Calculate the ionic strength of solution that 0.040 kg^-1 in K_a[Fe(CN)_s] (aq),

Question

Calculate the ionic strength of solution that 0.040 kg^-1 in K_a[Fe(CN)_s] (aq), 0.030 mol kg^-1 in KCI(aq) and 0.050 mol kg^-1 in NaBr(aq). Draw the T and P diagrams of partial miscible liquids a) with upper critical temperature (UCT) b) with lower critical temperature (LCT) c) with both UCT and LCT. At 2.257 K and 1.00 bar total pressure water is 1.77 per cent dissociated at equilibrium by way of the reaction 2H_2O = 2H_2 + O_2 (all are gasses). Calculate K. Devise cells in which the following are the reactions and calculate the standard cell potential in each case. a) Zn(s) + CuSO_4 (aq) rightarrow ZnSO_4(aq) + Cu(s) b) ZAgCI(s) + H_2(g) rightarrow 2 HCI(aq) + 2 Ag(s) A second order reaction of the A + 2B rightarrow P was carried out in a solution that was initially 0.050 mol dm^-3 in A and 0.030 mol dm^-3 in B. After 1.0 h the concentration is A had fallen to 0.0100 mol dm^-3 a) calculate the rate constant and b) what is the half-life of the reactants? The reaction mechanism for the renaturation of the double helix from its strands A and B A + B unstable helix (fast) Unstable helix rightarrow stable double helix (slow) Deduce the rate law. Consider the base-catalysed reaction AH + B = BH^+ A (fast) A^- + AH rightarrow product Deduce the rate law. The enzyme catalyzed conversion of a substrate at 25 degree C has a Michaelis constant of 9.0 times 10^5 mol dm^-3 and a maxiumum velocity of 2.24 times 10^-5 mol dm^-3 a^-1 when the enzyme concertation is 1.60 times 10^-9 mol dm^-3 calculate K_eat and eta. Is the enzyme "catalytically perfect"?

Explanation / Answer

Kcat = Vmax/ [E] tot

        = 2.25*10^-5 mol/dm^3 / (1.6*10^-9) =14062.5

n = Kcat/Km = 14062.5/ 9*10^5 = 0.016

This is not a catalytically perfect enzyme

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