What volume would 32.0 g of NO2 gas occupy at 3.12 atm and 18.0 Celsius

Volume of 32.0 g at 3.12 atm and 18.0 is 5.33 L.

Explanation:

Molar mass of = 46.0055 g/mol

So, 32.0 g of = mol of = 0.696 mol of

Let's assume behaves ideally.

Then    , where P, V, n, R and T represents pressure, volume, number of moles, gas constant and temperature (in kelvin scale) respectively.

Here = 3.12 atm, = 0.696 mol, R = 0.0821 L.atm/(mol.K) and T = (273+18.0)K = 291 K

So,      

        or,

        or,

So, volume of 32.0 g at 3.12 atm and 18.0 is 5.33 L.

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.