Chemistry, 15.11.2019 18:31 Kimberlytdb

Volhard titration was used to find out the mass% of i-in a 0.6532-g sample. a 50.00-ml of 0.05539 m silver nitrate solution was added to the sample and the precipitate was allowed to form. back titration of the remaining silver required 33.12 ml of 0.05233 mpotassium thiocyanate. determinethe mass% of i-(atomic weight 126.9 amu) in the sample.

Answers: 2

Show answers

Other questions on the subject: Chemistry

Chemistry, 22.06.2019 13:00, devontemiles8868

One of the hopes for solving the world's energy problem is to make use of the fusion reaction 21h +31h --> 42he + 10n + energy how much energy is released when 1 mol of deuterium is fused with 1 mol of tritium according to the above reaction? the masses of the atoms and the neutrons are as follows: 21h = 2.0140 amu 31h = 3.01605 amu 42he = 4.002603 amu 10n = 1.008665 amu. the speed of light is 2.9979 x 108 m/s.

Answers: 1

continue

Chemistry, 22.06.2019 14:30, Kiaraboyd9366

Select all of the statements which are true. electrons are located in shells or orbits around the atom. electrons orbit slowly around the atom. electrons travel in one flat path around the nucleus of an atom. the valence of an atom is determined by the number of electrons in the atom's outermost shell.

Answers: 1

continue

Chemistry, 22.06.2019 18:00, kingamir

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]

Answers: 1

continue

Chemistry, 22.06.2019 21:50, BookandScienceNerd

Answer the questions about this reaction: nai(aq) + cl2(g) → nacl(aq) + i2(g) write the oxidation and reduction half-reactions: oxidation half-reaction: reduction half-reaction: based on the table of relative strengths of oxidizing and reducing agents (b-18), would these reactants form these products? write the balanced equation: answer options: a. 0/na -> +1/na+1e- b. nai(aq) + cl2(g) → nacl(aq) + i2(g) c. +1/na+1e- -> 0 /na d. -1/2i -> 0/i2+2e- e. no f. 4nai(aq) + cl2(g) → 4nacl(aq) + i2(g) g. 2nai(aq) + cl2(g) → 2nacl(aq) + i2(g) h. 4nai(aq) + 2cl2(g) → 4nacl(aq) + 2i2(g) i. nai(aq) + cl2(g) → nacl(aq) + i2(g) j. 0/cl2+2e -> -1/2cl- k. yes

Answers: 1

continue

You know the right answer?

Volhard titration was used to find out the mass% of i-in a 0.6532-g sample. a 50.00-ml of 0.05539 m...