CHEMISTRY COMMUNITY

Phan Tran 1K

The Ka/Kb can also help you find the concentrations of H+ of OH- ions depending on whether the reaction is acidic or basic, which can then help you find pH.

Phan Tran 1K

Essentially, the question is telling you that there's a k before the catalyst was added, let's call it k1, and a k after, k2. 1000k1=k2. From here, you can plug in the Arrhenius equation and the givens in order to find the original activation energy. There's a confusing part where you have to take e...

Phan Tran 1K

K's unit changes depending on the order of the reaction because the reaction rate will always have to be some variation of M/s, and k compensates for the extra M's that gets added to the number when the reaction goes up in order.

Phan Tran 1K

I think you do include water if it's not in liquid phase and if the H or O is one of the things being oxidized/reduced (this is rare but can happen), although I'm not really sure.

Phan Tran 1K

They're basically asking you to give the ratio of N2 to H2, NH3 to H2, and NH3 to N2

Phan Tran 1K

First, balance everything but Hydrogen and electrons. Then write down how many Hydrogens you need to balance the equation on the side that needs the Hydrogens. Then replace the Hydrogens with H2O molecules and add the same number of OH- molecules to the opposite side of the equation. In effect, this...

Phan Tran 1K

An fast way to quickly identify what elements in a reaction could be being oxidized is to pinpoint anything that is not oxygen or hydrogen, since it is really rare for these two to have their oxidation numbers change. This usually leaves only two elements you have to calculate the oxidation numbers ...

Phan Tran 1K

We use platinum in a cell diagram when either the cathode or the anode lacks a solid that can be used to conduct electrons from one side of the beaker to the other. This is because platinum is inert and will not react with anything else in the reaction.

Phan Tran 1K

The best way to find the oxidation state of a transition metal is to use the oxidation states of the other molecules in the compound witht he transition metal.

Phan Tran 1K

There are a few rules that I memorized when assigning oxidation numbers, but they're fairly complicated 1.) Group 1 elements are always +1, Group 2 elements are always +2 2.) The oxidation numbers have to equal the total charge of the ion or molecule 3.) Hydrogen is +1 unless it is bonded to a metal...

Phan Tran 1K

What I tried doing is to have H+ ions in the oxidation equation too. When I tried this I got C2H5OH ---> 6e- +C2H4O + 2H+
The problem with this is that even though the atoms add up, it doesn't balance the charges out in the equations, and I can't figure that part out.

Phan Tran 1K

I think it's deltaS that can be zero. Not too sure if S itself can be zero, maybe at the hypothetical subzero?

Phan Tran 1K

I'm working through problem 8.99 in the 6th edition and I'm confused about how to apply finding the enthalpy of formation. The problem is long, and parred down to the relevant information it reads: A technician carries out the reaction 2 SO2(g) + O2(g) S 2 SO3(g) at 25 C and 1.00 atm in a constant-p...

Phan Tran 1K

For reaction enthalpies of combustion, its products-reactants, while it is the other way around for reaction enthalpies of formation. I just kinda memorize this.

Phan Tran 1K

Some problems will give you the amount of a substance and ask how much heat it gives off, and this is when J/mol comes into play. Other than that I don't think it matters if they're just asking for enthalpy.

Phan Tran 1K

-PΔV = -ΔnRT is used to calculate w, which can be calculated using the equation w=-P*ΔV. You need to calculate w in order to calculate ΔU. I don't have the seventh edition, so is there anymore information that the book gives you? Like whether the reaction is reversible or irreversible, or whether T/...

Phan Tran 1K

If it's a work problem, we're probably going to be dealing with gases. Heat problems, however, can have things in basically any state. There's a ton of homework problems, for example, that involve heating a piece of solid metal.

Phan Tran 1K

A good way to remember the difference between extensive and state property is that the distance between your house and the supermarket is a state property; no matter how you get to the supermarket, that distance will always be the same, it's the state the distance will always be in. The path you tak...

Phan Tran 1K

A good way to remember the difference between extensive and state property is that the distance between your house and the supermarket is a state property; no matter how you get to the supermarket, that distance will always be the same, it's the state the distance will always be in. The path you tak...

Phan Tran 1K

The external pressure in the equation is really just the normal pressure outside of the system. When work is done by the system and pushes the piston up, this displaces the particles on the other side of the piston and thus results in the energy change. The external pressure changes so little when t...

Phan Tran 1K

You can also check for this by finding the percent protonization by taking the concentration you get and dividing it bby the original concentration of your acid/base. If it's less than 5% then treating x as being so small that it is insignificant is valid.

Phan Tran 1K

Solving autoprotolysis should be the same as other weak acids/bases questions, using an ICE table and a pKa/pKb of 10^-7.

Phan Tran 1K

A reaction favoring either side of an equation is really just saying will the reaction proceed in the forwards or the backwards direction. You know this by calculating Q and then comparing it to K. Since in these quotients are a fraction with products on the top and reactants on the bottom, Q being ...

Phan Tran 1K

If the reaction is in water, then whatever molecule that isn't the hydronium or hydroxide ion is the conjugate acid/base, depending on whether the molecule before the reaction is an acid or a base. Acids have conjugate bases and bases have conjugate acids.

Phan Tran 1K

Don't quote me on this, but I vaguely recall from high school chemistry that temperature measures the activity level of the particles in a substance while heat measures the energy they give off. I'm probably wrong since my memory is shoddy, and I'm sure we'll learn about this more in detail later on...

Phan Tran 1K

If you really want to be sure you can always just check the percent ionization after you solve the problem. If the concentration of the conjugate base/acid divided by the initial concentration of the acid/base is less than 5%, then you're good.

Phan Tran 1K

Adding inert gases will only affect the equilibrium if the reaction is in a closed system. They add more particles, which increases the overall pressure and affects the other gases's partial pressure.

Phan Tran 1K

Wait, if compression changes which side of the equation is favored, would that eventually change the Kc too, since Kc tells you which side of the reaction is favored? I'm not really sure, can someone clarify?

Phan Tran 1K

Keep in mind that aqueous substances are used in the equilibrium constant, but not liquid substances.

To add on, think of adding more liquid to a system with an aqueous substance like increasing the volume of a gaseous substance. It decreases the concentration of aqueous substances and affects Kc

CHEMISTRY COMMUNITY

Search found 29 matches

Re: Ka Kb = Kw

Re: 15.69 6th Edition

Re: Units of k

Re: Water in a cell diagram

Re: 15.1

Re: Half Reactions in a Basic Solution

Re: Oxidation State

Re: Cell Diagrams (Using Platinum)

Re: Oxidation States

Re: Oxidation States

Re: problem 14.1 6th edition

Re: S=0

Application of Standard Enthalpy of Formation

Re: 4.15 (7th edition)

Re: Units

Re: 4D7

Re: Studying gases

Re: Extensive

Re: Extensive

Re: Piston Problem in class

Re: ignoring x

Re: autoprotolysis on Test 1?

Re: Equilibrium Favoring

Re: Conjugate base

Re: How temperature affects K

Re: ICE Tables [ENDORSED]

Re: Inert gases

Re: compression vs pressure

Re: Effect of water