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- The calcium ion binds to a certain protein to form a 1:1 complex. The following data was obtained in an experiment: Total Ca2+/uM 60 120 180 240 480 Ca2+ bound to Protein/uM 31.2 51.2 63.4 70.8 83.4 Determine graphically the dissociation constant of the Ca2+ - protein complex. The protein concentration was kept at 96uM for each run. (1uM = 1 x 10-6 M.)Calculate the pK at 25 degrees C of the following reaction using the data in the table. Use R = 8.314 J/mol/K. 4A +2B → 1C + 3D + 4E ΔHf0(kJ/mol) S0(J/mol/K) A 18.05 19.84 B 3.31 12.02 C -13.73 18.57 D -9.71 10.09 E -8.96 19.45 Remember, pK = -log(K) Report the answer to two decimal points.Calculate the entropy of activation for a collision between two structureless particles at 450 K, taking M = 92 g mol−1 and σ = 0.45 nm2.
- The table below provides data for the enthalpy and entropy of formation of compounds A and B at standard conditions (298 K) Compound DHfo (kJ mol-1) Sfo(J K-1mol-1) A –135.2 189.2 B –157.6 192.1 (b) A reversible isomerization reaction converts reactant A to product X. Calculate the Gibbs energy change at non-standard conditions (310 K) if the concentration of A is 2 x 10-4M and that of X is 3 x 10-6 M. Comment on whether the reaction is spontaneous at these conditions. The equilibrium constant Keq = 0.05. Assume standard temperature is 298K.Use the data below to determine the Michaelis-Menton constant [in mM] of a certain enzyme-catalyzed reaction. v = 0.152 mM/s at [S] = 0.334 mM v = 0.176 mM/s at [S] = 0.450 mMFor an aqueous solution saturated in both AgCl and AgI, at 1 bar and at 298 K. The equilibrium constants for this system are given as follow: Ksp(AgCl) = 1.8*10^-10, Ksp(AgI) = 8.5*10^-17, and Kw = 1.0*10^-14. How many formalities must be given to get unique values for all the concentrations?
- Reaction KspKsp ΔH°ΔH° ΔS°ΔS° FeCO3(s)⇄Fe2+(aq)+CO32−(aq)FeCO3(s)⇄Fe2+(aq)+CO32−(aq) 3×10−113×10−11 <0<0 >0>0 MnCO3(s)⇄Mn2+(aq)+CO32−(aq)MnCO3(s)⇄Mn2+(aq)+CO32−(aq) 2×10−112×10−11 <0<0 >0 The table above lists the equilibrium constants and changes in thermodynamic properties for the dissolution of FeCO3 and MnCO3 at 25°C. The two-particle diagrams below represent saturated solutions of each compound at equilibrium. (see attached image) a.) The particle diagrams best represent that ΔH°<0ΔH°<0 because the ions from both compounds are solvated by water molecules. b.) The particle diagrams best represent that ΔH°<0ΔH°<0 because both compounds produce about the same amount of CO32−CO32− ions from the dissolution. c.) The particle diagrams best represent that ΔS°>0ΔS°>0 because both compounds produce a very small amount of ions from the dissolution. d.) The particle diagrams best represent that the molar solubility is greater for…Calculate the molar Gibbs energy of substrate to enzyme binding, E + S ® ES, DG, at 25 °C if theequilibrium binding constant, K, at this temperature is 104 mol 1 L, [S] = 0.01 mol L-1, [E] =0.001 mol L-1.and [ES] = 0.001 mol L-1. R = 8.314 J mol1 k-1. Select the closest value.Please solve this ASAPPPPPPPPPPPPP. Disregard the answer already provided and resolve it thanks! In that same format. Thanks
- Find the variance (F) for the following equilibria: (i) 2 CaSO4(s) ↔ 2 CaO(s) + 2 SO2(g) + O2(g) (2) (ii) CuSO4·3H2O(s) ↔ CuSO4·H2O(s) + 2 H2O(g) (2) (iii) NH4Cl(s) ↔ HCl(g) + NH3(g), when the equilibrium is approached by starting only with the solid.Thermodynamics Quantities for Selected Substances at 298.15 K (25⁰C) Substance ∆H⁰f (kJ/mol) ∆G⁰f (kJ/mol) S (J/K-mol) Carbon C2H2(g) 226.7 209.2 200.8 C2H4(g) 52.30 68.11 219.4 Hydrogen H2(g) 0 0 130.58 Oxygen O2(g) 0 0 205.0 H2O(l) -285.83 -237.13 69.91 1. What is the value of ∆S⁰ for the catalytic hydrogenation of ethene to ethane: C2H4(g) + H2(g) → C2H6(g) + 2H2O(l) in J/K?Given below (c): Standard gibb's free energies (∆Gf0 kJ mol-1 ): UO2 = -962.7 UO22+ = -953.5 U4+ = -579.1 Fe2+ = -78.9 Fe(OH)3 ferrihydrite = -692.07 Mn2+ = -288.1 MnO2 pyrolusite = -465.1 HS- = 12.1 H+ = 0 H2O = -237.1 S0 = 0 Given: U(VI) as uraninite; UO2 (where Mn2+ = reductant; MnO2 pyrolusite = product): ∆ Gr0 = -21.3 KJ/mol E0 (emf) = 0.110 V n = 2 F = 96.42 QUESTION: Calculate Eh equation below to calculate at different pH: – Eh = E0 + (RT/nF) * lnK For U(VI) as uraninite; UO2 (where HS- = reductant; S0 = product): UO22+ + Hs- ---- > UO2 + S + H+ [A] pH 3 [B] pH 7