Given a 5.81g sample of h2 (fm =2.02), contained in a volume of 3.90l. how many atm of pressure would it have if the temperature was 64 degrees celsius?

The answer to your question is: P = 20.4 atm

Explanation:

Data

mass = 5.81 g

FM = 2.02

vol = 3.9 L

P = ?

T = 64°C = 337°K

R = 0.082 atm L / mol°K

Formula

               P = nRT / V

Process

              2.02 g  1 mol

              5.81 g  x

               x = (5.81 x 1) / 2.02

               x = 2.88 mol

             P = (2.88)(0.082)(337) / 3.9

             P = 20.4 atm

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.