Which element requires the least amount of
energy to remove the most loosely held electron
from a gaseous atom in the ground state?
<h3>Answer-</h3><h3>Na</h3>
Answer:
113.8g
Explanation:
Statement of problem: mass of 1.946mole of NaCl
Given parameters:
Number of moles of NaCl = 1.946mole
Unknown: mass of NaCl
Solution
To find the mass of NaCl, we apply the concept of moles which expresses the relationship between number of moles and mass according to the equation below:
Number of moles = 
To find the molar mass of NaCl:
the atomic mass of Na = 23g
atomic mass of Cl = 35.5g
Molar mass of NaCl = (23 + 35.5) = 58.5gmol⁻¹
Mass of NaCl = Number of moles x molar mass of NaCl
Mass of NaCl = 1.946 x 58.5 = 113.8g
An ionization suppressor is an alkali metal capable of preventing ionization, which can be used in atomic spectroscopy to determine matter composition.
<h3>What is ionization?</h3>
Ionization refers to the phenomena capable of converting neutral atoms/molecules to electrically charged atoms/ions.
Ionization is a process by which radiation (e.g., alpha, beta, gamma rays) can pass energy to inert matter.
Some examples of ionization suppressors include salts of alkali metals (for example, potassium), which can be used in atomic spectroscopy to determine matter composition.
Learn more about ionization here:
brainly.com/question/1445179
Cu(NO3)2>NO2+CuO+O2 balanced: 2Cu(NO3)2=4NO2+2CuO+O2