The number of electrons present in 4.2 gram of methane is

Explanation:

One CH4 molecules = 10 electrons ( Carbon 6e + 4Hydrogen 4e)

A mole of any substance witll have = 6.023*10^23 atom/ molecuels,

One mole of any substnce is equal is nothing but its moalr mass

Molar mass of CH4 = 16.04

It mean,

16.04  g of CH4 has = 6.023*10^23 molecules of methane

but

4.2 gram of CH4   = 6.023*10^23 * 4.2 /16.04 = 1.58*10^23 molecuels of CH4

So one molecule has =10 electrons

1.58*10^23 molecuels will have = 1.58*10^23  * 10 = 1.58*10^24

Hope this help you

answerca air mass;

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.