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leena1996
 one year ago
The following data were obtained in an experiment on inversion of cane sugar in the presence of 0.1N HCl.
Time(min) 0 60 120 180 360 infinity
Angle of
Rotation 13.1 11.6 10.2 9.0 5.87 3.8
After what time would you expect a zero reading in Polarimeter........its urgent can someone help me fast!
leena1996
 one year ago
The following data were obtained in an experiment on inversion of cane sugar in the presence of 0.1N HCl. Time(min) 0 60 120 180 360 infinity Angle of Rotation 13.1 11.6 10.2 9.0 5.87 3.8 After what time would you expect a zero reading in Polarimeter........its urgent can someone help me fast!

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chmvijay
 one year ago
Best ResponseYou've already chosen the best response.0WOW Very interesting question i Love it :)

leena1996
 one year ago
Best ResponseYou've already chosen the best response.0@chmvijay will u plz solve it.....

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0this is a theoretical experiment and not an actual experiment, right?

leena1996
 one year ago
Best ResponseYou've already chosen the best response.0yup it is theorotical just use chemical kinetics and answer this @nincompoop

chmvijay
 one year ago
Best ResponseYou've already chosen the best response.0hope @nincompoop answers this :)

leena1996
 one year ago
Best ResponseYou've already chosen the best response.0me too hopind the same......lets see...its not supposed to be very difficult .... it is class 12 question....ISC board @chmvijay

leena1996
 one year ago
Best ResponseYou've already chosen the best response.0ok here's the only hint formula\[k =\frac{ 2.303 }{ t }\log_{10}\frac{ r_{0}r _{\infty} }{ r_{t}r _{\infty} } \]

leena1996
 one year ago
Best ResponseYou've already chosen the best response.0@nincompoop @chmvijay @Frostbite i got to go back to work.......if you guys get the answer plz do reply here........thanks! Bye will be back soon

chmvijay
 one year ago
Best ResponseYou've already chosen the best response.0ok fine me to leaving to BED will see it hope @nincompoop solves this :)

leena1996
 one year ago
Best ResponseYou've already chosen the best response.0I guess i will have to ask this to my instructor...!

electrokid
 one year ago
Best ResponseYou've already chosen the best response.2in the formula \[k={2.303\over t}\log_{10}\frac{r_0r_\infty}{r_tr_\infty}\] k = rate of reaction (interms of change in optical activity) t = time r = angles of rotation from this, we calculate the rate of reaction \[ k={2.303\over120}\log_{10}\frac{13.1(3.8)}{10.2(3.8)}\\ k=1.57\times10^{3} \] use the equation again, to find "t" when r_t=0 rearranging, the terms, \[ t={2.303\over 0.00157}\log_{10}\frac{13.1(3.8)}{0(3.8)}\\ \Large\boxed{t=950\,{\rm min}\\ t=15\,{\rm hr}\;50\,{\rm min}} \]

leena1996
 one year ago
Best ResponseYou've already chosen the best response.0@electrokid Thanks a lot!!! It was great help....
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