P Question // Q 10) Find the value of the equilibrium constant expression. Fe3+ SCN- FESCN2+ 1.3 x 10-3 1.3 x 10-3 C -1.80 x 10-4 -1.80 x 10-4 +1.80 x 10-4 E 1.1 x 10-3 1.1 x 10-3 +1.80 x 10-4
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- Consider the mixing of 25.00 mL of 0.0500 M Pb(NO3)2 with 25.00 mL of 0.0500 M NaCl and the equilibrium reaction, PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq) Ksp = 1.20 x 10-5 What is the molar concentration of Pb2+ in the resulting mixture? (0.500, 0.025, 0.0125 or 0.100) What is the molar concentration of Cl- in the resulting mixture? (0.500, 0.025, 0.0125 or 0.100) What is the ion product (Q)? Will a precipitate form? (yes or no)We did an experiment called "An Equilibrium Constant". .2M Fe(NO3)3 .001 M NaSCN .1M HNO3 (in .1 M HNO3) (in .1 M HNO3) __________________________________________________________ Blank. 10 mL 0 mL Dilute to 25mL 1 10 mL 1 mL Dilute to 25 mL 2 10 mL 2 mL Dilute to 25 mL 3. 10 mL 3 mL Dilute to 25 mL 4. 10 mL 4 mL Dilute to 25 mL 5. 10 mL 5 mL Dilute to 25 mL _______ We put differing amounts of .001 M NaSCN solution (shown in column 2), mixed in 10 mL of .2 M Fe (NO3)3 solution with all above trials. Then we diluted every trial to 25 mL…What is [Cu2+] at equilibrium, given CuS, Ksp = 1.27 x 10^-36? A. 1.27 x 10^-36 MB. 6.35 x 10^-37 MC. 1.13 x 10^-18 MD. 1.61 x 10^-72 ME. There is not enough information in this prompt to answer this question.
- Calculate the equilibrium constant (Kc) for the following reaction: CH3COOH(aq) + OH-(aq)) CH3COO-(aq) + H2O(l) The Ka of acetic acid is 1.75 x 10-5; Kw for auto-ionization of water = 1 x 10-14 Hint: First write the equilibrium expressions for both species (reactants) separately then apply manipulation rules of K and find the answer)For the reaction PCl5(g)⇌ PCl3(g) + Cl2(g), the equilibrium constantKc = 1.1 x 10-2 at 400 K. What is the equilibrium constant for the reactionPCl3(g) + Cl2(g)⇌ PCl5(g) at 400 K?For the reaction PCl5(g) <=> PCl3(g) + Cl2(g), the equilibrium constantKc = 1.1 * 10-2 at 400 K. What is the equilibrium constant for the reactionPCl3(g) + Cl2(g) <=> PCl5(g) at 400 K?
- How I find the equilibrium constant? I was able to get K to be 0.001789 by having Ba and AsO = to S. I got 1.1e-13= [3S]^3 [2S]^2 so S=0.001789 then I plugged that in to find my Ba but got it wrong.Consider the mixing of 25.00 mL of 0.0500 M Pb(NO3)2 with 25.00 mL of 0.0500 M NaCl and the equilibrium reaction, PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq) Ksp = 1.20 x 10-5 What is the molar concentration of Pb2+ in the resulting mixture? What is the molar concentration of Cl- in the resulting mixture3 (i) In the Haber process for the production of ammonia, the following reaction occurs: N2 (g) + 3H2 (g) 2NH3 (g) DH is negative If the equilibrium concentrations for all the reactants and products at 6000C are: [N2] = 0.40 mol/dm3, [H2] = 1.20 mol/dm3 and [NH3] = 0.20 mol/ dm3 Construct an expression for Kc and calculate the numerical value of the equilibrium constant, Kc (ii) At 5000C Kc = 0.062 mol-2dm6. Compare this value to the one you calculated in 3(i) above and state whether the yield of ammonia is greater at 5000C or 6000C and briefly explain your choice.
- Determine the equilibrium concentration of WHITE (g). WALTER (g) ⇌ 2 WHITE (g) KC = 0.21 [WALTER]eq = 0.0033 M Group of answer choices: 0.026 M 1.2 × 10–1 M 8.3 × 10–2 M 6.9 × 10–3 M 1.44 MEqual quantities of 0.010M0.010M solutions of an acid HAHA and a base BB are mixed. The pHpH of the resulting solution is 9.4. Write the equilibrium equation for the reaction between HAand B. Express your answer as a chemical equation. Identify all of the phases in your answer. If Ka for HA is 9.0×10−5, what is the value of the equilibrium constant for the reaction between HA and B? Express your answer using one significant figure. Keq= What is the value of Kb for B? Express your answer using one significant figure. Kb=For the reaction 2HI(g)⇌H2(g)+I2(g) the equilibrium constant Kc=1.8×10−2 at 700 K. A. What is the equilibrium constant for the reaction H2(g)+I2(g)⇌2HI(g) at 700 K? (Express your answer using two significant figures.)