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vera_ewing
 one year ago
Chem
vera_ewing
 one year ago
Chem

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taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.0I know at stp the most ideal would be O2 and the C2H4 but I'm not for sure when pressure is high

vera_ewing
 one year ago
Best ResponseYou've already chosen the best response.0Would the answer be D. C?

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1Have you studies van der waals equation and constants?

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1Do you see a trend in the constants of these substances?

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1What trend do you see and what do you think it means?

vera_ewing
 one year ago
Best ResponseYou've already chosen the best response.0The bonds are decreasing from A to B? So I think that D. C would have ideal behavior at high pressures.

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1What do you mean that the bonds are decreasing?

vera_ewing
 one year ago
Best ResponseYou've already chosen the best response.0A has more bonds than B, C, and D.

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1That is true. Are you saying that oxygen gas will be most ideal at high pressures?

vera_ewing
 one year ago
Best ResponseYou've already chosen the best response.0Well I think Ne would have ideal behavior at high pressures. Or OO. Am I correct?

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1Yes, I get one of those. Do you have the van der waals constants for each?

vera_ewing
 one year ago
Best ResponseYou've already chosen the best response.0No. But I think it's OO, not Ne.

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/deviation5.html Why do you think it is OO?

vera_ewing
 one year ago
Best ResponseYou've already chosen the best response.0I think because it has one bond...

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1Why does that make it more ideal at high pressures do you think?

vera_ewing
 one year ago
Best ResponseYou've already chosen the best response.0Maybe because the one bond makes it more stable and it's between O and O?

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1Hm, let's look at the constants for O2 and Ne and see if we can draw any conclusions.

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1What values of constants do you think would make the van der waals equation simplify down to the ideal gas law?

vera_ewing
 one year ago
Best ResponseYou've already chosen the best response.0I'm not really sure...

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1If the an^2/V^2 term went to 0, what do you think would happen to van der waals?

vera_ewing
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure. Can you please explain?

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1If that term went to zero, then van der waals would simplify todw:1438902988176:dw

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1And what if the nb term went to 0?

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1Take a guess.dw:1438903094158:dw

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1No, only the nb term would go to zero.dw:1438903183657:dw

vera_ewing
 one year ago
Best ResponseYou've already chosen the best response.0Ohh okay! So it's Ne?

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1Right, because the van der waals constants are closer to 0.

vera_ewing
 one year ago
Best ResponseYou've already chosen the best response.0Ah, okay! Thank you so much for explaining it!

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1You're welcome :)
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