Hello there,
The answer to your question is:
C. Zinc
Hope this answer helps you.
Answer:
The reactive nucleophile is Ketone.
Explanation:
In organic chemistry, The process of acid - catalyzed aldol condensation starts from when ketone (or any aldehyde) is converted to an -enol, after which it attacks another ketone/aldehyde that has already been activated by parbonyl oxygen protonation.
The process of this is that first of all the ketone undergoes tautomerization to form -enol. Thereafter, the other carbonyl will undergo protonation which makes the carbon activated towards attack. Now, the nucleophilic enol will be added to the carbonyl in a [1,2]-addition reaction and we will now use deprotonation to obtain the neutral Aldol product.
Now, since only the ketone can produce an -enol, thus it is the nucleophile as aldehydes are better electrophiles
The compound contains an ester functional group.
An ester is a carbonyl (C=O) group with an alkyl (R) group on one side and an alkoxy (OR) group on the other.
We write the <em>condensed structural formula</em> of an ester as R(C=O)OR or RCOOR.
Ooooh boy alright. So, this may or may not be a limited reactant problem so we need to first find out of it is.
First, how many moles of each substance are there
the molar mass of BCl3 is <span>117.17 grams so 37.5 g / 117.17 is ~ .32 mol.
The molar mass of H2O is 18.02 so 60 / 18.02 is ~ 3.33 mol.
Now, for every 1 mole of BCl3, there are 3 moles of HCl created. Therefore, BCl3 can create ~ .96 moles.
For every 3 moles of H2O, there are 3 moles of HCl created. Therefore, HCl can create ~3.33 moles.
But, there is not enough BCl3 to support that 3.33 moles, only enough for .96 moles, therefore BCl3 is the limiting reactant. Now, to answer the question, simply multiply .96 moles by the molar mass of HCl.
.96 x 36.46 = ~35 g</span>
∆H ° rxn =-2855.56 kJ
<h3>Further explanation</h3>
Given
ΔHf CO₂ = -393.5 kJ/mol
ΔHf H₂O = -241.82 kJ/mol
ΔHf C₂H₆ = - 84.68 kJ/mol
Reaction
2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(g)
Required
ΔHrxn=
Solution
<em>∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants) </em>
∆H ° rxn = (4.-393.5+6.-241.82)-(2.-84.68)
∆H ° rxn = (-1574-1450.92)-(-169.36)
∆H ° rxn =-3024.92+169.36
∆H ° rxn =-2855.56 kJ
