Answer:
131.5 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
First, we will calculate the standard enthalpy of the reaction (ΔH°).
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)
) - 1 mol × ΔH°f(CaCO₃(s)
)
ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)
ΔH° = 179.2 kJ
Then, we calculate the standard entropy of the reaction (ΔS°).
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)
) - 1 mol × S°(CaCO₃(s)
)
ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)
ΔS° = 160.2 J/K = 0.1602 kJ/K
Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.
ΔG° = ΔH° - T × ΔS°
ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K
ΔG° = 131.5 kJ
Explanation:
As it is known that molarity is the number of moles present in a liter of solution.
Mathematically, Molarity = 
As it is given that molarity is 0.10 M and volume is 10.0 ml. As 1 ml equals 0.001 L. Therefore, 10.0 ml will also be equal to 0.01 L.
Hence, putting these values into the above formula as follows.
Molarity =
0.10 M = 
no. of moles = 0.001 mol
As molar mass of KCN is equal to 65.12 g/mol. Therefore, calculate the mass of KCN as follows.
No. of moles = 
0.001 mol = 
mass = 0.06152 g
Thus, we can conclude that 0.06152 grams of KCN are in 10.0 ml of a 0.10 M solution.
Answer:
62.15 m (squared)
Explanation:
We know that the Area of the square is 5.5 x 4= 22
so 90 degree to 30 degree is divide by 3
so u divide 22 by 3 and get 7.3 wich is the semi circle thing
now 7.3 x 5.5 = 40.15
22 + 40.15 = 62.15 m (squared)
Look im not really sure but I think this is the answer
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