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Aleks [24]
3 years ago
12

Describe the octet rule in terms of noble-gas configurations and potential energy

Physics
2 answers:
Wittaler [7]3 years ago
8 0
Elements will gain or lose electrons to form a noble gas configuration. Atoms that meet the octet rule are stable because their valence electrons have a relatively low potential energy. Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.


SashulF [63]3 years ago
5 0

Answer:

Octet rule is the ability of an element to have 8 electrons in its outer shell. This makes the element very stable.

Noble gases are also called stable gases due to their configuration. Noble gases configuration is the ability of an electron to lose or gain electrons so that they can obtain the electronic configuration that noble gases have.

Explanation:

Octet rule in terms of noble-gas configurations and potential energy can be described as the situation which is referred to rule of thumb whereby electrons combine together so that they have eight electrons in their outermost or valence shells which similar to the electronic configuration of noble gases.

Noble gases when they follow the Octet rule have a very low potential energy.

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A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.
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The final equilibrium temperature of the system is T = 12.48^oC

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In the following question we are provided with

Mass of the ice M_{i} = 129 g = 0.129 kg

Mass of the steam M_s = 19 g = 0.019 kg

Initial temperature is  T_i = 0°C

Temperature of  steam  T_s = 100°C

Following the change of state of water in the question

 The energy required by ice to change to water is mathematically given as

          Q_A = M_iL_f

Where L_f is a constant known as heat of fusion  and the value is 334*10^3 J/kg

           Q_A = 0.129 *334 *10^3  = 43086 J

The energy been released when the steam changes to water is mathematically given as

            Q_B = M_s * L_v

           Where L_v is a constant known as heat of vaporization and the value is 2256*10^3J/kg

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         The energy released when the temperature of water decrease from 100°C to 0°C is

                 Q_C = M_s *C_water (100°C)

Where C_{water} is the specific heat of water which has a value 4186J/kg \cdot K

                  Q_C = 0.019 *4186*100 = 7953.4

Looking at the values we obtained we noticed that ]

             Q_B + Q_C > Q_A

What this means is that the ice will melt

bearing in mind the conservation of energy

     looking the way at which water at different temperature were mixed according to the question

     Heat lossed by the vapor   = heat gained by ice

        Q_B + M_s *C_{water}(100-T) = Q_A + M_i C_{water} T

                                               T = \frac{Q_B+M_s *C_{water}(100^oC)-Q_A}{(M_s *C_{water})+(M_i*C_{water})}

                                               T = \frac{42864+7953.4-43086}{(0.019+0.129)(4186)}

                                              T = 12.48^oC

       

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