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ella [17]
3 years ago
15

Calculate how many grams of each product are formed when 4.05g of h2o is used. 2h2o=2h2+o2

Chemistry
1 answer:
NISA [10]3 years ago
3 0

Answer:

Explanation:

Given data:

Mass of water used = 4.05 g

Mass of each product produced = ?

Solution:

Chemical equation:

2H₂O   →  2H₂ + O₂

Number of moles of water:

Number of moles = mass/ molar mass

Number of moles = 4.05 g/ 18 g/mol

Number of moles = 0.225 mol

Now we will compare the moles of water with hydrogen and oxygen.

                             H₂O            :              H₂

                               2               :               2

                           0.225           :           0.225

                           

                           H₂O             :              O₂

                             2                :               1

                           0.225           :            1/2×0.225 = 0.113 mol

Mass of hydrogen:

Mass = number of moles × molar mass

Mass = 0.225 × 2 g/mol

Mass = 0.45 g

Mass of oxygen:

Mass = number of moles ×  molar mass

Mass = 0.113 mol × 32 g/mol

Mass = 3.616 g

                     

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Both salt products are water soluble.

2) With CuSO₄;

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Again both Salt products are water soluble.

3) With K₂SO₄;

                  Cu(NO₃)₂  +  K₂SO₄   →  CuSO₄  +  2 KNO₃
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4) With K₂S;

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Result:
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The volume density of atoms for a bcc lattice is 5 x 1026 m-3. Assume that the atoms are hard spheres with each atom touching it
ollegr [7]

Explanation:

It is known that for a body centered cubic unit cell there are 2 atoms per unit cell.

This means that volume occupied by 2 atoms is equal to volume of the unit cell.

So, according to the volume density

        5 \times 10^{26} atoms = 1 [tex]m^{3}

        2 atoms = \frac{1 m^{3}}{5 \times 10^{26} atoms} \times 2 atoms

                     = 4 \times 10^{-27} m^{3}

Formula for volume of a cube is a^{3}. Therefore,

           Volume of the cube = 4 \times 10^{-27} m^{3}

As lattice constant (a) = (4 \times 10^{-27} m^{3})^{\frac{1}{3}}

                                   = 1.59 \times 10^{-9} m

Therefore, the value of lattice constant is 1.59 \times 10^{-9} m.

And, for bcc unit cell the value of radius is as follows.

                 r = \frac{\sqrt{3}}{4}a

Hence, effective radius of the atom is calculated as follows.

                 r = \frac{\sqrt{3}}{4}a

                   = \frac{\sqrt{3}}{4} \times 1.59 \times 10^{-9} m

                   = 6.9 \times 10^{-10} m

Hence, the value of effective radius of the atom is 6.9 \times 10^{-10} m.

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