In the following reaction

2 VO43- + 3 Zn + 16 H+→ 2 V2+ + 3 Zn2+ + 8 H2O

the initial rate of disappearance of VO43- was found to be 0.56 M/s. What is the initial rate of appearance of Zn2+?

-0.56 M/s

0.37 M/s

0.56 M/s

0.84 M/s

1.12 M/s

0.84 M/s

correct it is 0.84

goodluck cuz i had to trust this answer and it was right.

To find the initial rate of appearance of Zn2+, we'll use the coefficients from the balanced chemical equation you provided:

2 VO43- + 3 Zn + 16 H+ → 2 V2+ + 3 Zn2+ + 8 H2O

The stoichiometry tells us that the ratio of the rate of disappearance of VO43- to the rate of appearance of Zn2+ is 2:3. Therefore, we can set up the following proportion:

(0.56 M/s) / (x M/s) = 2 / 3

To solve for x, the initial rate of appearance of Zn2+, we can cross multiply and solve for x:

(0.56 M/s) * (3 / 2) = x

x = 0.56 M/s * 3 / 2

x = 0.84 M/s

Therefore, the initial rate of appearance of Zn2+ is 0.84 M/s.