Carbohydrate metabolic enzymes bind d-glucose specifically. d-glucose has an estimated caloric value of 1 k. cal per 4 grams of carbohydrate.
based on the methods used to convert the energy of d-glucose into a useable form, what would you estimate the caloric value of l-glucose to be using the same method?

a) 1 k. cal per 4 grams
b) 1 k. cal per 2 grams
c) 1 k. cal per 8 grams
d) 0 k. cal per 4 grams
e) 2 k. cal per 8 grams

D) 0 k.cal per 4 grams

Explanation:

Genarally, glucose can be classified into two enantiomers such as d-glucose and l-glucose. The d-glucose is the most common sugar that bodies of living organisms use as source of energy. However, l-glucose is an organic compound and it is one of the aldohexose monosaccharides. It is the l-isomer of glucose and commonly refer to as a low-calorie sweetener. The l-glucose is  relatively indistinguishable in taste from d-glucose but cannot be used as a source of energy. Therefore, in the given problem:

if d-glucose has an estimated caloric value of 1 k.cal per 4 grams of carbohydrate, then the caloric value of l-glucose will be 0 k.cal per 4 grams.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

c8h10 is your answerr