Explanation:
Reversible reactions that happen in a closed system eventually reach equilibrium. At equilibrium, the concentrations of reactants and products do not change. But the forward and reverse reactions have not stopped - they are still going on, and at the same rate as each other.
Answer:
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Explanation:
Metallic bonds involve attraction between electrons and positively charged metal ions. The metals are ionized and electrons form a sea of valence electrons. These loosely bound electrons surround the nuclei of the metals.
The presence of this sea of electrons explains the fact that metals conduct electricity and heat due to the free valence electrons.
Due to the nature of the bonding between metal atoms,metals are malleable and ductile.
Due to the strong electrostatic interaction between metal ions and electrons, the metallic bond is very strong and is very difficult to break thereby accounting for the greater strength of metals as the size of the metallic ion decreases.
Answer : Electron P has greater energy difference than the Electron N.
Explanation :
Wavelength range of violet light = 400 - 500 nm
Wavelength range of orange light = 600 - 700 nm
The Planck's equation is,

where,
E = energy of light
c = speed of light
= wavelength of light
According to the Planck's equation, wavelength and energy follow inverse relation. As the wavelength increases, energy decreases.
From the given spectrum, the wavelength of violet light is less. We conclude that When electron P gives violet light on transition it means that energy difference between the energy level was high.
From the given spectrum, the wavelength of orange light is more. We conclude that When electron N gives orange light on transition it means that energy difference between the energy level was low.
So, Electron P which gives violet light on transition has greater energy difference than the Electron N.
Answer:

Explanation:
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In this case, given the masses of copper, zinc and lead, it is possible to compute the moles via their atomic masses first:

Now, we compute the atomic percentages as shown below:

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