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>
Answer:
To get number of neutrons, you must have the mass number and atomic number of that atom.

atomic number is number of protons
a region around a magnetic material or a moving electric charge within which the force of magnetism acts.
Answer:
Explanation:
Combustion reaction is given below,
C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)
Provided that such a combustion has a normal enthalpy,
ΔH°rxn = -1270 kJ/mol
That would be 1 mol reacting to release of ethanol,
⇒ -1270 kJ of heat
Now,
0.383 Ethanol mol responds to release or unlock,
(c) Determine the final temperature of the air in the room after the combustion.
Given that :
specific heat c = 1.005 J/(g. °C)
m = 5.56 ×10⁴ g
Using the relation:
q = mcΔT
- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT
ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005
ΔT= 836.88 °C
ΔT= T₂ - T₁
T₂ = ΔT + T₁
T₂ = 836.88 °C + 21.7°C
T₂ = 858.58 °C
Therefore, the final temperature of the air in the room after combustion is 858.58 °C