Answer:
See explanation for details
Explanation:
ammonium sulfate and barium nitrate Answer
Ba2+(aq) + SO4^2-(aq) ----------> BaSO4(s)
lead(II) nitrate and sodium chloride Answer
Pb2+(aq) + 2Cl-(aq) -----------> PbCl2(s)
copper(II) chloride and sodium hydroxide Answer
Cu2+(aq) + 2OH-(aq) -------------> Cu(OH)2(s)
There can only be a net ionic reaction when a precipitate is formed as the two solutions are mixed together.
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.
Answer:
Density = 11.4 g/cm³
Explanation:
Given data:
Density of lead = ?
Height of lead bar = 0.500 cm
Width of lead bar = 1.55 cm
Length of lead bar = 25.00 cm
Mass of lead bar = 220.9 g
Solution:
Density = mass/ volume
Volume of bar = length × width × height
Volume of bar = 25.00 cm × 1.55 cm × 0.500 cm
Volume of bar = 19.4 cm³
Density of bar:
Density = 220.9 g/ 19.4 cm³
Density = 11.4 g/cm³
