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- The value of K for aniline, C6H5NH₂, is 7.40×10-10. Write the equation for the reaction that goes with this equilibrium constant. It is not necessary to include states such as (aq) or (1). +(a) If the molar solubility of Nd₂(CO3)3 at 25 °C is 1.00e-07 mol/L, what is the Ksp at this temperature? Ksp = (b) It is found that 8.90e-06 g of Y₂(CO3)3 dissolves per 100 mL of aqueous solution at 25 °C. Calculate the solubility- product constant for Y₂(CO3)3. Ksp = (c) The Ksp of Ag₂C₂04 at 25 °C is 5.40e-12. What is the molar solubility of Ag2C₂04? solubility mol/LWrite equilibrium constant expressions for the following reactions in terms of concentration: (a) 2 CH₂Cl₂(g)CH₂(g) + CCl4(g) K = (b) Co(s) + H₂O(g) Co0(s) + H₂(g) K= (c) HCIO(aq) + H₂O(2) H30*(aq) + CIO (aq) K =
- The value of K, for isoquinoline, C,H,N, is 2.50×109. Write the equation for the reaction that goes with this equilibrium constant. It is not necessary to include states such as (ag) or (1).Consider the following reaction: CH4(g) + 2 H2S(g) = CS2(g) + 4 H2(g) A reaction mixture initially contains 0.50 M CH4 and 0.75 M H2S. If the equilibrium concentration of H2 is [H2]eq 0.44 M, find the equilibrium constant (Kc) for the reaction.(a) If the molar solubility of Mg3(PO4)2 at 25 oC is 6.26e-06 mol/L, what is the Ksp at this temperature?Ksp = (b) It is found that 0.00336 g of Ag2C2O4 dissolves per 100 mL of aqueous solution at 25 oC. Calculate the solubility-product constant for Ag2C2O4.Ksp = (c) The Ksp of Sc(OH)3 at 25 oC is 2.22e-31. What is the molar solubility of Sc(OH)3?solubility = mol/L
- Write the concentration equilibrium constant expression for this reaction. CH3CO2C2H5(aq)+H2O(l)→CH¸CO₂H(aq)+C2H5OH(aq) ☐starting with 1.2 M A and 0.25 M B, the equilibrium concentrations of A is found to be 0.75. 2A=B. (a) what is the equilibrium constant for this reaction.(b) what is the effect of catalyst on the equilibrium concentration of A? (c) If you assume that the reaction is aqueous, how is the equilibrium affected by the addition of the solvent (what happns when you double the solvent but not the solutes?)Calcium carbonate is sparingly soluble in water. CaCO:(s) = Ca²"(aq) + CO;²(aq) Kp= 3.4 x 109 (a) Calculate the solubility of CaCO; in sea water containing 0.01 mol L-ª of Ca²* ions. (b) Given the additional information below and using the Henderson Hasselbalch equation, what is the major form of carbonate in sea water and why? pH sea water = 8.1 H2CO: pKai = 6.35 HCO; pK2 = 10.33 Write down the relevant equilibria and explain why the actual solubility of calcium carbonate (4.6 x 10-7 mol L-') is higher than the value you determined in part (a)?
- Help me pleaseCoral structures found in the Great Barrier Reef are composed of calcium carbonate, CaCO3, and are under threat of dissolution due to ocean acidification. Consider the following equilibrium reaction equation. CaCO3(s) + CO2(aq) + H2O(l) ⇌ Ca2+(aq) + 2HCO3–(aq) Write the expression for the equilibrium constant, Kc for this reaction. Predict whether the pH of the ocean will increase or decrease as a result of a decrease in the partial pressure of carbon dioxide in the atmosphere (circle your answer). Calculate the molar solubility (s) of calcium carbonate in water at 25 °C when Ksp = 4.5 ´ 10–9.A solution of sodium chlorite is prepared using 0.38 mol L1 NaOH as the solvent. The pK, of chlorous acid is 2. The concentration of sodium chlorite after mixing, before any reaction takes place is 0.24 mol L1. Calculate the equilibrium concentrations of all acid species: [ HCIO2 ), [ CIO2 1, the pH and the pOH in the solution. Some assumptions may not work, on the other hand, if you do not make the correct assumptions first, the quadratic formula will give you the wrong answer. [ HCIO2 ] - mol L1, [ CIO2 ] = mol L-1, pH = pOH