The standard reduction potential for the reduction of ruo4−(aq) to ruo42−(aq) is +0.59 v.
you may want to reference(pages 860 - 867) section 20.4 while completing this problem.

use appendix e in the textbook. which of the following substances can oxidize ruo42−(aq) to ruo4−(aq) under standard conditions?

a) clo3−(aq)
b) cr2o72−(aq)
c) ni2+(aq)
d) pb2+(aq)
e) i2(s)

a) ClO₃⁻(aq)

b) Cr₂O₇²⁻(aq)

Explanation:

The ability of a species to act as an oxidizing agent depends on the standard reduction potential (E°red). The higher the E°red, the more able it is to be an oxidizing agent.

We want to oxidize RuO₄²⁻(aq) to RuO₄⁻(aq). The inverse reduction has a E°red = +0.59 V. Then, we need an species with a higher standard reduction potential.

Let's consider the following standard reduction potentials.

a) ClO₃⁻(aq)  / ClO₂(g)    E°red = 1.18 V

b) Cr₂O₇²⁻(aq)/Cr³⁺(aq)   E°red = 1.33 V

c) Ni²⁺(aq) /Ni(s)              E°red = -0.25 V

d) Pb²⁺(aq) /Pb(s)            E°red = -0.13 V

e) I₂(s)/I⁻(aq)                   E°red = 0.54 V

ClO₃⁻(aq) and Cr₂O₇²⁻(aq) can oxidize RuO₄²⁻(aq) under standard conditions.

space exploration has led to the development and technological improvement of many everyday items. (if this answer is wrong i will find the right answer) : )

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o