The oxidation and reduction equations for the following reaction are Pb(s) + Cu2+(aq) → Pb2+(aq) + Cu(s)

I.
Oxidation reaction

Pb (s) > Pb2 + (aq) + 2e Pb is oxidizing agent

Reduction reaction

Cu2 +(aq) + 2e- > Cu (s) Cu2 + is reducing agent

II.
Oxidation reaction

Pb (s) > Pb2 + (aq) + 2e Pb is oxidized

Reduction reaction

Cu2 + (aq) + 2e- > Cu (s) Cu2 + is reduced

III.
Oxidation reaction

Cu2 + (aq) + 2e- > Cu (s)

Reduction reaction

Pb (s) > Pb2 + (aq) + 2e

IV.

Oxidation reaction

Cu (s) + 2e- > Cu2+ (s)

Reduction reaction

Pb2+ (aq) > Pb (s) + 2e

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

la respuesta es si te gusta hacer lo que siempre haces cuando nunca ves a alguien en el espejo porque sabes que no siempre están ahí.

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).

water at room temperature is flowing through a pipe with pressure taps located 10 ft apart along one section of the pipe, which are connected to a mercury (s=13.6) manometer.