Write the equation for the Equilibrium constant, K, for the equilibrium between [CoCl4] 2– and [Co(H2O)6] 2+ .
The reaction happens in accordance with the Le-Chatelier's principle. The main species involved in the reaction is cobalt(II)chloride Or CoCl2, which usually comes in forms of hexahydrate, CoCl2. 6H2O. When dissolved in water both CoCl2. 6H2O will produce the complex ion [Co(H2O) 6]2+.

Write the equation for the Equilibrium constant, K, for the equilibrium between [CoCl4] 2– and [Co(H2O)6]...
Consider the equilibrium mixture: [Co(H2O)6]2+ (aq) + 4Cl- (aq) <--> [CoCl4]2- (aq) + 6H2O (l) What would happen to the color if you added NaCl? (more blue, more pink, no color change) What component of the NaCl would influence the equilibrium in this way?
1. Addition of chloride ion, Cl-(aq) to the cobalt complex, [Co(H2O)6]2+ causes the formation of a second cobalt complex, [CoCl4]2- and H2O(l). Write a balanced net ionic equation for this reaction. 2. Re-write the balanced equation from above this time including heat in the equation as either a reactant or product.
Co(H2O)6]2+, [CoCl4]2- Equilibrium (common ion) 1.Effect of concentrated HCl. Place about 10 drops of 1.0 M CoCl2 in a 75 mm test tube. 2.Add drops of concentrated HCl (Caution: Avoid inhalation and skin contact). This solution is found in the hood. Place the test tube in a test tube rack. 3.Remove and hold the coin stopper while you use it. Place coin stopper on bottle when done) until a color change occurs. 4.Slowly add water to the system and stir....
Aqueous solutions of Co2+ salts are pale pink. They contain the complex ion [Co(H2O)6]2+. Addition of concentrated HCl results in a profound colour change to dark blue, due to formation of [CoCl4]2–. The main absorption bands of [Co(H2O)6]2+ and [CoCl4]2– occur at 19400 cm–1 (e ~ 10 M–1 cm–1) and 14700 cm–1 (e ~ 1000 M–1 cm–1), respectively. (i) Explain the difference in the transition energies between the two complexes. (ii) Using a qualitative molecular orbital diagram, explain why the...
The equilibrium constant Kp for the reaction C(s)+H2O(g)⇌CO(g)+H2(g) is 2.44 at 1000 K. What are the equilibrium partial pressures of H2O, CO, and H2 if the initial partial pressures are PCO= 1.30 atm, and PH2= 1.40 atm? Part A What is the equilibrium partial pressure of H2O? What is the equilibrium partial pressure of CO? What is the equilibrium partial pressure of CO?
The equilibrium constant Kp for the reaction C(s)+H2O(g)⇌CO(g)+H2(g) is 2.44 at 1000 K. What are the equilibrium partial pressures of H2O, CO, and H2 if the initial partial pressures are PCO=1.15 atm, and PH2= 1.45 atm? Part A What is the equilibrium partial pressure of H2O? Part B What is the equilibrium partial pressure of CO? Part C What is the equilibrium partial pressure of H2?
Coordination numbers for the following complex ions: a) [Ti(H2O)6]3+ b) [Ti(EDTA)]- c) [TiCl4]- d) [Co(H2O)6]2+ e) [Co(EDTA)]2- f) [CoCl4]2- g) [Zn(H2O)6]2+ h) [Zn(EDTA)]2-
Solve these two questions plz
ASAP
6. Consider the chemical equation and equilibrium constant for the synthesis of ammonia at 25 °C: Ng(g) + 3 H2(g-2NH3(g) K=3.7 x 108 Calculate the equilibrium constant for the following reaction at 25 C: 7. Given the equilibrium constants for the first and second reactions, calculate the equilibrium constant for CO2 (g) + H2(g) CH3OH(g) the third reaction. K, = 1.00× 105 co(g) + H2O(g) 2co(g) + H2(g) K2 3.74x 103
Be sure to answer all parts. The equilibrium constant K, for the reaction Hz() + CO2() H2O(g) + CO(g) 1s 4.2 at 1650°C. Initially 0.76 mol H2 and 0.76 mol CO2 are injected into a 4.5-L flask. Calculate the concentration of each species at equilibrium. Equilibrium concentration of H2: Equilibrium concentration of CO2: Equilibrium concentration of H20: Equilibrium concentration of CO:
< Question 6 of 11 > Write the equilibrium-constant expression for the reaction shown in terms of [SO], [CO,1. [CS], and 250,(g) + CO,(g) = CS,() + 30,() Kc, which is sometimes symbolized as K or Keq, denotes that the equilibrium constant is expressed usis concentrations. For this question, K. means the same thing as K and Keq K=