(a) For a reaction `ABrarrP` , the rate law is given by, `r=k[A]^(1//2)[B]^(2).` What is the order of this reaction? (b) A first order reaction is fou

Correct Answer - B `(k_(T_(2)))/(k_(T_(1)))=(mu)^(DeltaT//10)` `(10^(-3))/(k_(T_(1)))=(mu)^(10//10)=2` `k_(T_(1))=(10^(-3))/(2)=0.5xx10^(-3)=5xx10^(-4)"min"^(-1)`

Correct Answer - A For zero order reaction, x=kt `:. (a)/(2)=kxxt_(1//2),i.e.,t_(1//2)=(a)/(2k)" ".....(i)` For frist order reaction, `t_(1//2)=(log_(e)2)/(k)" ".....(ii)` From eqs. (i) and (ii), `(a)/(2k)=(log_(e)2)/(k)` `a=log_(e)4M`

Correct Answer - C Let in 8 coupons S, U, R, F appears `x_(1), x_(2), x_(3), x_(4)` times. Then `x_(1) + x_(2) + x_(3) + x_(4) = 8`, where `x_(1), x_(2),...

`t_(25%)=20` min `k=(2.303)/(t)log((a)/(a-x))" "[{:(ato"Initial conc."),(xto" reacted"),(a-xto" remaining"):}]` `k=(2.303)/(20)log((cancel(100))/(cancel(75)))["Div. by"25]` `k=(2.303)/(20)(log4-log3)` `implies(2.303)/(20)xx(0.6021-0.4771)` `impliesk=(2.303)/(20)xx0.125s^(-1)` `t_(75%)=(2.303)/(k)log((cancel(100))/(cancel(25)))` `=(2.303xx20)/(2.303xx0.125)xx0.6021=t_(75%)=96.34sec`.

Correct Answer - `BE=(hc)/(lambda)-(1)/(2)Mu^(2)-(1)/(2)Mv^(2)`

Correct answer is (B) 3

1. Interest rate 2. Expansion of industrial sector

1. Newton’s first law shows an equivalence between the ‘state of rest’ and ‘state of uniform motion along a straight line.’  2. Newton’s first law of motion defines force as a...

Correct Answer - `(a) 2 cm//sec (b) 4 cm^(2)//sec`,Increase

Lenz’s law :  Whereas, Lenz’s law states that, the direction of the induced emf is such that the change is opposed. Ampere’s law :  Ampere’s law describes the relation between the induced magnetic...