A student ran the following reaction in the laboratory at 380 K: CH4(g) + CCl4(g) 2CH2Cl2(g)...
A student ran the following reaction in the laboratory at 380 K: CH4(g) + CCl4(g) 2CH2Cl2(g) When she introduced 4.26×10-2 moles of CH4(g) and 5.94×10-2 moles of CCl4(g) into a 1.00 liter container, she found the equilibrium concentration of CCl4(g) to be 5.11×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc =
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A student ran the following reaction in the laboratory at 380 K:
CH4(g) + CCl4(g) 2CH2Cl2(g)...
A student ran the following reaction in the laboratory at 293 K: 2CH2Cl2(g) CH4(g) + CCl4(g)...
A student ran the following reaction in the laboratory at 293 K: 2CH2Cl2(g) CH4(g) + CCl4(g) When she introduced 6.91×10-2 moles of CH2Cl2(g) into a 1.00 liter container, she found the equilibrium concentration of CCl4(g) to be 3.19×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc=?
A student ran the following reaction in the laboratory at 283 K: 2CH2Cl2(g) ->CH4(g) + CCl4(g) ...
A student ran the following reaction in the laboratory at 283 K: 2CH2Cl2(g) ->CH4(g) + CCl4(g) When she introduced 7.70×10-2 moles of CH2Cl2(g) into a 1.00 liter container, she found the equilibrium concentration of CCl4(g) to be 3.59×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc =
A student ran the following reaction in the laboratory at 294 K: 2CH2Cl2(g) ____> CH4(g) +...
A student ran the following reaction in the laboratory at 294 K: 2CH2Cl2(g) ____> CH4(g) + CCl4(g) When she introduced 7.04×10-2 moles of CH2Cl2(g) into a 1.00 Liter container, she found the equilibrium concentration of CCl4(g) to be 3.25×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction.
1. The equilibrium constant, Kc, for the following reaction is 42.2 at 288 K. 2CH2Cl2(g) <----->...
1. The equilibrium constant, Kc, for the following reaction is 42.2 at 288 K. 2CH2Cl2(g) <-----> CH4(g) + CCl4(g) When a sufficiently large sample of CH2Cl2(g) is introduced into an evacuated vessel at 288 K, the equilibrium concentration of CCl4(g) is found to be 0.140 M. Calculate the concentration of CH2Cl2 in the equilibrium mixture. _____ M 2. A student ran the following reaction in the laboratory at 295 K: 2CH2Cl2(g) <-------> CH4(g) + CCl4(g) When she introduced 8.46×10-2 moles...
A student ran the following reaction in the laboratory at 280 K: 2CH2Cl2(g) CH(g) + CC14(8)...
A student ran the following reaction in the laboratory at 280 K: 2CH2Cl2(g) CH(g) + CC14(8) When she introduced 8.40x10-2 moles of CH,CL2(g) into a 1.00 liter container, she found the equilibrium concentration of CCI,(g) to be 3.93x10-2 M. Calculate the equilibrium constant, K, she obtained for this reaction. K=
A student ran the following reaction in the laboratory at 425 K: PCl5(g) --> PCl3(g) +...
A student ran the following reaction in the laboratory at 425 K: PCl5(g) --> PCl3(g) + Cl2(g) When she introduced 4.59 moles of PCl5(g) into a 1.00 liter container, she found the equilibrium concentration of Cl2(g) to be 3.94×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction.
The equilibrium constant, Kc, for the following reaction is 9.52x10-2 at 350 K: CH4(8) + CC14(8)...
The equilibrium constant, Kc, for the following reaction is 9.52x10-2 at 350 K: CH4(8) + CC14(8) — 2CH2Cl2(g) Calculate the equilibrium concentrations of reactants and product when 0.305 moles of CH4 and 0.305 moles of CCl4 are introduced into a 1.00 L vessel at 350 K. [CH] = [CC14] = [CH2Cl2] = A student ran the following reaction in the laboratory at 531 K: COC12(8) CO(g) + Cl2(8) When she introduced 1.87 moles of COCl2(g) into a 1.00 liter container,...
A student ran the following reaction in the laboratory at 1185 K: 2SO2(g) + O2(g) 2SO3(g)...
A student ran the following reaction in the laboratory at 1185 K: 2SO2(g) + O2(g) 2SO3(g) When she introduced 8.07×10-2 moles of SO2(g) and 7.94×10-2 moles of O2(g) into a 1.00 liter container, she found the equilibrium concentration of O2(g) to be 5.86×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc=?
A student ran the following reaction in the laboratory at 324 K: 2NO(g) + Br2(g) 2NOBr(g)...
A student ran the following reaction in the laboratory at 324 K: 2NO(g) + Br2(g) 2NOBr(g) When she introduced 0.137 moles of NO(g) and 0.119 moles of Br2(g) into a 1.00 liter container, she found the equilibrium concentration of Br2(g) to be 6.92×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction.
A student ran the following reaction in the laboratory at 225 K: 2NOBr(g) 2 2NO(g) +...
A student ran the following reaction in the laboratory at 225 K: 2NOBr(g) 2 2NO(g) + Brz(g) When she introduced 0.198 moles of NOBr(g) into a 1.00 liter container, she found the equilibrium concentration of Br2(g) to be 1.89x10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc =
A student ran the following reaction in the laboratory at 280 K: 2CH2Cl2(g) CH(g) + CC14(8) When she introduced 8.40x10-2 moles of CH,CL2(g) into a 1.00 liter container, she found the equilibrium concentration of CCI,(g) to be 3.93x10-2 M. Calculate the equilibrium constant, K, she obtained for this reaction. K=
The equilibrium constant, Kc, for the following reaction is 9.52x10-2 at 350 K: CH4(8) + CC14(8) — 2CH2Cl2(g) Calculate the equilibrium concentrations of reactants and product when 0.305 moles of CH4 and 0.305 moles of CCl4 are introduced into a 1.00 L vessel at 350 K. [CH] = [CC14] = [CH2Cl2] = A student ran the following reaction in the laboratory at 531 K: COC12(8) CO(g) + Cl2(8) When she introduced 1.87 moles of COCl2(g) into a 1.00 liter container,...
A student ran the following reaction in the laboratory at 225 K: 2NOBr(g) 2 2NO(g) + Brz(g) When she introduced 0.198 moles of NOBr(g) into a 1.00 liter container, she found the equilibrium concentration of Br2(g) to be 1.89x10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc =
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