Answer:
the correct option is option c, I and IV
Explanation:
Atomic emission spectroscopy is used to determine quantity of element in a sample.
Atomic emission spectroscopy based on occurring of atomic emission when a valence electron from higher energy orbital comes back to lower energy orbital.
Light intensity emitted by a flame, plasma, arc of particular wavelength are used to excite the valence electron.
The intensity of atomic emission lines are proportional to number of atoms present in the excited state.
Emission intensity is affected by temperature of the excitation source and the efficiency of atomization.
Increase in temperature does not affect the ground state population.
Therefore, statements I and IV are correct.
So, the correct option is c.
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It can be found that 337.5 g of AgCl formed from 100 g of silver nitrate and 258.4 g of AgCl from 100 g of CaCl₂.
<u>Explanation:</u>
2AgNO₃ + CaCl₂ → 2 AgCl + Ca(NO₃)₂
We have to find the amount of AgCl formed from 100 g of Silver nitrate by writing the expression.
= 337.5 g AgCl
In the same way, we can find the amount of silver chloride produced from 100 g of Calcium chloride.
It can be found as 258.4 g of AgCl produced from 100 g of Calcium chloride.
Electron structure of sodium:
₁₁Na: 1s²2s²2p⁶3s¹
