The half-life for the viral inactivation if in the beginning `1.5 %` of the virus is inactivated per minute is (Given : The reaction is of first order)

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 )

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 )

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.

The half life of a first order reaction is

A viral preparation was inactivated in a chemical bath. The inactivated in a chemical bath. The inactivation process was found to the first order in virus concentration and at the beginning of the experiment 2.0% of the virus was found to be inactivated per minute The k for the inactivated per minute The k for the inactivated process is (assume that there is no change in rate in the interval)

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 ]