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Text Elements

moles of acid

moles of base

molarity of acid

molarity of base

volume of acid

volume of base

(54)(.10)(1) = (125)(1)(x) (54)(.10)/125 = .0432

(25)(.050)(1) = (345)(1)(x) (25)(.050)/345 = .00362

(50)(.5)(1) = (125)(2)(x) (50)(.5)/125/2 = .1

(.360)(2)(x) = (25)(.1)(2) (25)(.1)(2)/.360/2 = 6.944

(30)(1)(x) = (.1)(48)(1) (.1)(48)/30/1 = .16

(25)(1)(x) = (15)(2)(1) (15)(2)/25 = 1.2

(10)(2)(x) = (135)(1)(2) (135)(2)/10/2 = 13.5 = 14

(1.5)(1) = (20)(2.5)(3) (20)(2.5)(3)/1.5 = 100

(25)(.050)(2) = (.50)(1)(x) (25)(.050)(2)/.50 = 5

(7.5)(.020)(1) = (15)(1)(x) 7.5*.02*1/15=.04

(.010)(15)(1) = .15 (7.5)(.020)(4)=.06

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