Problem on MM equation
How to obtain relation between Vm and Km,given k(sec^-1) = Vmax/mg of enzyme x molecular weight x 1min/60 sec S* = 4.576(log K -10.753-logT+Ea/4.576T).
Expert
According to M.M Equation, V = Vmax x {S]/Km + {S] Since Vmax = ksec^-1 {Et}V = Ksec^-1{Et} x {S}/Km + {S}Substituting the given value of Ksec^-1 in the above equation,we get V = Vmax x mol.wt x 1min/60 sec x {S}/mg of enzyme / Km +{S]V = Vmax x mol.wt x {S}/mg of enzyme/Km +{S}Substituting the given value of {S}V = Vmax x mol.wt x 4.576(log K -10.753-logT+Ea/4.576T)/mg of enzyme/ Km + 4.576(log K -10.753-logT+Ea/4.576T)
What do you mean by the term hydra? Briefly define it.
Two tanks which contain water are connected to each other through a valve. The initial conditions are as shown (at equilibrium):
Can someone please help me in getting through this problem. 2.0 molar solution is acquired, when 0.5 mole solute is dissolved in: (i) 250 ml solvent (ii) 250 g solvent (iii) 250 ml solution (iv) 1000 ml solvent
explain the process of relative lowering of vapour pressure
The second law states that dS ≥ (dQ/T), where dS = dQ/T for a reversible process and dS > dQ/T for an irreversible process. a. Show that since dW12 = -dW21 (dWreverse = -dWforward) for a r
Which of the following solutions will have a lower vapour pressure and why? a) A 5% aqueous solution of cane sugar. b) A 5% aqueous solution of urea.
Which of the following would have the maximum osmotic pressure (assume that all salts are 90% dissociated): (a) Decimolar aluminium sulphate (b) Decimolar barium chloride (c) Decimolar sodium sulphate (d) A solution obtained by mix
arrange in order of basicity,pyridine,pipyridineand pyorine
Describe how dipole attractions, London dispersion forces and the hydrogen bonding identical?
Help me to go through this problem. 10 grams of a solute is dissolved in 90 grams of a solvent. Its mass percent in solution is : (a) 0.01 (b) 11.1 (c)10 (d) 9
18,76,764
1925312 Asked
3,689
Active Tutors
1415089
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!