CHEMISTRY COMMUNITY

Jim Brown 14B Lec1

I don't have 7th edition but check to see if there's any mL because that would account for the discrepancy.

Jim Brown 14B Lec1

A smaller equilibrium constant means at equilibrium the reactants are more favored, or it is less likely to decompose. This means that the lower the equilibrium constant, the more stable.

Jim Brown 14B Lec1

It is either stated explicitly or has the smaller k value

Jim Brown 14B Lec1

Use the formula t1/2= [A]/2k to solve for k, then modify the formula for the other proportions.
eg.
t1/16= 15[A]/16k
t1/4= 3[A]/4k
etc

Jim Brown 14B Lec1

It is included in the unique rate law but in the overall rate law it is only included if it is greater than zero. If the reactant is zero order then it will be to the power of zero and equal to 1.

Jim Brown 14B Lec1

If it doesn't specify then it doesn't matter. It's usually Kc.

Jim Brown 14B Lec1

If there's a change in pressure/volume, the entropy is dependent on Temperature. If there's a change in Temperature, the entropy is dependent on volume. You've gotta do one before the other or else there's just no calculation (that a 14B student knows how to do, idk about actual chemists).

Jim Brown 14B Lec1

Thank you! That clears up a lot :)

Jim Brown 14B Lec1

Q: A balloon expands from 3.3L at 298K to 9.2 x 105 L at 333K. What is ΔS ?
can someone post steps on how to solve?

Jim Brown 14B Lec1

P: atm
V: L
n: mol
R: J/k.mol
T: K

Jim Brown 14B Lec1

It depends on the constants and what needs to cancel. Both values should be provided so you just have to look at the question. It should usually be 8.314 J/k.mol

Jim Brown 14B Lec1

What day is your discussion, the answers are different depending on the day.

Jim Brown 14B Lec1

It will say in the question. Isothermal, same T, moles have to change. Increased Temperature, T changes.

Jim Brown 14B Lec1

Use the Boltzman constant when finding statistical Entropy
∆S=K•lnW

Jim Brown 14B Lec1

Heat capacity

Jim Brown 14B Lec1

Jim Brown 14B Lec1 wrote:
Jim Brown 14B Lec1 wrote:I think you need the bond enthalpy for ethane, then use it to heat water to 100º C and to reach its ∆Hvap

Enthalpy of combustion*

nvm this [censored] is rough idk how to do it :'(

Jim Brown 14B Lec1

Jim Brown 14B Lec1 wrote:I think you need the bond enthalpy for ethane, then use it to heat water to 100º C and to reach its ∆Hvap

Enthalpy of combustion*

Jim Brown 14B Lec1

I think you need the bond enthalpy for ethane, then use it to heat water to 100º C and to reach its ∆Hvap

Jim Brown 14B Lec1

can someone explain 6(d)? I tried using the additive equation for standard enthalpy of a reaction but I don't think I'm inputting the correct value for it or if that's even a possible way to get the answer We know the total enthalpy needs to equal 42, so add the enthalpies of formation for all the ...

Jim Brown 14B Lec1

Yes you can use this for a change in T

Jim Brown 14B Lec1

Anyone got the supplied equations? :)

Jim Brown 14B Lec1

A negative Delt. H means a -q (heat loss)

Jim Brown 14B Lec1

The problem will tell you whether it is constant temperature or pressure, either explicitly or through the other information it gives (i.e. ethalphy indicates constant pressure)

Jim Brown 14B Lec1

8.9 An ideal gas in a cylinder was placed in a heater and gained 5.50 kJ of energy as heat. If the cylinder increased in volume from 345 mL to 1846 mL against an atmospheric pressure of 750. Torr during this process, what is the change in internal energy of the gas in the cylinder?

Jim Brown 14B Lec1

Those are used for Entropy calculation at constant Volume or Pressure. Lavelle didn't explain why other than for 3/2 is a result of there being 3 dimensions and velocity being "1/2"

Jim Brown 14B Lec1

The reaction is not isolated and takes place in a heat reservoir. This way, even if the reaction is exothermic, the overall temperature will remain the same at the beginning and the end of the reaction.

Jim Brown 14B Lec1

2015 Chem class coming in CLUTCH

Jim Brown 14B Lec1

2015 Chem class coming in CLUTCH

Jim Brown 14B Lec1

My handwriting is terrible but

Jim Brown 14B Lec1

You only have to take the bond enthalpy of the bonds broken and the new bonds formed

Jim Brown 14B Lec1

Both use J/kg

Jim Brown 14B Lec1

If Gibbs Free Energy is negative, the reaction is spontaneous (exergonic). If Gibbs is positive, the reaction is nonspontaneous (endergonic).
The change in Gibbs is equal to the change in enthalpy minus the change in entropy times T.

Jim Brown 14B Lec1

There are 8 common strong bases worth committing to memory. It consists of Alkali and Alkaline Earth metals paired with Hydroxide Ions. LiOH - lithium hydroxide NaOH - sodium hydroxide KOH - potassium hydroxide RbOH - rubidium hydroxide CsOH - cesium hydroxide Ca(OH)2 - calcium hydroxide Sr(OH)2 - s...

Jim Brown 14B Lec1

In Lecture Lavelle said we should just commit them to memory. The staircase that is similar to that of the metalloids can be a good guiding line for which oxides this includes.

Jim Brown 14B Lec1

All Brönstead acids/bases are Lewis Acids/Bases, but something can be considered a lewis acid even if it might not qualify as a Brönstead acid. This is because a Brönstead acid needs to donate a proton (needs to contain hydrogen) while a Lewis Acid only needs to be able to accept the lone pair.

Jim Brown 14B Lec1

This is only possible with transition metals. Transition metals lose electrons more easily, and they do not lose them based on completing an octet because they utilize the d-orbital.

Jim Brown 14B Lec1

You've just gotta memorize which elements bond to make amphoteric compounds. The list is similar (but not identical) to the list of metalloids.

Jim Brown 14B Lec1

Pi-bonds do not allow free rotation, if the molecule is rotated, the pi-bond will break. This results in rigidly structured molecules.

Jim Brown 14B Lec1

A single bond and a double bond are both 1 region of electron density. Double bonds will determine the different hybridizations and the shape of the molecule, but even with these added it still follows the same rules as the single region of electron density in the VSEPR model.

Jim Brown 14B Lec1

: sodium chloride

Jim Brown 14B Lec1

There aren't different bond angles, all the angles are the same. You just have to memorize the general rules for the specifics of each shape and apply it to the various molecules that follow that structure.

Jim Brown 14B Lec1

Not sure but it's pretty easy. Its the character A followed by X with a subscript of the number of bonds with another atom, then E with a subscript of the number of lone pairs.

Jim Brown 14B Lec1

Rosha Mamita 1E wrote:Yes, though the shape will be different and would not be called linear

This is true for most electron geometries and will result in a bent shape, but you can still get a linear shape when there are 5 areas of electron density and 3 of them are lone pairs

Jim Brown 14B Lec1

Yes, a trigonal bipyramidal electron geometry with 3 lone pairs will give you a linear shape with 180º angles

Jim Brown 14B Lec1

You want to try and minimize the formal charge

Jim Brown 14B Lec1

Molecular geometry takes into account only the atoms surrounding the central atom. This includes shapes like bent, trigonal pyramidal, square planar, etc. Electron-pair geometry is based on the number of electron density fields. This includes only the basic geometries (linear, trigonal planar, tetra...

Jim Brown 14B Lec1

It depends on the number of electron fields. See how the T-shaped has five and the square planar has 6? It's worth committing to memory.

Jim Brown 14B Lec1

Electron affinity increases across a period and decreases down a group.

Jim Brown 14B Lec1

The Noble gasses are nonreactive. Neon has a greater effective charge, but its octet is complete, so it won't pick up any electrons, where chlorine will in order to complete the octet.

Jim Brown 14B Lec1

Variable valency just refers to chemicals (typically [if not always?] a metal) that form multiple ions with different charges. (Chemicals like iron (II) and iron (III), tin (III) and tin (IV), etc.)

Jim Brown 14B Lec1

: salttime

Jim Brown 14B Lec1

l corresponds to the orbital (s, p, d, f, etc.) Orbitals increase by two for each new shape (s has 1, p has 3, d has 5, etc) The l quantum number 2 represents a p orbital, so there are 3 subshells. The l quantum number 3 represents a d orbital, so there are 5 subshells. The l quantum number 1 repres...

Jim Brown 14B Lec1

Ryberg's constant is 3.29e+15

Jim Brown 14B Lec1

Use wavelength to calculate E then work backwards from total Energy. Set up the problem E=Rh(1/(n[final])^2 - 1/(n[initial])^2). Divide everything by hR. You end up with .888 = 1/(n[final])^2 - 1/(n[initial])^2. Test values of integers and you'll end up with a change from 1 to 3.

Jim Brown 14B Lec1

Convert wavelength to energy by using the energy formula E=hν. Because νλ is equal to the speed of light this can be rewritten as E=hC/λ. You now have a formula that you can plug in values for and get the energy for one the emission of one atom (3.37 e-17). Next, convert the mass of sodium into atom...

Jim Brown 14B Lec1

I asked Lavelle, he said a pdf is fine. The textbook is to help you learn so whatever works for you.

Jim Brown 14B Lec1

Is water wet? @Dr. Lavelle

Jim Brown 14B Lec1

The "R" is a constant like "h", "C" or Avogadro's number. It was derived through experimentation.

Jim Brown 14B Lec1

I meant like if I made a post during week zero (24th-30th) does that count as points for week one?

Jim Brown 14B Lec1

Do posts made during week zero count as posts for week one?

Jim Brown 14B Lec1

The 4S orbital is located closer to the nucleus, so the energy level is higher and will fill out first. It's easy to tell which orbitals have higher energy levels by looking at the organisation of the elements filling those orbitals on the periodic table.

Jim Brown 14B Lec1

The method I use to solve limiting reactant problems is by converting the reactants' mass into moles. I don't think this is impossible because once you have converted the reactants into moles, you compare the two. The reactant with less moles is the limiting reactant. I believe you can solve these ...

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