1
JEE Main 2019 (Online) 10th April Morning Slot
+4
-1
A bacterial infection in an internal wound grows as N'(t) = N0 exp(t), where the time t is in hours. A does of antibiotic, taken orally, needs 1 hour to reach the wound. Once it reaches there, the bacterial population goes down as $${{dN} \over {dt}} = - 5{N^2}$$. What will be the plot of $${{{N_0}} \over N}$$ vs. t after 1 hour?
A
B
C
D
2
JEE Main 2019 (Online) 9th April Morning Slot
+4
-1
The given plots represent the variation of the concentration of a reactant R with time for two different reactions (i) and (ii). The respective orders of the reactions are :
A
0,1
B
1,0
C
0,2
D
1,1
3
JEE Main 2019 (Online) 8th April Evening Slot
+4
-1
For a reaction scheme $$A\buildrel {{k_1}} \over \longrightarrow B\buildrel {{k_2}} \over \longrightarrow C$$,

if the rate of formation of B is set to be zero then the concentration of B is given by :
A
$${k_1}{k_2}[A]$$
B
$$\left( {{{{k_1}} \over {{k_2}}}} \right)[A]$$
C
$$({k_1} + {k_2})[A]$$
D
$$({k_1} - {k_2})[A]$$
4
JEE Main 2019 (Online) 8th April Morning Slot
+4
-1
For the reaction 2A + B $$\to$$ C, the values of initial rate at diffrent reactant concentrations are given in the table below. The rate law for the reaction is :

[A] (mol L-1) [B] (mol L-1) Initial Rate (mol L-1s-1)
0.05 0.05 0.045
0.10 0.05 0.090
0.20 0.10 0.72
A
Rate = k[A][B]2
B
Rate = k[A]2[B]2
C
Rate = k[A]2[B]
D
Rate = k[A][B]
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