The hest off tion of waer 15. 16 Base Conjugate acid of the base D 18. att base
ID: 147747 • Letter: T
Question
The hest off tion of waer 15. 16 Base Conjugate acid of the base D 18. att base ofthe acid C at pH 11 17 20 Calculate the OH ion concentration (M) in a solution having the pH value of4 10-M b.06 x 104 c.1040 1.2x10" e. 104 21. The energy which is required to start the reaction to an equilibrium state is named: a standard free energy b. froe energy of the reaction c. activation energy d. estropy e. enthalpy 22. The maximun useful work that a reaction could perform at constant temperature and pressure, and depends on the displacement of the system from equilibrium is called: a free energy b. entropy c.enthalpy d heat content e. first law of The measure of the randomness or orderliness of the energy and matter in a system is: a free energy e. high energy 23. b. entropy c. enthalpy d. standard free energy 24. Gh e is hydrolyzed by glucose-6-phosphatase and the free glucose is released. At equilibrium at 25 C and at pH 7 we have [glucose-6-phosphate] = 5x10"M and [glucose] = [Pi] = 99.95 x 10" ML Calculate the equilibrium constant Keq' for the reaction of hydrolysis a. 0.1 b.199.8 c. 300 d. 0.01 e. 168.5Explanation / Answer
12. pKa of a weak base = - logpKa
= - log 3.1 X 10-6
= 5.51
Answer is option (d)
13. Molarity of water = Number of moles of solute / Volume of solution in litres
= 1/(18) / 1/(1000)
= 1000g / 18 g
Answer is option (d)
14. Heat of vaporization of water is amount of heat required to convert liquid form of water to vapor form. It is 540 cal/g.
Answer is option (b)
15. Acid A is HA
16. Base B is H2O
17. Conjugate acid is H3O+
18. Conjugate base is A-
19. pKa of NH3 is 9.25. Hence at pH 11 ammonia would predominate.
Answer is option (a)
20. pH + pOH = 14
4 + pOH = 14
pOH = 10
pOH = -logOH-
10 = -logOH-
OH- = 10-10
Answer is option (c)
21. Answer is option (c) Activation Energy.
22. Answer is option (a) Free Energy.
23. Answer is option (b) Entropy.
24. Equilibrium constant (Keq) = [Concentration of Products]a / [Concentration of Reactants]b
Glucose-6-Phosphate Glucose + Phosphate
Keq = 99.95 X 10-3 X 99.95 X 10-3 / 5 X 10-5
= 199.8
Answer is option (b)
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