Uppose the reaction between naphthalene and 2-bromo-2-methylpropane did not run to completion, so that the limiting reagents were not consumed. how would the two TLC plates have looked in your first and second run in this case?

A)there wouuld have been two spots in both cases

B)There would have been three spots in both cases

C)There would have been four pots in both cases

D)There would have been no distinct spots but only streaks.

C)There would have been four pots in both cases

Explanation:

In an incomplete reaction of naphthalene and 2-bromo-2-methylpropane, we started with 2 spots (one each for the two starting materials) and we should end up with 3 spots If spots will appear clearly, one for naphthalene, one for the alkyl halide and one for the product. However, this is possible to have same number of spots initially and finally when we consider 2 spots each for naphthalene and 2-bromo-2-methylpropane then after reaction we have 2 spots of products as each 1 spot of reactants decreases. Finally making initial 4 and final 2+1+1 =4.

So, the answer is (C)

answer: -

reaction:

na + cl ⇌ nacl

balanced chemical equation:

na + 2cl ⇌ nacl2

sodium + clorine goes to form sodium chloride (table salt).

the product is nacl2.

and the reactants are na and cl.

(note: if you don’t know what ⇌ means, read below: -

⇌(double head arrow) means reversible reaction).

b is the correct answer

mark me brainliest and hope this