The substance ammonia has the following properties: normal melting point: 195.4 K normal boiling point: 239.8 K triple point: 5.9×10-2 atm, 195.3 K critical point: 111.5 atm, 405.5 K At temperatures above 405.5 K and pressures above 111.5 atm, NH3 is a . NH3 does not exist as a liquid at pressures below atm. NH3 is a at 5.90×10-2 atm and 249.5 K. NH3 is a at 1.00 atm and 236.0 K. NH3 is a at 24.6 atm and 185.6 K.

i'm wiling to try and you : )

the answers are

c and c

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.