The ore contains 55.4% calcium phosphate (related to the mineral apatite) so the amount of Ca3(PO4)2 is 55.4%x=1000g so x=1000/0.554= 1.805kg. Now for the % of P in this amount of calcium phosphate, use all the masses of the elements in Ca3PO4= Ca=40.078 x 3= 120.23 and (PO4)2= (30.974+64)2=189.95 (NB oxygen is 16 mass x 4 =64) so the total mass is 310.2 and we have 61.95 of P (Pmass x 2) so 61.95/3102.= 0.19 or 19% P. So of the 1.805 x 0.19= 0.34kg of phosphorus.
2 × (atomic mass of Ag) + (atomic mass of Cl (
Because they can look at the fossils that they left behind and see what kind of animal it was and what happened to it if another animal is inside of the fossil or if the bones are misplaced
Talking about ions, a cation has a positive charge. That means a positive charge is gained by an atom that gives out electrons.
Answer:
Empirical Formula = NH₄NO₃ (Ammonium Nitrate)
Solution:
Step 1: Calculate Moles of each Element;
Moles of N = %N ÷ At.Mass of N
Moles of N = 35.0 ÷ 14
Moles of N = 2.5 mol
Moles of O = %O ÷ At.Mass of O
Moles of O = 59.96 ÷ 16
Moles of O = 3.7475 mol
Moles of H = [100% - (%N + %O)] ÷ At.Mass of H
Moles of H = [100% - (35.0 + 59.96)] ÷ 1.008
Moles of H = [100% - 94.96] ÷ 1.008
Moles of H = 5.04 ÷ 1.008
Moles of H = 5 mol
Step 2: Find out mole ratio and simplify it;
N H O
2.5 5 3.7475
2.5/2.5 5/2.5 3.7475/2.5
1 2 1.5
Multiply Mole Ratio by 2,
2 4 3
Result:
Empirical Formula = N₂H₄O₃
Or,
Empirical Formula = NH₄NO₃
This empirical formula is also a Molecular Formula for Ammonium Nitrate a well known Fertilizer and often misused in the formation of Explosives.
