Given the following: i) n20(g) 1/2 02(g) 2 no(g) ii) n2(g) 02(g) 2 no(g) ke= 4.1 x 10-31 ke= 1.7 x 10-13 find the value of the equilibrium constant for the following equilibrium reaction: n2(g) 1/202(g) n20(g) a) 7.0 x 10-44 b) 4.2 x 1017 c) 2.4 x 10-18 d) 1.6 x 109 e) 2.6 x 10-22

10-14. (a) when 100.0 ml of weak acid ha were titrated with 0.093 81 m naoh, 27.63 ml were required to reach the equivalence point. find the molarity of ha. (b) what is the formal concentration of a- at the equivalence point? (c) the ph at the equivalence point was 10.99. find pk. for ha. (d) what was the ph when only 19.47 ml of naoh had been added?

Is heavenly controller dating someone?

Answer asap need it by wednesday morning carry out the following calculations on ph and ka of from data. i. calculate the ph of 0.02m hcl ii. calculate the ph of 0.036m naoh iii. calculate the ph of 0.36m ca(oh)2 iv. calculate the ph of 0.16m ch3cooh which has ka = 1.74 x 10-5 mol dm-3 v. calculate ka for weak acid ha which has a ph of 3.65 at 0.30m concentration vi. calculate the ka of a solution made by mixing 15.0 cm3 0.2m ha and 60.0 cm3 0.31m a-. [ph= 3.80] vii. calculate the ph of a solution made by mixing 15.0 cm3 0.1m naoh and 35.0 cm3 0.2m hcooh. [ka = 1.82 x 10-4 m]