As we know,
1 D = 3.34 × 10⁻³⁰ C.m
So,
1.44 D = ?
Solving for 1.44 D,
= (3.34 × 10⁻³⁰ C.m × 1.44 D) ÷ 1 D
1.44 D = 4.80 × 10⁻³⁰ C.m
Dipole Moment is given as,
Dipole Moment = q × r
Solving for q,
q = Dipole Moment / r ------ (1)
Where,
Dipole Moment = 4.80 × 10⁻³⁰ C.m
r = 163 pm = 1.63 × 10⁻¹⁰ m
Putting values in eq. 1,
q = 4.80 × 10⁻³⁰ C.m / 1.63 × 10⁻¹⁰ m
q = 2.94 × 10⁻²⁰ C
As,
1.602 × 10⁻¹⁹ C = 1 e⁻
So,
2.94 × 10⁻²⁰ C = X e⁻
Solving for X,
X = (2.94 × 10⁻²⁰ C × 1 e⁻) ÷ 1.602 × 10⁻¹⁹ C
= 0.183 e⁻
Result:
So one element is containing + 0.183 e⁻ while the other element is containing - 0.183 e⁻.
The answer would be A, 100g since it’s equal
radiation is energy that comes from a source. (travels through the speed of light)
Answer:
32
Explanation:
0.56dm of gas has 0.8g
22.4dm of gas Z has a mass of 22.4/0.56X0.8
the relative molecular mass of gas Z is 32
<span> Au</span>₂(SeO₄)₃
O = -2 × 4 = -8
Se = + 6
So,
(+6 - 8) = -2
Means (SeO₄) contains -2 charge, Now multiply -2 by 3
-2 ₓ 3 = -6
Means,
Au₂ + (-6) = 0
Au₂ = +6
Or,
Au = 6 / 2
Au = +3
Result:
Au = +3
Se = +6
O = -2
Ni(CN)₂
Cyanide (CN⁻) contains -1 charge,
So,
N = -3
C = +2
Then,
Ni + (-1)₂ = 0
Ni - 2 = 0
Or,
Ni = +2
Result:
N = -3
C = +2
Ni = +2
