Answer:
2-methyl-2-pentyl-1,3-dioxolane
Explanation:
In this case, we have two reactions:
First reaction:
1-heptyne + mercuric acetate -------> Compound A
Second reaction:
Compound A + HOCH2CH2OH -------> Compound C
<u>First reaction</u>
In the first reaction, we have as a main functional group a triple bond. We have to remember that mercuric acetate in sulfuric acid will produce a ketone. The carbonyl group (C=O) would be placed in the most substituted carbon of the triplet bond (in this case, carbon 2). With this in mind, we will have as a product: heptan-2-one. (See figure 1).
<u>Second reaction</u>
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In this reaction, we have as reagents:
-) Heptan-2-one
-) Ethylene-glycol 
-) Sulfuric acid 
When we put ethylene-glycol with a ketone or an aldehyde we will form a cyclic acetal. In this case, this structure would be formed on carbon 2 forming 2-methyl-2-pentyl-1,3-dioxolane. (See figure 2).
I hope it helps!
C. both a physical and chemical change occur.
Question options:
1. Insoluble
2. Soluble
3. Solvable
Answer:
Soluble
Explanation:
The solubility is a chemical property referring to the ability of the solute to dissolve in a solvent and form a homogeneous mixture (solution). It is measured in terms of the maximum amount of solute dissolved in a solvent at equilibrium.
<u>Therefore, a substance is soluble in a particular liquid if it is able to dissolve in a solvent.</u>
Answer:
At the burner temp. and pressure, 18.85 litres of air is needed to completely combust each gram of propane
Explanation:
The combustion stoichiometry is as follows:
C₃H₈ + 5O₂ = 4 H₂O + 3CO₂ The molecular weights (g/mol) are:
MW 44 5x32 4x18 3x44
So each gram of propane is 1/44 = 0.02272 mol propane
and will need 5 x 0.02272 = 0.1136 mol oxygen
At 0.21 mol fraction oxygen in air, 0.1136 / 0.21 = 0.54 mol air is needed to burn the propane.
At the low pressure in the burner we can use the Ideal Gas Law
PV=nRT, or V = nRT/P
P = 1.1 x 101325 Pa = 111457 Pa
T = 195°C + 273 = 468 K
R = 8.314
and we calculated n = number of moles air = 0.54 mol
So V m³ = 0.54 x 8.314 x 468 / 111457 = 0.0188 m³ = 18.85 litres air.
