The structure of the alkyl bromides used in a malonic ester synthesis of ethyl 2-methyl-4-pentenoate are as drawn in the attached image.
<h3>Ethyl 2-methyl-4-pentenoate by Malonic ester synthesis.</h3>
The malonic ester synthesis is a chemical reaction characterized by the alkylation of diethyl malonate or a similar ester of malonic acid at the carbon alpha (directly adjacent) to both carbonyl groups, and subsequently converted to a substituted acetic acid.
Hence, it follows from the structure of Ethyl 2-methyl-4-pentenoate that the alkyl bromides used are Ethyl bromide and methyl bromide.
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brainly.com/question/17237043
Explanation:
As the total concentration is given as 1.2 mM. And, it is also given that salt present in the solution is NaCl.
As sodium chloride is an ionic compound so, when it is added to water then it will dissociate into sodium and chlorine ions as follows.

So, it means in total there will be formation of 2 ions when one molecules of NaCl dissociates.
Therefore, concentration of chlorine ions will be calculated as follows.
Concentration of
ions =
= 0.6 mM
Thus, we can conclude that the concentration of chloride ions is 0.6 mM.
Answer:
single displacement reaction
Explanation:
coefficient if Ag is 2
after it is balanced
The product of the complete combustion of any fuel (in this case, acetylene) are indeed water and carbon dioxide.
Balancing the combustion reaction,
C2H2 +(5/2) O2 --> 2CO2 + H2O
The number of moles of C2H2 will be,
(12 g) x (1 mole/26 g) = 6/13 mole
Then, the number of moles of O2 is,
(12 g) x (1 mole/32 g) = 3/8 mole
Therefore the limiting reaction is the O2. Getting the amount of CO2 and H2O produced from balancing,
CO2 = (3/8 moles) x (2 moles CO2/ 5/2 mole O2)(44 g/ 1 mole) = 52.8 g
H2O = (3/8 moles) x (1 mole / 5/2 mole O2)(18 g / 1 mole) = 2.7 g
Answer:
0.39 moles
Explanation:
To find how many moles are in 50.0 g of CaC₂O₄ you divide the grams of the sample by the molar mass of the compound;
=0.39 mol
The grams cancel out and you are left with moles!
I hope this help ^-^