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.
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Given that 1 mole contains 6.02x10^23 molecules, 3.0x10^23 is just around half a mole. Then we check the number of moles for each choice:
A. This is approximately half a mole, since the molar mass of Br2 is 159.8 g/mol.
B. He has a molar mass around 4 g/mol, so this is 1 mole.
C. H2 has a molar mass of 2.02 g/mol, so this is 2 moles.
D. Li has a molar mass of around 6.97 g/mol, so this is around 2 moles.
Therefore the only choice that fits is A. 80 g of Br2.
11. I would say physical because the color of the item is changed and the texture and density is changed aswell.
