-OH is elctron donating -C=-N is electron withdrawing -O-CO-CH3 is electron withdrawing -N(CH3)2 is electron donating -C(CH3)3 is electron donating -CO-O-CH3 is electron withdrawing -CH(CH3)2 is electron donating -NO2 is electrong withdrawing -CH2
<u>Answer:</u> The chemical equations are given below.
<u>Explanation:</u>
The chemical equation for the reaction of lead nitrate and sodium hydroxide follows:

By Stoichiometry of the reaction:
1 mole of aqueous solution of lead nitrate reacts with 2 moles of aqueous solution of sodium hydroxide to produce 1 mole of solid lead hydroxide and 2 moles of aqueous solution of sodium nitrate.
The chemical equation for the reaction of lead hydroxide and hydroxide ions follows:
![Pb(OH)_2(s)+2OH^-(aq.)\rightarrow [Pb(OH)_4]^{2-}(aq.)](https://tex.z-dn.net/?f=Pb%28OH%29_2%28s%29%2B2OH%5E-%28aq.%29%5Crightarrow%20%5BPb%28OH%29_4%5D%5E%7B2-%7D%28aq.%29)
By Stoichiometry of the reaction:
1 mole of lead hydroxide reacts with 2 moles of aqueous solution of hydroxide ions to produce 1 mole of aqueous solution of tetra hydroxy lead (II) complex
Hence, the chemical equations are given above.
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In the above equation, the product formed is :

_____________________________

The empirical formula : C₆H₂Cl₂O
The molecular formula : C₁₂H₄Cl₄O₂
<h3>Further explanation</h3>
Given
The percentage composition
Required
The empirical formula and the molecular formula
Solution
The mol ratio of the components :
C : H : Cl : O
=44.76/12 : 1.25/1 : 44.05/35.5 : 9.94/16
=3.73 : 1.25 : 1.241 : 0.621 divide by 1.241
= 3 : 1 : : 1 : 0.5 x 2
= 6 : 2 : 2 : 1
The empirical formula : C₆H₂Cl₂O
(Empirical formula)n=molecular formula
(C₆H₂Cl₂O)n=321.97
(160.986)n=321.97
n=2
(C₆H₂Cl₂O)₂=C₁₂H₄Cl₄O₂
The molecular formula : C₁₂H₄Cl₄O₂