Answer:
-1.94 * 10^-18 J
Since the electron moved from a higher to a lower energy level (n = 3 → n = 1) it is an emission.
Explanation:
From Rydberg equation;
E = -RH(1/n^2final - 1/n^2initial)
For a transition from n = 3 → n = 1
RH = 2.18 * 10^-18 J
E = -(2.18 * 10^-18) (1/1^2 - (1/3^2)
E = -(2.18 * 10^-18) (1-1/9)
E= -(2.18 * 10^-18) (8/9)
E = -1.94 * 10^-18 J
Because the resulting hybridized orbitals are more stable
A) energy is converted from one form to another
Answer:
0.33 mol/kg NH₃
Explanation:
Data:
b(NH₃) = 0.33 mol/kg
b(Na₂SO₄) = 0.10 mol/ kg
Calculations:
The formula for the boiling point elevation ΔTb is
i is the van’t Hoff factor — the number of moles of particles you get from a solute.
(a) For NH₃,
The ammonia is a weak electrolyte, so it exists almost entirely as molecules in solution.
1 mol NH₃ ⟶ 1 mol particles
i ≈ 1, and ib = 1 × 0.33 = 0.33 mol particles per kilogram of water
(b) For Na₂SO₄,
Na₂SO₄(aq) ⟶ 2Na⁺(aq) + 2SO₄²⁻(aq)
1 mol Na₂SO₄ ⟶ 3 mol particles
i = 1 and ib = 3 × 0.10 = 0.30 mol particles per kilogram of water
The NH₃ has more moles of particles, so it has the higher boiling point.
For this problem we can use half-life formula and radioactive decay formula.
Half-life formula,
t1/2 = ln 2 / λ
where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days
Hence,
8.04 days = ln 2 / λ
λ = ln 2 / 8.04 days
Radioactive decay law,
Nt = No e∧(-λt)
where, Nt is amount of compound at t time, No is amount of compound at t = 0 time, t is time taken to decay and λ is radioactive decay constant.
Nt = ?
No = 1.53 mg
λ = ln 2 / 8.04 days = 0.693 / 8.04 days
t = 13.0 days
By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg
Hence, mass of remaining sample after 13.0 days = 0.499 mg
The answer is "e"
