in a given homologous series of hydrocarbons, the boiling point generally increases as the size of the molecules increases. the
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Answer:
The option closest to the percentage yield is option;
d. 81.1
Explanation:
The given chemical equation of the reaction is presented as follows;
CaO (s) + H₂O (l) → Ca(OH)₂
The mass of CaO in the experiment, m = 2.00 g
The volume of water with which the CaO was reacted = Excess volume of water
Number of moles = Mass/(Molar mass)
The mass of Ca(OH)₂ recovered, actual yield = 2.14 g
The molar mass of CaO = 56.0774 g/mol
The number of moles of CaO in the reaction, n₁ = 2.00 g/(56.0774 g/mol ≈ 0.036 moles
The molar mass of Ca(OH)₂ = 74.093 g/mol
The number of moles of Ca(OH)₂ in the reaction, n₂ = 2.14 g/(74.093 g/mol) ≈ 0.029 moles
From the given chemical reaction, one mole of CaO reacts with one mole of H₂O to produce one mole of Ca(OH)₂
Therefore, 0.036 moles of CaO will produce 0.036 moles of Ca(OH)₂
Mass = Number of moles × Molar mass
The mass of 0.036 moles of Ca(OH)₂ ≈ 0.036 moles × 74.093 g/mol = 2.667348 grams
∴ The theoretical yield of Ca(OH)₂ = 2.667348 grams
The percentage yield = (2.14 g)/(2.667348 grams) × 100 = 80.23%
Therefore, the option which is closest is option d. 81.1.
The names of these constitutional isomers are cyclopentane, methylcyclobutane, ethylcyclopropane, and 1,1-dimethylcyclopropane.
Constitutional isomers have the same molecular formula. This means that they have the same number of carbon and hydrogen atoms, with a general formula of:
When we consider the possible structures, the first option is cyclopentane, the simplest five-membered ring. When we reduce the ring to four carbons, we have the methylcyclobutane and when we reduce it to three carbons, we can arrange the other two carbons in an ethyl group or we can put two methyl groups on a single carbon atom to get the final two constitutional isomers.
You can learn more about constitutional isomers here:
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