Answer:
c
Explanation:
I just learned all this in chemistry too lol
got 100 on the test last week so its good
Given
Mass of K2O2 = 1.0 g
Molar mass of K2O2 = 110.0 g/mol
Calculate the # moles of K2O2
# moles = mass of K2O2/molar mass
= 1.0 g/110.0 g.mol-1 = 0.0091 moles
1 mole of K2O2 contains 2*6.023*10^23 atoms of oxygen
Therefore, 0.0091 moles of K2O2 will correspond to:
= 0.0091 moles * 2* 6.023*10^23 atoms/1 mole
= 1.096 * 10^24 atoms of oxygen
<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
I think it is calcium carbonate, from sedimentary rocks and such
