CHEMISTRY COMMUNITY

Dabin Kang 1B

The reaction quotient (Q) is similar to the equilibrium constant (K), but K is the constant when the reaction is in equilibrium while Q is for any stage of the reaction except when it is at equilibrium.

Dabin Kang 1B

KP = PPCl3 PCl2 / PPCl5
KP (PPCl5 / PCl2) = PPCl3 (PCl2 / PPCl5)(PPCl5 / PCl2)
KP PPCl5 / PCl2 = PPCl3

Dabin Kang 1B

Because the reactants are being used to make the product, the equilibrium concentration of the reactants would be less than the initial while the product's equilibrium concentration would be higher than its initial. Since we don't know the amount of reactants used to make the product, we would put i...

Dabin Kang 1B

They will probably be given.

Dabin Kang 1B

P4 is nonpolar because the 4 P atoms are arranged about a center of symmetry and they have the same electronegativity.

Dabin Kang 1B

Yes, but you should still try plugging in both numbers just to make sure. :)

Dabin Kang 1B

We skipped molecular orbital theory so I don't think we have to do this problem.

Dabin Kang 1B

viewtopic.php?p=2547&sid=0d36d6a0a83ae99de2b5592ede5e800f#p2547

Dabin Kang 1B

Dabin Kang 1B

Electronegativity is directly proportional to effective nuclear charge, so when electronegativity is high, the nuclear charge is high as well. The high nuclear charge strongly attracts the electrons and pulls them in, decreasing the energy of the orbitals.

Dabin Kang 1B

One exception to the octet rule would be when atoms exceed their octet. Starting from phosphorus in period 3, atoms can utilize their d subshell in bonding.

Dabin Kang 1B

In molecules like $SO_{4}^{2-}$ in which sulfur exceeds its octet, the remaining electrons would go into the d orbitals, which should be filled in the same way the s and p orbitals are filled using the Pauli exclusion principle and Hund's rule.

Dabin Kang 1B

I suppose the arrangement containing the lone pairs is called the electron geometry and the actual shape of the molecule is called the molecular geometry. This is a great simulation that shows both the electron and molecular geometry: https://phet.colorado.edu/sims/html/molecule-shapes/latest/molecu...

Dabin Kang 1B

The ground state electron configuration of an element is the same as the regular configuration in which the electrons are in the lowest possible energy state. For example, the ground state electron configuration of oxygen is 1s2 2s2 2p4. The excited state electron configuration shows when an electro...

Dabin Kang 1B

This isn't a complete explanation but here's a list of what I have observed from looking at these graphs. The radial distribution function equals zero when ever the radius is equal to zero, signifying that the probability of finding an electron at the nucleus is zero. It can be seen that radii of th...

Dabin Kang 1B

The electron configuration of nitrogen is 1s2 2s2 2p3, which shows that each p-orbital contains one electron. However, the electron configuration of oxygen is 1s2 2s2 2p4, which shows that the px orbital contains two electrons. This leads to electron-electron repulsion which causes instability and m...

Dabin Kang 1B

Na has a smaller first ionization energy than Al because Al has a greater nuclear charge than Na, which causes the electrons to be more attracted to the center. Also, ionization energy tends to decrease down a group and increase across a period. The first ionization energy is the amount of energy it...

Dabin Kang 1B

Oh, I thought you were asking about part C. You would use $\lambda =\frac{h}{p}$ to solve for the wavelength. How did you get E=vh/λ ?

Dabin Kang 1B

The question doesn't ask for the wavelength of an ejected electron. It asks for the wavelength of the photon that caused the ejection of an electron.

Dabin Kang 1B

Find the molar mass of the unknown element M by subtracting the mass of the hydrogen and the oxygen atoms from the given molar mass of the molecule: 74.10 - 2(1.008 + 15.999) = 40.086 g/mol This molar mass is the closest to the molar mass of Calcium, so we can assume that M is calcium. Calcium sulfi...

Dabin Kang 1B

Use E\left ( photon \right )=E\left ( threshold \right )+E_{k} Find the energy of the photon: E\left ( photon \right )=E\left ( threshold \right )+\frac{1}{2}m_{e}v_{e}^{2} E\left ( photon \right )=\left ( 1.66\times 10^{-17} J\right )+\frac{1}{2}\left...

Dabin Kang 1B

You can also convert them to fractions and multiply by the denominator.

For example, you can convert 1.666 to $\frac{5}{3}$ and multiply by 3 to get 5.

CHEMISTRY COMMUNITY

Search found 22 matches

Re: Reaction Quotient

Re: Example from Pg 134 in Course Reader

Re: Change

Re: Ka and Kb values

Re: Question 4.39 [ENDORSED]

Re: Positive Concentrations

Re: HW Problem 4.77

Re: Homework Problem 4.43

Re: Post All Chemistry Jokes Here

Re: Relationship between Electronegativity and Orbital Energy

Re: Octet Rule [ENDORSED]

Re: The use of hidden d orbital

Re: Naming Compounds

Re: Ground state vs. excited state

Re: Graphs on page 44 of the textbook

Re: Question 2.81

Re: ionization energy

Re: Question about 1.33

Re: Question about 1.33

Re: Problem E.15 in Fundamentals

Re: Question about 1.33 Part C

Re: Whole Number Coefficients