Answer:
A. ΔG° = 132.5 kJ
B. ΔG° = 13.69 kJ
C. ΔG° = -58.59 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.
ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)
where,
n: moles
ΔH°f: standard enthalpy of formation
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))
ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)
ΔH° = 178.3 kJ
We can calculate the standard entropy of the reaction (ΔS°) using the following expression.
ΔS° = ∑np . S°p - ∑nr . S°r
where,
S: standard entropy
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))
ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)
ΔS° = 160.6 J/K. = 0.1606 kJ/K.
We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.
ΔG° = ΔH° - T.ΔS°
where,
T: absolute temperature
<h3>A. 285 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ
<h3>B. 1025 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ
<h3>C. 1475 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ
<h3>
Answer:</h3>
The centripetal acceleration is 26.38 m/s²
<h3>
Explanation:</h3>
We are given;
- Mass of rubber stopper = 13 g
- Length of the string(radius) = 0.93 m
- Time for one revolution = 1.18 seconds
We are required to calculate the centripetal acceleration.
To get the centripetal acceleration is given by the formula;
Centripetal acc = V²/r
Where, V is the velocity and r is the radius.
Since time for 1 revolution is 1.18 seconds,
Then, V = 2πr/t, taking π to be 3.142 ( 1 revolution = 2πr)
Therefore;
Velocity = (2 × 3.142 × 0.93 m) ÷ 1.18 sec
= 4.953 m/s
Thus;
Centripetal acceleration = (4.953 m/s)² ÷ 0.93 m
= 26.38 m/s²
Hence, the centripetal acceleration is 26.38 m/s²
I’m pretty sure the answer is Barium. I hope it helps.
Markovnikov rule, in organic chemistry, a generalization, formulated by Vladimir Vasilyevich Markovnikov in 1869, stating that in addition reactions to unsymmetrical alkenes, the electron-rich component of the reagent adds to the carbon atom with fewer hydrogen atoms bonded to it, while the electron-deficient component ...
Answer:
Correct option is B)
According to de-Broglie,
λ=mvh=6.023×10232×5×104cm/sec6.62×10−27ergsec=4×10−8cm=4Ao
