Suppose a helium-3 nuclide transforms into a helium-4 nuclide by absorbing a proton and emitting a positron. Complete the nuclea
1 answer:
<u>Answer:</u> The nuclear equation for the conversion of He-3 nuclide to He-4 nuclide is given above.
<u>Explanation:</u>
Nuclear reaction are defined as the reactions in which nucleus of an atom is involved.
Positron emission is defined as the emission process in which positron particle is emitted. In this process, a proton gets converted to neutron and an electron neutrino particle.
The chemical equation for the reaction of He-3 with a proton follows:
Hence, the nuclear equation for the conversion of He-3 nuclide to He-4 nuclide is given above.
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Ni(OH)₂ ⇄ Ni⁺² + 2 OH⁻
Ksp = [Ni⁺²][OH⁻]² = S (2S)² = 4S³
where S is molar solubility.
at pH = 10
[H⁺] = 10⁻¹⁰
[H⁺][OH⁻] = 10⁻¹⁴
so [OH⁻] = 10⁻⁴ M
Ksp = S [10⁻⁴ + 2S]²
Ksp is very small so the molar solubility of OH⁻ will be very small
so (10⁻⁴ + 2S) is about 10⁻⁴
so Ksp = S x 10⁻⁸
S =
= 6 x 10⁻⁸ M
Ksp = [Ni⁺²][OH⁻]² = S (2S)² = 4S³
where S is molar solubility.
at pH = 10
[H⁺] = 10⁻¹⁰
[H⁺][OH⁻] = 10⁻¹⁴
so [OH⁻] = 10⁻⁴ M
Ksp = S [10⁻⁴ + 2S]²
Ksp is very small so the molar solubility of OH⁻ will be very small
so (10⁻⁴ + 2S) is about 10⁻⁴
so Ksp = S x 10⁻⁸
S =