When 25.0 mL of a solution of 0.200 M NalOz was added to acidified iodide ions, the iodine produced reacted with 20.3 mL of sodium thiosulfate (Na2S2O3). Calculate the concentration of sodium thiosulfate
solution given the REDOX equations below.
IO3 + 51° + 6H → 312 + 3H2O
12 + 2S2O32- → 21° +S4062

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current

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.