What is an electrolyte?

Both white phosphorus and red phosphorus are made up only of phosphorus atoms.Does this mean they are both pure substances? Coul

choli [55]

True, both of the red and white phosphorus is a pure substance.

Both red and white phosphorus is made of phosphor element but they have a different structure which causes them to have a different color. This phenomenon called allotropes when a chemical element has two or more different form.They are not compounds as they are only made of phosphorus. A compound should be made by at least two different elements.

4 0

3 years ago

What is the oh of a solution with a ph of 9.8

Veseljchak [2.6K]

It would be 4.2, hope this helps.

4 0

3 years ago

If you stay at 90 degrees Celsius and a gas in a liquid solution has a pressure of 2 atm, what solubility should it have if it h

grigory [225]

Answer:

im not sure

Explanation:

5 0

3 years ago

A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are

VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell = 0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

a)

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

b)

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

c)

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

8 0

3 years ago

In an electroplating process, copper (ionic charge +2e, atomic weight 63.6 g/mol) is deposited using a current of 10.0 A. What m

salantis [7]

Answer : The mass of copper deposit is, 1.98 grams

Explanation :

First we have to calculate the charge.

Formula used : $Q=I\times t$