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3 years ago

What happens to the valence electrons of a metal and a nonmetal when they form an ionic bond

Physics

1 answer:

shusha [124]3 years ago

7 0

Answer:

Explanation:

Ionic bonds are also known as electrovalent bonds.

For ionic bonds to form, there is transfer of valence electrons from the metal to the non-metal. Metals are electropositive and would easily allow electrons to leave.

The goal of bonding is to achieve the octet configuration. An atom can lose or gain electrons to achieve this.

For example, to form NaCl, Na would loose its 1 valence electron. Cl would gain the Valence electron and complete it's octet with it.

This transfer of electron creates the ionic bond.

The metal which is less electronegative becomes positively charged. The non metal being more electronegative would become negatively charged.

Ionic compounds are polar compounds.

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Consider an isolated system, by which we mean a system free of external forces. If the sum of the particle energies in the syste

Airida [17]

Answer:

The system's potential energy is -147 J.

Explanation:

Given that,

Energy = 147 J

We know that,

System is isolated and it is free from external forces.

So, the work done by the external forces on the system should be equal to zero.

$W=0$

We need to calculate the system's potential energy

Using thermodynamics first equation

$\Delta U=W-\Delta E$

Put the value into the formula

$\Delta U=0-147$

$\Delta U=-147\ J$

Hence, The system's potential energy is -147 J.

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3 years ago

The coefficient of linear expansion of copper is 17 x 10^-6 K-1. A block of copper 30 cm wide, 45 cm long, and 10 cm thick is he

schepotkina [342]

Answer:

The change in volume is $6.885\times 10^{- 5}\$

Solution:

As per the question:

Coefficient of linear expansion of Copper, $\alpha = 17\times 10^{- 6}\ K^{- 1}$

Initial Temperature, T = $0^{\circ}$ = 273 K

Final Temperature, T' = $100^{\circ}$ = 273 + 100 = 373 K

Now,

Initial Volume of the block, V = $30\times 45\times 10\times 10^{- 6}\ m^{3} = 0.0135\ m^{3}$

$V' = V(1 + \gamma \Delta T)$

$\gamma = 3\alpha$

$V' = V(1 + 3\alpha \Delta T)$

where

V' = Final volume

$V' - V= 0.0135\times 17\times 10^{- 6} \times (T' - T))$

$\Delta V= 0.0135\times 3\times 17\times 10^{- 6} \times (373 - 273)) = 6.885\times 10^{- 5}\$

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