forms of a chemical element that are made up of equal numbers of protons and electrons but different numbers of neutrons.
Generally, the isotopes of a chemical element have the same chemical properties because of their atomic number but different physical properties due to their atomic weight (mass number).
The two isotopes of nitrogen are nitrogen-14 and nitrogen-15.
Given the following data;
Relative abundance of N-14 = 99.63%
Atomic mass of N-14 = 14.003
Relative abundance of N-15 = 0.37%
Atomic mass of N-15 = 15.000
The atomic mass is;
14.003 × (99.63/100) + 15.000 × (0.37/100)
Atomic mass = 14.003 × (0.9963) + 15.000 × (0.0037)
Atomic mass = 13.9512 + 0.0555
Atomic mass = 14.0067 amu.
<em>Therefore, the atomic mass of nitrogen is 14.0067 amu. </em>
