Molecular mass= (14.01∗1)+(1.008∗3)
14.01+3.024=17.03g/mol
If acetone has a density of 0.7857
the mass in grams of point A is 22.4 g and the volume at point B is 8.32 mL.
<h3>What is acetone?</h3>
Acetone is known as a chemical substance that is usually found in the environment but can also be produced artificially. Acetone is a polar organic product that interacts very well with water molecules, generating dipole-dipole relationships.It is colorless with a distinctive smell and taste, we find it in products known as <u>cleaning and personal care products</u>, but we can also use it as a solvent for substances.
Also in the environment in <u>plants, trees and in volcano emissions or in forest fires</u>, it does not become <em>toxic</em> in low doses but if it is exposed to an individual in high doses it can become <em>fatal</em>.
In the statement we can find that acetone has a density of 0.7857
.
Therefore, we can confirm that if acetone has a density of 0.7857
the mass in grams of point A is 22.4 g and the volume at point B is 8.32 mL.
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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
Answer:
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Explanation: