The gaining of electron by an atom results in the formation of anion shown by the negative charge on the atom whereas lose of electron results in the formation of cation shown by positive charge on the atom. The atom lose or gain electron to complete their octet and get stable in nature.
The chlorine atom will gain an electron and form chloride anion with one negative charge on it. The chloride ion is more stable in nature compared to the chlorine atom due to complete octet of chloride ion by gaining of electron.
Electronic configuration of chlorine atom is:
![[Ne]3s^{2}3p^{5}](https://tex.z-dn.net/?f=%5BNe%5D3s%5E%7B2%7D3p%5E%7B5%7D)
By gaining of one electron, electronic configuration of chloride ion is:
![[Ne]3s^{2}3p^{6}](https://tex.z-dn.net/?f=%5BNe%5D3s%5E%7B2%7D3p%5E%7B6%7D)
Thus, the equation that shows the formation of the chloride ion from a neutral chlorine atom is:
Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
The answer is CH3CH2CH2CHO
<span>The number next to the simbol of the element ions (as a superscript) means the number of charges of the ion. For example N (+),, where (+) is a superscript means that the charge of the ion is 1+. S(2-), where (2-) is a superscript, means that the charge of the ion is (2-). OH (-), where (-) is a superscript, means that the charge of OH ion is (1-) . </span>
The answer I got was : 52.2
