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. The rate constant is the reaction rate divided by the concentration
terms.
Explanation:
The rate constant can be determined from the rate law because it is the reaction rate divided by the concentration terms. I hope I could help! :)
Answer:
Chicken nuggets
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Answer:
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Answer:
The answer to your question is 27 g of Al
Explanation:
Data
mass of Al = ?
moles of Al₂O₃ = 0.5
The correct formula for the product is Al₂O₃
Balanced chemical reaction
4Al + 3O₂ ⇒ 2Al₂O₃
Process
1.- Calculate the molar mass of the product
Al₂O₃ = (27 x 2) + (16 x 3)
= 54 + 48
= 102 g
2.- Convert the moles of Al₂O₃ to grams
102 g ---------------- 1 mol
x ---------------- 0.5 moles
x = (0.5 x 102) / 1
x = 51 g of Al₂O₃
3.- Use proportions to calculate the mass of Al
4(27) g of Al --------------- 2(102) g of Al₂O₃
x --------------- 51 g
x = (51 x 4(27)) / 2(102)
x = 5508 / 204
x = 27 g of Al
