Answer:
Even-number fatty acids such as palmitate undergoes complete β-oxidation in the liver motochondria to CO₂ because the product, acetyl-CoA can enter the TCA cycle.
Oxidation of odd-number fatty acids such as undecanoic acid yields acetyl-CoA + propionyl-CoA in their last pass. Propionyl-CoA requires additional reactions including carboxylation in order to be able to enter the TCA cycle.
The reaction CO2 + propionyl-CoA ----> methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, a biotin-containing enzyme, which is inhibited by avidin. Palmitate oxidation however, does not involve carboxylation.
Explanation:
Even-number fatty acids such as palmitate undergoes complete β-oxidation in the liver motochondria to CO₂ because their oxidation product, acetyl-CoA, can enter the TCA cycle where it is oxidized to CO₂.
Undecanoic acid is an odd-number fatty acid having 11 carbon atoms. Oxidation of odd-number fatty acids such as undecanoic acid yields a five -carbon fatty acyl substrate for their last pass through β-oxidation which is oxidized and cleaved into acetyl-CoA + propionyl-CoA. Propionyl-CoA requires additional reactions including carboxylation in order to be able to enter the TCA cycle. Since oxidation is occuring in a liver extract, CO₂ has to be externally sourced in order for the carboxylation of propionyl-CoA to proceed and thus resulting in comlete oxidation of undecanoic acid.
The reaction CO2 + propionyl-CoA ----> methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, a biotin-containing enzyme. The role of biotin is to activate the CO₂ before its tranfer to the propionate moiety. The addition of the protein avidin prevents the complete oxidation of undecanoic acid by binding tightly to biotin, hence inhibiting the activation and transfer of CO₂ to propionate.
Palmitate oxidation however, does not involve carboxylation, hence addition of avidin has no effect on its oxidation.
Explanation:
The valency of the element is the measure of the combining power of the element with the other atoms when the element forms compounds or molecules.
<u>Thus, valency has to known to find the how the elements have been combined and how many electrons have been lost, gain or shared.</u>
Given that:
Element A has 3 valence electrons and element B has 2 valence electrons. To find the ionic compound, the valency of the cations and the anions are interchanged and are written in subscripts. Thus,
A B
3 2
Cross multiplying, we get the formula : 
<u>Hence, 3 is the subscript for Element B.</u>
Explanation:
The initial concentrations for a mixture :
Acetic acid at equilibrium = 0.15 M
Ethanol at equilibrium = 0.15 M
Ethyl acetate at equilibrium = 0.40 M
Water at equilibrium = 0.40 M
Initially:
0.15 M 0.15 M 0.40 M 0.40 M
At equilibrium
(0.15-x)M (0.15-x) M (0.40+x) M (0.40+x) M
The equilibrium constant is given by expression
![K_c=\frac{[CH_3CO_2C_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3CO_2C_2H_5%5D%5BH_2O%5D%7D%7B%5BCH_3COOH%5D%5BC_2H_5OH%5D%7D)
Solving for x:
x = 0.0333
The equilibrium concentrations for a mixture :
Acetic acid at equilibrium = (0.15-x)M = (0.15-0.033) M = 0.117 M
Ethanol at equilibrium = (0.15-x)M = (0.15-0.033) M = 0.117 M
Ethyl acetate at equilibrium = (0.40+x)M = (0.40+0.033) M = 0.433 M
Water at equilibrium = (0.40+x)M = (0.40+0.033) M = 0.433 M
Answer:
The mass percent of a solution of 7.6 grams sucrose in 83.4 grams of water is 8.351 %.
Explanation:
Given,
Mass of Sucrose = 7.6 grams
Mass of Water = 83.4 grams
In this solution, Sucrose is solute and water is the solvent.
Mass percent of a solution can be calculated using the formula,
Mass percent = (Mass of Solute/Mass of Solution)(100)
As sucrose is solute, mass of solute = 7.6 grams
As the solution contains both Sucrose and Water,
Mass of solution = 7.6 grams + 83.4 grams = 91 grams
Substituting the values, Mass percent = (7.6/91)(100) = 8.351 %.
