Write a nuclear equation for the alpha decay of 23892U.23892U → 0−1e + 23893Np23892U → 10n + 23792U23892U → 42He + 23490Th23892U
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Answer:
C₃H₅O₂
4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O
Explanation:
The reaction can be expressed as:
CₓHₓOₓ + nO₂ → CO₂ + H₂O
Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so <u>the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂</u>:
All of the hydrogens atoms in the compound ended up becoming H₂O, so <u>the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O</u>:
Because the compound is composed only by C, H and O, <u>the mass of O in the compound can be calculated by substraction</u>:
0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O
In order to determine the empirical formula, we calculate the moles of each component:
- mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C
- mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H
- mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O
Then we divide those values by the lowest one:
0.0236 mol C ÷ 0.0141 = 1.67
0.0354 mol H ÷ 0.0141 = 2.51
0.0141 mol O ÷ 0.0141 = 1
If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.
- The reaction is:
4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O