<u>Given:</u>
% Al = 35.94
% S = 64.06
<u>To determine:</u>
Empirical formula of a compound with the above composition
<u>Explanation:</u>
Atomic wt of Al = 27 g/mol
Atomic wt of S = 32 g/mol
Based on the given data, for 100 g of the compound: Mass of Al = 35.94 g and mass of S = 64.06 g
# moles of Al = 35.94/27 = 1.331
# moles of S = 64.06/32 = 2.002
Divide by the smallest # moles:
Al = 1.331/1.331 = 1
S = 2.002/1,331 = 1.5 ≅ 2
Empirical formula = AlS₂
Answer:
high concentrations of molecular is to a lower concentration of monoculars
Answer:
200J
Explanation:
This is because 500J - 300J = 200J
I hope this helps!!!
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