<span>If a mole of aluminum weighs 26.98 grams, that means 1 atom of aluminum weighs = (26.98 g/mole) / (6.023 x 10^23 atoms/mole) = 4.479 x 10^-23 grams,
</span>so, it is not possible because 1 atom weighs that much we calculated which is <span>almost 100 times more than the amount you mentioned</span>
Answer 1) : The density of the hot air inside the balloon can be found out by using ideal gas equation;
PV = nRT;
As n is number of moles and in gases, number of moles along with mass per mole is equal to the density of the gas.
If the moles in the gas are more the density will be more.
here, density (ρ) = mass (m) / volume (V); substituting in the ideal gas equation we get,
ρ = mP / RT
Answer 2) ρ (hot air) = ρ (cold air) X
Here according to the formula because T(hot air) >T(cold air),
So, the density of hot air greater than the density of cold air.
The relationship between the ρ (h) = ρ(c) X
Answer:
ΔG°rxn = -72.9 kJ
Explanation:
Let's consider the following reaction.
HCN(g) + 2 H₂(g) → CH₃NH₂(g)
We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression:
ΔG°rxn = ΔH° - T.ΔS°
where,
ΔH° is the standard enthalpy of the reaction
T is the absolute temperature
ΔS° is the standard entropy of the reaction
ΔG°rxn = -158.0 KJ - 387 K × (-219.9 × 10⁻³ J/K)
ΔG°rxn = -72.9 kJ
So here's how you find the answer:
Given: (rate constants)
K₁ = 4.0 x 10⁻⁴ M⁻¹s⁻¹.
T₁ = 25.0 C = 293 K.
k₂ = 2.6 x10⁻³ M⁻¹s⁻¹.
T₂ = 42.4 C = 315.4 K.
R = 8,314 J/K·mol.
Use the equation:
ln(k₁/k₂) = Ea/R (1/T₂ - 1/T₁).
Transpose:
Ea = R·T₁·T₂ / (T₁ - T₂) · ln(k₁/k₂)
Substitute within the given transposed equation:
<span>Ea = 8,314 J/K·mol · 293 K · 315.4 K ÷ (293 K - 315.4 K) · ln (4.0 x 10⁻⁴ M⁻¹s⁻¹/ 2.6 x 10⁻³ M⁻¹s⁻¹).
</span>
Continuing the solution we get:
<span>Ea = 768315 J·K/mol ÷ (-22,4 K) * (-1.87)
</span>
The value of EA is:
<span>Ea = 64140.58 J/mol ÷ 1000 J/kJ = 64.140 kJ/mol.</span>
Answer:
the partial pressure of Xe is 452.4 mmHg
Explanation:
Dalton's law of partial pressures says that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases.
The partial pressures can be calculated with the molar fraction of the gas, in this case, Xe.
Molar fraction of Xe is calculated as follows:


Then, 0.29 is the molar fraction of Xe in the mixture of gases given.
To know the parcial pressure of Xe, we have to multiply the molar fraction by the total pressure:
Partial Pressure of Xe=1560mmHg*0.29
Partial Pressure of Xe=452.4mmHg