the given reaction is reverse to the basicity reaction of amine.
so that,
equilibrium constant = 1/Kb
= 1/(4.2*10^-4)
= 2.38*10^3
answer: e
What is the equilibrium constant for the following reaction? CH_3 NH^+_3 + H_2 O equivalentto CH_3...
Write products for the following acid-base reaction () CH_3 OH + H_2 SO_4 rightarrowoverleftarrow CH_3 OH + NA^+ NH^-_2 rightarrowoverleftarrow CH_3 NH^+_3 CL^- + Na^+ OH^- rightarrowoverleftarrow + rightarrowoverleftarrow
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What is the pH of a 0.34 M solution of methylamine (CH_3 NH_3, K_, - 4.4 times 10^-4) at 25 degree C? A) 1.92 B) 5.56 C) 0.47 D) 12.08 E) 13.53 What the pOH of a 0.033 M HI solution? A) 10.59 B) 3.41 C) 1.48 D) 15.48 E) 12.52 Which of the following is not capable of acting as an acid? A) SO^2-_3 B) H_2O C) H_2 O D) H_2 SO_3 E) HSO^-_3 For the reaction Br_2 (g)...
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Be sure to answer all parts. The equilibrium constant K_c for the reaction H_2(g) + Br_2(g) = 2HBr(g) is 2.180 Times 10^6 at 730DegreeC. Starting with 5.20 moles of HBr in a 12.0-L reaction vessel, calculate the concentrations of H_2, Br_2, and HBr at equilibrium. [H_2] = M [Br_2] = M [HBr] = M
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