CHEMISTRY COMMUNITY

505801516

E refers to the cell's potential when it is under non-standard conditions, whereas E* is when the potential of the cell is at its standard positions, and is also not proportional to the amount of the substance.

505801516

you first have to know that in order to find [C] you have to compare experiment 1 and experiment 4. this should show you that the order of [C] is 0 since the rates of 1 and 4 are both independent from the concentration of C. To find [A] you do the same thing with experiments 1 and 2, which should le...

505801516

the rate of both a forward and reverse reaction are increased with the presence of a catalyst, since the catalyst allows for another path to be made available to the reaction allowing it to progress, which in turn lowers the activation energy.

505801516

when you calculate the change in the gibbs free energy, you are calculating the amount of work that a system is able to do. the system is doing this work on its surroundings. if both the gibbs freww energy, and the delta G max of a system are equal, that is where we get -nFE

505801516

Elizabeth Harty 1A wrote:How do you know if there are sigma bonds available for rotation?

pi bonds are the bonds that are not able to rotate, but sigma bonds do have that ability

505801516

you use a comma instead of a line to separate Fe since it is in the same aqueous state and of the same species.

505801516

a unique reaction rate entails the rat being the same for the concentrations of both the products and the reactants when that reaction rate is multiplied by the coefficient that goes with it. there is no need to specify a species when you are writing a reaction rate.

505801516

since any free element's oxidation number is 0, the oxidation number of O3 would be 0

505801516

when you are solving it in terms of pre-equilibrium you really only have to make sure that the steps prior are the ones that are at equilibrium. you continue to do this until there are only reactant concentrations left in the rate equation.

505801516

the way that you can tell if a species is a catalyst is by recognizing that it is first consumed and then produced. this would entail a species being a rectant in the first step and a product in the second one.

505801516

The mechanism proposed for the oxidation of iodide ion, I−, by the hypochlorite ion, ClO−, in aqueous solution is shown. ClO−(aq)+H2O(l)⇌HClO(aq)+OH−(aq)fast in both directions I−(aq)+HClO(aq)→HIO(aq)+Cl−(aq) slow HIO(aq)+OH−(aq)→IO−(aq)+H2O(l) fast Complete the rate law for the formation of IO− imp...

505801516

Calculate the activation energy, Ea, in kilojoules per mole for a reaction at 74.0 ∘C that has a rate constant of 0.290 s−1 and a frequency factor of 9.26×1011 s−1

this question got me stuck. can anyone explain it?
.

505801516

The rate of the given reaction is 0.180 M/s.

A+3B⟶2C

What is the relative rate of change of each species in the reaction?

i got this quesiton correct on the homework, but i am not entirely sure why. can anyone help to walk me through it?

505801516

could someone please explain how to do this. i really dont undersatand
What would the potential of a standard hydrogen electrode (S.H.E.) be under the following conditions?

[H+]=0.53 MPH2=5.0 atmT=298 K

505801516

on number 14 on the homework (Complete this table relating the values of E∘cell and ΔG∘ to K.) i figured out the table but im not sure i completely understnd why. could someone explain it to me?

505801516

Identify the oxidizing and reducing agents in the skeletal (unbalanced) reaction. Then, balance the reaction, including the phase (solid, liquid, etc.) of each species. The reaction takes place in basic aqueous solution. Pb(OH)2−4(aq)+ClO−(aq)⟶PbO2(s)+Cl−(aq) i am confused on how to do the balanced ...

505801516

I'm super lost on this question
The ΔHvap
of a certain compound is 48.75 kJ·mol−1
and its ΔSvap
is 73.70 J·mol−1·K−1.

What is the boiling point of this compound?

505801516

Can someone kind of just sum up the steps on this one? I keep thinking I have it and then ending up with very different, incorrect answers. Calculate the standard enthalpy, entropy, and Gibbs free energy at 298 K for the given reaction, using the data in the table of thermodynamic properties. NH4Cl(...

505801516

Consider a process in which an ideal gas is compressed to one-fourth of its original volume at a constant temperature.

Calculate the entropy change per mole of gas.

I'm confused about this question. Can anyone explain it to me?

505801516

Unless otherwise specified, assume that the temperature is 298K (25 C). This is what I do when it's not given and it always works.

505801516

I would say that doing as many practice problems as you can is the most helpful. Professor Lavelle posts textbook ones each week, and I've found these to be super useful! I would also say to watch ANY optional video/explanation he posts, they are always helpful. And when in doubt, go to the step-up ...

505801516

Elizabeth Harty 1A wrote:How do you know if there are sigma bonds available for rotation?

pi bonds don't/can't rotate, but sigma bonds can.

505801516

Im confused on this question on the homework A constant‑volume calorimeter was calibrated by carrying out a reaction known to release 1.51 kJ of heat in 0.400 L of solution in the calorimeter (q=−1.51 kJ) , resulting in a temperature rise of 3.98 ∘C . In a subsequent experiment, 200.0 mL of 0.40 M H...

505801516

Im confused about the second part of #12 on the homework
Suppose a 3.805 g
sample of a second compound, compound B, is combusted in the same calorimeter, and the temperature rises from 23.49 ∘C
to 26.82 ∘C.

What is the heat of combustion per gram of compound B?

505801516

I'm confused on this question
Automobile airbags contain solid sodium azide, NaN3,
that reacts to produce nitrogen gas when heated, thus inflating the bag.

2NaN3(s)⟶2Na(s)+3N2(g)

Calculate the value of work, w,
for the system if 10.5 g NaN3
reacts completely at 1.00 atm
and 22 ∘
C.

505801516

Kw is only used for water. The subscripts are what I use to remember: w for water, c for concentrations, and p for pressure.

505801516

Ka is the dissociation constant for weak acids and Kb is the dissociation constant for weak bases. They are both calculated in the same way that you would calculate Kc.l

505801516

since strong acids will fully dissociate, they release more ions of H+. when more H+ ions are released, this gives you a lower pH value.

505801516

505968894 wrote:Is a negative pH possible?

even though the scale is usually 0-14, it is possible for pH to be negative.

505801516

Phoebe 1H wrote:Also, what does amphiprotic mean?

it means that it is able to react as both an acid and a base since it can donate and accept a proton

505801516

its unique because the reaction rate is the same for each concentration of products and reactions when it gets multiplied by the stoich coefficient.

505801516

No, ICE tables are just used for calculating the equilibrium for gases.

505801516

Yes! In class he said that he allows us to download the book on up to 3 different devices for 4 years of use, so you can download it multiple ways!

505801516

K is the reactants and products of the reaction when it is at equilibrium. Q is the proportion at any given time during the reaction, so it is not the value at equilibrium.

CHEMISTRY COMMUNITY

Search found 34 matches

Re: difference between E and Ecell naught

Re: Achieve week 8-10 #7

Re: Forward v Reverse Catalyst

Re: - n F E

Re: chelating ligands [ENDORSED]

Re: Short Notation

Re: Unique Reaction Rates [ENDORSED]

Re: oxidation number

Re: pre-equilibrium reaction mechanism

Re: WK 8-10 HW #18

homework question

question on homework

review question

homework confustion

homework question

homework question

Achieve Question

Week 6-7 Homework Question

Week 6-7 Homework Question [ENDORSED]

Re: Question 4D7

Re: Rusty on High School Chem [ENDORSED]

Re: chelating ligands [ENDORSED]

HW #18

HW #12

HW #14

Re: Kw

Re: Ka and Kb

Re: pH of weak acids vs strong acids

Re: Acids and Bases

Re: Acids and Bases

Re: Unique Reaction Rates [ENDORSED]

Re: ICE tables: solids

Re: Textbook Inquiry [ENDORSED]

Re: Q and K