Consider the following system at equilibrium where h° = -18.8 kj/mol, and kc = 10.5 , at 350 k. 2 ch2cl2 (g) ch4 (g) + ccl4 (g) when 0.21 moles of ch2cl2 (g) are removed from the equilibrium system at constant temperature:

i don't see a text, but i am gonna assume this *orbit* the curved path, usually elliptical, described by a planet, satellite, spaceship, etc., around a celestial body, as the sun.

c. electrical charges of ionic bonds.

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.