Condensation is the process of watervapor in the air is changing into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds. Condensation is also the opposite of evaporation

8 0

3 years ago

Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el

Likurg_2 [28]

Answer: 281 hours

Explanation:-

1 electron carry charge= $1.6\times 10^{-19}C$

1 mole of electrons contain= $6.023\times 10^{23}$ electrons

Thus 1 mole of electrons carry charge= $\frac{1.6\times 10^{-19}}{1}\times 6.023\times 10^{23}=96500C$

$Cu^{2+}+2e^-\rightarrow Cu$

$96500\times 2=193000Coloumb$ of electricity deposits 1 mole or 63.5 g of copper

0.0635 kg of copper is deposited by 193000 Coloumb

11.5 kg of copper is deposited by= $\frac{193000}{0.0635}\times 11.5=34952756$ Coloumb

$Q=I\times t$

where Q= quantity of electricity in coloumbs = 34952756 C

I = current in amperes = 34.5 A

t= time in seconds = ?

$34952756 C=34.5A\times t$

$t=1013123sec=281hours$

Thus it will take 281 hours to plate 11.5 kg of copper onto the cathode if the current passed through the cell is held constant at 34.5 A.

8 0

3 years ago

Fill in the blank! (for 15 pts)

Artist 52 [7]

In order:

alkali

reactive

shiny

electricity

soft

low

good luck!

4 0

2 years ago

A chemist adds 185.0mL of a 2.6 M iron(II) bromide FeBr2 solution to a reaction flask. Calculate the millimoles of iron(II) brom

RUDIKE [14]

Answer:

0.48 moles

Explanation:

The bromide has a molarity of 2.6M.

This simply means that in 1dm^3 or 1000cm^3 of the solution, there are 2.6 moles.

Now, we need to get the number of moles in 185ml of the bromide. It is important to note that the measurement ml is the same as cm^3.

We calculate the number of moles as follows.

If 2.6mol is present in 1000ml

x mol will be present in 185 ml.

To calculate x = (185 * 2.6) ÷ 1000

= 0.481 moles = 0.48 moles to 2 s.f

7 0

3 years ago

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