K(s) + 1/2 Cl2(g) → KCl(s) ΔH = -437 kJ/mol The elements K and Cl react directly to form the compound KCl according to the equation above. Refer to the information above and the table to answer the questions that follow.

Cl2(g) + 2e- → 2Cl-(g)

Which of the following expressions is equivalent to ΔHo for the reaction represented above?

Process ΔH° (kj/mol)
K(s) → K(g) v
K(g) → K+(g) + e^- w
Cl2(g) → 2Cl(g) x
Cl(g)+ e^- → Cl^-(g) y
K+(g) + Cl(g)^- → KCL z

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]