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jeka57 [31]
3 years ago
8

Which statement correctly describes the reactivity of noble gases, according to the octet rule? Their atoms have eight electrons

in their valence shells, so noble gases are very reactive. Their atoms have one electron in their valence shells, so noble gases are very reactive. Their atoms have one electron in their valence shells, so noble gases are very unreactive. Their atoms have eight electrons in their valence shells, so noble gases are very unreactive.
Physics
1 answer:
Kipish [7]3 years ago
5 0

Answer:

The correct option is : Their atoms have eight electrons in their valence shells, so noble gases are very unreactive.

Explanation:

The octet rule state that atoms tend to complete their last energy levels with eight electrons, and that this configuration make them very stable and unreactive.

Noble gases are characterized as unreactive atoms, and this is associated with the fact that they have a complete valence shell, it means that they have eight electrons on it (they follow the octet rule).

Atoms with less electrons on their valence shells tend to react with another atom, forming bonds, to complete their valence shells (with eight electrons).

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madreJ [45]

Answer:

Part a)

E = 0

Part b)

E = 6.77 \times 10^7 N/C

Part c)

Electric field inside the conductor is again zero

E = 0

Part d)

E = 8.52 \times 10^6 N/C

Explanation:

Part a)

conducting sphere is of radius

R = 2 cm

so electric field inside any conductor is always zero

So electric field at r = 1 cm

E = 0

Part b)

Now at r = 3 cm

By Gauss law

E = \frac{kq}{r^2}

E = \frac{(9\times 10^9)(6.77 \muC)}{0.03^2}

E = 6.77 \times 10^7 N/C

Part c)

Again when we use r = 4.50 cm

then we will have

Electric field inside the conductor is again zero

E = 0

Part d)

Now at r = 7 cm

again by Gauss law

E = \frac{kQ}{r^2}

E = \frac{(9\times 10^9)(6.77\mu C - 2.13\mu C)}{0.07^2}

E = 8.52 \times 10^6 N/C

5 0
4 years ago
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Answer:

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Explanation:

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Answer:

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Explanation:

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Answer:

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Answer:

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