CHEMISTRY COMMUNITY

Aileen Nguyen 3N

You can see that A does not work because the intermediates do not cancel out the cumulative reactants for B should be 2CLO and O2 since one of the O2's are cancelled out So the two valid ones are B&C Student B: 2ClO; O2 Student C: 2ClO; O2 you can tell that the second steps are the rate limititi...

Aileen Nguyen 3N

[Current] = [Initial] + [Change in initial] [O2] = [O2 initial] + [change in initial O2] Flip both equations: 1 / [O2] = 1 / ([O2 initial] + [change in initial O2]) Since Q = [Products] / [Reactants] = 1 / [O2], plug in: Q = 942.23 = 1 / ([.00117] + [Change in initial O2]) Rearrange and solve using ...

Aileen Nguyen 3N

n is based on the total number of electrons transferred in the redox reaction. Fe2+ goes to Fe, meaning that two electrons are transferred per Fe, and there's a molar coefficient of 2, so in the net reaction we transfer 4 electrons.

Aileen Nguyen 3N

Carbon Monixide is an Electorphile. The C=O bond is very polar and the carbonyl carbon is very positive. The carbonyl bond is very polar. There is a partial positive charge on the carbon and a partial negative charge on the oxygen, because oxygen is more electronegative than carbon. This charge sepa...

Aileen Nguyen 3N

To relate pH to the equations we learned, E° has to be 0 when pH 7 since pH=7 when concentration of H+ and OH- are equal to 1.0 x 10^-7 mol.L^-1 when k=1 E°is 0 take ∆G°=-nFE°and ∆G°=RTlnK and then you get E°= RT/nF (lnK) at pH=7: K= [H+]/[OH-]= 1.0 x 10^-7 mol.L^-1/1.0 x 10^-7 mol.L^-1= 1 and E = 0...

Aileen Nguyen 3N

There are a lot of students in my BL that are in 14C currently and many that haven't taken 14B in over a year and they're doing well!

Aileen Nguyen 3N

Happy Valentine's Day!

Aileen Nguyen 3N

Great joke!
However, you're supposed to post all chemistry jokes here:
viewtopic.php?f=156&t=4291

Aileen Nguyen 3N

You would use E˚cell = E˚cathode - E˚anode with E˚ at the electrodes being the the electron potential for the half reaction. 2H+ (aq, 1.0 M) +2e- ----> H2 (g, 1 atm) E˚cathode = 0 V H2 (g, 1 atm) ----> 2H+ (aq, .075 M) +2e- E˚anode = 0 V 2H+ (aq, 1.0 M) +2e- + H2 (g, 1 atm) ----> 2H+ (aq, .075 M) +2...

Aileen Nguyen 3N

Enjoy!

Aileen Nguyen 3N

alcohol is not a problem it's a solution

Aileen Nguyen 3N

1) if it is near an anion which has a high charge density (small and high charge)
2) if the cation is large since its electrons will be further away from the protons in its nucleus

Aileen Nguyen 3N

Acids tend to accept electrons, and bases tend to donate electrons. Acids contain H+ ions. They tend to grab electrons and form hydrogen gas and since metals don't have a tight grip on their electrons. They give them up and the solid metal gradually disappears. The reaction is Bases contain OH- ions...

Aileen Nguyen 3N

Al3+ forms a complex with water as ligands and since Al3+ is a Lewis acid AlCl3 gets substituted by Al(H2O)6

Aileen Nguyen 3N

A chem moderator said to omit 12.117 :)

Aileen Nguyen 3N

It would specify if the problem wanted the solution to be in bars

Aileen Nguyen 3N

As2O3 is amphoteric because it can react with either acid or base.

Ex. Reaction with Acid
As2O3 we have: As2O3 + 6HCl --> 2AsCl3 + 3H2O

Reaction with base:
NaOH + As2O3 + 3H2O --> 2Na[As(OH)4]

Aileen Nguyen 3N

Reactants are strongly favored so we push the reaction as far to the left as possible and then start from the new initial conditions Pressures (bar) 2 HCl(g) -> H2(g) + Cl2(g) Original 2.0 1.0 3.0 New initial 4.0 0.0 2.0 Change -2x +x +x Final 4.0-2x +x 2.0-x K= PH2 PCl2 / PHCl2 3.2 x 10 ^-34 = x(2....

Aileen Nguyen 3N

You would use molarity as the units when calculating equilibrium concentrations!

Aileen Nguyen 3N

Aileen Nguyen 3N

Accurately portrays my love life

Aileen Nguyen 3N

HClO is an oxoacid and in oxoacids hydrogen has to be bonded with oxygen.

Aileen Nguyen 3N

A photon checks into a hotel and is asked if he needs an help with his luggage.
He replies " No, I'm traveling light."

Aileen Nguyen 3N

3D is only lower in energy than 4s after the 4s orbital is occupied

Aileen Nguyen 3N

I solved this problem using the Rydberg equation instead of the bohr frequency condition.

You solved the n1=1 so for a line that's 102.6 nm you use the formula v=c/(lambda) = 2.922 x10^15 s^-1

Then use the Rydberg to solve for n2 which would give you n2=3

Aileen Nguyen 3N

Do you have 11 protons? Cause your sodium fine.

CHEMISTRY COMMUNITY

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