MAl₂(SO₄)₃·xH₂O:
(mAl×2) + (mS×3) + (mO×12) + (mH₂O×x)
(27×2)+(32×3)+(16×12)+(x×18) = 342 + 18x [g]
mAl₂: 27×2 = 54 [g]
54g ---------- 13,63%
342+18x ---- 100%
0,1363(342+18x) = 54
46,6146 + 2,4534x = 54
2,4534x = 7,3854
x ≈ 3
>>> Al₂(SO₄)₃·3H₂O <<<<
:)
Answer:
I think it the carbon 14 which is a half life of 5730 because in 4.5billion years ago I don't think any human being existed on earth I'm not sure I hope that I helped you and good luck .
Answer:
a. 0.5 mol
b. 1.5 mol
c. 0.67
Explanation:
Fe3+ + SCN- -----> [FeSCN]2+
a. The ratio of the product to Fe3+ is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of Fe3+ was used. Leaving 0.5 mol remaining at equilibrium
b. The ratio of the product to SCN= is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of SCN- was used. Leaving 1.5 mol remaining at equilibrium
c. KC = 0.5/(0.5*1.5) = 0.67
