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n200080 [17]
2 years ago
14

A compound is 48.6 % C, 8.2 % H, and 43.2 % O by mass. Calculate the empirical formula of this compound.

Chemistry
1 answer:
maria [59]2 years ago
6 0
Answer:
C2H3O
Explanation
:
C. H. O
48.6. 8.2. 43.2
r.a.m 12. 1. 16
no.of 48.4÷12. 8.2÷1. 43.2÷16
moles
4.05÷2.7. 8.2÷2.7. 2.7÷2.7
2. 3. 1
Therefore C2H3O
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If 3.0 liters of 0.50 m nacl (aq) is mixed with 9.0 liters of 0.2777 m nacl (aq), what is the final concentration of the resulti
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First, we shall calculate the total number of moles present in the final solution.
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When 10.0 grams of ch4 reacts completely with 40.0 grams of o2 such that there are no reactants left over, 27.5 grams of carbon
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Moles of O₂ = mass of O₂/molar mass of O₂ = 40/32 = 1.25

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both CH₄ and O₂ are present in the exact required quantity.

Hence, moles of CO₂ formed = moles of CH₄ reacted = 0.625

Mass of CO₂ formed = moles of CO₂ x Molar mass of CO₂ = 0.625 x 44 = 27.5g

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