Asecond reaction mixture was made up in the following way:
10 ml 4.0 m acetone + 20 ml 1.0 m hcl + 10 ml 0.0050 m i2 + 10 ml h2o
a. what were the initial concentrations of acetone, h+ ion, and i2 in the reaction mixture?
(acetone) m; (h+) m; (i2)0 m
b. it took 120 seconds for the i2 color to disappear from the reaction mixture when it occurred at the
same temperature as the reaction in problem 2. what was the rate of the reaction?

A. We can calculate the initial concentrations of each by the formula:

initial concentration ci = initial volume * initial concentration / total mixture volume

where,

total mixture volume = 10 mL + 20 mL + 10 mL + 10 mL = 50 mL

ci (acetone) = 10 mL * 4.0 M / 50 mL = 0.8 M

ci (H+) = 20 mL * 1.0 M / 50 mL = 0.4 M               (note: there is only 1 H+ per 1 HCl)

ci (I2) = 10 mL * 0.0050 M / 50 mL = 0.001 M

B. The rate of reaction is determined to be complete when all of I2 is consumed. This is signified by complete disappearance of I2 color in the solution. The rate therefore is:

rate of reaction = 0.001 M / 120 seconds

rate of reaction = 8.33 x 10^-6 M / s

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.