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Chemistry 7 Online
OpenStudy (anonymous):

The vapour density of partially dissociated iodine at a certain temperature was found to be 80. What is its degree of dissociation?

OpenStudy (anonymous):

do you got any formulas to work with?

OpenStudy (anonymous):

No idea

OpenStudy (anonymous):

yea i'm sorry i don't think i will be able to help with this one

OpenStudy (anonymous):

Can u call some one else

OpenStudy (anonymous):

.587?

OpenStudy (anonymous):

yes u r correct @Ishaan94

OpenStudy (anonymous):

@Ishaan94 hw do u get it

OpenStudy (anonymous):

Vapor density of dissociation vapor = 127. Let x be the degree of dissociation. 80/120 = 1/(1+x) => x = .587 I hope you can get the equation and number moles at equilibrium.

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