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Alisiya [41]
2 years ago
11

How many atoms are centered on the (100) plane for the fcc crystal structure?

Chemistry
1 answer:
Setler [38]2 years ago
5 0

Here, we have to get the number of atoms present in the 100 plane of the FCC crystal lattice.

There will be 2 atoms in 100 plane of FCC crystal lattice.

In the face centered crystal (FCC) lattice there are atoms at each corner of the cube and each are shared by 4 another atoms. And an atom is present at the face of the crystal.

For the 100 plane of the Miller indices the intercepts are a, ∞, ∞ or 2a, ∞, ∞.

Thus, for the 4 atoms of the corner at the cube shared by 4 other atoms will contribute, 4 × \frac{1}{4} = 1 and the un-shared atoms at the face will contribute another 1, which make the total atom 1 + 1 = 2.

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A hot air balloon is filled to 1250 m3 at 27 C. At what temperature will the balloon be filled to 1600 m3 if the pressure remain
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384.2 K

Explanation:

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Fynjy0 [20]

Answer:

Rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

Explanation:

According to equation   2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

                     ⇒ -\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}  -----------------------------(1)

    Given that the rate of disappearance of oxygen = -\frac{d[O_{2} ]}{dt} = 3.64 x 10⁻³ M/s

             So the rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = ?

from equation (1) we can write

                                   \frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]

                                ⇒ \frac{d[SO_{3}] }{dt} = 2 x 3.64 x 10⁻³ M/s

                                ⇒ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

7 0
3 years ago
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