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leena1996

  • 2 years 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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  1. chmvijay
    • 2 years ago
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    WOW Very interesting question i Love it :)

  2. leena1996
    • 2 years ago
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    @chmvijay will u plz solve it.....

  3. nincompoop
    • 2 years ago
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    this is a theoretical experiment and not an actual experiment, right?

  4. leena1996
    • 2 years ago
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    yup it is theorotical just use chemical kinetics and answer this @nincompoop

  5. chmvijay
    • 2 years ago
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    hope @nincompoop answers this :)

  6. leena1996
    • 2 years ago
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    me too hopind the same......lets see...its not supposed to be very difficult .... it is class 12 question....ISC board @chmvijay

  7. leena1996
    • 2 years ago
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    *hoping

  8. leena1996
    • 2 years ago
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    ok here's the only hint formula\[k =\frac{ 2.303 }{ t }\log_{10}\frac{ r_{0}-r _{\infty} }{ r_{t}-r _{\infty} } \]

  9. leena1996
    • 2 years ago
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    @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

  10. chmvijay
    • 2 years ago
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    ok fine me to leaving to BED will see it hope @nincompoop solves this :)

  11. nincompoop
    • 2 years ago
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    I can't… LOL

  12. nincompoop
    • 2 years ago
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    @electrokid

  13. leena1996
    • 2 years ago
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    I guess i will have to ask this to my instructor...!

  14. electrokid
    • 2 years ago
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    in the formula \[k={2.303\over t}\log_{10}\frac{r_0-r_\infty}{r_t-r_\infty}\] 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, \[ t={2.303\over 0.00157}\log_{10}\frac{13.1-(-3.8)}{0-(-3.8)}\\ \Large\boxed{t=950\,{\rm min}\\ t=15\,{\rm hr}\;50\,{\rm min}} \]

  15. leena1996
    • 2 years ago
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    @electrokid Thanks a lot!!! It was great help....

  16. chmvijay
    • 2 years ago
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    @electrokid nice :)

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