A mixture of two dye molecules, blue and red, was tested by TLC analysis, where the blue dye is more polar than the red. The spotted plate was supposed to be eluted with a mixture of 30% hexane and 70% acetone. However, one student mistakenly used a 70% hexane and 30% acetone solvent mixture instead. The expected change in the Rf value(s) for the two molecules would be: (Select all that apply)

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]