Answer:
1) HNO3/H2SO4, 2) CH3CH2CH2Cl/AlCl3
Explanation:
Benzene is a stable aromatic compound hence it undergoes substitution rather than addition reaction.
When benzene undergoes substitution reaction, the substituent introduced into the ring determines the position of the incoming electrophile.
If I want to synthesize m-nitropropylbenzene, I will first carry out the nitration of benzene using HNO3/H2SO4 since the -nitro group is a meta director. This is now followed by Friedel Craft's alkykation using CH3CH2CH2Cl/AlCl3.
 
        
             
        
        
        
Answer:
a. Minimum 1.70 V
b. There is no maximum.
Explanation:
We can solve this question by remembering that the cell potential is given by the formula
ε⁰ cell = ε⁰ reduction -  ε⁰  oxidation
Now the problem states the cell must provide at least 0.9 V and that the reduction potential of the  oxidized species  0.80 V, thus
ε⁰ reduction -  ε⁰  oxidation ≥  ε⁰ cell 
Since ε⁰  oxidation is by definition the negative of ε⁰ reduction , we have
ε⁰ reduction - ( 0.80 V )  ≥  0.90 V
⇒ ε⁰ reduction  ≥ 1.70 V
Therefore,
(a) The minimum standard reduction potential is 1.70 V
(b) There is no maximum standard reduction potential since it is stated in the question that we want to have a cell that provides at leat 0.9 V
 
        
             
        
        
        
Answer:
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Explanation:
 
        
             
        
        
        
Oxygen has to be involved when methanol is ignited
        
             
        
        
        
It will lose them and become stable