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Yakvenalex [24]
3 years ago
6

Using the potential given below, calculate the activity of cl^- in 1M KCl solution.

Chemistry
1 answer:
MissTica3 years ago
3 0

Answer:

Explanation:

E° (calomel electrode) = 0.268 v

E( calomel electrode, 1M KCl) = 0.280 v

For calomel electrode

E = E^0 _( Hg^+/Hg ) - \frac{.059}{2} \times log( a_{cl^-} )^2

.280 = .268 - \frac{.059}{2} \times log( a_{cl^-} )^2

.012=  - \frac{.059}{2} \times log( a_{cl^-} )^2

log( a_{cl^-})^2= - .4067

a_{cl^-_}  = 0.626 .

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1.- From the balanced equation, we know that 1 mol of NH₄NO₃ produces 1 mol of N₂O so, we use proportions to find the number of moles produced when 1.40 moles of NH₄NO₃ react.

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ELECTRON NUMBER: Electron Number is equal to the atomic number.

NEUTRON NUMBER: It is Mass Number - Atomic Number.

CHARGE: It is due to the addition or removal of the electron. + charge when electron removed and - charge when electron is added.

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    ELECTRON NUMBER: 38

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