A25.0 ml sample of an acetic acid solution is titrated with a 0.175 m naoh solution. the equivalence point is reached when 26.6 ml of the base is added. the concentration of acetic acid in the sample was m.

.186 M

Explanation:

Remember, the equivalence point is reached when the number of moles of the two reactants is the same.

1. For this problem, obtain the number of moles of NaOH by multiplying the concentration and volume given.

2. Once obtain, the number of moles must be the same for the weak acid.

3. With step two in mind, simply solve for molarity by dividing the value obtained and 25 mL.

Calculation.

1. moles of NaOH = (.175)(.0266) = 4.66e-3

2. moles of Acetic acid = 4.66e-3

3. Concentration of acetic acid: (4.66e-3)/(.025) = .186 M

the answer is either b or c

8 m/s because for him to get the same amount of energy as ray is putting out, bobby, who weighs half as much as ray, needs to run twice as fast.