CHEMISTRY COMMUNITY

Joseph Hsing 2C

jeffrey van 2I wrote:Thanks, so just to confirm, I would just divide the value of Kw (1.0E-14) by the given Ka, right?

Yup this is correct in order to find Kb.

Joseph Hsing 2C

Which section is this question from? I can't find it.

Joseph Hsing 2C

I think you would need Ka and Kb values since most of the given compounds are weak acids and bases.

Joseph Hsing 2C

Only thing you missed was to balance the equation, put a 3 in front of HClO. Your final answer would then be missing a cube root.

Joseph Hsing 2C

For liquids omit any solvents because it will be in excess so there would be no point tracking their molar concentration. Both solids and liquids are excluded because their effective concentrations will stay constant throughout the reaction.

Joseph Hsing 2C

I don't know the full context of the question but just from reading what you typed, I'm guessing there's some confusion on sig figs: 0.025, 1.6, and 1.7 all have 2 sig figs. Make sure to include all non-zero digits left of the decimal in your sig fig count.

I hope this helped!

Joseph Hsing 2C

With two x values, there's always one value that just doesn't make sense with the numbers you have, like if the value is greater than your inital molar concentrations.

Joseph Hsing 2C

Pressure is associated with gasses, and something can only have a concentration if it is dissolved in a solvent.

Joseph Hsing 2C

The pH of the diluted solution is given so set that equal to -log[H30+] and then solve for [H30+]. I got a value of 5.62 * 10^-14. Now you can use Kw = [H30+][OH-] = 10^-14, substitute and solve for [OH-] and that is the molar concentration of hydroxide ions of the diluted solution of 0.18M. Next, u...

Joseph Hsing 2C

CH3COOH should be an acid.

Joseph Hsing 2C

Look for the electronic geometry of a molecule which is basically counting how many areas of electron density. This number will tell you directly how many hybridized orbitals it is using. So 2 regions will use sp hybridized, 3 will use sp2, ..., 6 will use sp3d2.

Joseph Hsing 2C

This is because oxygen has one more electron than a half-full orbital as in nitrogen with 3 of 6 p filled. So the energy needed to remove that outlier electron on oxygen will be lower despite the normal trend of ionization energy increasing down a period.

Joseph Hsing 2C

I think more oxygens means the strength of the bond to hydrogen will decrease because of more resonance, basically more electron delocalization away from the hydrogen because more electronegative atoms are present.

Joseph Hsing 2C

If an attached ligand is monodentate then count 1 towards the coordination number for each, and for bidentates count 2 towards the coordination number of the central atom, etc. Hope this helped!

Joseph Hsing 2C

The coordination number of a central atom refers to the number of atoms, ions, or molecules bonded to it, while the steric number is the number of bonds and lone pairs.

Joseph Hsing 2C

To figure out if a ligand is polydentate, look at the molecular structure and look if there are atoms like carbon between the donor atoms. Spacing is the key factor.

Joseph Hsing 2C

Yes you're correct for accounting for both the ligands inside and outside of the coordination sphere.

Joseph Hsing 2C

Yup for oxidative number, we take into account the atoms and ions inside and outside of the coordination sphere.

Joseph Hsing 2C

If you can imagine the Lewis structure in your head from the molecular formula, then you can check for resonance and as a result delocalized pi bonds. I would recommend just drawing out the different structures just to be safe though and more practice.

Joseph Hsing 2C

It probably has to do with the relative stability of the different ligands and their bonding. I'm assuming water is less stable than some of the compounds you listed when bonded with a metal ion.

Joseph Hsing 2C

There exists tetradentates, pentadentates, and octadentates. I'm guessing there is a cap to how many atoms a ligand can bind to, but I'm not sure on that exact number.

Joseph Hsing 2C

The 3 possible structures of C3H4 include one ring structure and two linear structures. Since all three of these have different atom connectivities, we cannot say they are resonance structures of each other.

Joseph Hsing 2C

The carbons on the end have 3 regions of electron density which means that they have 3 sp2 orbitals and 1 p orbital. The carbons in the middle, whose amount is determined by x only has 2 regions of electron density which means they have 2 sp hybrid orbitals and 2 p orbitals. This means that in H2CC...

Joseph Hsing 2C

Could someone explain the rule of when x is even or odd in H2C(C)xCH2 that determines if the terminal hydrogens lie in the same plane? More specifically why exactly does this happen? I know from the answer that odd number of carbon atoms will result in pi bonds that are perpendicular and an even num...

Joseph Hsing 2C

The only shapes we will have to know will be the ones directly covered in lecture! He won't test us on material that he hasn't discussed.

Joseph Hsing 2C

We probably will since the lectures are prerecorded, but only Lavelle can confirm this!

Joseph Hsing 2C

This sulfite ion in VSEPR model notation is AX3E, so the molecular geometry will be trigonal pyramidal.

Joseph Hsing 2C

An ion by definition will have lost or gained electrons, so this can result in a change of the number of electron densities, therefore ions tend to be angular shaped. For OClO there are 3 regions of electron so the electron domain geometry will be trigonal planar, but since only 2 of the 3 positions...

Joseph Hsing 2C

We will probably be given a couple of choices of Lewis structures and then asked what is the correct one. Since it's on CCLE, there won't be a drawing feature I'm guessing.

Joseph Hsing 2C

Palladium is more stable if it completes its d-orbital as opposed to filling the s-orbital and leaving 2 empty electrons in the d-orbital.

Joseph Hsing 2C

Lavelle never mentioned diagonal relationships in lecture, so I don't think so. Focus on horizontal and vertical trends. :)

Joseph Hsing 2C

Yes, you're right, electron repulsion from the lone pair pushes the other fluorine bonds which contribute to molecular shape which in turn makes an overall molecule polar or nonpolar. Hope this helped.

Joseph Hsing 2C

Organic molecules have carbon-hydrogen bonds and these are made by biological systems, inorganic molecules don't have these bonds. Line structures is referring to the rings and lines formed by carbon chains.

Joseph Hsing 2C

I think one factor to focus on for determining boiling points is examining how strong the force of attractions that are present and how many there are between the atoms in a molecule. For example, more Van der Waals forces results in more energy needed to melt or boil a compound and thus increases t...

Joseph Hsing 2C

Midterm 2 covers all material after MT1 to the end of Outline 3 (Chemical Bonds)

Joseph Hsing 2C

The less energy that is needed for an atom to lose an electron, the greater the number of bonds that atom can participate in. Hope this helped :)

Joseph Hsing 2C

Yup, atomic radius increases due to the extra shell of electrons.

Joseph Hsing 2C

I believe that all group 13 elements are electron deficient and act like Lewis acids. Not entirely sure on the second question but I think electron deficient just means the atom doesn't have enough electrons to fill an octet, but compounds with boron and aluminum are exceptions to the octet rule.

Joseph Hsing 2C

Nonmetals will typically gain electrons and metals will typically lose electrons. If an ion gains e- it will have a negative charge and if it loses e- it will have a positive charge. Hope this helps!

Joseph Hsing 2C

Pressing 2nd then comma is also a shortcut for *10^ this helps save a lot of time for me.

Joseph Hsing 2C

I second what Hannah says! Make sure you put an extra pair of parentheses in the denominator when using your calculator.

Joseph Hsing 2C

Convert the given wavelength of light to energy using E=hc/lambda, then concert joules to eV with the conversion factor above has stated. After finding the difference in eV convert from eV/atom to kJ/mol!

Joseph Hsing 2C

Rydberg formula is used to calculate the wavelength of spectral lines, while de Broglie describes the wavelength of any particle. There are 3 different numbers I believe when using Rydberg and the last one not mentioned was 2.178 x 10^-18 J for energy calculations. Look for what the question calls f...

Joseph Hsing 2C

I feel that the actual math operations behind Shrodinger's equation are well beyond the scope of a general chem class. I think we should focus on the conceptual idea behind what the equation stands for including that electrons can be described by a wave function which then describes electron orbitals.

Joseph Hsing 2C

I'm not entirely sure, but I feel in Chem there are rarely absolutes as most things are viewed in comparison to one another. I know it is possible for a noble gas to receive an electron, just very very unlikely, so we can't say zero affinity.

Joseph Hsing 2C

Since you are finding the wavelength of the threshold energy needed to eject electrons, then you can disregard kinetic energy.

Set work function equal to h*frequency to find frequency.

Next, use c=lambda*frequency to solve for wavelength.

Hope this helps :)

Joseph Hsing 2C

I think the most important concept to understand is the basic trends of ionization energy and atomic radius as you move down a period or group.

Joseph Hsing 2C

Could someone clarify if we do 5 posts per week in terms of the participation points or a total of 50 at the end of the quarter?

Joseph Hsing 2C

I believe ionization energy increases as you move up a group because the valence electrons are getting closer and closer to the nucleus, thus needing more energy to counteract the attraction of electrons with the nucleus. Basically, less shielding effect.

Joseph Hsing 2C

This isn't a mass percent problem, it only consists of dimensional analysis. So, for part A the process of solving the question would be: 3.00g CaBr2 x (1 mole CaBr2 / 199.197g) x (1 mole Ca2+ / 1 mole CaBr2) You would repeat the same steps for part C, just make sure to convert kg to g first. Hopef...

Joseph Hsing 2C

The question asked to calculate the moles of: a) Ca2+ ions in 3.00 g of CaBr2 c) F- ions in 25.2 kg of UF6 I seem to be getting the calculations involving ions wrong. Is there something I possible could be not considering and could someone walk through these questions? I'm finding the molar mass of ...

Joseph Hsing 2C

You can find the molar mass of a Helium atom on the periodic table: 4.002 g/mol. Since we are using Joules with Planck's constant, convert grams to kilograms, so 4.002 x 10^-3 kg. And make sure to multiply by Avogadro's number so 6.002 x 10^23 He atoms/mol in this question so that you get the value ...

Joseph Hsing 2C

Got it thanks everyone! ^^

Joseph Hsing 2C

First, make sure you convert the given wavelength to meters so it would be 486.1 x 10^-9 m. Use v=c/lambda to convert to frequency. Plug frequency and n1=2 into Rydberg Equation to find n2. You should get n2=4.

Joseph Hsing 2C

How many decimals are we expected to use for constants like the speed of light and Planck's constant for the midterm?

Joseph Hsing 2C

After calculating the work function by subtracting max KE from energy of the photon, you should then divide total energy by the work function. The answer should be the max number of electrons ejected!

Joseph Hsing 2C

For me, the most important aspect of studying is being an active questioner and try to think of every aspect of a question and bring them up during discussions or this forum. Still working on trying to find the right balance of textbook questions that works for me.

Joseph Hsing 2C

Ah I see I wasn't paying attention to the decimal. Thank you!

Joseph Hsing 2C

I could be wrong but when the prof was calculating the wavelengths if the first example wouldn't the answer be 551 x 10^-7 instead of 551 x 10^-9?

Search found 59 matches