Calculate the cell potential, the equilibrium constant, and the free-energy change for:
1 answer:
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Answer:
C₁₀H₂₀O
Explanation:
The molecular formula must be . The combustion reaction will occur between the fuel and oxygen gas:
+ O₂ → CO₂ + H₂O
For Lavoisier law, the mass of the reactants must be equal to the mass of the products (mass conservation):
10.0 + mO₂ = 28.16 + 11.53
mO₂ = 29.69 mg
Supposing that all the oxygen and the menthol were consumed, let's calculate the number of moles of the compounds, knowing, for the Periodic Table, that:
MC = 12 g/mol, MO = 16 g/mol, MH = 1 g/mol
MCO₂ = 12 + 2x16 = 44 g/mol
MH₂O = 2x1 + 16 = 18 g/mol
MO₂ = 2x16 = 32 g/mol
n = mass (g)/molar mass
nCO₂ = 0.02816/44 = 6.4x10⁻⁴ mol
nH₂O = 0.01153/18 = 6.4x10⁻⁴ mol
nO₂ = 0.02969/32 = 9.3x10⁻⁴ mol
The molar number is proportional in the molecule, so, in CO₂, the number of C is 6.4x10⁻⁴ mol, and of O is 1.28x10⁻³. All the carbon of the methol will be in CO₂, and all the H will be in H₂O. The number of moles of O will be the difference of moles in H₂O and the O₂, then:
nC = 6.4x10⁻⁴ mol
nH = 2x6.4x10⁻⁴ = 1.28x10⁻³ mol
nO = (2x6.4x10⁻⁴ + 6.4x10⁻⁴) - (2x9.3x10⁻⁴) = 6.0x10⁻⁵
The empirical formula is the molecule formula with the small subscripts numbers, which represent the number of moles of the atoms in the molecule. So, let's divide all the number of moles for the small on 6.0x10⁻⁵.
nC = (6.4x10⁻⁴)/(6.0x10⁻⁵) = 10
nH = (1.28x10⁻³)/(6.0x10⁻⁵) = 20
nO = (6.0x10⁻⁵)/(6.0x10⁻⁵) = 1
So, the empirical formula of methol is C₁₀H₂₀O.
Answer:
B
Explanation:
Please take a look at the picture attached for the drawings and structures.
C2H4 is a alkene (C-C double bond). When steam (water) is added, it turns into an alcohol, where the double bond breaks and a (-OH) functional group is attached to one of the Carbons. In this case, C2H4 ethene is turned into ethanol.
When an alcohol undergoes oxidation, primary alcohols turn into aldehyde (-CHO) or carboxylic acids (-COOH). Secondary alcohol turns into ketone. Ethanol is a primary alcohol. And since it later reacts with propanol, it can only form carboxylic acid when it oxidizes. The product in this reaction is ethanoic acid.
Carboxylic acid reacts with alcohol to form an ester (-COO-). the -COOH group from acid combines with the -OH group from alcohol to form an ester bond -COO- while eliminating H2O. Therefore, when propanol undergoes esterification with ethanoic acid, propyl ethanoate is produced. It is the answer of B.
