CHEMISTRY COMMUNITY

KVu 3G

I believe Professor Lavelle mentioned that it is derived from not only simplifying the constants, but also multiplying by a factor which converts lnQ to logQ

KVu 3G

I believe it's because the carbons on the parent chain aren't counted for the substituent chain

KVu 3G

You know that the substance is a first order, since the amount of the substance does not change the half life (50 days regardless of how much you start with.) Since you know it's a first order, I believe you can simply keep dividing in half until you reach 3g OR Use half life = 0.693/k and solve for...

KVu 3G

I believe you can use an integrated rate law to solve for initial concentration if you aren't already given it. You can also use the integrated rate law to draw a graph which tells you the order of the reaction. For example, if the graph of ln[A] vs. t is a straight line, you know you have a first o...

KVu 3G

a refers to the coefficient in a general equation $aA+ bB \rightarrow ...$

The term -1/a defines how fast A is being consumed relative to the other reactants and products in the equation. For example B may not be consumed at the same rate A does, it's rate of consumption is dependent on b.

KVu 3G

n is the number of moles of electrons transferred. When the redox reaction between zinc and copper are balanced there are 2 moles of electrons transferred.

$Zn \rightarrow Zn^{2+} + 2e^{-}$
$Cu^{2+} + 2e^{-}\rightarrow Cu$

KVu 3G

I believe the enthalpy of atomization is the enthalpy change associated with breaking apart a molecule into it's individual substances.

For example, the enthalpy of atomization for H2O would be the sum of the bond enthalpies to split the molecule into it's individual atoms.

KVu 3G

You find w using w=-P * delta V which can be derived from the ideal gas equation. PV=nRT P\Delta V = \Delta n RT -P\Delta V =- \Delta n RT Substitute -P\Delta V for - \Delta n RT gives you w=-\Delta nRT . New equation is \Delta U=q + (-\Delta nRT) where q is given, R is a constant, T=298K, a...

KVu 3G

Yes it is true, I like to think of this in terms of a reaction profile. For example, say the reactants have a lower energy than their products. That means that the reaction is endothermic and requires energy. However, once you switch them, and go from products to reactants, you're now going from a h...

KVu 3G

A video explaining how to fill molecular orbital diagrams

KVu 3G

Correct me if I'm wrong but, molecular shape is determined by the orientation of the atoms and regions of electron density. With resonance, the bonds are shifted around, but the number of electron density regions remains the same, so the orientation of the atoms and the molecular shape of the atoms ...

KVu 3G

For elements such as Chromium or Copper, they will have a filled or half filled d orbital. Electrons prefer to be in either a half filled or full d-orbital because it is more stable due to symmetry. For example, Copper will have an electron configuration of [Ar]3d 10 4s 1 because it is more stable w...

CHEMISTRY COMMUNITY

Search found 12 matches

Re: Nerst Equation?

Re: Tert- and Sec- Prefixes

Re: Quiz #2 Question #1 on Thursday

Re: Integrated vs. Differential Rate Laws

Re: Changing Differential Rate Law to Integrated Rate Law

Re: Example 14.3 moles confusion? (Not HW question, example.

Re: Enthalpy of atomization

Re: Homework 8.49 question

Re: Question regarding Enthalpy

Video: Molecular Orbital Diagrams and Configurations

Re: Shape

Re: Chromium, Copper, etc.