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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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<h3>What is Potential and Kinetic energy?</h3>

Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.

mass of rod, mab = 2.4kg

mass of rod, mbc = 4kg

conservation of energy

T_{1}  + V_{1} = T_{2}  + V_{2}

h_{ab}  = h_{bc}  = 0.18m

potential energy at position 1,

V1 = m_{ab} gh_{ab}  + m_{bc} gh_{bc}

V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18

V1 = 11.30112

kinetic energy T1 at position 1 is zero

potential energy at position 2 is zero

K.E at position 2,

T_{2} = \frac{1}{2} l_{ab} w^{2}_{ab} +  \frac{1}{2} m_{bc} v^{2}_{G} +  \frac{1}{2} lw^{2}_{bc}

l_{ab} =\frac{m_{ab} l^{2}_{ab}  }{3}

= 1/3 *4 * (0.36)²

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l =\frac{m_{bc} l^{2}_{bc}  }{12}

= 1/12 *4 * (0.6)²

=0.12kg m²

on putting the values in above equation we get,

T₂ = 1.0667vb²

0 + 11.30112 = 1.0667vb² + 0

vb = 3.2549 m/s

to learn more about Kinetic and potential energy go to - brainly.com/question/18963960

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