CHEMISTRY COMMUNITY

garrett_perozich

Why is the substituent given priority in regards to naming? In other words, why wouldn't it be 3-(3-chlorocycloheptnyl)-2-methylpropan-1-ol? I thought that the functional group always receives the lowest number.

garrett_perozich

No, you wouldn't. As long as you specified within your work that that was what K was equal to, then you should be fine.

garrett_perozich

Is 4,4-dimethyl-2-hexanol the same thing as 4,4-dimethylhexan-2-ol? In other words, can we number the location of the alcohol before the "ol" and have it represent the same molecule as if we numbered before the longest carbon chain?

garrett_perozich

For 1B, how would a student know to use Cv for step 2? I understand that you would either use Cv or Cp since the problem told you to treat it as an ideal gas, but what doesn't make sense to me is why you'd use Cv if they volume was not constant (it changes from 3.0 to 3.5L). Also, the pressure isn't...

garrett_perozich

On quiz two, we had a reaction with two steps:

Step 1: this was the slow step and used a catalyst
Step 2: this was a fast step and produced the same catalyst

So my question is, why would the catalyst be a part of the rate law equation if it is produced as soon as it is used up?

garrett_perozich

If you draw it out, you will see why it gets a partial positive charge. Look at the bonds. Our middle carbon (the one you're talking about with the positive charge) is bonded to a carbon on the left (1 bond), + a carbon on the right (now we have 2 bonds), and then a hydrogen (that's 3 bonds). But ou...

garrett_perozich

The difference between these two is that, in problem 34, we are given the products, and as we can see, our hydroxide stays as a hydroxide. So, this would only be one step, since our O is still bonded to our H. In 26, we have to first break our HI up and attach our H to the carbon, but we now have an...

garrett_perozich

The molecules would need to collide again for this next step to happen.

garrett_perozich

When you look at the activation energies, you are looking at the energy needed to move from its previous "state" to the new transition state (TS). So, for Ea1, it is from the reactants "axis", to the top of TS1. For Ea2, though, we have to look at the difference between the inter...

garrett_perozich

When drawing a reaction profile given the equation, for example, CH3CHCH2 + HBr, how do we know whether the final Gibbs free energy will be lower or higher than the initial? We were asked to do this twice on the homework for chapter 4, once for Gibbs, and the other for potential, and both times I as...

garrett_perozich

Curved arrows are used to represent the transfer of a pair of electrons. So while it may be used for a bond, in which case it would represent the transfer of the two electrons that are being shared between the atoms for that particular bond, it may also be used on a lone pair of electrons, in which ...

garrett_perozich

I tell chemistry jokes periodically, but usually there is no reaction.

garrett_perozich

If you take a look at page 59 in the course reader, we aren't really "ignoring" the products. As seen at the top of the page, the rate that the reactants decrease is equal to the rate that the products increase (with the opposite sign). Also, it is easier to study an initial reaction when ...

garrett_perozich

Hi, The book actually doesn't divide by 100. We know that K1 is 10x greater than K2. So that means that K1 = 10*K2. Now if we do K1/K1, substituting 10*K2 for K1, we get 10*K2/K2, or 10, since the K2's cancel. Now, that value is a constant for our equilibrium value when solving for our Gibbs Free En...

garrett_perozich

#8 on Quiz 1 Preparation says, "For reaction: 2C+2H2=C2H4, DeltaH= +52.3 kJ/mol and DeltaS= -53.07 J/K*mol at 298 K. At what temperature will this reaction be spontaneous?" Initially, I thought the formula needed to solve for the temperature was DeltaG= DeltaH - T*DeltaS but then I saw th...

CHEMISTRY COMMUNITY

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