Answer:
59.077 kJ/mol.
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>
k₂ = 3k₁ , Ea = ??? J/mol, R = 8.314 J/mol.K, T₁ = 294.0 K, T₂ = 308.0 K.
ln(3k₁/k₁) = (Ea / 8.314 J/mol.K) [(308.0 K - 294.0 K) / (294.0 K x 308.0 K)]
∴ ln(3) = 1.859 x 10⁻⁵ Ea
∴ Ea = ln(3) / (1.859 x 10⁻⁵) = 59.077 kJ/mol.
In chemistry, pH<span> is a numeric scale used to specify the acidity or basicity of an aqueous solution. It is approximately the negative of the base 10 logarithm of the concentration of hydronium ions. We calculate as follows:
pH = -log [H3O+]
pH = -log[</span><span>5.45 × 10–5 M]
pH = 4.3</span>
i think it is 8. I might be wrong.
Answer:
Here's what I get
Explanation:
A. Initial observation
Gary's shell had slime and an odour.
B. Independent variable
The independent variable is the one that the experimenter changes.
There are two independent variables: the rubbing with seaweed and the drinking of Dr. Kelp.
C. The dependent variable
The dependent variable is the amount of slime and odour.
D. The conclusion
Sponge Bob can conclude that rubbing the shell with seaweed and drinking Dr. Kelp removes the slime and odour.
However, this was a poorly designed experiment. He doesn't know if it is the seaweed or the Dr. Kelp that gives the result or if he must use both together. He should change only one independent variable at a time.
Answer:
V = 5.17L
Explanation:
Mass of gas = 8.7g
T = 23°C = (23 + 273.15)K = 296.15K
P = 1.15 atm
V = ?
R = 0.082atm.L / mol.K
From ideal gas equation
PV = nRT
P = pressure of the gas
V = volume of the gas
n = no. Of moles
R = ideal gas constant
T = temperature of the gas
no of moles = mass / molar mass
Molar mass of Chlorine = 35.5g / mol
No. Of moles = 8.7 / 35.5
No. Of moles = 0.245 moles
PV = nRT
V = nRT / P
V = (0.245 * 0.082 * 296.15) / 1.15
V = 5.9496 / 1.15
V = 5.17L
The volume of the gas is 5.17L
