The inactivation rate of a viral preparation is proportional to the amount of virus. In the first minute after preparation, 10% of the virus is inactivated. The rate constant for viral inactivation is __________ `xx 10^(-3) "min"^(-1)`. (Nearest integer)
[Use : ln 10 = 2.303 , `log_(10) 3= 0.477`,
Property of logarithm : `logx^y = y log x` ]

The inactivation of a viral preparation in a chemical bath is found to be first order reaction. If in the beginning, 2.5% of the virus is inactivated per minute, calculate the rate constant of the viral inactivation.

For a certain first order reaction 32% of the reactant is left after 570s. The rate constant of this reaction is "_________" xx 10^(-3) s^(-1) . (Round off to the Nearest Integer ) . [ Given : log_(10) 2 = 0.301, ln 10 = 2.303 ]

The inactivation of a viral preparation in a chemical bath is found to be a first order reaction. The rate constant for the viral inactivation if in beginning 1.5% of the virus is inactivated per minute is (Given : In (100)/(98.5)=0.01511 )

If the rate constant for a second order reaction is 2.303 xx 10^(-3) s^(-1) then the time required for the completion of 70% of the reaction is (log3=0.48)

N_(2)O_(5(g)) rarr 2NO_(2(g)) + (1)/(2)O_(2(g)) . In the above first order reaction the initial concentration of N_(2)O_(5) is 2.40 xx 10^(-2) mol L^(-1) at 318 K. The concentration of N_(2)O_(5) after 1 hour was 1.60 xx 10^(-2) mol L^(-1) . The rate constant of the reaction at 318K is ______ xx 10^(-3) "min"^(-1) . (Nearest integer) [Given: log3= 0.477, log 5= 0.699]

The rate constant of a reaction with a virus is 3.3xx10^(-4)S^(-1) . Time required for the virus to become 75% inactivated is

For the first order reaction A to 2 B ,1 mole of reactant A gives 0.2 moles of B after 100 minutes. The half life of the reaction is ...... min. (Round off to the nearest integer). [Use ln 2 =0. 69, ln 10 = 2. 3 Properties of logarithms : ln x ^(y) = y ln x, ln ((x)/(y)) = ln x - ln y] Round off to the nearest integer)

Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at 25^(@)C . After 9h , the fraction of sucrose remaining if f. The value of log_(10)((1)/(f)) is _________ xx 10^(-2) . (Rounded off to the nearest integer ) [Assume , ln 10 = 2.303, ln 2= 0.693 ]

If 6^(3 - 4x) 4^(x+5) = 8 (Given log_(10) 2 = 0.301 and log_(10) 3 = 0.477 ), then which one of the following is correct ?