#5: Determine the empirical formula of a compound that contains 15.5 g phosphorus and 20.0 g oxygen. I will give Brainliest to best answer + SHOWS WORK!! fake answers will be reported and deleted. <3

Given that

Vapour density V.D = 59

- Moleculat mass

MM =

2X VD = 2 XS9

= 118 u

Mass ti of

40. 68%

Mass / of H 5.08.2

Mass % of 0 100 - 140.68 +5.08)

= 54.24%

Now,

=

Mass 7.

Element

z =

y = 3.39

y X 2 =

1 x2= 2

С

40.68

12

= 3.39 3.39

3.39

P

5.08

1

5.08 5.08 15 1.5X2 = 3

3.39

54.24

16

3.39

3.39

= 1

, 1 2 = 2

O

3.39

=

59.

. Empherical formula , EF = C₂ H₂ O₂

Empherical mass, EM

Molecular mass

n. factor

Empherical mass

= 2.

59

6. -

3. Molecular formula = (EF) are = (Calg Oz]2

عوار)

= C4 H604

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.