Respuesta :
Explanation:
The given data is as follows.
Thickness = 0.28 mm = [tex]0.28 \times \frac{1}{10}[/tex] cm = 0.028 cm
Area = 0.40 [tex]m^{2}[/tex] = [tex]0.40 \times 10^{4} cm^{2}[/tex] = 4000 [tex]cm^{2}[/tex]
As, it is known that volume = area × thickness
So, Volume = [tex]4000 cm^{2} \times 0.028 cm[/tex]
= 112 [tex]cm^{3}[/tex]
As density is mass divided by volume. So, mass of chromium will be calculated as follows.
Density = [tex]\frac{mass}{volume}[/tex]
7.20 [tex]g/cm^{3}[/tex] = [tex]\frac{mass}{112 cm^{3}}[/tex]
mass = 806.4 g
As, mass of 1 mole of chromium is 52 g. So, number of moles in 806.4 g of chromium will be as follows.
No. of moles = [tex]\frac{mass}{molar mass}[/tex]
= [tex]\frac{806.4 g}{52 g}[/tex]
= 15.50 mol
In chromate ion, ([tex]CrO^{2-}_{4}[/tex]) charge on Cr is +6. It means that 6 electrons are needed to reduce [tex]Cr^{+6}[/tex] into Cr.
As, 1 mole of [tex]Cr^{+6}[/tex] ions require 6 moles of electrons. Therefore, moles of electrons for 15.50 mol will be calculated as follows.
6 × 15.50 mol = 93.04 mol
To calculate number of electrons we multiply number of moles by Avogadro's number as follows.
[tex]93.04 mol \times 6.02 \times 10^{23}[/tex]
= [tex]560.13 \times 10^{23}[/tex]
= [tex]5.6 \times 10^{25}[/tex] electrons
There is magnitude of [tex]6.241 \times 10^{18}[/tex] times the charge on an electron is equal to 1 coulomb.
Hence, number of coulombs will be as follows.
No. of coulombs = [tex]\frac{5.6 \times 10^{25}}{6.241 \times 10^{18}}[/tex]
= [tex]0.897 \times 10^{7}[/tex] C
or, = [tex]8.97 \times 10^{6}[/tex] C
Thus, we can conclude that [tex]8.97 \times 10^{6}[/tex] C are required to plate a layer of chromium metal with given data.
