CHEMISTRY COMMUNITY

ishaa Diwakar 4E

You will need to know about the neutralization with a strong acid and a strong base. You will probably have to determine whether the solution will be acidic or basic given the amount of acid and base and the volume of solution.

ishaa Diwakar 4E

Acid rain forms from the reaction between CO2, SO2 and NO2 with H2O in the atmosphere. These reactions form the acids H2CO3, H2SO4, and HNO3, which release H+ ions in solution to create acidity. The main way to reduce acid rain is to reduce the emission of CO2, SO2 and NO2 into the atmosphere.

ishaa Diwakar 4E

When the complex is negative, you would name the metal with the ending -ate. If it is positive or neutral, the metal name will remain unchanged.

ishaa Diwakar 4E

This counts as one sigfig. Only 0s after a number after the decimal will be significant, like 0.200 will have three sigfigs.

ishaa Diwakar 4E

If a compound is considered amphoteric, it is able to act as an acid or a base. For bronsted pairs, you could react the compound with water to see if it is able to act as both. For lewis pairs, you could probably see how the compound reacts with a known acid and a known base.

ishaa Diwakar 4E

You look at the charges of the ligands and the overall charge of the compound. If the compound happens to be bound to something else, you would use those charges to determine the charge of the compound.

ishaa Diwakar 4E

A super acid is an acid that has a H+ concentration greater than 1.0 mol. L-1, which is equivalent to a pH of 0.

ishaa Diwakar 4E

pKa is the -log[Ka]. Similarly to pH, it is a simplified way of writing concentration using the -log equation as an operator. We don't need to know how to calculate pKa for this class.

ishaa Diwakar 4E

You would need to react each of the given salts with water to determine the acid and base. For part a, for example: NaC6H5O + H2O --> NaOH + C6H5OH. I'm not sure if this question asks you to label the acid/base as strong/weak, but if it does, you should be able to figure out at least one from each p...

ishaa Diwakar 4E

Yes, exactly. The sig figs from any other numbers in the question should be used to dictate the number of sig figs after the decimal place in your pH answer.

ishaa Diwakar 4E

Because the requirements for a H-bond to form are that there is an H bonded to a N, O or F, and that there is another N, O or F nearby that is charged or has a lone pair. Since the lone pair is negatively charged (electron), it can attract the partial positive charge of the hydrogen.

ishaa Diwakar 4E

Water acts as a bronsted acid because it can donate a proton: H2O --> H+ + OH-. It acts as a base because it can also accept a proton:
H2O + H+ --> H3O+.

ishaa Diwakar 4E

We also probably need to know how to name coordination compounds and to understand the concept of chelating ligands.

ishaa Diwakar 4E

The main characteristic is that they have the potential to donate two protons (H+) since there are two H atoms per molecule. For example, with H2SO4, the first equation would be: H2SO4 + H2O --> HSO4- + H3O+ , and the second equation would be: HSO4- + H2O --> SO4 2- + H3O+

ishaa Diwakar 4E

I am pretty sure the only givens are those we have seen on the midterm and tests. You may be able to ask about abbreviations as long as they are not on the periodic table/have been explained in class. It would be best to memorize ion charges, but worst case you can calculate the formal charges of po...

ishaa Diwakar 4E

It could be a typo, but it might be signifying that p is 'equivalent' to -log, rather than 'equal' to -log. The difference would be that the use of p (like in pKA) is the same as the use of -log with [KA], yet saying p = -log would be inaccurate since -log is acting as an operator.

ishaa Diwakar 4E

It is also important to notice whether there are single or double bonds around the atoms with lone pairs, since only single bonds can rotate to form multiple ligands. For example, NH2-CH2-CH2-NH2 has single bonds attaching the Ns, so it can form a shape like \__/, which can be bidentate.

ishaa Diwakar 4E

In this case, the question tells you that this molecule is an acid, and this will usually happen if they are asking about an amphoteric molecule. However, you can determine whether it is an acid or base depending on the molecule. With -COOH molecules, they will usually either be neutral charge or ne...

ishaa Diwakar 4E

I'm not completely sure either, but it might just be traditional as you said, because pH and pOH are just opposite scales.

ishaa Diwakar 4E

Not really, but if you can remember the common oxidation states for the main transition metals, it can help you narrow down the options. Along with this, if you can recognize which ligands are neutral and memorize the charges of other common ligands, it will be easier to figure out the metal's oxida...

ishaa Diwakar 4E

Bronsted acids are proton donors and the bases are proton acceptors, while Lewis acids are lone pair acceptors and the bases are lone pair donors.

ishaa Diwakar 4E

The main difference between strong acids and weak acids is that strong acids dissociate completely in solution while weak acids do not (the same concept applies for bases). Therefore, strong acids/bases will have higher H+/OH- concentrations in solution (respectively). Because of this, strong acids ...

ishaa Diwakar 4E

I don't remember the exact number, but this question confused me too. It's supposed to be the total number of potential H-bonding sites on the molecule, and this can be for H-bond sites with ANY molecule, not only more of the same molecule. So, it will be the number of Hydrogens bonded to N,O, and F...

ishaa Diwakar 4E

A bronsted acid is a proton donor, and the base is a proton acceptor. A Lewis acid is a lone pair acceptor while the base is a lone pair donor. The "opposite" part is just that the "donor" and "acceptor" titles for acids and bases are the opposite for the two categories.

ishaa Diwakar 4E

H2CO3 and H2SO4 are considered diprotic, meaning they usually lose two protons as anions. So, when they lose only one, they are able to lose one more and act as an acid, even with a negative charge.

ishaa Diwakar 4E

Also, often molecules with different electron domain and VSEPR geometries will be polar, since a lone pair will be in place of the ED and will prevent the dipoles from cancelling. An example of this is PF3.

ishaa Diwakar 4E

I'm pretty sure the test doesn't have anything from the midterm. It will most likely focus on polarity, IMFs, VSEPR, and then else everything we cover until the test.

ishaa Diwakar 4E

I think you are right, but also the main reason that the Hydrogen bonds are stronger is due to electronegativity (EN). Since Hydrogen bonds are formed between H (with the lowest EN) and N,O, or F (with the three highest ENs), the EN difference is the greatest and therefore the dipole is strongest. T...

ishaa Diwakar 4E

Also, once the light passes the threshold energy, it can eject electrons. Since the energy of the light is directly correlated to its frequency, increasing the frequency will only cause the energy of the emitted electron to increase. Therefore, the graph you draw should show a horizontal line (at an...

ishaa Diwakar 4E

Since lone pairs take up more space, the bond angles adjacent to the lone pairs will increase and the opposing angles will decrease.

ishaa Diwakar 4E

Not exactly. In a covalent bond, the electrons are shared when the orbitals from each atom overlap. So when the anion distortion occurs, the cation's positive charge attracts the anion's electrons towards it and the orbitals overlap, which is why it is similar to a covalent bond.

ishaa Diwakar 4E

The p-block atoms with more protons will have a higher nuclear charge than those in the s-block because each electron is affected by all of the protons in the atom. This means that if an atom has 8 protons and 8 electrons, for example, each electron will individually experience the nuclear pull from...

ishaa Diwakar 4E

I think it is because the anion distortion is caused by the cation (which transfers its electrons) attracting the anion's electrons towards itself. This may cause the electrons to be almost shared between the atoms, like in a covalent bond.

ishaa Diwakar 4E

Also, many molecules with resonance structures have charges, so they are likely to form ionic bonds. For example, a lot of polyatomic ions have resonance.

ishaa Diwakar 4E

This is because in Bi+3, it would lose the three electrons in the 6p block first, while in Tl+3, it would lose its one electron in the 6p block and then the other two electrons from the 6s block since it only has one electron in the 6p block.

ishaa Diwakar 4E

I think that the molecule with the strongest bonds would be the one where the C and X atoms have the greatest difference in electronegativity.

ishaa Diwakar 4E

To add on to this, covalent bonds are stronger than ionic bonds in biological systems, since molecules will often be dissolved in water. Otherwise, ionic bonds are usually stronger.

ishaa Diwakar 4E

I think those are the main exceptions, and also this same rule will apply to elements in the same position in the 4d section.

ishaa Diwakar 4E

This is because the 3d and 4s blocks are at very similar levels in terms of electron shells. Metals in the d block will very often form oxidation states with charges greater than 2+, although they will lose electrons from the 4s orbital first.

ishaa Diwakar 4E

I think this applies only to the 3d and 4d rows, since they do not include the f block.

ishaa Diwakar 4E

Atoms with low ionization energy will usually have higher electronegativity. Therefore, the atom will be less likely to lose its own electrons and more likely to attract electrons from another atom. Electron affinity, in my experience, is usually given. If not, you can calculate it using other values.

ishaa Diwakar 4E

This is because the first 1 is referring to the number of electron shells, which correlates to the first row of the periodic table. The second 1 is referring to a p-orbital. This could not exist since none of the elements with only 1 electron shell can contain a p-orbital.

ishaa Diwakar 4E

Yes, since each excited electron will emit a very specific wavelength of light based on the electron level jump, which correlates to a line on the spectrum.

ishaa Diwakar 4E

The equation most likely would not be applied to photons, since they do not have a mass or momentum.

ishaa Diwakar 4E

The equation E=hv is not exactly proven by the photoelectric effect, but it shows the positive relationship between energy of the photon and frequency. The photoelectric effect proves (or shows) that photons have particle-like properties, since increasing the intensity of light did not eject electro...

ishaa Diwakar 4E

If the energy is equal to the work function, the electron would not be ejected from the metal, therefore it would have 0 kinetic energy.

ishaa Diwakar 4E

Yes, it is D, because there is one electron that will be excited by every one photon (1:1 ratio).

ishaa Diwakar 4E

This relates to the photons emitted by the electrons after they drop from energy levels. Since they are discrete, they can only emit specific wavelengths, and therefore each atom has a unique emission spectrum.

ishaa Diwakar 4E

I think he mainly focused on Balmer and Lyman, but there are also the Paschen, Brackett, and Pfund series. The Balmer series corresponds to the visible light spectrum, the Lyman series corresponds to UV light, and the last three correspond to infrared light. They are differentiated by wavelength.

ishaa Diwakar 4E

The grouping is also most likely determined by the wavelengths of the emitted photons, which are grouped based on UV light (Lyman) and visible light (Balmer).

ishaa Diwakar 4E

So the Lyman series (for UV) will correlate to any energy level drop that includes the drop from n=2 to n=1, since that is the largest energy level difference. The Balmer series (for visible light) will correlate to the energy level drop that includes that from n=3 to n=2, since that is the second l...

ishaa Diwakar 4E

Yes, I think that the Angstrom is a more commonly used unit in this situation, so you can use it as if it is an SI unit.

ishaa Diwakar 4E

Since you have the molar masses and the percentage composition, you just convert the percent composition to grams and divide by the molar mass. Then, using the molecular formulas, you can write a chemical reaction and balance it.

ishaa Diwakar 4E

When we start learning the Quantum unit, we will be using more of pico- and nano- units. Also, in biology, micro- units are very commonly used.

ishaa Diwakar 4E

I actually learned this method in high school too! It is actually similar to one of the methods Lavelle taught, where he told us to solve for the amount of products each molar amount of the reaction would form. This method basically just removes the second step of solving for the products.

ishaa Diwakar 4E

Lavelle says that we are doing reduction reactions later in the quarter, so don't worry about it now because you won't find this problem on the test.

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