The given compound is Aluminum sulfate, Al2(SO4)3:
Molar masses:
Aluminum = 27 g/mol
Sulfur = 32 g/mol
Oxygen = 16 g/mol
The total molar mass is 342 g/mol
The ratio by mass of the elements:
Aluminum = 27*2/342
= 0.16
Sulfur = (32*3)/342
= 0.28
Oxygen = (16*12)/342
= 0.56
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The amount remaining is obtained using the half life data. The total amount remaining after one hour is equal to 1/( e^ (ln2/(20/60))*1). The final answer is 0.125 grams.
"There are three rules on determining how many significant figures are in a number: Non-zero digits are always significant. Any zeros between two significant digits are significant. A final zero or trailing zeros in the decimal portion ONLY are significant."
Answer:
see explanations
Explanation:
4NH₃(g) + 5O₂(g) => 4NO(g) + 6H₂O(g)
Ci(NH₃) = 3.5mole/4L = 0.875M
Cf(NH₃) = 1.6mole/4L = 0.400M
Rate-1 => Δ[NH₃]/Δt = |(0.400M - 0.875M)/3min| = 0.158M/s
Rate-2 => 6(Δ[NH₃]/Δt) = 4(Δ[H₂O]/Δt) => 6/4(0.158M/s) = 0.237M/s
Rate-3 => 5(Δ[NH₃]/Δt) = 4(Δ[O₂]/Δt) => 5/4(0.158M/s) = 0.237M/s
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NOTE: When setting up comparative rate expressions for a given reaction, set the rates expressions as equal then swap coefficient values. Then solve for rate of interest and substitute givens.
example: for NH₃ and H₂O
- set rates expressions equal => Δ[NH₃]/Δt = Δ[H₂O]/Δt
- then swap and insert coefficients from given rxn ...
- solve for rate of interest ...
4NH₃(g) + 5O₂(g) => 4NO(g) + 6H₂O(g)
=> 6(Δ[NH₃]/Δt) = 4(Δ[H₂O]/Δt)
=> Δ[H₂O]/Δt = 6/4(Δ[NH₃]/Δt) = 6/4(0.237M/s) = 0.237M/s
