in order to solve a series, the root test is applied and I have this limit
## \lim_{n \rightarrow +\infty} \sqrt[n] {\left| {\frac {i^{\frac { n} {3}}} { \frac {2n} {3} +1}} \right| } ##
I don't understand why at the second step the numerator becomes 1, cannot recall why it becomes 1, that is...
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ok now i got it, yesterday i was too tired to see it 😅. Basically what I do is.
##8,5*10^{-2}-x+x+x= 8,5*10^{-2}+x=n_{tot}##
so
## n_{tot}=\frac{P_{tot*V}}{RT}##=0,125mol
##8,5*10^{-2}+x=0,125mol##
##X=0,125-8,5*10^{-2}=0,04 mol##
now I just plug in Kp equation and done
Yes I've done it: but i am not understanding how to get x, usually Kc or Kp or partial pressures, or molar fractions are given, but here ?
sorry for posting a pic btu i did not know how to draw it here
I am stuck, i have compute the total moles as :
## n_{tot}=\frac{P_{tot*V}}{RT}##=0,125 mol
##Kp= \frac{P_{tot}\frac {n_{PCL_3}}{n_{tot}}*P_{tot}\frac {n_{CL_2}}{n_{tot}}}{P_{tot} \frac {n_{PCL_3}}{n_{tot}}}=\frac {P_{tot}}{n_{tot}}*\frac {n_{PCL_3}n_{CL_2}}{n_{PCL_5}}##
From here, can't go...
ok, i was neglecting the methanol, basically i was considering it as non volatile solute, in that case I should not consider it, but here it must be taken into account. thanks
Isn't the formula to be applied ##P= \chi_{solvent}P^o _{solvent}##?
in which ##P^o _{solvent}##= 18.7mmHg and
##\chi_{solvent}= \frac{mol_{solvent}}{mol_{solvent}+mol_{solute}}=\frac{9.0 mol}{9.0 mol+0,50 mol}##
but seems doesn't leads to the expected result
thank you now it makes more sense, all works perfectly.
Are we sure is like that ? ##CH_2= CH_2 + H_2O_{(l)} -> ## so this is a way of writing that or is just a typo? never seen before such writing
I've also tried like this
##\Delta H°_{ethanol}=-277\frac{KJ}{mol}##
##\Delta H°_{H_2O_{(l)}}=-285,8 \frac{KJ}{mol}##
##\Delta H°_R=\sum \Delta H°_{products}-\sum \Delta H°_{reagents}##=
##\Delta H°_R= (-277\frac{KJ}{mol})-(-285,8 \frac{KJ}{mol})= 8,8##
but i think, being positive si not the...
corrected but still doesn't match -1,5*10^3 KJ :confused:
from table the standard enthalpy of ethanol is ##\Delta H°=-277 \frac{KJ}{mol}##
the grams of ethanol are
##2000 cm^3*0,79\frac {g}{cm^3}=1580g##
and the moles are
##\frac{1580g}{(24,02+16,00+6,06)\frac g {mol}}= 34,3##mol
By...
from table the standard enthalpy of ethanol is ##\Delta H°=-277 \frac{KJ}{mol}##
the grams of ethanol are
##0,002 cm^3*0,79\frac {g}{cm^3}=1,58*10^{-3}g##
and the moles are
##\frac{1,58*10^{-3}g}{(24,02+16,00+6,06)\frac g {mol}}= 3,43*10^{-5}##mol
By proportionality, having already the...