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.

b) mutualism

hope this : )

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.