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mylen [45]
3 years ago
6

The hypothetical elements shown here (figures a–d) do not include hydrogen or helium. Which element would you expect to bond cov

alently with an oxygen atom to form a two-atom molecule?
Chemistry
1 answer:
Marizza181 [45]3 years ago
4 0
The answer is "Figure a" i did it on plato lol
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How many coulombs are required to plate a layer of chromium metal 0.28 mm thick on an auto bumper with a total area of 0.40 m2 f
Rufina [12.5K]

Explanation:

The given data is as follows.

Thickness = 0.28 mm = 0.28 \times \frac{1}{10} cm = 0.028 cm

Area = 0.40 m^{2} = 0.40 \times 10^{4} cm^{2} = 4000 cm^{2}

As, it is known that volume = area × thickness

So,             Volume = 4000 cm^{2} \times 0.028 cm      

                                = 112 cm^{3}

As density is mass divided by volume. So, mass of chromium will be calculated as follows.

                   Density = \frac{mass}{volume}  

            7.20 g/cm^{3} = \frac{mass}{112 cm^{3}}      

                      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 = \frac{mass}{molar mass}                            

                                     = \frac{806.4 g}{52 g}

                                     = 15.50 mol

In chromate ion, (CrO^{2-}_{4}) charge on Cr is +6. It means that 6 electrons are needed to reduce Cr^{+6} into Cr.

As, 1 mole of Cr^{+6} 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.

               93.04 mol \times 6.02 \times 10^{23}

                      = 560.13 \times 10^{23}

                     = 5.6 \times 10^{25} electrons

There is magnitude of 6.241 \times 10^{18} times the charge on an electron is equal to 1 coulomb.

Hence, number of coulombs will be as follows.

              No. of coulombs = \frac{5.6 \times 10^{25}}{6.241 \times 10^{18}}

                                           = 0.897 \times 10^{7} C

or,                                        = 8.97 \times 10^{6} C

Thus, we can conclude that 8.97 \times 10^{6} C are required to plate a layer of chromium metal with given data.

4 0
2 years ago
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