CHEMISTRY COMMUNITY

Amanda Lin 2I

Catalysts are regenerated. They are used up, then produced.

Amanda Lin 2I

The rate limiting step is the slowest step, and it determines the rate of the overall reaction.

Amanda Lin 2I

The one with a higher, more positive Ecell value is usually reduced while the one with a lower, more negative Ecell value is usually oxidized.

Amanda Lin 2I

If there is, it would be rare because the molecules would need to collide simultaneously.

Amanda Lin 2I

Something is an ideal gas if it does not take up volume and has no intermolecular attractive forces.

Amanda Lin 2I

A catalyst increases k, but it does not affect K because the ratio of products to reactants remains the same.

Amanda Lin 2I

The exponent is the order of the reaction and is not related to the coefficient.

Amanda Lin 2I

Elementary reactions are the steps that make up the overall reaction.

Amanda Lin 2I

K and Cl dissociate in water, so you just need to work with the metal ions.

Amanda Lin 2I

I think either way is fine as long as the anode is on the left and the cathode is on the right.

Amanda Lin 2I

Because the pH is higher than the pKa, CH3COOH will be deprotonated, forming CH3COO-, which has a charge of -1.

Amanda Lin 2I

ΔG° = ΔH° - TΔS°
ΔH° and ΔG° would be the most similar when ΔS° is closest to 0. Going from a solid to another solid would have the smallest change in entropy, so A is the correct answer.

Amanda Lin 2I

If there is no conducting solid, then you use an inert electrode.

Amanda Lin 2I

For concentration cells both sides have the same components, just different concentrations.

Amanda Lin 2I

The amount of energy that leaves the system is -w.
wmax = ΔG = -nEF

Amanda Lin 2I

I don’t think the ions in the salt bridge can run out.

Amanda Lin 2I

Without the salt bridge, the cathode side will become too negative, and the electrons will not flow

Amanda Lin 2I

Fe3+ and Fe2+ are not separated by | because there is no phase change.

Amanda Lin 2I

The oxidation number is zero because it only consists of one element.

Amanda Lin 2I

ΔG° is the Gibbs free energy at standard conditions. You can find ΔG using ΔG=ΔG°+RTlnQ.

Amanda Lin 2I

For an adiabatic system there is no heat transfer so q=0 and ΔU=w.

Amanda Lin 2I

If the products side has moles of gas than the reactants side, then the reaction has done expansion work.

Amanda Lin 2I

This equation can be used to calculate the change in entropy at a constant temperature.

Amanda Lin 2I

You just count the coefficients to determine the equilibrium shift.

Amanda Lin 2I

An increase in pressure causes a decrease in entropy. An increase in volume causes an increase in entropy. An increase in temperature causes an increase in entropy.

Amanda Lin 2I

Because there is already more disorder at a higher temperature, there will not be as great of a change in entropy.

Amanda Lin 2I

It’s a conversion factor. 1 L.atm=101.325 J.

Amanda Lin 2I

There’s a constants and equations sheet given, and it’s also posted on the class website.

Amanda Lin 2I

Standard molar entropy is the entropy of 1 mole of substance at 1 atm.

Amanda Lin 2I

You would use w=-PΔV, solve for w, then plug w into ΔU=q+w.

Amanda Lin 2I

Heat and work are not state properties because they depend on the path taken to reach that state.

Amanda Lin 2I

If the system does not exchange matter or energy with its surroundings, it is isolated.

Amanda Lin 2I

The standard state of a substance is its phase at 25°C and 1 atm.

Amanda Lin 2I

I believe molar heat capacity is the amount of heat needed to increase the temperature of 1 mole by 1°C, and specific heat capacity is the amount of heat needed to increase the temperature of 1 gram by 1°C.

Amanda Lin 2I

Reactions can still occur within the system without exchanging matter or energy with its surroundings.

Amanda Lin 2I

Yes, they are different. Also, if the pressure is increased by adding an inert gas, there is no change in concentration.

Amanda Lin 2I

If K is less than 10^-3, and the percent ionization is less than 5%, then you can approximate.

Amanda Lin 2I

Ka is the acid dissociation constant. Kb is the base dissociation constant.

pKa = -log(Ka)
pKb = -log(Kb)

Amanda Lin 2I

When you reduce a reactant, the equilibrium shifts toward the reactants in order to maintain the ratio of products to reactants.

Amanda Lin 2I

The actual solution is 250 mL. You would need to calculate the molarity of the actual solution and plug it into -log(x).

Amanda Lin 2I

Q is calculated like K and is a ratio, so it would be not have units.

Amanda Lin 2I

Because the reaction is endothermic, it favors the formation of the product, X.

Amanda Lin 2I

Endothermic reactions favor the formation of products.

Amanda Lin 2I

Increasing the concentration of reactants will cause more products to be formed in order to maintain equilibrium. The ratio of products to reactants does not change.

Amanda Lin 2I

Yes, K does not change.

A decrease in volume with more moles of gas on the reactant side will cause the reaction to produce more products.
A decrease in volume with more moles of gas on the product side will cause the reaction to produce more reactants.

Amanda Lin 2I

For part c, Cl2 is more stable than F2 because it has a smaller equilibrium constant.

Amanda Lin 2I

I don't think we will have to memorize the conversions between units of pressure. If we do have to convert them, they should be on the constants and equations sheet.

Amanda Lin 2I

You can simplify the expression if it is a cubic function and K is small.

Amanda Lin 2I

I think it was K<10^-4.
K<10^-3 is used to determine that equilibrium favors the reactants.

Amanda Lin 2I

Removing some of the product will cause the reaction to make more product in order to reach equilibrium again.

Amanda Lin 2I

Iron bound to a porphyrin ligand forms a heme complex, which is a component of myoglobin.

Amanda Lin 2I

It lowers the energy of the molecule, and the molecule has alternating single and double bonds.

Amanda Lin 2I

You use ferrate if the compound is an anion. You use iron if the compound is neutral or a cation.

Amanda Lin 2I

The affinity of CO binding to hemoglobin is stronger than the affinity between O2 and hemoglobin.

Amanda Lin 2I

The coordination sphere includes the central metal cation that the ligands attach to.

Amanda Lin 2I

en is ethylenediamine or NH2CH2CH2NH2. It is on the chart that Dr. Lavelle gave us for naming coordination compounds.

Amanda Lin 2I

He talked about cisplatin or cis-diammine-dichloro-platinum(II).

Amanda Lin 2I

A coordinate covalent bond is when a lone pair of electrons from one atom is shared with another atom.

Amanda Lin 2I

A complex consists of a central metal ion that ligands attach to through coordinate covalent bonds.

Amanda Lin 2I

A complex consists of a central metal ion that ligands attach to through coordinate covalent bonds. A coordination compound consists of at least one complex. However, I have also seen these two used interchangeably.

Amanda Lin 2I

It forms a ring when two or more of the atoms of the ligand bind to a central metal atom.

Amanda Lin 2I

A coordination sphere is a central atom with ligands bonded to it.

Amanda Lin 2I

A polydentate ligand is a ligand that has multiple atoms with lone pairs that can be used to bond to a central metal atom.

Amanda Lin 2I

I think ligands are listed in alphabetical order.

Amanda Lin 2I

Look at the arrangement of the amine groups. Only b would allow for the two amine groups to bond to the same metal center.

Amanda Lin 2I

Dipole-dipole is the interaction between the partially positive end of a polar molecule and the partially negative end of another polar molecule.

Amanda Lin 2I

That is probably the most efficient way because limiting reactant problems often ask for the theoretical yield of the product.

Amanda Lin 2I

Yes, however, nonpolar molecules can have temporary dipoles.

Amanda Lin 2I

Oxygen with two bonds has a formal charge of 0.
Nitrogen with three bonds has a formal charge of 0.
Carbon with four bonds has a formal charge of 0.

Amanda Lin 2I

505106414 wrote:Can someone explain the exception for oxygen and nitrogen? It was on the midterm and I got it wrong.

Nitrogen has 3 e- in its p orbital. Oxygen has 4 e- in its p orbital, which results in an electron pair. The electron-electron repulsion from this pair causes a lower ionization energy.

Amanda Lin 2I

A molecule with a dipole moment is polar. A molecule with no dipole moment is nonpolar.

Amanda Lin 2I

Hydrogen bonds with atoms with high electronegativity, such as N, O, and F.

Amanda Lin 2I

An atom is polarizable if it has a larger radius and less electronegativity.

Polarizing power is the ability of a cation to distort an anion by pulling electrons away from the anion. Polarizability is the ability to pull those electrons toward the anion.

Amanda Lin 2I

Ionization energy is minimum amount of energy needed to remove an electron from a neutral atom. Electronegativity is a measure of an atom's tendency to attract a pair of electrons. Both ionization energy and electronegativity generally increase as you go left to right on the periodic table and as yo...

Amanda Lin 2I

Dissociation energies are always positive because energy is absorbed in order to break the bond.

Amanda Lin 2I

The negative charge would be on the electronegative atom, and because oxygen is more electronegative than chlorine, it is more stable if oxygen has a formal charge of -1.

Amanda Lin 2I

You would divide it by Avogadro's constant.

Amanda Lin 2I

They are delocalized because they move between two atoms instead of staying with one atom.

Amanda Lin 2I

I would say just read the question carefully. Identify what it's asking you to find and see what information is already given.

Amanda Lin 2I

I believe it goes around the one with the overall charge.

Amanda Lin 2I

Gallium has 3 valence electrons, so it is more likely to lose those 3 electrons than to gain 5 electrons.

Amanda Lin 2I

The ground state electron configuration for Ru is [Kr]4d^7 5s^1. However, because it is asking for Ru3+, you would need to remove 3 electrons, leaving you with [Kr]4d^5.

Amanda Lin 2I

I find the number of valence electrons for the element. Then, I subtract the value of the positive charge from the number of valence electrons.

Amanda Lin 2I

1 Angstrom = 10^-10 m
Angstroms are the units for measuring the atomic radius.

Amanda Lin 2I

The s sublevel has 1 orbital, so it can have a max of 2 e-. The p sublevel has 3 orbitals, so it can have a max of 6 e-. The d sublevel has 5 orbitals, so it can have a max of 10 e-. The f sublevel has 7 orbitals, so it can have a max of 14 e-.

Amanda Lin 2I

Wave functions are the mathematical representation of orbitals.

Amanda Lin 2I

If the position is more precise, the momentum is more uncertain. If the momentum is more precise, the position is more uncertain. The equation calculates this uncertainty.

Amanda Lin 2I

1p cannot exist because the n=1 shell only has s orbitals.

Amanda Lin 2I

Paired electrons have opposite spins while parallel electrons have the same spin.

Amanda Lin 2I

The wavelength would need to be in meters, but the mass of the neutron can stay in kg because that is the SI unit for mass.

Also, 1 J = 1 kg*m^2*s^-2, so the units would cancel out leaving you with m/s for your final answer.

Amanda Lin 2I

I think it covers chemical bonding as well.

Amanda Lin 2I

Sig figs should be applied to the final answer because rounding during your calculations could affect your results.

Amanda Lin 2I

I believe elements can only have line spectrums because electrons have specific energy levels. When electrons go down an energy level after being excited, they release a specific wavelength of light.

Amanda Lin 2I

The mass of a proton and a neutron is 1.673*10^-27 kg, the velocity is 2.75*10^5 m/s, and Planck's constant is 6.626*10^-34 J.s. Plug in to De Broglie's equation: (6.626*10^-34 J.s) / ((1.673*10^-27 kg)(2.75*10^5 m/s)) = 1.44*10^-12 m The wavelength of the proton and the neutron are the same because...

Amanda Lin 2I

The velocity has to be in m/s instead of km/hr.

(125 km/hr)(1000 m/1 km)(1 hr/3600 s) = 34.72 m/s

(6.626*10^-34)/(275*34.72) = 6.94*10^-38 m

Amanda Lin 2I

Isabella Dal Porto 1I wrote:Why is the Rydberg equation only applicable towards hydrogen atoms?

The Rydberg equation only works for atoms with one electron. If there is more than one electron, the formula will produce incorrect results.

Amanda Lin 2I

Yes, you would use a 100 g sample, so there would be 63.15 g C, 5.30 g H, and 31.55 g O. You would then convert these to moles.

Amanda Lin 2I

Rounding for sig figs is usually done at the end. If you round during the calculations, it could affect your final answer.

Amanda Lin 2I

For problems asking to calculate the percent yield, the actual yield is usually provided in the prompt.

Amanda Lin 2I

There is an order to the elements in chemical formulas due to the Hill System. Carbon comes first, next is hydrogen, and then it is the rest of the elements alphabetized. There are also exceptions to this rule.

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