1) a student dissolved 0.1916 g of an unknown diprotic acid in 100 ml of distilled water. the acid was then titrated with 0.1025m naoh solution. the second equivalence point showed the sharpest change in ph, and so it was used to determine the molar mass of the unknown acid. the volume of naoh needed to reach the equivalence point was 27.5 ml.

a. calculate the number of moles of naoh used in the titration to reach the second equivalence point.

b. calculate the number of moles of diprotic acid, based on the fact that we are examining the second equivalence point.

c. calculate the molar mass of the diprotic acid.

a.

b.

c.

Explanation:

a. To calculate the number of moles of NaOH we need the following definition of molarity:

b. As we have a diprotic acid, it means that when the second equivalence point is reached, two moles of NaOH have reacted by one mol of acid, therefore the relation we use to determine the moles of acid is:

c. Finally, we determine the molar mass as follows:

idk what is it? is this a question that you know the answer or u dont

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally