claculate the Ionic % and covalent % in HCL
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answer is 17%, ihave taken some data to make it clear...
dipole(meu)=e * d,
pecent ionic =
actual dipole/calculated dipole*100..
mark as brainliest please!!!
dipole(meu)=e * d,
pecent ionic =
actual dipole/calculated dipole*100..
mark as brainliest please!!!
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AnjaliRoy00:
karnatka
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Answered by
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Hii Ma'am !!!!
ws will first cal dipole moment
(mu)=e*d
e=1.6*10^-19C
d=1.275*10^-10
mu=(1.6*10^-19)*(1.275*10^-10)
=2.04*10^-29
and we know actual dipole moment of hcl is 0.34*10^-29
so %ionic= (0.34*10^-29)/(2.04*10^-29)
=17%
Hope it helps
ws will first cal dipole moment
(mu)=e*d
e=1.6*10^-19C
d=1.275*10^-10
mu=(1.6*10^-19)*(1.275*10^-10)
=2.04*10^-29
and we know actual dipole moment of hcl is 0.34*10^-29
so %ionic= (0.34*10^-29)/(2.04*10^-29)
=17%
Hope it helps
ma'am plz mark this ans brainliest if you like it...plzzz
bro , I think I have replied it earlier than u,,,
but it depends in the asker whom it mark brainliest
good try bro
nice job...
Thanks a lot ma'am for marking brainliest.
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