If you had a 0.650 l solution containing 0.0120 m of zn2+(aq), and you wished to add enough 1.34 m naoh(aq) to precipitate all of the metal, what is the minimum amount of the naoh(aq) solution you would need to add? assume that the naoh(aq) solution is the only source of oh−(aq) for the precipitation.

11.6 mL

Explanation:

First we need to calculate the number of moles of Zn2+ present in the solution:

As the charge of ion zinc is 2+ and the charge of hydroxide is 1-, we need double moles of NaOH:

As we have the concentration and the moles, we can calculate the volume:

c. heat could easily travel through the molecules in the air and escape/enter the flask. removing the air slows down the heat transfer.  

the formula for the conversion of temperature from degree celsius to kelvin is:

the sample of water at is converted to kelvin temperature by using the above formula as:

on comparing the two samples of water at or we can say that at and at respectively. the difference between the temperatures of two samples of water is:

hence, one can say that the sample of water at is greater than the sample of water at by .