If a solution of 20mL of 0.050M K+ is added to 80mL of 0.50M ClO4- will a precipitate form and what is the value of Qsp? For KClO4, Ksp = 1.07 x 10-2

Q< K hence a precipitate will not form.

Explanation:

First convert the concentration to molL-1

For number of moles of K^+ = 0.05×20/1000 = 1×10^-3 moles

If we have 1×10^-3 moles in 20cm3

Then in 1000cm^3 we have 1×10^-3 moles×1000/20= 0.05 M

For ClO4^-= 0.50×80/1000= 0.04 moles

If we have 0.04 moles in 80cm3

Then in 1000cm^3 we have 0.04 moles×1000/80= 0.5 M

Q= [K^+] [ClO4^-]

Q= [0.05] [0.5]

Q= 0.025= 2.5×10-2

Q< K hence a precipitate will not form.

1. 1 mole of anything contains 6.022x10^23 pieces  

d  

2. a  

3. c, number  

5. 1.24x10^24atoms / 6.022x10^23atoms/mole = 2moles h2 atoms  

6. fe2br3 mass = 2moles x mass fe + 3 moles x mass br  

molar mass = grams / number of moles = g/mole  

to find moles = g / molar mass  

to find mass = g = molar mass / moles