Answer:
The correct option is C. 21900.3. I calculated 21945 J, which makes option C closely aligned with my result.
Explanation:
Data
mass = 150 g
initial temperature T1 = 10°C
final temperature T2 = 45°C
Cw = 4.18 J/g°C
Formula
Q = mCΔT = mC(T2 - T1)
Substitution
Q = (150)(4.18)(45 - 10)
Simplification
Q = (150)(4.18)(35)
Result
Q = 21945 J
N₀ signifies the quantity of C-14 atoms per kg of carbon in the original sample at time = 0 seconds, when the carbon composition matched that in today’s atmosphere. As time progresses to ts, the number of C-14 atoms per kg declines to N, due to radioactive decay. λ indicates the decay constant.
Hence, we have N = N₀e - λt, which is the equation for radioactive decay. Rearranging gives us N₀/N = e λt, or In(N₀/N) = - λt, which becomes equation 1.
The sample contains mc kg of carbon, leading to an activity measured as A/mc decay per kg. The variable r represents the initial mass of C-14 in the sample at t=0 relative to the total mass of carbon which is calculated as [(total number of C-14 atoms at t = 0) × ma] / total mass of carbon. Thus, N₀ equates to r/ma, which becomes equation 2.
The activity of the radioactive element is directly related to the atom count at the moment. The activity equation A = dN/dt = λ(N) indicates that: A = λ₁(N × mc). Rearranging provides N = A / (λmc), represented in equation 3.
By integrating equations 2 and 3, we can solve for t yielding
t = (1/λ) In(rλmc/m₀A).
Based on the equation:
ΔG = ΔH - TΔS = 0
It follows that ΔS = ΔH/T
So, ΔS = n*ΔHVap / Tvap
- where n represents the number of moles calculated as mass/molar mass
For a mass of 24.1 g
and a molar mass of 187.3764 g/mol
substituting gives:
∴ n = 24.1 / 187.3764g/mol
= 0.129 moles
The molar enthalpy of vaporization, ΔHvap, is 27.49 kJ/mol
The temperature in Kelvin, Tvap = 47.6 + 273 = 320.6 K
After substitution, we compute ΔS, the change in entropy:
∴ΔS = 0.129 mol * 27490 J/mol / 320.6 K
= 11 J/K
Answer: The change in enthalpy will be -13.
Explanation:-
Endothermic reactions absorb heat, while exothermic reactions release heat. In the case of an endothermic reaction, the change in enthalpy is represented as positive, whereas for an exothermic reaction, it is negative.
When one mole of A combines with one mole of B to form three moles of C
So the stoichiometric ratio being halved also results in the enthalpy for the reaction being halved.
Thus, for this reaction:
The resulting change in enthalpy is -13.
