CHEMISTRY COMMUNITY

ZainAlrawi_1J

Lyndon sees his crush and panics, accidentally dropping 382.7 mg of HCl and 147.1 mg CaO into a 1L flask. After filling that flask with water up to the 1L mark, what is the pH of the resulting solution? How do you go about deriving the equation HCl + Cao --> CaCl2 + H2O from this starting to point? ...

ZainAlrawi_1J

The oxidation number is the charge of a particular atom or molecule.
The coordination number is the number of separate points at which ligands attach to a central atom

ZainAlrawi_1J

Refer to table 6C.7 in the book to clear up this confusion on how to determine acid strengths, I have it attached here: https://i.imgur.com/IFnknpN.png

ZainAlrawi_1J

Both of these are oxoacids, both oxoacids have the same number of Oxygen atoms, so you take a look at the electronegativity of the attached group. H2C is more electronegative than HGe (C is more electronegative than Ge), therefore it is the stronger acid.

ZainAlrawi_1J

You use -ate with negatively charged atoms.

ZainAlrawi_1J

Both 'cyano' and 'cyanido' refer to CN- and both are accepted for this class. Please refer to the following document that Dr. Lavelle emailed us: https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14A/NamingCoordinationCompounds.pdf I would personally stick to the traditional naming schem...

ZainAlrawi_1J

You would count a double bond as a single point, except in the case of molecules with polydentate qualities, that are able to bind to the central metal in more than one place. These molecules could contribute more attachments to the bonding number.

ZainAlrawi_1J

It is important to first note that both CH3COOH and HCOOH are Carboxylic Acids, meaning that they both contain the COOH group, and as such we can use the information summarized in Table 6C.7 of the textbook in order to understand how to derive which of the two is the stronger acid. For carboxylic ac...

ZainAlrawi_1J

Can someone explain how the hybridization of the lone pairs in part b of this question are derived? I have never seen a question like this before and I'm not even sure that we have went over it.

ZainAlrawi_1J

A single bond is a sigma bond.
A double bond consists of one sigma bond, one pi bond
A triple bond consists of one sigma bond, two pi bonds.

ZainAlrawi_1J

There must be a side by side overlap between orbitals in order for them to form double or triple bonds, this specific arrangement, wherein orbitals are bonded side by side as opposed to end to end, is referred to as pi bonding.

ZainAlrawi_1J

Bronsted acids usually start with a Hydrogen "H" atom, while Bronsted bases usually end in an Oxygen and Hydrogen "OH"

ZainAlrawi_1J

Cisplatin helps in mitigating cancerous cell mutation, it is an important example to know the structure and significance of for this course.

ZainAlrawi_1J

When in an acid loses a hydrogen atom in a chemical reaction, the formed compound is called the conjugate base.

ZainAlrawi_1J

The more electronegative an atom, the weaker the acids it will form.

ZainAlrawi_1J

2 charges of 1 or -1 is usually preferable.

ZainAlrawi_1J

Electronegativity is correlated with shorter bond lengths, which are also stronger bonds by virtue of their compactness

ZainAlrawi_1J

Yes, we have to understand the naming algorithm for coordination compounds.

ZainAlrawi_1J

Bond order is defined as the difference between the # of bonds and the # of antibonds

ZainAlrawi_1J

Bond order is defined as the difference between the # of bonds and the # of antibonds

ZainAlrawi_1J

Bent molecules have a lone pair.

ZainAlrawi_1J

Pi bonds exist in all double and triple bonds

ZainAlrawi_1J

Strongest to weakest:

1. Ion-Ion
2. Ion-Dipole
3. Hydrogen Bond
4. Dipole-Dipole
5. London Dispersion

ZainAlrawi_1J

We consider unbonded electron pairs as well as single, double, and triple bonds to just be individual "regions of electron density" for the sake of VSEPR models

ZainAlrawi_1J

Yes, a triple bond has 1 sigma bond and 2 pi bonds.

ZainAlrawi_1J

The number of bonding pairs and lone pairs attached to a central atom determines its shape, I would recommend looking up a VSEPR chart as it shows how intuitive it is.

ZainAlrawi_1J

The shape and as such the bond angles do influence the strength of interactions and is one of the reasons molecular shape is valuable.

ZainAlrawi_1J

Hydrogen atom covalently bonded to Nitrogen, Oxygen, or Fluorine.

ZainAlrawi_1J

Draw the most stable structure (formal charges in mind) in order to accurately predict the shape

ZainAlrawi_1J

Yes, any discrepancy in electronegativity causes dipole moments

ZainAlrawi_1J

Polarizability increases toward down and to the left on the periodic table, while polarizing power increases up and to the right on the periodic table.

ZainAlrawi_1J

Yes, elements with higher electronegativity are more likely to get electrons as they are closer to reaching a stable configuration.

ZainAlrawi_1J

The magnetic quantum number ranges from the values -l to +l

ZainAlrawi_1J

All elements with access to the d orbital follow the same exception as Xenon as they can hold more electrons because of their access to the orbital.

ZainAlrawi_1J

All elements with access to the d orbital follow the same exception as Xenon.

ZainAlrawi_1J

For the purposes of the course so far, we don't need to distinguish between them, though it is very good to know this for future courses

ZainAlrawi_1J

In increasing strength:
Single bonds < Double bonds < Triple bonds.

ZainAlrawi_1J

There's unfortunately no extra credit opportunities for 14A :(

ZainAlrawi_1J

The second test will cover content from only after the midterm, check the syllabus for more info.

ZainAlrawi_1J

Just sorted it out by doing the formal charges for the atoms in both possible structures, unfortunately had to the formal charges for all the oxygen atoms to get a clear answer but its definitely obvious now. Thanks everyone!

ZainAlrawi_1J

As in part (c) of question 2C.5, the double bond is drawn between the O and N atoms, why is this the case? Why isn't it drawn between the Cl and O atoms?

ZainAlrawi_1J

Use the equation ΔpΔx ≥ h/4π

wherein
Δp = momentum, also equal to m*v (mass times velocity)
Δx = position
h = planck's constant

ZainAlrawi_1J

Elements with d orbitals (period 3 and beyond), can have expanded valence shells.

ZainAlrawi_1J

Resonance describes any lewis structure with more than one way of being drawn without altering the structure of the drawn molecule.

ZainAlrawi_1J

The ionization energy is the amount of energy that is required to remove an electron from an atom, while an atom's electron affinity is the amount of energy released or spent when an electron is added to an atom.

ZainAlrawi_1J

It decreases a the distance between the electrons increases further away from the nucleus, as such decreasing their attraction.

ZainAlrawi_1J

The formula sheet is available to print out on Lavelle's site.

ZainAlrawi_1J

If we need to work with any actual numerical values, then yes, they will be given.

ZainAlrawi_1J

We would only have to calculate the differences in electronegativity between already given values, as in we should be provided with either an electronegativity table or the relevant exact figures for a given problem.

ZainAlrawi_1J

Base it purely off the number order,
3d is n = 3
4s is n = 4
4 > 3
3 is at a lower energy level

ZainAlrawi_1J

Delocalized bonds are not tied down to a single nucleus in the same way localized bonds are. Delocalized bonds occur in every example of resonance bonding, as they contain electrons that are free to move about in the structure and thus allowing different lewis structures to be draw representing the ...

ZainAlrawi_1J

This is an "exception to the exception" that is the Aufbau Principle and I don't think we're quite required to know. There is a more concise explanation on this link: https://socratic.org/questions/57b1630911ef6b4fbd986ed8

ZainAlrawi_1J

One proton has a greater affect on radius than one electron, hence why as you go from left to right across a period, atomic radius tends to decrease.

ZainAlrawi_1J

Any 5 problems off the Chemical Bonds section on the syllabus are fair game

ZainAlrawi_1J

The ground state describes an atom at its lowest possible energy level. An atom is at an excited state when the energy level of its electron is at a higher energy level than its ground state. You can tell them apart by looking at the electron configuration, as any electron configuration in which the...

ZainAlrawi_1J

Yes, we'll receive the values and units for all the constants we'd need as we did on the front page of Test #1.

ZainAlrawi_1J

For the purposes of this class we don't need to know much beyond what he went over in lecture about what an electron's spin is. However, you could liken the spin of an electron to a the spinning of our earth, as it goes in a single direction, spins on an axis, and it is a property that describes its...

ZainAlrawi_1J

Orbitals are not a real part of an atom, it is essentially a mathematical way of computing an atom an electron's wave-like and complex pattern of behavior.

ZainAlrawi_1J

As far as this class is concerned, always use the noble gas configuration when applicable unless otherwise indicated. It saves a lot of time and is definitely the "proper" way of doing it.

ZainAlrawi_1J

For elements that are considered exceptions, you write 4s --> 3d because 4s has less energy, but most of the time it you would just do 3d --> 4s

ZainAlrawi_1J

h (plack's constant) is usually given in J*s, which means mass is most simply computed in kilograms, and length in meters as J=kg×m^2

ZainAlrawi_1J

De Broglie does not apply to electromagnetic radiation as it is focused on objects with mass and momentum. All others (that we have learned at least) are EM equations

ZainAlrawi_1J

No, do not use molar masses as part of figuring out how many significant figures should be used. Only use given usable values in the equation.

ZainAlrawi_1J

The De Broglie equation is used to describe the wave properties of matter, and as such the equation is reserved for things that have mass, such as electrons. In practice, De Broglie is used when we are solving for an object's wavelength, and it has mass. This is in contrast to what you would use whe...

ZainAlrawi_1J

Since you have 1.26 moles of NH3 and 1.33 mol of O2, then you can already deduce that NH3 is the limiting reactant. You can now use the mole ratio of 4 NH3 for every 4 NO molecules to deduce how many grams of NO are produced, and then use the same process to figure out how many grams of H2O, by usin...

ZainAlrawi_1J

114 and 114. have the same number of sig figs.
114 and 114.0 do not have the same number of sig figs.

114 has 3 significant figures.
114.0 has 4 significant figures.

ZainAlrawi_1J

Capital M denotes molarity, which is moles of solute per liter of solution.

ZainAlrawi_1J

E1. Step 1: 1 mol Ag * (6.022*10^23 Ag atoms / 1 mol Ag) = 6.022*10^23 atoms of Ag Step 2: Multiply 144 by 2 to get 288 (convert radius to diameter), and then multiply by the number of atoms of Ag. --> 288 * 6.022*10^23 = 1.73*10^26pm Step 3: Convert our answer from picometers (pm) to meters (m). --...

ZainAlrawi_1J

The question is written confusingly but has a rather straight forward answer, it is more logic-based than it is chemistry. You have 9 atoms of Gallium at 70g/mol (total of 9 * 70 = 630). You want to figure out how many atoms of Astatine, which is given as 210 g/mol, would be equal to this. Therefore...

ZainAlrawi_1J

For some reason I can't seem to get the same answers as the textbook, anyone care to do a step by step? E.25 (a) Determine the number of KNO 3 formula units in 0.750 mol KNO 3. (b) What is the mass (in milligrams) of 2.39 ? 10 20 formula units of Ag2SO4? (c) Estimate the number of NaHCO 2 formula un...

ZainAlrawi_1J

In order to find out how many atoms of oxygen are in 5.15 g of Epsom salts we have to find out how many moles of magnesium sulfare heptahydrate (246.48 g/mol) are present in the sample. O = (5.15 g / 246.48 g/mol) * (11 mol O-atoms/1 mol MgSO4·7H2O) * (6.022x10^23 atoms/mol) = 1.38 x 10^23 atoms of ...

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