<span>First divide the ionization energy by Avogadro's number to get the energy per atom of potassium;
</span>419 kj/mol / 6.023 x 10²³
= 4.19 x 10⁵ / 6.023 x 10²³ = 6.96 x 10⁻¹⁹
E = hc/λ
where lambda (λ<span>) is the wavelength, h is Planck's constant, c is the speed of light
</span>E = 6.96 x 10⁻¹⁹ j/atom<span>
h =</span>6.63x10⁻³⁴<span> Js
c = 3 x 10</span>⁸ m/s
λ = ?
λ = hc/E = (6.63x10⁻³⁴ x 3 x 10⁸ ) / 6.96 x 10⁻¹⁹ = 285.8nm = 286nm
<span>The longest wavelength of light capable of this ionization is 286nm.</span>
Answer:
Option a.
0.01 mol of CaCl₂ will have the greatest effect on the colligative properties, because it has the biggest i
Explanation:
To determine which of the solute is going to have a greatest effect on colligative properties we have to consider the Van't Hoff factor (i)
These are the colligative properties:
ΔP = P° . Xm . i → Lowering vapor pressure
ΔT = Kb . m . i → Boiling point elevation
ΔT = Kf . m . i → Freezing point depression
π = M . R . T → Osmotic pressure
Van't Hoff factor are the numbers of ions dissolved in the solution. For nonelectrolytes, the i values 1.
CaCl₂ and KNO₃ are two ionic solutes. They dissociate as this:
CaCl₂ → Ca²⁺ + 2Cl⁻
We have 1 mol of Ca²⁺ and 2 chlorides, so 3 moles of ions → i = 3
KNO₃ → K⁺ + NO₃⁻
We have 1 mol of K⁺ and 1 mol of nitrate, so 2 moles of ions → i = 2
Option a, is the best.
dim? im not so fluent in this but i did research yesterday
Metals are all in the center of the periodic table and are all made up of metals
Co2
Explanation:
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l)