A solution containing 0.14M Pb2+, 1.5×10−6M Pb4+, 1.5×10−6M Mn2+, 0.14M MnO−4, and 0.83M HNO3 was prepared. For this solution, the balanced reduction half‑reactions and overall net reaction shown can occur. 5[Pb4++2e−↽−−⇀Pb2+] ∘+=1.690 V
2[MnO−4+8H++5e−↽−−⇀Mn2++4H2O] ∘−=1.507 V
5Pb4++2Mn2++8H2O↽−−⇀5Pb2++2MnO−4+16 H+

A) Determine ∘cell , Δ∘ , and for this reaction. B) Calculate the value for the cell potential, cell , and the free energy, Δ , for the given conditions. C) Calculate the value of cell for this system at equilibrium. D) Determine the pH at which the given concentrations of Pb2+, Pb4+, Mn2+, and MnO−4 would be at equilibrium.

stupid i wont let you get a chance

explanation:

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

it is b that is the answer.