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2. K. is 3.8 x 10 at 1000K for the equilibrium I8)21g.) Starting with 0.0456 moles...
2. 21(g) the value of K-3.76x10sat 1000K. If initially LI-10 M,00 sOCOM For the reaction I...
2. 21(g) the value of K-3.76x10sat 1000K. If initially LI-10 M,00 sOCOM For the reaction I is heated to 1000K a. Which way does ig) hws the reaction need to shift to reach equilbrium (eft toward reactants or right toward products? b. Construct an ICE table describing the change in terms of a positive parameter x Initial 1.00 0.00 Change Equilibrium c. Solve for the change x and obtain the equilibrium concentration of la
9. The equilibrium constant(K.) for the gaseous dissociation of Iodine molecules to lodine atoms (shown below) 12(8...
9. The equilibrium constant(K.) for the gaseous dissociation of Iodine molecules to lodine atoms (shown below) 12(8) 21(8) is 3.76 x 10-3 at 1000K. If 0.15 mole of 12(g) is placed in a 12.3-L flask at 1000 K, what are the concentrations of 12(g) and I(g) when the reaction comes to equilibrium?
The equilibrium constant, K, for the following reaction is 1.20 times 10^-2 at 500 k. PCl_5...
The equilibrium constant, K, for the following reaction is 1.20 times 10^-2 at 500 k. PCl_5 (g) irreversible PCl_3 (g) + Cl_2 (g) At equilibrium mixture of the three gases in a 1.00 L. flask at 500 K contains 0.203 M PCl_5, 4.93 times 10^-2 M PCl_5 and 4.93 times 10^-2 M Cl_2. What will be the concentration of the three gases once equilibrium has been reestablished, if 2.81 times 10^-2 mol of PCl_3 (g) is added to the flask?...
A Q=K: the reaction mixture is at equilibrium O T OM B. Q<K; the reaction will...
A Q=K: the reaction mixture is at equilibrium O T OM B. Q<K; the reaction will shift towards the reactants C. Q>K; the reaction will shift towards the products → KA А е е D. Q>K; the reaction will shift towards the reactants MOL (E. Q<K; the reaction will shift towards the products. M280. 00 3. For the reaction M SES00.0 3 to noiriwa H2 (9)+CO2(9) <==> H20 (9) +CO (9) Kc = 0.798 at 320°C. 0.492 moles of H2...
At 35°C, K = 2.3 x 10-5 for the reaction 2 NOCI(9) – 2 NO(g) +...
At 35°C, K = 2.3 x 10-5 for the reaction 2 NOCI(9) – 2 NO(g) + Cl (9) Calculate the concentrations of all species at equilibrium for each of the following original mixtures. a. 2.4 moles of pure NOCl in a 2.0-L flask [NOCI) - t t M t [NO] - M t [Cl] - M b. 1.0 mole of NOCI and 1.0 mole of NO in a 1.0-L flask [NOCI] - M [NO] - M (Cl) - M c....
At 35°C, K =2.0 x 10-8 for the reaction 2 NOCI(9) = 2 NO(g) + Cl2(9)...
At 35°C, K =2.0 x 10-8 for the reaction 2 NOCI(9) = 2 NO(g) + Cl2(9) Calculate the concentrations of all species at equilibrium for each of the following original mixtures. a. 2.2 moles of pure NOCI in a 2.0-L flask [NOCI) = M [NO] = | М [Cl] = M b. 1.0 mole of NOCI and 1.0 mole of NO in a 1.0-L flask [NOCI) = M [NO] = M (Cl2] = 1 M c. 2.0 mole of NOCl...
1) The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g)...
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...
consider the reaction A+B<----->C+D, which has an equilibrium constant K, equal to 3.4 x 10^2 1....
consider the reaction A+B<----->C+D, which has an equilibrium constant K, equal to 3.4 x 10^2 1. If one begins a reaction by placing 0.6000 moles of A in 1.0 L container as well as 0.150 moles of B, what will be the equilibrium concentrations of A, B, C, and D? 2 Once the reaction in problem one reaches equilibrium, some additional B is injected into the flask from an outside source. LeChatelier's principal says the reaction will be a) unchanged...
explain 21,22,23 UTILIT In this solution is a. 2.6 x 10-1° M b. 3.8 x 10...
explain 21,22,23
UTILIT In this solution is a. 2.6 x 10-1° M b. 3.8 x 10 M c. 9.58 d. 4.42 e. none of these 21. How many moles of pure NaOH must be used to prepare 1.0 L of a solution that has pH = 12.26? a. 8.2 x 10-16 mol b. 0.018 mol c. 5.5 x 10-13 mol d. 1.74 mol e. none of these 22. Calculate the pH of a 0.045 M HCl solution. K a. 1.23...
The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) +...
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 0.329 M HI, 4.41×10-2 M H2 and 4.41×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 2.54×10-2 mol of H2(g) is added to the flask? [HI] = M [H2] = M [I2] = M
2. 21(g) the value of K-3.76x10sat 1000K. If initially LI-10 M,00 sOCOM For the reaction I is heated to 1000K a. Which way does ig) hws the reaction need to shift to reach equilbrium (eft toward reactants or right toward products? b. Construct an ICE table describing the change in terms of a positive parameter x Initial 1.00 0.00 Change Equilibrium c. Solve for the change x and obtain the equilibrium concentration of la
9. The equilibrium constant(K.) for the gaseous dissociation of Iodine molecules to lodine atoms (shown below) 12(8) 21(8) is 3.76 x 10-3 at 1000K. If 0.15 mole of 12(g) is placed in a 12.3-L flask at 1000 K, what are the concentrations of 12(g) and I(g) when the reaction comes to equilibrium?
The equilibrium constant, K, for the following reaction is 1.20 times 10^-2 at 500 k. PCl_5 (g) irreversible PCl_3 (g) + Cl_2 (g) At equilibrium mixture of the three gases in a 1.00 L. flask at 500 K contains 0.203 M PCl_5, 4.93 times 10^-2 M PCl_5 and 4.93 times 10^-2 M Cl_2. What will be the concentration of the three gases once equilibrium has been reestablished, if 2.81 times 10^-2 mol of PCl_3 (g) is added to the flask?...
A Q=K: the reaction mixture is at equilibrium O T OM B. Q<K; the reaction will shift towards the reactants C. Q>K; the reaction will shift towards the products → KA А е е D. Q>K; the reaction will shift towards the reactants MOL (E. Q<K; the reaction will shift towards the products. M280. 00 3. For the reaction M SES00.0 3 to noiriwa H2 (9)+CO2(9) <==> H20 (9) +CO (9) Kc = 0.798 at 320°C. 0.492 moles of H2...
At 35°C, K = 2.3 x 10-5 for the reaction 2 NOCI(9) – 2 NO(g) + Cl (9) Calculate the concentrations of all species at equilibrium for each of the following original mixtures. a. 2.4 moles of pure NOCl in a 2.0-L flask [NOCI) - t t M t [NO] - M t [Cl] - M b. 1.0 mole of NOCI and 1.0 mole of NO in a 1.0-L flask [NOCI] - M [NO] - M (Cl) - M c....
At 35°C, K =2.0 x 10-8 for the reaction 2 NOCI(9) = 2 NO(g) + Cl2(9) Calculate the concentrations of all species at equilibrium for each of the following original mixtures. a. 2.2 moles of pure NOCI in a 2.0-L flask [NOCI) = M [NO] = | М [Cl] = M b. 1.0 mole of NOCI and 1.0 mole of NO in a 1.0-L flask [NOCI) = M [NO] = M (Cl2] = 1 M c. 2.0 mole of NOCl...
explain 21,22,23
UTILIT In this solution is a. 2.6 x 10-1° M b. 3.8 x 10 M c. 9.58 d. 4.42 e. none of these 21. How many moles of pure NaOH must be used to prepare 1.0 L of a solution that has pH = 12.26? a. 8.2 x 10-16 mol b. 0.018 mol c. 5.5 x 10-13 mol d. 1.74 mol e. none of these 22. Calculate the pH of a 0.045 M HCl solution. K a. 1.23...
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