Triple equality ( ) is used to
equivalency or showing two substances equivalent. It used for molar
mass calculations.
Ca3(PO4)2 : The metal cation is Ca2+ and anion formula is PO43-
We Used value of the anion charge as the subscript for the cation and use the value of the cation charge as the subscript for the anion : to get compound formula is Ca3(PO4)2.
so, can write Triple equality ( ) for
Ca3(PO4)2 :
3 Ca2+ 2
PO43-
or alternatively 1 mol of
Ca3(PO4)2
3 moles of Ca2+
1 mol of
Ca3(PO4)2
2 moles of PO43-
what is the ksp expression for a solution of Ca3(PO4)2?
Excess Ca3(PO4)2 is added to a beaker of water. At equilibrium, the PO4 concentration is 1.8 X 10^-6 M. What is the Ksp for Ca3(PO4)2? Please show all work
Calculate the solubility of Ca3(PO4)2 in a solution containing 0.0539 M Na3PO4. Ksp=2.0*10^-29 for Ca3(PO4)2
In a purely competitive market, there is a triple equality. What is it? The triple equality can be broken up into two equations that each signify a difference efficiency. What are these efficiencies? Why are they important?
Which of the following compounds will be soluble in water: NaNO3, Ca3(PO4)2, CuCl2, and Fe2S3? Ca3(POa)2 and Fe2S3 CuCl2 and Fe2S3 NANO3, Ca3(PO4a)2, and Fe2S3 NaNO3 and CuCl2 and Ca3(PO4)2 ONaNO3
What mass of P4 is possible from the reaction of 45.0g of Ca3(PO4)2, 70.0 of SiO2 and 18.0g of C? The balanced reaction is: 2 Ca3(PO4)2 (s) + 6 SiO2 (s) + 10 C(s) ---> P4 (s) + 6 CaSiO3 (l)+ 10 CO(g)
What is the maximum mass of P4 that can be obtained from 56.2 g Ca3(PO4)2, 37.3 g SiO2 and excess carbon? The chemical equation for the reaction is given below. 2 Ca3(PO4)2(s) + 6 SiO2 + 10 C(s) ⟶ P4(g) + 6 CaSiO3(l) + 10 CO(g) Give your answer in grams with three significant figures. Molar masses, in g mol-1: Ca3(PO4)2, 310.18 SiO2, 60.09 P4, 123.88
the crc handbook gives the solubility of ca3(po4)2 at 25 c as .0012g/100ml . determine the molar solubility of ca3(po4)2 . this is the molarity of saturated solution . and determine ksp of ca3(po4)2
---Percent yield of Ca3(PO4)2 from CaCL2 Mass of beaker + CaCL2 = 342.231g, Mass of beaker = 341.852g, Mass of filter paper + Ca3(PO4)2 ppt = 0.726, Mass of filter paper = 0.506 1. Show the calculation for theoretical yield of Ca3(PO4)2 2. Show the calculation for percent yield of Ca3(PO4)2
Question 24 1 pts Write the solubility product constant expression for calcium phosphate, Ca3(PO4)2 O [Ca2+][PO43-M[Ca3(PO4)2] O Ca2+1*[PO43-12 O [Ca2+][PO43-1 O [Ca2+12[PO43-13