Answer to Question #14180 in Other Chemistry for benjamin nuyles
mL of solution. b) How many moles of KMnO4 are present in 75.0 mL of a 0.0850 M solution? c) How many
millimeters of 11.6 M HCl solution are needed to obtain 0.105 mol of HCl?
(b) this is like (a) but in reverse. start with 75.0 mL and convert to L (1L/1000 mL) then multiply by the molarity (0.0850 mol/L). this way "L" cancels out and you are left with your answer in mol.
(c) start with 0.105 mol. multiply by the molarity (1 L/11.6 mol). mol will cancel out and you will be left with your answer in L. then convert from L to mL using the conversion factor, (1000 mL/ 1L)
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