CHEMISTRY COMMUNITY

Samuel Tzeng 1B

I'm pretty sure it should be given

Samuel Tzeng 1B

I think you just have to know if it's pure or not

Samuel Tzeng 1B

He hasn't explicitly stated when we need to submit the final yet

Samuel Tzeng 1B

It means the catalyst is not consumed during the reaction

Samuel Tzeng 1B

Yes at higher temperatures, molecules have higher kinetic energy

Samuel Tzeng 1B

The SI units for time is seconds

Samuel Tzeng 1B

The total entropy is the entropy of the surroundings plus the entropy of the system

Samuel Tzeng 1B

I think different types of catalysts use different methods to lower the activation energy

Samuel Tzeng 1B

No a zero order reaction won't be influenced by a catalyst

Samuel Tzeng 1B

The reduction table in the back of the book shows you which metals will be reduced

Samuel Tzeng 1B

The instantaneous rate is the rate at any given time, the rate is given by the tangent line at that time

Samuel Tzeng 1B

The oxidation number change in a chemical reaction show what is being reduced and oxidized

Samuel Tzeng 1B

the rate constant is always positive

Samuel Tzeng 1B

it's the slowest step in a reaction

Samuel Tzeng 1B

You'll likely need to be given the rate law or experimental data

Samuel Tzeng 1B

Increasing the size of the metal does not change the cell potential

Samuel Tzeng 1B

If it's a cathode solution (like on test 2) then diluting it will decrease cell potential

Samuel Tzeng 1B

You should know that enzymes can act as catalysts

Samuel Tzeng 1B

0.44 is not the rate constant but the actual rate so you would multiply the number of oxygen's by 0.44

Samuel Tzeng 1B

It relates the rate constant k to activation energy Ae and temperature T

Samuel Tzeng 1B

You'll have to find the order of the individual reactants then add them up

Samuel Tzeng 1B

1/[A] vs time

Samuel Tzeng 1B

You will most likely not need to determine which step is the slow step

Samuel Tzeng 1B

Probably during dicussion

Samuel Tzeng 1B

Since the equation has 3 moles of O2 on the reactant side, the rate at which oxygen reacts will be 3 times the rate of reaction
Similarly, since the equation has 2moles of H2O on the product side, the rate at which water forms will be 2 times the rate of reaction

Samuel Tzeng 1B

it is the rate constant of the reverse reaction

Samuel Tzeng 1B

Isothermal systems have constant temperature, this does not mean the same as reversible

Samuel Tzeng 1B

if delta G is negative the reaction is spontaneous

Samuel Tzeng 1B

finding the temperature at delta G=0 gives the minimum temperature for the reaction to be spontaneous

Samuel Tzeng 1B

q = deltaH at constant pressure

Samuel Tzeng 1B

use q=mCdeltaT when you're given the mass and use q=nCdeltaT when you're given the moles

Samuel Tzeng 1B

If work is done on the system it is positive, if the system is doing work its surroundings work is negative

Samuel Tzeng 1B

P is pressure, V is volume, n is # of moles of gas, R is the gas constant, T is temperature

Samuel Tzeng 1B

Exothermic

Samuel Tzeng 1B

Pressure in an open beaker is always constant since there isn't a volume to drop or increase pressure

Samuel Tzeng 1B

I don't think so.

Samuel Tzeng 1B

if temp increases then the reaction will shift to the left, if temp decreases then the reaction will shift to the right

Samuel Tzeng 1B

I believe they'll give partial credit if you use the right steps

Samuel Tzeng 1B

pKa + pKb = 14, then use pKb to solve for Kb

Samuel Tzeng 1B

the reaction will shift left

Samuel Tzeng 1B

I'm pretty sure the thermodynamics section

Samuel Tzeng 1B

A stronger acid will have a lower pKa

Samuel Tzeng 1B

The test is during your discussion session on week 3

Samuel Tzeng 1B

Lewis acids are proton acceptors while Bronsted acids are proton donors

Samuel Tzeng 1B

It helps us predict how a reaction will react when a physical change occurs

Samuel Tzeng 1B

Yes they're inversely proportional

Samuel Tzeng 1B

Writing (n/V)RT is the same as nRT/V

Samuel Tzeng 1B

If the ratio of products to reactants is larger than that of the equilibrium constant, then Q>K

Samuel Tzeng 1B

Since the activity of solids and liquids are equal to 1, they aren't included in the calculation of Q and K, this leaves only gases and aqueous substances

Samuel Tzeng 1B

If all other conditions remain the same, the equilibrium constant won't change

Samuel Tzeng 1B

If the forward reaction is favored, then reactants will have a negative change and products will have a negative change, and vice versa when the reverse reaction is favored

Samuel Tzeng 1B

If it has the structure binding site-spacer-spacer-binding site then it is typically chelating

Samuel Tzeng 1B

Elements with the third energy shell and on since the d-orbital can also have electrons

Samuel Tzeng 1B

You don't need to add anything for positive ions or neutral molecules

Samuel Tzeng 1B

they can act as both an acid and a base

Samuel Tzeng 1B

pi bonds in double and triple bonds don't hybridize

Samuel Tzeng 1B

HNO3 is stronger since it’s conjugate base NO3- is more stable than the conjugate base of HNO2

Samuel Tzeng 1B

Conjugate acids are bronsted bases with an H+
Conjugate vases are bronsted acids without an H+

Samuel Tzeng 1B

You can represent end to end by drawing 2 infinity signs next to each other and represent side to side by drawing two 8’s next to each other

Samuel Tzeng 1B

I believe so

Samuel Tzeng 1B

Yes the concentration of H+ would be double that of H2SO4

Samuel Tzeng 1B

The transition metal cation comes before the anion

Samuel Tzeng 1B

9C.1: Coordination Complexes
9C.2: The Shapes of Complexes

Samuel Tzeng 1B

chrisleung wrote:Yes, they are two names for the same geometry, for molecules with configuration AX2E or AX3E

AX3E would be trigonal pyramidal, AX2E2 would be bent

Samuel Tzeng 1B

The larger molecule will have stronger London forces

Samuel Tzeng 1B

NO is linear, I think it's slightly polar

Samuel Tzeng 1B

I don't think we would need to know this as it's either super uncommon or not possible

Samuel Tzeng 1B

it would be between the oxygen atoms of one molecule and the hydrogen of another

Samuel Tzeng 1B

since the C-H bonds and C-Cl bonds have differing polarities, the molecule will be polar and will have dipole-dipole interactions

Samuel Tzeng 1B

the dipole between O-H bonds will cause the H to be slightly negative, while the dipole between the Se-O bonds will make the O slightly positive, this causes the H atoms of one molecule to be attracted to the O atoms of another, creating dipole-dipole interactions

Samuel Tzeng 1B

London, Dipole-dipole, dipole-induced dipole, ion-dipole, h-bondings, ion-induced dipole

Samuel Tzeng 1B

Dipole moments cause parts of a molecule to be charged, this leads to positively charged parts of one molecule being attracted to negatively charged parts of another, leading to intermolecular forces.

Samuel Tzeng 1B

I think they're the same thing

Samuel Tzeng 1B

CH4 and CCl4 won't have dipole-dipole interactions since they don't have permanent dipole moments

Samuel Tzeng 1B

dipole moments occur when electrons are unequally shared, this can affect intermolecular forces as it makes parts of the molecule positively/negatively charged

Samuel Tzeng 1B

Look at the difference in electronegativity between atoms and look for cases of hydrogen bonding

Samuel Tzeng 1B

It shows that they're bonded through ionic bonds rather than covalent bonds

Samuel Tzeng 1B

a) [Ar]
b) [Ar]3d^10 4s^2
c) [Kr]4d^5
d) [Ar]3d^10 4s^2

Samuel Tzeng 1B

a) [Ar]3d^10
b) [Xe]4f^14 5d^10 6s^2
c) [Ar]3d^10;
d) [Xe]4f^14 5d^10

Samuel Tzeng 1B

N can't be double bonded to Cl since Cl is a halogen and halogens don't form double bonds

Samuel Tzeng 1B

Br has three bonds (one with each F) and two pairs of lone electrons, while the F's have 3 pairs of lone electrons

Samuel Tzeng 1B

E represents lone pairs on the central atom

Samuel Tzeng 1B

Since double bonds are stronger, they pull the atoms closer together and are therefore shorter.

Samuel Tzeng 1B

a.2 bonding pairs, 2 lone pairs
b.2 bonding pairs, 4 lone pairs
c.3 bonding pairs, 2 lone pairs
d.5 bonding pairs, 1 lone pair

Samuel Tzeng 1B

a.Mg3As2
b.In2S3
c.AlH3
d.H2Te
e.BiF3

Samuel Tzeng 1B

a.[Ar]
b.[Kr]5s^2 4d^10
c.[Xe]
d.[Ar]3d^8

no unpaired electrons for all

Samuel Tzeng 1B

E=pc=mc^2 only applies to em radiation

E=mv^2/2 is used for kinetic energy of an object

Samuel Tzeng 1B

Electrons are paired if they're in the same orbital and parallel if they're in separate orbitals and have the same spin

Samuel Tzeng 1B

I used the ones from the chem website

Mass of neutron = 1.674 927 × 10^−27 kg

Mass of proton = 1.672 622 × 10^−27 kg

Samuel Tzeng 1B

6 – 8pm, Wednesday November 6

Samuel Tzeng 1B

1 eV = 1.602 × 10^-19 J

Samuel Tzeng 1B

wavelength = h/(mv)

if wavelength is shorter than 10^-15m it is undetectable

Samuel Tzeng 1B

since the speed of light is always the same c = 3.0*10^8 and c=wavelength * frequency, as wavelength decreases frequency must increase and the same idea applies to energy since frequency and energy are directly related

Samuel Tzeng 1B

find number of joules from the light
32J/s * 2.0s = 64J

then divide joules by the energy per photon
64J/(4.7*10^-19) = 1.4*10^20

finally divide by Avogadro's #

1.4*10^20/(6.022*10^23) = 2.3*10^-4 mol photon

Samuel Tzeng 1B

I think it's reasonable as long as it's lower than the speed of light

Samuel Tzeng 1B

E(photon)=1.66*10^-17+KE(electron)= 1.66*10^-17J + 0.5 * (9.11*10^-31kg) * (3.6*10^6m/s)^2 = 2.25*10^-17J

then you can use the energy of the photon to find the wavelength using E=hv

Samuel Tzeng 1B

I don't think so

Samuel Tzeng 1B

If your experiment is inaccurate yet precise it may indicate that there is a problem with the devices/methods you are using to conduct your experiment

Samuel Tzeng 1B

Darren Nguyen 1F wrote:
Samuel Tzeng 1H wrote:Molar mass of XA = total mass - mass of H2O

How do I get total mass?

It's the molar mass of HA + XOH = (250+125)

Sorry I meant total molar mass - molar mass of H2O

Samuel Tzeng 1B

Start off by finding moles of tin and oxygen

Sn: 1.50g Sn / 118.71g/mol = .01264 mol Sn
O: (28.35g - 26.45g - 1.50g) / 16.00g/mol = 0.025 mol O

With this you can tell the empirical formula is SnO2

The oxide would be Tin(IV) Oxide since each oxygen has a -2 charge so the tin must have a +4 charge

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