What is the [OH^-] if the pH is 4.900

Achemist wants to extract copper metal from copper chloride solution. the chemist places 0.50 grams of aluminum foil in a solution containing 0.75 grams of copper (ii) chloride. a single replacement reaction takes place. (ii) chloride. a single replacement reaction takes place. which statement explains the maximum amount of copper that the chemist can extract using this reaction? a) approximately 0.36 grams, because copper (ii) chloride acts as a limiting reactant b) approximately 1.8 grams, because copper (ii) chloride acts as a limiting reactant c) approximately 0.36 grams, because aluminum acts as a limiting reactant d) approximately 1.8 grams, because aluminum acts as a limiting reactant

Straightforward questions answered in the powerpoint slidesreaction: heating the starting materials under refluxwhat does it mean to heat under reflux? why do we choose water as the reflux solvent? what are boiling chips used for? why do we put a condenser on top of the reaction? why do we add heat and let the reaction stir for 30 minutes? why do we add sulfuric acid to the reaction after it cools as opposed to when it’s still hot? separation: filtration of precipitatewhy don’t we do an aqueous and organic extraction in the separatory funnel? why do you rinse the salicylic acid on the filter with ice cold water? purification: recrystallization of salicylic acid (no hot filtration needed)what is the difference in the amount of room temperature water vs. boiling water needed to dissolve the salicylic acid (assume a 1.2 gram yield of salicylic acid)? remember, in the lab if you need x ml of boiling water to dissolve a solid, then you should add a little more (definitely no more than 1.5 times the theoretical amount) to ensure it doesn’t recrystallize prematurely.analysis: melting point of salicylic acidwhat can you conclude if the melting point of the salicylic acid you just synthesized is 152-155oc and the 1: 1 mix of your product and “synthetic” salicylic acid is 151-154oc?

Find the mass, in grams, of 5.00*10^23 molecules of f2