Mannoheptulose is a sugar found in avocados. If each C-C bond contains 76 kcal of energy and each C-H bond contains 91 kcal, how
many kcal of energy are available through the complete catabolism of this sugar
1 answer:
Answer:
Total Kcal energy produced in the catabolism of mannoheptulose = 1184 Kcal
Explanation:
The molecular formula of mannoheptulose is C₇H₁₄O₇.
The structure is as shown in the attachment below.
Number of C-C bonds present in mannoheptulose = 6
Number of C-H bonds present in mannoheptulose = 8
Since the each C-C bond contains 76 Kcal of energy,
Amount of energy present in six C-C bonds = 6 * 76 = 456 Kcal
Also, since each C-H bond contains 91 Kcal of energy;
amount of energy present in eight C-H bonds = 8 * 91 = 728 Kcal
Total Kcal energy produced in the catabolism of mannoheptulose = 456 + 728 = 1184 Kcal
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Given that:
- The heat of vaporization = 22.44 kJ/mol = 22440 J/mol
- normal boiling point which is the initial temperature = 0.4° C = (273 + (-0.4))K = 272.6 K
- volume = 250 mL = 0.250 L
- Mass of butane = 0.8 g
- the final temperature = -22° C = (273 + (-22)) K = 251 K
The first step is to determine the vapor pressure at the final temperature of 251K by using the Clausius-Clapeyron equation. This is following by using the ideal gas equation to determine the numbers of moles of butane gas. After that, the mass of butane present in the liquid is determined by using the relation for the number of moles.
Using Clausius-Clapeyron Equation:
where;
P1 and P2 correspond to the temperature at T1 and T2.
∴
replacing the values into the given equation, we have;
As such, at -22° C; the vapor pressure = 0.427 atm
Now, using the ideal gas equation:
PV = nRT
where:
- P = Pressure
- V = volume
- n = number of moles of butane
- R = universal gas constant
- T = temperature
∴
Making (n) the subject of the formula:
We all know that the standard molecular weight of butane = 58.12 g/mol
∴
Using the relation for the number of moles which is:
mass = 0.00518 mole × 58.12 g/mol
mass = 0.301 g
∴
The mass of butane in the flask = 0.301 g
But the mass of the butane present as a liquid in the flask is
= 0.8 g - 0.301 g
= 0.499 g
In conclusion, the mass of the butane present as a liquid in the flask is 0.499 g
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