Search found 30 matches

by Megan Gianna Uy 3L
Tue Dec 04, 2018 3:02 pm
Forum: Hybridization
Topic: P orbital
Replies: 7
Views: 66

Re: P orbital

to know how many p orbitals there are count the regions of electron density.

1 region: s
2 regions: sp
3 regions: sp2
4 regions: sp3
5 regions: sp3d
by Megan Gianna Uy 3L
Tue Dec 04, 2018 3:00 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: polarity
Replies: 11
Views: 97

Re: polarity

A molecule that is a dipole (meaning positively charged on one side and negatively charged on the other) is polar. A molecule that is nonpolar is, as previously stated, symmetrical.
by Megan Gianna Uy 3L
Tue Dec 04, 2018 2:57 pm
Forum: Naming
Topic: Naming a Compound
Replies: 2
Views: 32

Naming a Compound

What are the rules to naming a compound and in what order do you write the element names?
by Megan Gianna Uy 3L
Thu Nov 29, 2018 11:52 am
Forum: Determining Molecular Shape (VSEPR)
Topic: AX2E2
Replies: 14
Views: 175

Re: AX2E2

Molecules with the molecular formula AX2E2 have four regions of electron density so its electron geometry would be tetrahedral. Its molecular geometry would be bent with a bond angle of <<109.5.
by Megan Gianna Uy 3L
Thu Nov 29, 2018 11:41 am
Forum: Determining Molecular Shape (VSEPR)
Topic: Radicals
Replies: 2
Views: 46

Re: Radicals

Typically, questions will not give you a radical. Most of the time the questions will take away or add an electron so that all electrons are either bonded or in a lone pair.
by Megan Gianna Uy 3L
Thu Nov 29, 2018 11:38 am
Forum: Biological Examples
Topic: CH 17 6TH EDITION HW 17.31d
Replies: 4
Views: 58

Re: CH 17 6TH EDITION HW 17.31d

sodium bisoxalato(diaqua)ferrate(III)

sodium: Na
bisoxalato: (C2O4)2
diaqua: (OH2)2
ferrate(III): Fe

Na[Fe(OH2)2(C2O4)2]
by Megan Gianna Uy 3L
Tue Nov 20, 2018 5:21 pm
Forum: Lewis Structures
Topic: Midterm question
Replies: 8
Views: 89

Re: Midterm question

Since resonance is the blending of structures, each N-O bond is an average of the double and single bond shown on the lewis structure. Therefore you would take the average of the two bonds. (140pm+120pm)/2 = 130pm
by Megan Gianna Uy 3L
Tue Nov 20, 2018 5:16 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Hund's rule vs Pauli's
Replies: 3
Views: 91

Re: Hund's rule vs Pauli's

Pauli exclusion principle:
1. no more than 2e- per orbital
2. if two electrons are in the same orbital then spin paired

Hund's Rule:
1. due to electron repulsion electrons in the same shell occupy different orbitals
2. with parallel spin
by Megan Gianna Uy 3L
Tue Nov 20, 2018 5:11 pm
Forum: Sigma & Pi Bonds
Topic: What makes up a sigma bond?
Replies: 1
Views: 50

Re: What makes up a sigma bond?

A sigma bond consists of two orbitals, each with 1e- interacting end-to-end to for 1sigma bond. Sigma bonds allow bound atoms to rotate, resulting sigma bonds have e- density with cylindrical symmetry around internuclear axis.
by Megan Gianna Uy 3L
Thu Nov 15, 2018 9:27 pm
Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
Topic: 6th edition 4.67
Replies: 2
Views: 49

Re: 6th edition 4.67

The higher valence shell, the lower the ionization energy. This is because the more inner shells there are, the more the valence electrons get shielded from the electrostatic attraction of the nucleus. Electrons close to the nucleus have a higher ionization energy than electrons further away.
by Megan Gianna Uy 3L
Thu Nov 15, 2018 9:25 pm
Forum: Formal Charge and Oxidation Numbers
Topic: Formal Charge
Replies: 10
Views: 199

Re: Formal Charge

If an element is in row 3+ then it can have an expanded octet. However, formal charge is still important because it makes sure your lewis structure is in its most stable form, making it easier for you to identify the shape of the molecule.
by Megan Gianna Uy 3L
Thu Nov 15, 2018 9:18 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Trigonal Planar
Replies: 5
Views: 52

Re: Trigonal Planar

Whether a structure has a single bond or multiple bonds do not matter in case of determining bond angles. Instead, look at the number of regions of electron density.
by Megan Gianna Uy 3L
Thu Nov 08, 2018 1:49 pm
Forum: Resonance Structures
Topic: Resonance Structures
Replies: 3
Views: 96

Re: Resonance Structures

Resonance structures are generally more stable since they are lower in energy.
by Megan Gianna Uy 3L
Thu Nov 08, 2018 1:45 pm
Forum: Lewis Structures
Topic: Mystery Element
Replies: 4
Views: 90

Re: Mystery Element

Typically, the center element would be the element that is the least electronegative. Or sometimes, the question will give you a specific element to use.
by Megan Gianna Uy 3L
Thu Nov 08, 2018 1:33 pm
Forum: Octet Exceptions
Topic: Octet
Replies: 6
Views: 101

Re: Octet

Exceptions to the octet rule include:
1. If the element is the the 3rd or more period, it can have more than an octet since it can use the d-subshell.
2. Radicals: H, He, Li, Be can have less than an octet
by Megan Gianna Uy 3L
Fri Nov 02, 2018 11:33 am
Forum: Lewis Structures
Topic: Formal Charge Tricks
Replies: 5
Views: 85

Re: Formal Charge Tricks

To find the formal charge use the formula FC = V (L+ S/2).
Where FC=formal charge
V= number of valence electrons
L= number of lone electrons that arent bonded
S= number of shared electrons that are being bonded
by Megan Gianna Uy 3L
Fri Nov 02, 2018 11:29 am
Forum: Photoelectric Effect
Topic: Photoelectric Experiment - Chem Test 2
Replies: 3
Views: 128

Re: Photoelectric Experiment - Chem Test 2

To find 5 moles of ejected electron, you would multiply your answer by plancks constant and multiply that by 5.
by Megan Gianna Uy 3L
Fri Nov 02, 2018 11:22 am
Forum: Octet Exceptions
Topic: The center atom
Replies: 20
Views: 314

Re: The center atom

In this case, oxygen is in the middle since the formal charge with oxygen in the middle is 0; whereas if chlorine is the the middle the formal charge would be +1.
by Megan Gianna Uy 3L
Thu Oct 25, 2018 11:12 pm
Forum: Quantum Numbers and The H-Atom
Topic: 1D.23 b and d, 7th edition
Replies: 1
Views: 56

Re: 1D.23 b and d, 7th edition

the ml value corresponds to the orientation of the orbital,
for example, the p orbital can have an ml value -1, 0, +1
which corresponds with the px, py, and pz orbitals.
Therefore, when the ml value is specified, there can only be one orbital.
by Megan Gianna Uy 3L
Thu Oct 25, 2018 11:03 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Units
Replies: 8
Views: 155

Re: Units

The Heisenberg Uncertainty Equation is deltap x delta x = h/4pi,
where deltap =m x deltav

m= mass (kilograms)
deltav = distance(meters)/time(second)
deltax = distance(meters)
h= planck's constant (kg x m^2/s)
by Megan Gianna Uy 3L
Thu Oct 25, 2018 9:49 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: electron config. 3d and 4s
Replies: 5
Views: 144

Re: electron config. 3d and 4s

3d is a lower energy level because it is in the 3rd shell, whereas the 4s electrons are in the 4th shell. This means that the 4s electrons are outer electrons, further from the nucleus, and are shielded by the 3d electrons. Also, when removing an electron, the electron would be removed from the 4s o...
by Megan Gianna Uy 3L
Fri Oct 19, 2018 10:50 am
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: H bar
Replies: 5
Views: 75

Re: H bar

He has the equation sheet posted on his website with all the equations he's giving us for the tests. In this case the equation on the sheet is (delta p)(delta x)>=h/(4pi), so you dont need to know the value of hbar since hbar =h/2pi.
by Megan Gianna Uy 3L
Fri Oct 19, 2018 10:44 am
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Nodal Planes
Replies: 3
Views: 80

Re: Nodal Planes

Nodes are regions of electron absense, since psi=0 at those regions. So, in the cone (nodal shape) there aren't any electrons there.
by Megan Gianna Uy 3L
Fri Oct 19, 2018 10:40 am
Forum: DeBroglie Equation
Topic: De Broglie equation
Replies: 3
Views: 63

Re: De Broglie equation

c=(lambda)(frequency) and E=h(frequency)
Therefore, E=(hc)/lambda
ALSO, E=mc^2, where m*c=p, so E=pc
Thus, pc=(hc)/lambda,
Therefore, lambda=h/p, and p=momentum(m*v)
So, lambda=h/(m*v)
by Megan Gianna Uy 3L
Thu Oct 11, 2018 9:15 pm
Forum: Photoelectric Effect
Topic: photoelectric effect
Replies: 2
Views: 59

Re: photoelectric effect

One of the equations that Lavelle wrote on his board was that E(photon)-E(energy to remove e)=E(excess). So basically, the total energy of the photon is equal to the energy required to remove an electron (threshold energy) plus the kinetic energy of the photon (excess energy). Another equation that ...
by Megan Gianna Uy 3L
Thu Oct 11, 2018 8:53 pm
Forum: Properties of Light
Topic: HW 1A.9 [ENDORSED]
Replies: 4
Views: 53

Re: HW 1A.9 [ENDORSED]

In order to find the energy, use the formula E=hv, since the frequency is given.
by Megan Gianna Uy 3L
Thu Oct 11, 2018 3:05 pm
Forum: Properties of Electrons
Topic: Help with question 1.13
Replies: 2
Views: 46

Re: Help with question 1.13

for the first part, use the rydberg formula where v=R( (1/n sub1^2)-(1/n sub2^2)) and plug that into c=lambda*v. Therefore you get c=lambda*(R( (1/n sub1^2)-(1/n sub2^2))). 3.00x10^8=(3.29x10^15)*((1/4^2)-(1/2^2))*lambda. lambda = 4.86x10^-7m = 486nm for part b, this belongs to the balmer series sin...
by Megan Gianna Uy 3L
Wed Oct 03, 2018 1:41 pm
Forum: Empirical & Molecular Formulas
Topic: Confusion on number of moles
Replies: 4
Views: 113

Re: Confusion on number of moles

Depending on the chemical equation, CuCl2H24H20 and Cl2 probably had a 1:2 molar ratio, respectively. Thus, you would have to multiply, .0417molsCuCl2H24H20 x (2molCl2/1molCuCl2H24H20), leaving you with .0834molsCl2.
by Megan Gianna Uy 3L
Wed Oct 03, 2018 1:34 pm
Forum: Empirical & Molecular Formulas
Topic: Writing formula from diagram
Replies: 2
Views: 65

Re: Writing formula from diagram

For exercises F1 I just counted the number of carbons, hydrogens, and oxygens that I saw on the diagram, using the given code. For F1 I got C10H16O.
by Megan Gianna Uy 3L
Wed Oct 03, 2018 1:27 pm
Forum: Limiting Reactant Calculations
Topic: Post-assessment problem, neutralizing acids
Replies: 2
Views: 60

Re: Post-assessment problem, neutralizing acids

For this problem, what I did was to first convert 1kg CaCO3 into moles of CaCO3 using its molar mass. 1000g/100.09g = 9.99molCaCO3. I assumed that CaCO3 was the limiting reagent since that was the only given information. Then, I converted 9.99molCaCO3 to moles of CO2. Since the balanced equation sho...

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