CHEMISTRY COMMUNITY

Kevin Tang 4L

I don't know which part you are stuck on but I can work through some of part c: First lets find the oxidation states: Cr3+ goes to CrO42- . 3+ charge for Cr3+ and +6 charge for Cr in CrO42- This means this is oxidized MnO2 goes to Mn2+. +4 charge for Mn in MnO2 and 2+ charge in Mn2+. This is reduced...

Kevin Tang 4L

I believe in lecture he said that Mercury will only be used as an inert conductor if it present in the reaction initially but I"m not entirely sure, sorry. But I believe Platinum should be used if an inert conductor is needed unless Mercury is part of the reaction. This is correct. When in dou...

Kevin Tang 4L

Usually the one with the more positive E value, the reduction potential, the greater chance that it is going to be reduced, and thus will be the cathode. The one with the lower E value will most likely be the one oxidized, being the anode.

Kevin Tang 4L

Jonathan Christie 1I wrote:What would be the reason to use Hg over Pt?

The only case you use Hg over Pt is if Hg is involved in the re-dox reaction. Like for example Hg2+ 2e- -> Hg(l). Then the Hg(l) can be used as the electrode because it is already part of the overall reaction. In this case you don't need to add Pt(s).

Kevin Tang 4L

Platinum is an inert electrode. It is used when theres is no metal present.

Mercury(l) is one non-solid that is an electrode so when it is used in a re-dox reaction, you don't need to use Platinum. (Ignore this part if you don't understand).

Kevin Tang 4L

Because the reactants are used up, thus the negative sign and the products are created, thus being positive.

Hope this helps and correct me if I'm wrong. Thanks!

Kevin Tang 4L

Wouldn't a zero order reaction have the rate law be: rate=k?
It would still have a rate law, it would just be the constant k.

Kevin Tang 4L

Yes, because we know that [C] is a zeroth order reaction, it has no effect on the overall rate equation.
Rate=k[A][B]^2, C is not present, so it doesn't affect the rate, regardless of its initial concentration.

Kevin Tang 4L

ΔG = ΔH - TΔS represents the change in Gibbs free energy. Spontaneous reactions are exergonic, when ΔG is less than 0, showing a release in free energy. Reactions that are not spontaneous, on the other hand, are endergonic when ΔG is greater than 0.

Kevin Tang 4L

Using the equation ΔG = -RTlnK.
ln 1=0 thus when K=1, ΔG = 0

Kevin Tang 4L

When a metal is oxidized and gains electrons, it changes from solid form to aqueous. This change in state corrodes the metal.

Kevin Tang 4L

Jchellis 1I wrote:Was this covered in class? Or were we just expected to know it?

Don't recall if this particular type of problem was explicitly covered in class, but we should know this based on the information that Lavelle has given us. It is also in our homework which suggests we are expected to know it.

Kevin Tang 4L

How do I determine if a reaction is spontaneous or non spontaneous? E.g. is water melting spontaneous? How do I think through these different scenarios given? Spontaneous means the Gibbs Free Energy is less than 0 and not spontaneous means Gibbs Free Energy is greater than 0. Ice will melt spontane...

Kevin Tang 4L

If you want more practice for this type of question 14.25 in the 6th edition of the book is a good resource. It gets confusing for me because you have to reverse the E value and understand if the question is asking for the strength as a reducing or oxidizing agent and whether it is asking the streng...

Kevin Tang 4L

Why do we use Platinum instead of some other metal? I know we can use C(graphite) too.

Kevin Tang 4L

Yes, when delta H is positive and delta S is positive, it will be spontaneous at high temperatures and nonspontaneous at low temperatures. This is because when delta S is positive, -T Delta S will be negative. If the T in the equation is large enough, then the whole combined -T Delta S can be greate...

Kevin Tang 4L

Sorry but so Monday's (the 25th) lecture won't be included with the material for test 2? No, you can check his Chem 14B also for this information: "Test 2 covers all topics on Gibbs free energy and electrochemistry up to but not including the Nernst equation (which is all the new material cove...

Kevin Tang 4L

It is because it is an acidic solution. If it is a basic solution, you have to use OH-.

Hope this helps and correct me if I'm wrong.

Kevin Tang 4L

I believe a reducing agent is one that gets oxidized as it loses electrons. The oxidizing agent, on the other hand, is the one that gets reduced as it gains electrons.

Correct me if I'm wrong. Hope this helps.

Kevin Tang 4L

For #10, should both heat capacities (of ice and of water be the same), or should we use the heat capacity of ice on the left side of the equation. I remember at the review he used 4.184 for both, but shouldn't we take into account C of ice as well? Because the Ice is at 0C, it will melt into water...

Kevin Tang 4L

can someone explain the reasoning behind the three steps to solve for enthalpy of #12B So we are trynna calculate the DeltaHrxn(total) of the human body at 37C and we know the DeltaHrxn=-2756kJ at 200C First, we need to heat up the reactants C6H12O6 and 6O2 to 200 which will be your DeltaH1 Then we...

Kevin Tang 4L

Could someone explain for 4a how he got 5.0x10^3?? When I worked it out on my own I got -498 and all my elements cancelled out. Your answer is correct. He gave the answer DeltaHrxn= -5.0*10^5 Joules which is the same thing as -498kJ. Just make sure you use your sig figs as there are only 2 sig figs...

Kevin Tang 4L

Make sure you account for the DeltaH. Because we are only using 1/4 of the mols of reactants and get 1/4 of the mols of products, we have to divide the DeltaH of the reaction by 4. Thus the DeltaH of the reaction is -61.25kJ when 1.00 mol of CO2 produced in the reaction.

Kevin Tang 4L

I think you do use Cv,m for this because we are using the equation DeltaS=Cv,m(lnT2/T1).

Thus, for DeltaS for the change in temperature we do (3.73mol)(3/2)(8.3145 J*mol-1*K-1)(ln 348K/323K)

Kevin Tang 4L

(1)(219.2)(163)+(6)(29.4)(163) only gives you part of the answer as you need to heat it up from 37 to 200, then use the DeltaHrxn then cool it down from 200 to 37. DeltaH1=(1)(219.2)(163)+(6)(29.4)(163) DeltaH2=-2756kJ DeltaH3=(6)(37.1)(-163)+(6)(75.3)(-163) DeltaH1+DeltaH2+DeltaH3 Should give you t...

Kevin Tang 4L

lukezhang2C wrote:Are there answers to these problems?

He said he wouldn't be posting answers for this practice test.

Kevin Tang 4L

can someone explain how to do #5 For this problem, we have to find the change in entropy, delta S. To do this we want to change the Helium and Krypton to mols. We get 2.25mol He and 1.49mol Kr. Then we can plug them into the equation DeltaS=nRln(V2/V1). For the Helium, it is placed in a compartment...

Kevin Tang 4L

Ok, nevermind I realized my mistake. Breaking the bonds of the reactants releases 1083kJ of energy and we then have to use 1174kJ of energy to form the bonds to make the products. Thus there is a net negative delta H. So the delta H = -91kJ for the reaction My mistake was I was looking at the change...

Kevin Tang 4L

For the reaction H2O2->1/2 O2 + H2O, The energy needed to break the bonds on the left is 1083kJ and the energy need to break the bonds on the products side is 1174kJ. Why is the delta H = -91kJ and not 91kJ?

Kevin Tang 4L

Go to tomorrow's review session that starts at 12pm at CS50. The TA will review Thermochem and Thermodynamics if you are having trouble with that.
Doing all the assigned homework questions is also a great way to study!

Kevin Tang 4L

If there is both steam and water both at 100 degrees Celcius, steam will burn more because it has more heat energy. The energy of condensation (phase change from gas to liquid) from the vapor to water is what will cause the more severe burns.

Hope this helps.

Kevin Tang 4L

You can find information about this on page 423 in the 6th Edition of the book.

Kevin Tang 4L

How you do calculate the heat released by 5.025g of Kr(g) at .400 atm as it cools from 97.6 to 25 with a constant volume?

This is part b, with a constant pressure I got q=90.6 using q/(.06mol)(20.79J/K*mol)=-72.6K

Kevin Tang 4L

I believe the SI unit for energy is joules

Kevin Tang 4L

Looking through the textbook may help you a lot while doing this problem. Look at example 1 on page 264 of edition 6. At the bottom of the example, it says Related Exercises: 8.3 To solve this problem, we first need to get the area which is pi(r)^2. They give the diameter which is 3cm. 3/2=1.5cm whi...

Kevin Tang 4L

They shouldn't take off points if you use kJ unless they specifically say to use joules. But it is important to keep in mind that joules is the SI unit for energy.

Kevin Tang 4L

You can ignore x in an equilibrium equation when the percent deprotonation or protonation is less than 5%. This is because x will be negligible in the denominator related to the other initial value.

Hope this helps correct me if i'm wrong.

Kevin Tang 4L

Is the 5% the measurement of protonation or deprotonation?

Kevin Tang 4L

I realized that when I made my ICE table, I didn't do 2x for the right side which threw off the numbers for the rest of the numbers. After adjusting I got the correct answer.

Kevin Tang 4L

I realized that when I made my ICE table, I didn't do 2x for the right side which threw off the numbers for the rest of the numbers. After adjusting I got the correct answer.

Kevin Tang 4L

I am stuck on this problem. I did 17.4mg into .0174g and that into 3.95*10^-4 mols/.25L = 1.58*10^-3 M which is the value i have for x in the ice table. But then when i plug it into the equilibrium equation I get 3.944312*10^-9. What am I doing wrong?

Kevin Tang 4L

If we are not given Ka, can we just find Ka by using an ice table? I don't think it would be possible to find Ka value using an ice table if we don't know the variable x. We use the ice table so we can calculate the change in concentration knowing the Ka or Kb value. Instead, to find Ka value, you ...

Kevin Tang 4L

I have the 6th Edition of the book so the table number will be different. But for the 6th Edition, it is table 12.2, the Kb values. Look for methylamine CH3NH2. This will give you the Kb value and you can change this to its Ka value using 1x10^-14=[Ka][Kb]

Hope this helps.

Kevin Tang 4L

Although all these molecules are polyprotic, do we need to calculate the change in H3O+ value for the second deprotonation? It seems like the Ka2 value is too small.

When can we use approximations in this problem?

Kevin Tang 4L

Henry Dudley 1G wrote:What is the value of R?

The value of R is the gas constant, which I'm guessing will be a given value.

Kevin Tang 4L

The current concentration is denoted as Q, whereas the equilibrium concentration is denoted as K. Using the value K, we can tell if the reactants or the products are favored. If K>0, then the products are favored and if K<0, then the reactants are favored. If Q>K, that means that there are too many ...

Kevin Tang 4L

Solids have a concentration of 1, which doesn't change anything in the equilibrium concentration equation. For example, C(s)+ O(g) <-> CO(g), if Carbon was included in K because it was, it would be [CO]/[O]*1 which is just [CO]/[O]. You don't want to include the concentration of Carbon though, as it...

Kevin Tang 4L

Can anyone help me with the bond angle for number one ?? I have difficulty determining bond angles so for this problem what would it be and why? This angle will be slightly less than 120 degrees. This is because it is sp2 hybridized, and has a VSEPR formula of AX2E the shape is bent. The lone pair ...

Kevin Tang 4L

Leela_Mohan3L wrote:Thank you! Does the compound in number 31 have a net charge?

[Ni (NH3)5 NO2]^+2
Because the Nickel has a 3+ charge and the NO2 has a 1- charge, it brings it to a net charge of 2+.
NH3 has no charge.

I hope this helps.
Please correct me if I made a mistake. Thanks

Kevin Tang 4L

You should look at fundamentals J and Chapter 12 in the 6th edition book.

Hope this helps.

Kevin Tang 4L

Yeah, this is correct. The conjugate acids accept an H+ from the conjugate base.
The same can be done for the conjugate base, which loses an H+.

Kevin Tang 4L

Another common example of an Amphoteric Compound is HCO3-

This is amphoteric because it can lose the H+ to form CO3^2- and it can also gain an H+ to form H2CO3

It just means it can both act as an acid and a base.

Kevin Tang 4L

I don't seem to fully grasp the concept of pi bonds. I understand that double bonds have one sigma and one pi and that triple bonds have one sigma and two pi bonds. Pi bonds also cannot rotate because of its orientation, it extends out perpendicular down, and up and then connect. What I don't unders...

Kevin Tang 4L

The first bond you draw is sigma, any additional bonds are pi.

Single bonds are always sigma bonds. Double bonds have one sigma and one pi bond. Triple bonds have one sigma and two pi bonds.

Hope this helps and correct me if im wrong. Thanks.

Kevin Tang 4L

One way to remember is that there is a plus like shape on the partially positive side of the dipole:

(plus like shape on this side) +--------------> (points towards the partially negative side)

Kevin Tang 4L

Hydrogen bonds are bonds specifically between hydrogen and Nitrogen, Oxygen and Florine. This is because N, O, F have the highest electronegativity. This means that the partial charge is higher between Hydrogen and N,O,F. It is more polar than the bond between Hydrogen and Sulfur. Because it is less...

Kevin Tang 4L

They will not always have induced dipoles, but induced dipoles and momentary dipole (London Dispersion Forces) are the only intermolecular force that nonpolar molecules can have.

Please correct me if I'm incorrect. Hope this helps.

Kevin Tang 4L

First, we need to change the information given about the composition of this molecule into the molecular formula: 37.5% C * 32.04g/mol = 12.0g C / 12.0107g/mol = 1mol C 12.6% H * 32.04g/mol = 4.03g H / 1.007g/mol = 4mol H 49.9% O * 32.04g/mol = 16.0g O / 15.9994g/mol = 1mol O CH4O, which can also be...

Kevin Tang 4L

CH2O's central atom, Carbon, has two single bonds and one double bond, the first bond is sigma and one of the bonds in the double bond is a pi bond. So there are three sigma bonds and one pi bond. CH2O is sp2 hybridized and has a shape of trigonal planar. See 4.1, pg 109 in 6th Edition.

Kevin Tang 4L

None of them are radicals because all of them end in an even number. In increasing order, it goes CH2 (2-), CH3(-), CH4, CH2, CH3 (+), CH2 (2+), This is because of the molecular geometry. CH2 (2-) is bent, sp3 hybridized which means its going to have angle less than 120 (much lower because of two lo...

Kevin Tang 4L

I'm not exactly sure about ClO2+, but generally, you would want the formal charge for the central atom to be 0, and the outer atoms with the charge. Oh, I figured it out. The formal charge of the central atom, Cl, in ClO2+ is actually has a formal charge of +1 because there is going to be a positiv...

Kevin Tang 4L

The central atom does have three lone pairs of electrons that may give the impression that it looks trigonal bipyramidal, but the angle between the two F is 180degrees, which is linear.

Kevin Tang 4L

What should the formal charge for the central atom of ClO2+ be. Should it have a charge of one because it is positively charged overall. I don't quite understand which formal charge should go on what molecule.

Thanks

Kevin Tang 4L

are electron affinity and electronegativity related? They are related in the sense that they follow the same periodic trend. The highest electronegative and highest electron affinity are on the top right side of the periodic table. However, the noble gases are an exception in electron affinity as t...

Kevin Tang 4L

Adding on to Connie's post, the lowest ionization energy atom will be in the center (ionization energy and electronegativity follow the same trend) because the atoms on the outside are more electronegative/higher ionization energy and want electrons.

Kevin Tang 4L

Elements in period 3 have an empty d-orbital, so it does not have to obey the octet rule. Hydrogen, helium, beryllium, and lithium are exceptions to the octet rule because they form duplets. Radicals are also exceptions to said rule. If radicals are exception to this rule, can they have nine valenc...

Kevin Tang 4L

Question 6b is asking if a Helium atom or a "new" element GarBreadium have longer wavelengths. The answer key says the GarBreadium will have a longer wavelength but isn't it He because it is less heavy, therefore the denominator in de Broglie's equation \lambda =h/m\cdot v be smaller, cor...

Kevin Tang 4L

can someone further explain number 6c? I'm not sure how the wavelengths correspond with lower frequency Refer to the equation c=(lambda)(v), you can rewrite this as c/(lambda)=v or c/v=(lambda). You can see they are inversely related; when wavelength increases, the frequency decrease and vice versa.

Kevin Tang 4L

304981930 wrote:Jk final frequency I get 3.77 * 10^14. Why is this off

This is because you didn't add the KE with the threshold value. You only did 2.5x^-19 =hv
v=3.8x10^14
If you add the KE, you should get the answer 6.8*10^14 Hz

KE= 1/2mv^2

Kevin Tang 4L

For problem 9b, the electron configuration of Chromium is [Ar]3d5 4s1. I thought that the 4s orbital was filled before the 3d orbital. I know Copper and some other elements also follow this trend. Why do these elements only fill one of the two electrons in the 4s or 5s orbitals?

Kevin Tang 4L

Daisylookinland4B wrote:For #6, how are you supposed to convert the molar mass of GarBreadium to kg in order to use the de broglie equation?

You change it to kg you divide the molar mass of GarBreadium by Avogadro's constant, 6.02*10^23 which gives you 5.244*10^-24g then you divide again by 1000 to get the value in kg.

Kevin Tang 4L

It's kinda confusing that we use kg because all the other units we use don't use kilo. E.g. the SI unit for distance is meter, not kilometer; we use mols, not kilomols.

However, for mass, the SI unit is kilograms as other units use kg. E.g joules, which is measured in kg⋅m2⋅s−2.

Kevin Tang 4L

The Heisenberg Indeterminacy equation states that ΔP*ΔX must be greater than or equal to h/(4pi). The lower the uncertainty in position, the greater the uncertainty in momentum; the lower the uncertainty in momentum, the greater the uncertainty the position will be. However, my question is if we kno...

Kevin Tang 4L

The possible number for the magnetic quantum number, denoted m l , can be calculated knowing the value of l. It can be -l, -l+1 .... l-1, l. This number shows the orientation of the specific orbital. The difference between m l and n is that n tells us about energy and how far the electrons extend out.

Kevin Tang 4L

Madison is correct, The first quantum number (n) is a positive integer (0,1,2,3 .....) The second quantum number (l) is any positive integer that is less than n, thus (0,1,2,3..... n-1) The reason why (1,1,0) doesn't work is because n=1, which means the only value possible for l can be 0. If n=2, th...

Kevin Tang 4L

In order to do this problem, you need to convert keV (kiloelectronvolt) to joules.
There are 6.242x10^15 keV in 1 joule:

140.511keV*(1 joule/ 6.242x10^15)= 2.251x10^-14 joules

Kevin Tang 4L

I am having trouble with part b and d and I still don't know if this is right. I would appreciate if someone corrected me if I'm wrong. My understanding is that ml are the different orbitals in a subshell. Each individual orbital can only have a maximum of two electrons. So that means the answer can...

Kevin Tang 4L

As the mass increases, the denominator in the equation λ=h/mv increases which in turn decreases its wavelength because λ is a constant.

Do you know why 10^-18 specifically? Is it because it is the smallest wavelength that is detectable by most instruments?

Thanks

Kevin Tang 4L

Yes, everything with mass has a wavelength but because the baseball is relatively large compared to an electron, the wavelength is very short. In the de Broglie wavelength: λ=h/mv, when m increases, the wavelength becomes shorter (when the denominator is larger, the wavelength is smaller) Compare th...

Kevin Tang 4L

Metals exposed to the sun are affected by the photoelectric effect because the sun is emitting wavelengths of visible light, metals that have a threshold energy that is equal to or less than the energy of the sun's photons will scatter electrons. As for the weight, metals do lose a negligible amount...

Kevin Tang 4L

You are correct, 25.92x10^-3 is not correct scientific notation, it should be 2.592^-2.
The problem seems to be that when 25.92mg was converted into grams, 10^-3 was just added instead of also moving one decimal over.

Kevin Tang 4L

First, we need to find the percent composition of Potassium of the various molecules: K:39.098g/mol KCl:74.5513g/mol 39.098/74.5513g/mol = 52.44% K K2S:110.262g/mol 78.196(K2)/110.262g/mol= 70.92% K K3PO4:212.27g/mol 117.294/212.27g/mol= 55.26% K Then you use the percent composition to find how many...

Kevin Tang 4L

You just made a small mistake by using Nitrogen (14.007g/mol) rather than Oxygen (15.9994g/mol) which would explain the error. 63.15g C/12.011g/mol = 5.257 molC / 1.971=2.666*3=8 Carbon 31.55g N/ 15.994g/mol = 1.971 molN / 1.971=1*3=3 Nitrogen 5.30g H/1.008g/molH = 5.257 molH / 1.971=2.666*3=8 Hydro...

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