CHEMISTRY COMMUNITY

dmcclintick

Use the ideal gas law, PV=nRT

dmcclintick

First, figure out the half reactions. You have chromium and copper in the solid (and thus neutral) state, going to aqueous CrCl3 and CuSO4. So, the half reactions must be Cr 3+ + 3e - ----> Cr and Cu 2+ ----> Cu Then you look up their standard reduction potentials and you can see that the 1st is -.7...

dmcclintick

I believe it would be:

Al ----> Al₃₊ + 3e^-

3O₂ +12e^- ----->2O₃^6-

If you balance the electrons (4 times the first equation), you get

4Al + 3O₂ ---> 2Al₂O₃

dmcclintick

Oxygen is more electronegative and pulls the shared electrons toward themselves, leaving the Carbon electron deficient and thus makes it electrophilic.

dmcclintick

Question 44 asks for the IUPAC name for the following compound: (CH 3 CH 2 ) 2 NCH 2 CH 2 CH 3 The correct answer is N,N-diethylpropanamine, which makes sense, however, I drew it like this: CH 3 -CH 2 -CH 2 -N-CH 2 -CH 2 -CH 3 | CH 3 Which would give the name N-methyl N-propylpropanamine. Why is thi...

dmcclintick

I believe this is because the first way (in the Intro to O Chem Book), the IUPAC naming is used and in the Quiz 3 example the common name is used. As far as I know, Vinyl is not used for IUPAC naming. Because Vinyl must be the end of a chain, it can be considered a substituent for common naming, or ...

dmcclintick

Our numbering schemes could either be:

1,2,4

OR

1,3,4

Both options have 1 as the first number, so we move onto the second number. Option A has 2 and option B has 3 as the second number, so we are done, Option A is better for numbering.

dmcclintick

The double bond takes priority in this situation for the numbering, so we must count with the double bond. The parent chain in this example is 5, so the double bond is either at Carbon 4 or Carbon 1. Because the double bond takes priority, we count it at Carbon 1, which gives us 1-pentENE for double...

dmcclintick

Because the electrons in the bond are not static, the electrons move back and forth, sometimes being closer to one of the Bromines. This means that at any given moment in time one of the Bromines can have a slightly positive charge, making it an electrophile for that moment.

dmcclintick

Those are the same thing. It just depends on whether you subtract ln(initial) from ln(final) or the other way around. Notice the k is positive in one equation and negative in the other, but both are equal to T. Your approach should work just fine, as should the solution manual's.

dmcclintick

Because entropy is a state function, we can think of this problem as 2 separate steps: 1. The change in volume of the gas at fixed temperature and 2. The change in temperature of the gas at fixed volume To do the first part: deltaS = nRln(V2/V1) and the second part: deltaS = nCvln(T2/T1) Just add th...

dmcclintick

Your first mistake is that the E ^{\circ} value is .033V, not .33V. The second error is that you calculated RT/F to be .0025678233 when it is supposed to be .025678233. If you make these corrections, it should work as I just did it using those numbers and got the correct answer. I believe the lamina...

dmcclintick

Why is R used in the Nernst Equation when there are often no gasses involved (only aqueous solutions for example)? It seems interesting that the universal gas constant would be used for a reaction in which there are no gasses.

dmcclintick

It is used to allow the electrons to flow from the left to the right, and I believe platinum and graphite are used because they won't affect the continuous flow of electrons from the anode to the cathode

dmcclintick

G is a measure of the available "free" energy in a system. It is a relationship between enthalpy and entropy (G = H - TS). We know that the universe tends to greater disorder \Delta S , and using the above equation, we divide both sides by T and get \frac{\Delta G}{T} = \frac{\Delta H}{T} ...

dmcclintick

In work of expansion, w = -p \Delta V , the negative is needed because of the change in volume. If a gas does expansion work, it loses energy in order to do that said work, so we would expect w to be positive. However, in this situation, the volume is expanding so delta v (Vfinal - Initial) would be...

dmcclintick

Adiabatic system, from the course reader means there is no heat flow in or out of the system, which means q = 0. Isothermal means there is no temperature change. I believe that in a adiabatic system, which is insulated, a gas can do work, or have work done on it, (-/+w) but because there is no q, or...

dmcclintick

And also, why is $\Delta$ H a state function, while q is not? I understand that q_p is a state function but I do not understand why holding the pressure constant then makes it a state function.

dmcclintick

As an object or solution becomes hotter or cooler, does the heat capacity change? For instance, it seems like it would be much easier to heat 1 mol of water at room temperature 1 degree as opposed to 1 mol of water from 95 to 96 degrees.

CHEMISTRY COMMUNITY

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Homework Chapter 2 Question 44

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