Answer:
a. in pure water Solubility (x) = 1.26 x 10⁻⁴M
b. in 0.202M M⁺² Solubility (x) = 9.963 x 10⁻¹²M
The large drop in solubility is consistent with the common ion effect.
Explanation:
a. Solubility in pure water
Given: M(OH)₂ ⇄ M⁺² + 2OH⁻
I --- 0 0
C --- x 2x
E --- x 2x
Ksp = [M⁺²][OH⁻]² = (x)(2x)² = 4x³ => x = CubeRt(Ksp/4)
solubility in pure water = x = CubeRt(8.05 x 10⁻¹²/4) = 1.26 x 10⁻⁴M
b. Solubility in presence of 0.202M M⁺² as common ion.
Given: M(OH)₂ ⇄ M⁺² + 2OH⁻
I --- 0.202M 0
C --- +x +2x
E --- 0.202M + x 2x
≈ 0.202M
Ksp = [M⁺²][2x]² = (0.202)(2x)² = (0.202)(4x²) = 8.05 x 10⁻¹²
=> x = (8.05 x 10⁻¹²)/(0.202)(4) = 9.963 x 10⁻¹²M
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The answer is (2). You can think about this question in terms of the Bohr's model of the atom or in terms of quantum chemistry. In the Bohr model, electrons exist in discrete "shells," each respresenting a fixed spherical distance from the nucleus in which electrons of certain energy levels orbit the nucleus. The larger the shell (the greater the "orbit" radius), the greater the energy of the "orbiting" electron (I use quotations because electrons don't actually orbit the nucleus in the traditional sense, as you may know). Thus, according to the Bohr model, a third shell electron should be farther from the nucleus and have greater energy than an electron in the first shell.
The quantum model is differs drastically from the Bohr model in many ways, but the essence is the same. A larger principal quantum number indicates 1) greater overall energy and 2) a probability distribution spread a bit more outward.
The answer for the problem is explained below.
The option for the answer is "D".
<u><em>Therefore the energy of the light is 4.25 × 10^-19 J</em></u>
Explanation:
Given:
wavelength (λ) = 468 nm = 468×10^-9 m
speed of light (c) = 3.00 x 10^8m/s
Planck's constant is 6.626 x 10^-34J·s
To solve:
energy of light (E)
We know,
E =(h×c) ÷ λ
E = ( 6.626 x 10^-34 × 3.00 x 10^8) ÷ 468×10^-9
E = 4.25 × 10^-19 J
<u><em>Therefore the energy of the light is 4.25 × 10^-19 J</em></u>
