The molecular formula of a compound is C₂H₂O₄.
Take 100 grams of compound:
1) ω(C) = 26.7% ÷ 100% = 0.267
m(C) = ω(C) × m(compound)
m(C) = 0.267 × 100 g.
m(C) = 26.7 g.
n(C) = m(C) ÷ M(C).
n(C) = 26.7 g ÷ 12 g/mol.
n(C) = 2.22 mol; amount of carbon
2) ω(H) = 2.2 % ÷ 100% = 0.022
m(H) = 0.022 × 100 g.
m(H) = 2.2 g.
n(H) = 2.2 g ÷ 1 g/mol.
n(H) = 2.2 mol; amount of hydrogen
3) ω(O) = 71.1 % ÷ 100%.
ω(O) = 0.711
m(O) = 0.711 × 100 g
m(O) = 71.1 g
n(O) = 71.1 g ÷ 16 g/mol
n(O) = 4.4 mol; amount of oxygen
4) n(C) : n(H) : n(O) = 2.2 mol : 2.2 mol : 4.4 mol /2.2 mol.
n(C) : n(H) : n(O) = 1 : 1 : 2
M(CHO₂) = 45 amu; empirical formula
90 amu ÷ 45 amu = 2 CHO₂
More info about empirical formula: brainly.com/question/1873039
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Answer:
Boiling of a substance
Explanation:
When something is boiled, heat is released to the surrounding, making it an exothermic reaction (a type of chemical reaction)
I think transition metals
Answer:
21.6 g
Explanation:
The reaction that takes place is:
First we<u> convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:
- 9.6 g CH₄ ÷ 16 g/mol = 0.6 mol CH₄
- 64.9 g O₂ ÷ 32 g/mol = 2.03 mol O₂
0.6 moles of CH₄ would react completely with (2 * 0.6) 1.2 moles of O₂. As there are more O₂ moles than required, O₂ is the reactant in excess and CH₄ is the limiting reactant.
Now we <u>calculate how many moles of water are produced</u>, using the <em>number of moles of the limiting reactant</em>:
- 0.6 mol CH₄ *
= 1.2 mol H₂O
Finally we<u> convert 1.2 moles of water into grams</u>, using its <em>molar mass</em>:
- 1.2 mol * 18 g/mol = 21.6 g
<u>Answer:</u> The chemical equations are given below.
<u>Explanation:</u>
The chemical equation for the reaction of lead nitrate and sodium hydroxide follows:

By Stoichiometry of the reaction:
1 mole of aqueous solution of lead nitrate reacts with 2 moles of aqueous solution of sodium hydroxide to produce 1 mole of solid lead hydroxide and 2 moles of aqueous solution of sodium nitrate.
The chemical equation for the reaction of lead hydroxide and hydroxide ions follows:
![Pb(OH)_2(s)+2OH^-(aq.)\rightarrow [Pb(OH)_4]^{2-}(aq.)](https://tex.z-dn.net/?f=Pb%28OH%29_2%28s%29%2B2OH%5E-%28aq.%29%5Crightarrow%20%5BPb%28OH%29_4%5D%5E%7B2-%7D%28aq.%29)
By Stoichiometry of the reaction:
1 mole of lead hydroxide reacts with 2 moles of aqueous solution of hydroxide ions to produce 1 mole of aqueous solution of tetra hydroxy lead (II) complex
Hence, the chemical equations are given above.