Answer:
The rate law may be determined only by experiment.
Explanation:
For a reaction, A + B ---> C, the rate law can only be determined from experimental data. Chemists determine the rate of reaction by carefully observing the changes in the concentration of species as the reaction progresses.
Hence, the rate law is not determined by inspection of the chemical reaction equation, it must be obtained from the experimental data, hence the answer given.
Answer:
The amount of NaF produced is doubled.
(d) is correct option.
Explanation:
Given that,
A 2 mole sample of F₂ reacts with excess NaOH according to the equation.
The balance equation is
If the reaction is repeated with excess NaOH but with 1 mole of F₂
The balance equation is
Hence, The amount of NaF produced is doubled.
(d) is correct option.
Answer:
ΔH°c = -2219.9 kJ
Explanation:
Let's consider the combustion of propane.
C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(l)
We can find the standard enthalpy of the combustion (ΔH°c) using the following expression.
ΔH°c = [3 mol × ΔH°f(CO₂(g)) + 4 mol × ΔH°f(H₂O(l))] - [1 mol × ΔH°f(C₃H₈(g)) + 5 mol × ΔH°f(O₂(g))]
ΔH°c = [3 mol × (-393.5 kJ/mol) + 4 mol × (-285.8 kJ/mol)] - [1 mol × (-103.8 kJ/mol) + 5 mol × (0 kJ/mol)]
ΔH°c = -2219.9 kJ
Answer:
104.84 moles
Explanation:
Given data:
Moles of Boron produced = ?
Mass of B₂O₃ = 3650 g
Solution:
Chemical equation:
6K + B₂O₃ → 3K₂O + 2B
Number of moles of B₂O₃:
Number of moles = mass/ molar mass
Number of moles = 3650 g/ 69.63 g/mol
Number of moles = 52.42 mol
Now we will compare the moles of B₂O₃ with B from balance chemical equation:
B₂O₃ : B
1 : 2
52.42 : 2×52.42 = 104.84
Thus from 3650 g of B₂O₃ 104.84 moles of boron will produced.
Answer:
A = 679.2955 ppm
Explanation:
In this case, we already know that 64Cu has a half life of 12.7 hours. The expression to use to calculate the remaining solution is:
A = A₀ e^-kt
This is the expression to use. We have time, A₀, but we do not have k. This value is calculated with the following expression:
k = ln2 / t₁/₂
Replacing the given data we have:
k = ln2 / 12.7
k = 0.0546
Now, let's get the concentration of Cu:
A = 845 e^(-0.0546*4)
A = 845 e^(-0.2183)
A = 845 * 0.8039
A = 679.2955 ppm
This would be the concentration after 4 hours
