CHEMISTRY COMMUNITY

Yijia_Yang_3A

When a molecule is able to donate multiple protons. Usually happens when there are multiple H+.

Yijia_Yang_3A

take the negative log of H+ and do 10^-pH, then you will get H+

Yijia_Yang_3A

i think they are the same thing

Yijia_Yang_3A

Like previous answers, you need to compare pH and pKa. If pKa is lower, then it is negatively charged. If pKa is higher, it is neutral

Yijia_Yang_3A

I had a question about a problem from a UA Worksheet. It asks what the oxidation number for Na3[Co(OH)3Br3]. When we solve for the problem, should we assume Na3 will have a 3+ and is attracted to a coordination compound relating to that or should I go about it another way? I would say it is safe to...

Yijia_Yang_3A

I find it helpful to think of delocalized pi bond as something that would occur when 1). There are multiple resonance structures, and 2). There is at least one pie bond.

Yijia_Yang_3A

what dentate is CO3-2? Can't it be monodentate and bidentate bc metals can bond with the two oxygen ligands? correct me if I am wrong. Thank you! Also the same situation with oxalate... I think you are right. CO3 should be a polydentate since the two oxygens have a single bond and three lone pairs ...

Yijia_Yang_3A

CesarLec1 wrote:I believe see saw is not included because although it has 4 bonds in it, it is very uncommon for a molecule to take this shape as the tetrahedral or the square planar shape are much more desirable due to lower repulsions.

I am confused. Why then is square planar counted since it has two lone pairs?

Yijia_Yang_3A

It is Cobalt because there needs to be a transition metal cation in the sphere

Yijia_Yang_3A

same as the last post. when you multiply the total mass by the percentage of each element, you have to take what you end up with and divide that by its molar mass, which can be found on the periodic table.

Yijia_Yang_3A

I think so because the sapling hw asked us to. However I think it is helpful to know conceptually what they are when compared.

Yijia_Yang_3A

yes because it is the most stable state

Yijia_Yang_3A

a triple has 1 sigmas and 2 pi bonds. you wont be able to tell they are shown as straight lines

Yijia_Yang_3A

dipole is when one region of electronegativity is higher than another/others, reuslting in an uneven electron cloud.

Yijia_Yang_3A

You get AX3E as a result (one lone pair of electrons pushing down on the other three sorrounding elements), which translates into trigonal pyramidal

Yijia_Yang_3A

To answer your question brief: yes, the hydrogen atom needs to be bonded to N, or O, or F on both sides in order to form a H-bond, and the second molecule in which the hydrogen is bonded to needs to have a lone pair.

Yijia_Yang_3A

Always keep in mind that vapor pressure has an inverse relationship with boiling point.

Yijia_Yang_3A

ionic bonding is stronger than hydrogen bonding. Hydrogen bonding is a strong type of dipole-dipole, which ranks lower than ionic bonding.

Yijia_Yang_3A

I like to think that electronegativity is how hard it is to pull an electron away, and electron affinity as how likely an electron is able to join. Hope this is helpful.

Yijia_Yang_3A

It's negative because it's the amount of energy needed to be input to break the bonds. It wouldn't make sense to be positive because you would then have to give off energy. Hope it is helpful! :)

Yijia_Yang_3A

Mostly covalent bonds since you are trying to figure out whether it's polar or nonpolar. Also dipole refers to covalent bonded elements where electrons aren't equally shared. I guess you can do it for ionic bonds, but dipole definition wouldnt hold true.

Yijia_Yang_3A

nobel gases have full octet, thus they are not included for electronegativity purposes

Yijia_Yang_3A

The smaller the cation and the higher the charge, the greater the ability to distort the electron cloud. So, you want to compare these three elements in terms of size (electronegativity) and charge. We know Be2+ and Sr2+ have the same charge, so we find instead that Be is smaller than Sr, thus Be 2+...

Yijia_Yang_3A

Formal charges is low is when it's the most stable. If that's achieved, second rule is that more electronegative atoms are better off having a negative charge (closer to noble gas form).

Yijia_Yang_3A

I don't think there is any exceptions. The bigger the atom, the larger the LDF it experiences. So on the periodic table, LDF increases from left to right, top to down.

Yijia_Yang_3A

What i do is count the valence electron and subtract from it the number of dots and lines

Yijia_Yang_3A

I think we are only suppose to know the electron configuration of transition metals.

Yijia_Yang_3A

it means 4s is filled before 3d. But when you write it out, 3d is before 4s

Yijia_Yang_3A

As you go from left to the right side of the periodic table, nuclear charge increases, therefore there is more attraction to the center. once there is, the radii decreases

Yijia_Yang_3A

The expanded octet is more stable.

Yijia_Yang_3A

p is momentum, which is given by the multiplication of m and v, mass and velocity, respectively.

Yijia_Yang_3A

When you are trying to figure out wavelength, mass or speed, De Broglie is a useful option

Yijia_Yang_3A

dont trust atoms because they make up everything!

Yijia_Yang_3A

Such an inspiring story! I wish you the best of luck in the coming years!

Yijia_Yang_3A

true if the question is asked under the same condition, ie. same surface.

Yijia_Yang_3A

It is important to find the correct delta x in this question. It is given in the problem that the radius is .05nm, which translates into 5x10^-11 m. Since it said 1%, so multiply the radius by 1%, then 2 (as how the module was taught). Then the problem should get easy from there.

Hope it helps :)

Yijia_Yang_3A

10^-15 m! anything smaller than that has detectable wavelike property, bigger than that does not.

Yijia_Yang_3A

delta x in this case equals 1 (given in the question), and use delta=1 to solve for delta p in Heisenberg's equation. once you get p, divide that by 10^3 kg in turn you will get change in velocity. I believe whatever you'll get in change in v is a very small number, which will leave you defenseless ...

Yijia_Yang_3A

when energy is being emitted, it is negative; conversely, when energy is being absorbed, it is positive. energy itself isn't negative.

Yijia_Yang_3A

he said it was pure physics so i guess not

Yijia_Yang_3A

The energy gaps get smaller as it goes because as distance from nucleus goes, there is less force pulling the electron towards its nucleus. Hence, we understand that it requires more energy to pull the electron in.

Yijia_Yang_3A

Hi!

1. add the threshold energy and kinetic energy (which you obtained from solving a. and b.) together to find Etotal
2. use E=cV to solve for V.
3. use C=v x lambda to solve for wavelength

Hope this helps :)

Yijia_Yang_3A

1. find out Ek by plugging in the mass of the electron and the velocity of the electron respectively in equation Ek=(1/2)mV^2 2. since you obtained the work function (how much energy is needed to remove an electron from a zinc atom) from reading the question, all you need to do is add up the work fu...

Yijia_Yang_3A

first figure out the frequency by using the formula v=c/lambda, then plug in the frequency you get into the Rydberg formula. Because the wavelength falls into the UV spectrum, which in turns tells us that its initial energy should be 1, leaving n2 as the undefined variable. Plug everything in Rydber...

Yijia_Yang_3A

5 china

Yijia_Yang_3A

Yes, it ultimately gets too small to be counted as effective. Think about the concentration, since it keeps on getting diluted (90 times), whatever the content is gets too small to even exist because the volume of the solution gets too big.

Yijia_Yang_3A

The multiple choice question reads: during a summer camping weekend 4 moles of butane (C4H10) gas were used for cooking. Choose the right balanced equation for the combustion of 4 moles of butane gas. What is the net number of moles of gas produced? I was able to find the correct balanced equation: ...

Yijia_Yang_3A

Yes, usually the first steps of a problem is converting grams to moles. Simply divide the grams of the compound by how heavy one mole of this compound is, then you get the moles of this compound.

Yijia_Yang_3A

Often time when you are done with a problem, the end result you get should be in standard units. But honestly I think any unit should be fine since they are refer back to the same amount (assuming you have the correct results).

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