Answer:
that results in an alcohol and a carboxylic acid.
Explanation:
Saponification is a chemical reaction process of alkaline hydrolysis of esters(R'COOR group) by which soap is obtained.
For Example, when a base such as sodium hydroxide [NaOH] is used to hydrolyze an ester, the products are a carboxylate salt and an alcohol. Because soaps are prepared by the alkaline hydrolysis of fats and oils.
In a saponification reaction, alkaline hydrolysis of fats and oils with sodium hydroxide yields propane-1,2,3-triol and the corresponding sodium salts of the component fatty acids.
i.e Fat or oil + caustic alkali ⇒ Soap + propane-1,2,3-triol
As a specific example, ethyl acetate and NaOH react to form sodium acetate and ethanol:
The reaction goes to completion in the image below:
Answer:
3.09kg
Explanation:
First, let us write a balanced equation for the reaction. This is illustrated below:
2C8H18 + 25O2 —> 16CO2 + 18H2O
Molar Mass of C8H18 = (12x8) + (18x1) = 96 + 18 = 114g/mol
Mass of C8H18 from the balanced equation = 2 x 114 = 228g
Converting 228g of C8H18 to kg, we obtained:
228/1000 = 0.228kg
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Mass of CO2 from the balanced equation = 16 x 44 = 704g
Converting 704g of CO2 to kg, we obtained:
704/1000 = 0.704kg
From the equation,
0.228kg of C8H18 produced 0.704kg of CO2.
Therefore, 1kg of C8H18 will produce = 0.704/0.228 = 3.09kg of CO2
Answer:
1) The value of Kc : (C.) remains the same
2) The value of Qc : (A.) is greater than Kc
3) The reaction must : (B.) run in the reverse direction to reestablish equilibrium.
4) The number of moles of Br2 will : (B.) decrease
Explanation:
Value of concentration equilibrium constant Kc depends only on temperature. Since temperature remains constant, therefore, Kc remains constant. Decrease in volume means increase in pressure. Increase in pressure favors the side with less gaseous species. Hence, increase in pressure will favor the reverse reaction towards reactants.
Answer:
a) But-1-ene
b) E-But-2-ene
c) Z-But-2-ene
d) 2-Methylpropene
Explanation:
In this case, if we want to draw the <u>isomers</u>, we have to check the<u> formula </u>
in this formula we can start with a linear structure with 4 carbons. We also know that we have a double bond, so we can put this double bond between carbons 1 and 2 and we will obtain <u>But-1-ene.</u>
<u />
For the next isomer, we can move the double bond to carbons 2 and 3. When we do this can have two structures. When the methyl groups are placed on the same side we will obtain <u>Z-But-2-ene</u>. When the methyls groups are placed on opposite sides we will obtain <u>E-But-2-ene.</u>
<u />
Finally, we can use a linear structure of three carbons with a methyl group in the middle with a double bond, and we will obtain <u>2-Methylpropene.</u>
<u />
See figure 1 to further explanations.
I hope it helps!
<u />
<u />