Im no sure what it is i just love helping little boys out. at this age it gives me a thrill.
The reactant that will be the best reactant for a nucleophilic aromatic substitution is NO₂- NO₂. The correct option is b.
<h3>What is nucleophilic
aromatic substitution?</h3>
Nucleophilic aromatic substitution is a substitution process of nucleophile substance is substituted by halides in an aromatic ring. Aromatic compounds contain this type of substitution.
In option b, the compound is the one nitroxide group substituted by halogen, that is fluorine. The fluorine group is substituted in these given aromatic compounds.
Thus, the correct option is b, NO₂- NO₂.
To learn more about nucleophilic aromatic substitution, refer to the link:
brainly.com/question/28265482
#SPJ4
The question is incomplete. Your most probably complete question is given below:
NO₂F
NO₂- NO₂-F
CH₃-O
CH₃-O-F
Question #1
Potasium hydroxide (known)
volume used is 25 ml
Molarity (concentration) = 0.150 M
Moles of KOH used
0.150 × 25/1000 = 0.00375 moles
Sulfuric acid (H2SO4)
volume used = 15.0 ml
unknown concentration
The equation for the reaction is
2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l)
Thus, the Mole ratio of KOH to H2SO4 is 2:1
Therefore, moles of H2SO4 used will be;
0.00375 × 1/2 = 0.001875 moles
Acid (sulfuric acid) concentration
0.001875 moles × 1000/15
= 0.125 M
Question #2
Hydrogen bromide (acid)
Volume used = 30 ml
Concentration is 0.250 M
Moles of HBr used;
0.25 × 30/1000
= 0.0075 moles
Sodium Hydroxide (base)
Volume used 20 ml
Concentration (unknown)
The equation for the reaction is
NaOH + HBr = NaBr + H2O
The mole ratio of NaOH : HBr is 1 : 1
Therefore, moles of NaOH used;
= 0.0075 moles
NaOH concentration will be
= 0.0075 moles × 1000/20
= 0.375 M
The model most likely to predict the future event is the model of the weather systems. The model of the weather system provides much crucial information such as the temperature!
You need to calculate the molar mass for Al(CN)3 using the atomic weights for Al, C, and N given on the periodic table.
1 Al (26.98) + 3 C (3 x 12.01) + 3 N (3 x 14.01) = 105.04 g Al(CN)3 / mole
183 g Al(CN)3 x (1 mole Al(CN)3 / 105.04 g Al(CN)3) = 1.74 moles Al(CN)3
