Answer:
k₂ = 4.06 x 10⁻² s⁻¹.
Explanation:
- From Arrhenius law: <em>K = Ae(-Ea/RT)</em>
where, K is the rate constant of the reaction.
A is the Arrhenius factor.
Ea is the activation energy.
R is the general gas constant.
T is the temperature.
- At different temperatures:
<em>ln(k₂/k₁) = Ea/R [(T₂-T₁)/(T₁T₂)]</em>
k₁ = 5.8 × 10⁻³ s⁻¹, k₂ = ??? , Ea = 33600 J/mol, R = 8.314 J/mol.K, T₁ = 298.0 K, T₂ = 348.0 K.
- ln(k₂/5.8 × 10⁻³ s⁻¹) = (33600 J/mol / 8.314 J/mol.K) [(348.0 K - 298.0 K) / (298.0 K x 348.0 K)] = (4041.37) (4.82 x 10⁻⁴) = 1.9479.
- Taking exponential of both sides:
(k₂/5.8 × 10⁻³ s⁻¹) = 7.014.
∴ k₂ = 4.06 x 10⁻² s⁻¹.
Answer:
When dealing with the force of gravity between two objects, there are only two things that are important – mass, and distance. The force of gravity depends directly upon the masses of the two objects, and inversely on the square of the distance between them.
For 1 mole of any substance, there are 6.02214086 × 10^<span>23 particles (atoms, molecules, ions, etc. depending on the substance)
So, for 1 mole of water, there are </span>6.02214086 × 10^23 water particles, and so on
So, in 1 mole of Lithium, there are 6.02214086 × 10^23 Lithium atoms
Answer:
3.50*10^-11 mol3 dm-9
Explanation:
A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, Ag2C2O4, at 25°C. The measured potential difference between the rod and the SHE is 0.5812 V, the rod being positive. Calculate the solubility product constant for silver oxalate.
Ag2C2O4 --> 2Ag+ + C2O4 2-
So Ksp = [Ag+]^2 * [C2O42-]
In 1 L, 2.06*10^-4 mol of silver oxalate dissolve, giving, the same number of mol of oxalate ions, and twice the number of mol (4.12*10^-4) of silver ions.
So Ksp = (4.12*10^-4)^2 * (2.06*10^-4)
= 3.50*10^-11 mol3 dm-9
