The empirical formula : CH₃
<h3>Further explanation</h3>
Given
2.5 g sample
2.002 g Carbon
Required
The empirical formula
Solution
Mass of Hydrogen :
= 2.5 - 2.002
= 0.498
Mol ratio C : H :
C : 2.002/12 = 0.167
H : 0.498/1 = 0.498
Divide by 0.167 :
C : H = 1 : 3
Answer:
b) Phosphorus acid
Explanation:
To distinguish the type of acid of phosphorus with the oxidation state of +3, we need to be familiar with the chemical formula of each of the compounds:
Orthophosphoric acid H₃PO₄
Phosphorus acid H₃PO₃
Metaphosphoric acid HPO₃
Phyrophosphoric acid H₄P₂O₇
Now that we know the formula of the given compounds, the algebraic sum of all the oxidation numbers of all atoms in a neutral compound is zero:
Only phosphorus acid yielded an oxidation state of +3 for phosphorus in the compound.
H₃PO₃:
we know the oxidation state of H = +1
O = -2
The oxidation state of P is unknown. We can express this as an equation:
3(+1) + P + 3(-2) = 0
3 + P -6 = 0
P-3 = 0
P = +3
The basic units for density is
and any get of units that has those units in the proper place can be considered a density unit. The ones that has those specifically are A, B, E and F
Sir Joseph John Thomson OM PRS (18 December 1856 – 30 August 1940) was a British physicist and Nobel Laureate in Physics, credited with the discovery of the electron, the first subatomic particle to be discovered.
Explanation:
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