Practice OAT Questions 45 and 64

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Aletheia

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Hi,

Can someone please work out questions 45 and 64 for me on the practice OAT exam in the Chem section? Thanks!
 
In question 45, you are finding the new volume of gas after a temperature and pressure change. First, you start with the original volume (10 liters). The pressure goes from 1 atm to .5 atm. If you cut the pressure in half, you double the volume. This will be the 1/.5 term (.5/1 doesn't make sense because then you'd be cutting the volume in half along with pressure). The temperature (remember to work with kelvins, with 273 kelvins being equal to 0 degrees celcius) goes from 373 kelvins to 300 kelvins. Really, all you need to do is, again, find the order of the fraction (373/300 or 300/373). Well, you are lowering the temperature, which cools the gas and decreases volume. So the term must be 300/373 for the temperature decrease. The correct answer must be D.

In question 64, all you have to do is assign an oxidation value to Nitrogen on both the reactants side and the products side, and simply find out what the difference between those numbers is. On the reactants side, Nitrogen is found in the molecule NO and Oxygen has an oxidation value of -2 (virtually all the time) so Nitrogen must have a value of +2 because the molecule NO has no net charge. On the products side, Nitrogen is found in HNO3 and again, the Oxygens all have -2 and there are 3 of them, so you'll get -6 from the Oxygens. The Hydrogen will be a +1 because you aren't working with any elements that have extremely low electronegativity (i.e. the reactive metals, such as the alkali series). Therefore, you have a net of -5 from the elements other than Nitrogen, so for this to be a neutral molecule (as it has no net charge) you must give Nitrogen a +5 oxidation value here. Nitrogen started at +2 and went to +5 for a difference of 3, which is the correct answer.
 
Thanks! I see where I ran into trouble on those now. The first one confused me because I hadn't seen anything like that before in studying gas questions but thinking it through makes complete sense.
 
If this is wrong someone please correct it:

63:

E=Ered-Eox (it spells out Redox if you're having trouble remembering it)

From the final reaction we see the Cu--> stuff, thus our reduction potential is left as is for this half reaction. We see that Cl2 is a product in the final reaction but a reactant in the half cell reaction, thus we must switch the half cell reaction so it reads 2Cl- --> Cl2 + 2e. When we do this, the sign of the E changes to -1.36. Next, figure out which is reduced and which is oxidized. We can see that Cu gains electrons (LEO the lion says GER--gain electrons reduces) so it is reduced. We can also see that Cl2- loses electrons (lose electrons oxidizes). Therefore the answer is -1.36+.34!

I haven't reviewed my acids and bases yet so someone else can take those.