TLS Online TPP Program

#Question id: 8849


Humans can digest starch but not cellulose because

#SCPH28 | Zoology
  1. the monomer of starch is glucose, while the monomer of cellulose is galactose.
  2. humans have enzymes that can hydrolyze the beta (β) glycosidic linkages of starch but not the alpha (α) glycosidic linkages of cellulose.
  3. humans have enzymes that can hydrolyze the alpha (α) glycosidic linkages of starch but not the beta (β) glycosidic linkages of cellulose.
  4. humans harbor starch-digesting bacteria in the digestive tract.
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TLS Online TPP Program

#Question id: 15463

#SCPH05 I Biotechnology

A system in which there is exchange of energy but not of mass, is called a/an    ______                 system

TLS Online TPP Program

#Question id: 15464

#SCPH05 I Biotechnology

The quantity of heat required to evaporate 1 kg of a saturated liquid is called

TLS Online TPP Program

#Question id: 15465

#SCPH05 I Biotechnology

50 g benzaldehyde vapour is condensed at 179°C. What is the enthalpy of the liquid relative to the vapour?  (Standard latent heat for vaporisation for benzaldehyde is 38.40 kJ gmol-1)

TLS Online TPP Program

#Question id: 15466

#SCPH05 I Biotechnology

Water at 25°C enters an open heating tank at a rate of 10 kg h-1. Liquid water leaves the tank at 88°C at a rate of 9 kg h-1; 1 kg h-1 water vapour is lost from the system through evaporation. At steady state, what is the rate of heat input to the system?
h (liquid water at 25°C = 104.8 kJ kg-1
h (liquid water at 88°C = 368.5 kJ kg-1
h (saturated steam at 88°C = 2656.9 kJ kg-1)

TLS Online TPP Program

#Question id: 15467

#SCPH05 I Biotechnology

Fumaric acid is produced from Malic acid using Fumarase. Calculate standard heat of reaction for the following transformation:
C4H6O5 → C4H4O4 + H2O
Given: (Δhc°)malic acid= -1328.8 kJ/gmol 
   (Δhc°)Fumaric acid= -1334.0 kJ/gmol

TLS Online TPP Program

#Question id: 15468

#SCPH05 I Biotechnology

In downstream processing of gluconic acid, concentrated fermentation broth containing 20% (w/w) gluconic acid is cooled in a heat exchanger prior to crystallisation. 2000 kg h-1 liquid leaving an evaporator at 90°C must be cooled to 6°C. Cooling is achieved by heat exchange with 2700 kg h-1 water initially at 2°C. The final temperature of the cooling water is 50°C. 
Assume (heat capacity of gluconic acid is 0.35 cal g-1 °C-1).
h (liquid water at 90°C = 376.9 kJ kg-1)
h (liquid water at 6°C = 25.2 kJ kg-1)
h (liquid water at 2°C = 8.4 kJ kg-1 )
h (liquid water at 50°C = 209.3 kJ kg-1)
h (gluconic acid at 90°C=0)
How much heat is removed to the cooling water?