17.40 sec is the time will take to smell second perfume after diffusion takes place
Acc. to Graham's law of Diffusion
Diffusion of Gas inversely proportional to square root of its Molecular mass.
<u>rb (Perfume B)</u>
ra(perfume A)
=
its equation (1)
Give molar mass of Perfume A = 275 g/mol
molar mass of Perfume B= 205g/mol putting value in (1)
» 
<em> its eq (2)</em>
» Perfume B will defuse 1:16 times faster than perfume A.
Hence, perfume B will be first smelled by Person.
Sf Equal volume V of two goes diffuse in t1 and t2 sec. respectively ton

Now,, from eq(2) toto
1/1:15 = 
ta
Given tb =the smell of Perfume B as it diffuse faster
ta= 1.15 x 15 see
=ta2=17.40 sec
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it's SN1 reaction,
it's takin double step to complete,first formation of carbocation and then attack of reagents,so it's SN1!
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Because his first law states that an object with a net force of zero acting on it will remain at rest, if initially at rest, or it will maintain a constant velocity.
Answer:
d
Explanation:
the farther you are from the equater the dolder it is.