<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
cation means the primation in the air
Please help me! i only need one more to be the best score
Answer:
2.453.
Explanation:
<em>∵ pH = - log[H₃O⁺]
</em>
[H₃O⁺] = 0.00352 M.
<em>∴ pH = - log[H₃O⁺] </em>= - log(0.00352) = <em>2.453.</em>
1 mole ----------- 22.4 L ( at STP )
3.75 moles ----- ?
3.75 x 22.4 / 1 => 84.0 L
