Answer:
[H2]eq = 0.0129 M
[F2]eq = 1.0129 M
[HF]eq = 0.9871 M
Explanation:
∴ Ke = [HF]² / [H2]*[F2] = 1.15 E2
experiment:
∴ n H2 = 3.00 mol
∴ n F2 = 6.00 mol
∴ V sln = 3.00 L
⇒ [H2]i = 3.00 mol / 3.00 L = 1 M
⇒ [F2]i = 6.00 mol / 3.00 L = 2 M
[ ]i change [ ]eq
H2 1 1 - x 1 - x
F2 2 2 - x 2 - x
HF - x x
⇒ K = (x)² / (1 - x)*(2 - x) = 1.15 E2
⇒ x² / (2 - 3x + x²) = 1.15 E2 = 115
⇒ x² = (2 - 3x + x²)(115)
⇒ x² = 230 - 345x + 115x²
⇒ 0 = 230 - 345x + 114x²
⇒ x = 0.9871
equilibrium:
⇒ [H2] = 1 - x = 1 - 0.9871 = 0.0129 M
⇒ [F2] = 2 - x = 2 - 0.9871 = 1.0129 M
⇒ [HF] = x = 0.9871 M
Answer:
The atom is divisible particle and can be subdivided into smaller particles proton, neutron and electrons was not stated by John dalton.
Explanation:
The postulate of Dalton's atomic theory that atom is indivisible particle and can not be subdivided into smaller particles was later changed because atom can be divided into neutrons, protons and electrons.
Electron:
The electron is subatomic particle that revolve around outside the nucleus and has negligible mass. It has a negative charge.
Symbol= e⁻
Mass= 9.10938356×10⁻³¹ Kg
It was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.
Proton and neutron:
While neutron and proton are present inside the nucleus. Proton has positive charge while neutron is electrically neutral. Proton is discovered by Rutherford while neutron is discovered by James Chadwick in 1932.
Symbol of proton= P⁺
Symbol of neutron= n⁰
Mass of proton=1.672623×10⁻²⁷ Kg
Mass of neutron=1.674929×10⁻²⁷ Kg
All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron.
Answer:
C
Explanation:
will silver wires conduct electricity better than copper wires ?
Answer: The first isotope has a relative abundance of 79% and last isotope has a relative abundance of 11%
Explanation: Given that the average atomic mass(M) of magnesium
= 24.3050amu
Mass of first isotope (M1) = 23.9850amu
Mass of middle isotope (M2)=24.9858amu
Mass of last isotope(M3)= 25.9826amu
Total abundance = 1
Abundance of middle isotope = 0.10
Let abundance of first and last isotope be x and y respectively.
x+0.10+y =1
x = 0.90-y
M = M1 × % abundance of first isotope + M2 × % of middle isotope +M3 ×% of last isotope
24.03050= 23.985× x + 24.9858 ×0.10 + 25.9826×y
Substitute x= 0.90-y
Then
y = 0.11
Since y=0.11, then
x= 0.90-0.11
x=0.79
Therefore the relative abundance of the first isotope = 11% and the relative abundance of the last isotope = 79%
Answer:
4Fe + 3O₂ → 2Fe₂O₃
Explanation:
Fe → ²⁺
O → ²⁻
But Iron III is Fe³⁺
So we have Fe³⁺ and O²⁻, the formula for the oxide must be Fe₂O₃ so the equation can be:
4Fe + 3O₂ → 2Fe₂O₃
