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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!
 11 months ago
 11 months 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!
 11 months ago
 11 months ago

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chmvijayBest ResponseYou've already chosen the best response.0
WOW Very interesting question i Love it :)
 11 months ago

leena1996Best ResponseYou've already chosen the best response.0
@chmvijay will u plz solve it.....
 11 months ago

nincompoopBest ResponseYou've already chosen the best response.0
this is a theoretical experiment and not an actual experiment, right?
 11 months ago

leena1996Best ResponseYou've already chosen the best response.0
yup it is theorotical just use chemical kinetics and answer this @nincompoop
 11 months ago

chmvijayBest ResponseYou've already chosen the best response.0
hope @nincompoop answers this :)
 11 months ago

leena1996Best ResponseYou've already chosen the best response.0
me too hopind the same......lets see...its not supposed to be very difficult .... it is class 12 question....ISC board @chmvijay
 11 months ago

leena1996Best ResponseYou've already chosen the best response.0
ok here's the only hint formula\[k =\frac{ 2.303 }{ t }\log_{10}\frac{ r_{0}r _{\infty} }{ r_{t}r _{\infty} } \]
 11 months ago

leena1996Best 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
 11 months ago

chmvijayBest ResponseYou've already chosen the best response.0
ok fine me to leaving to BED will see it hope @nincompoop solves this :)
 11 months ago

leena1996Best ResponseYou've already chosen the best response.0
I guess i will have to ask this to my instructor...!
 11 months ago

electrokidBest ResponseYou've already chosen the best response.2
in 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, \[ t={2.303\over 0.00157}\log_{10}\frac{13.1(3.8)}{0(3.8)}\\ \Large\boxed{t=950\,{\rm min}\\ t=15\,{\rm hr}\;50\,{\rm min}} \]
 11 months ago

leena1996Best ResponseYou've already chosen the best response.0
@electrokid Thanks a lot!!! It was great help....
 11 months ago
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