Answer:
The heat of combustion is -25 kJ/g = -2700 kJ/mol.
Explanation:
According to the Law of conservation of energy, the sum of the heat released by the combustion reaction and the heat absorbed by the bomb calorimeter is equal to zero.
Qcomb + Qcal = 0
Qcomb = - Qcal
The heat absorbed by the calorimeter can be calculated with the following expression.
Qcal = C × ΔT
where,
C is the heat capacity of the calorimeter
ΔT is the change in temperature
Then,
Qcomb = - Qcal
Qcomb = - C × ΔT
Qcomb = - 1.56 kJ/°C × 3.2°C = -5.0 kJ
Since this is the heat released when 0.1964 g o quinone burns, the energy of combustion per gram is:
The molar mass of quinone (C₆H₄O₂) is 108 g/mol. Then, the energy of combustion per mole is:
Answer:
The correct answer is 1 glycogen degradation would slow down.
Explanation:
Glycogen is the principle storage polysaccharide present in the liver and muscle of human body.
Glycogen contain both alpha-1,4-glycosidic linkage and alpha -1,6-glycosidic linkage.During glycogenolysis some glucose residues are transferred from branch point of the glycogen to its end and thereafter a single glucose residue is linked to the branch point of glycogen by alpha-1,6-glycosidic linkage.
The alpha-1,6-glycosidic linked glucose of glycogen is finally get separated from glycogen by the catalytic activity of alpha-1,6-glycosidase enzyme in the final step of glycogenolysis.
According to the given question if there is no alpha-1,6-glycosidic linkage in the glycogen then glycogen degradation will slow down.
Answer:
I think <em><u>bar graph</u></em> is the appropriate way to present the data.
The physical properties of alkenes and alkynes are generally similar to those of alkanes or cycloalkanes with equal numbers of carbon atoms. Alkynes have higher boiling points than alkanes or alkenes, because the electric field of an alkyne, with its increased number of weakly held π electrons, is more easily distorted, producing stronger attractive forces between molecules.
