G-10 Assume the equilibrium constant K 1 for a reaction A + B 2 C. You...
I. Write the equilibrium constant expression, K for the following reaction Fe"(ag) FENCS(ag) SCN (ag) + CFENCS D 2. In part A of Experiment 34, it was assumed that all of the SCN added to the reaction mixture was converted to product (FeNCS). Which of the following best explains why that assumption was valid? A) The moles of Fe added equaled the moles of SCN- added. B) In part A, the SCN was the limiting reactant whereas in part B,...
equilibrium help!! What is the equilibrium constant for the reaction CO(g) + 3 H2(g) - CH4(g) + H2O(g) if at 20 °C the equilibrium molar concentrations are [CO] -0.613, [H2] = 1.839, (CH4) = 0.387, and [H20) - 0.387? Format Β Ι Ο The reaction for the formation of the diamine-silver ion is as follows: Ag (aq) + 2NH3(aq) + Ag(NH3)2(aq) a. Write the equilibrium constant expression for this reaction. for the above b. An experiment was carried out to...
Equilibria can be treated mathematically with the equilibrium constant, K. For the reaction Fe3+ + SCN − equilibrium reaction arrow FeSCN2+, the equilibrium constant expression is [FeSCN2+ ] K = [Fe3+ ] · [SCN − ] where, for example, [Fe3+ ] is the molar concentration (mol Fe3+ / L solution) present in an equilibrium mixture. At some temperature, a chemist found the following equilibrium concentrations. [Fe3+ ] = 8.17 ✕ 10−3 M, [SCN − ] = 8.60 ✕ 10−3 M,...
Section Name Experiment 23 Advance Study Assignment: Determination of the Equilibrium Constant for a Chemical Reaction 1. A student mixes 5.00 mL 2.00 X 10M Fe(NO), with 5.00 ml 2.00 x 10-M KSCN. She finds that in the equilibrium mixture the concentration of FeSCN is 1.40 x 10M. Find K for the reaction Fe(aq) + SCN (aq) FeSCN2(aq). Step 1 Find the number of moles Fe and SCN initially present. (Use Eq. 3.) (5.00 x103 LX (300X163) = (x 103...
Are my solutions correct? Please help to break these down or correct me. Thank you so much Ta Calculate the solubility of AgCN in a solution containing L0x10 M HCL. K, for AgCN -22x 10 K, for HCN- 6.2 x 101 and 22 x10 G.2 x 10-10 3.SS x03 u,3.58 xx o, 001 1.88 x 10-3 M) 7b. At a certain temperature, K 1.1 x 10' for the reaction, Fe(aq)+ SCN(a)FeSC (aq) Calculate the concentrations of Fe, SCN, and FeSCN"2...
prelab /postlab all questions please Questions 1. For the reaction at 20°C, NH3(aq) + H+ (aq) NH4(aq), the equilibrium constant is calculated to be K = to 4.5 x10%. CHAT] A. Write the Equilibrium expression for this reaction. Keq = ỞNH TH] B. From the size of the number for Keq, does the equilibrium lie to the left or to the right? Right 2. If the reaction between iron(III) ion and thiocyanate ion, Fe3+ (aq) + SCN (aq) → FeSCN2+(aq),...
1) The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.311 M HI, 4.18×10-2 M H2 and 4.18×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 2.85×10-2 mol of I2(g) is added to the flask? 2) The equilibrium constant, K, for the following reaction is 1.20×10-2 at...
Equilibrium Concentrations for a Simple Addition Reaction At a certain temperature K = 1.10x103 for the reaction: Fe3+(aq)SCN"(aq) FeSCN2+(aq) 5.00x102 mol of Fe(NO3)3 is added to 8.70x 10-1 L of 1.84 x 10-1 M KSCN. Neglecting any volume change and assuming that all species remain in solution--: Calculate the equilibrium concentration of Fe3+ (in mol/L) mol/L 1 pts 提交答案 Tries 0/8 Calculate the equilibrium concentration of SCN (in mol/L). mol/L 1 pts [提交答案 Tries 0/8 Calculate the equilibrium concentration of...
Data and Calculations: Determination of the Equilibrium Constant for a Chemical Reaction Method II Volume in mL 2.00 x 103 M Fe(NO) Volume in mL, Depth in mm Volume in ml. 2.00 x 103 M Method I Mixture Unknówn KSCN Water Absorbance Standard FESCNP 4mL 1 5.00 x 10 M 1,00 .227 3mL 2 5,00 202 x 10 M 2,00 90 x 10 M .304 3 5,00 3.00 2mL 955 x 104 M I ImL 4 5.00 4,00 19x 10...
(1). The equilibrium constant, K, for the following reaction is 1.32×10-3 at 565 K. NH4Cl(s) =NH3(g) + HCl(g) An equilibrium mixture in a 10.4 L container at 565 K contains 0.285 mol NH4Cl(s), 4.47×10-2 M NH3 and 2.95×10-2 M HCl. What will be the concentrations of the two gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 4.25 L? [NH3] = M [HCl] = M