Recent content by DottZakapa

yes

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

🤷‍♂️ results coincide, so i guess he is right

o ok, thank you, didn't know it. Thanks

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...

thanks