Search found 36 matches
- Tue Jun 12, 2018 5:37 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: 12.25
- Replies: 2
- Views: 375
Re: 12.25
thank you!
- Tue Jun 12, 2018 5:35 pm
- Forum: Bronsted Acids & Bases
- Topic: 12.65 e and f
- Replies: 3
- Views: 688
Re: 12.65 e and f
Determine whether an aqueous solution of each of the following salts has a pH equal to, greater than, or less than 7. Write a chemical equation to justify your answer. (a) NH4Br; (b) Na2CO3; (c) KF; (d) KBr; (e) AlCl3; (f) Cu(NO3)2. for e the solutions manual states that Al(H2O)6 3+ + H2O --> H3O+ +...
- Tue Jun 12, 2018 5:13 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Step Up
- Replies: 1
- Views: 334
Step Up
Hi,
Is the step up for today from 5-7 in Hedrick hall 125 still planned? I was just there and nobody was there. Thanks
Is the step up for today from 5-7 in Hedrick hall 125 still planned? I was just there and nobody was there. Thanks
- Tue Jun 12, 2018 3:18 pm
- Forum: Bronsted Acids & Bases
- Topic: 12.65 e and f
- Replies: 3
- Views: 688
12.65 e and f
Why does the solutions manual show Al(H2O)6 for e and Cu(H2O)6 for f?
- Tue Jun 12, 2018 2:38 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: 12.25
- Replies: 2
- Views: 375
12.25
I understand how to get the molarity of Ba(OH)2 and Ba2+ and OH-, but how do we find the molarity of H3O+?
- Wed Jun 06, 2018 10:40 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Wording of Questions Regarding Molecular Shape
- Replies: 2
- Views: 423
Re: Wording of Questions Regarding Molecular Shape
You may be confused between the difference of electron geometry and molecular geometry. When naming using the electron geometry, all the regions of electron density are included. So an atom that has two bonds and one lone pair would be trigonal planar since there are three regions of electron densit...
- Wed Jun 06, 2018 10:35 am
- Forum: Hybridization
- Topic: Hybridization
- Replies: 3
- Views: 424
Re: Hybridization
Yes, that's true. The idea of a hybridized orbital is that it is a way for us to describe and understand how bonding can happen when it appears that there are not enough unpaired electrons to form all of the bonds in a molecular structure. If we see that a molecular structure has 4 areas of electron...
- Tue Jun 05, 2018 7:29 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Ligand binding
- Replies: 4
- Views: 583
Ligand binding
I'm confused on how we know where a ligand can bind. Can someone help explain this?
- Thu May 31, 2018 8:43 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Molecular Geometry [ENDORSED]
- Replies: 1
- Views: 316
Re: Molecular Geometry [ENDORSED]
Electron geometry is the three dimensional configuration of a molecule by counting all the electron dense regions. This means lone pairs and bonds are considered equal. For example, if a central atom was surrounded by 4 bonds it would have the electron geometry tetrahedral. A central atom with 3 bon...
- Thu May 31, 2018 9:20 am
- Forum: Hybridization
- Topic: Hybridization and it's relation to valence electrons
- Replies: 3
- Views: 441
Re: Hybridization and it's relation to valence electrons
Yes. Carbon typical forms 4 singles bonds. Each of these single bonds will have a hypridization of 2sp3. This is because we need a better way to explain how carbon can form 4 bonds if we usually see the valence electrons written as 2s2 2p2.
- Thu May 31, 2018 9:18 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Problem 4.9
- Replies: 3
- Views: 454
Re: Problem 4.9
If you draw the lewis structure, you will see that I forms 1 bond with each Cl atom. After giving each Cl a full octet, there are still 4 electrons that are left over. Two lone pairs are added to I because it can have more than an octet. The lewis structure will now have 3 bonds and 2 lone pairs aro...
- Wed May 23, 2018 9:56 pm
- Forum: Resonance Structures
- Topic: Worksheet 7 #2
- Replies: 1
- Views: 402
Worksheet 7 #2
Can someone please attach a pic of the two resonance forms? I can only figure out one of them.
- Wed May 23, 2018 8:48 pm
- Forum: Polarisability of Anions, The Polarizing Power of Cations
- Topic: Dipole moments
- Replies: 3
- Views: 708
Re: Dipole moments
If a bond is polar, it has a dipole moment. A polar bond means that the two atoms have different electronegativities. Whichever has a larger electronegativity is partially more negative and the atom is partially more positive. This unequal sharing of electrons causes a dipole moment. To show this, y...
- Wed May 23, 2018 8:39 pm
- Forum: Lewis Structures
- Topic: Homework Problem 3.65
- Replies: 3
- Views: 440
Re: Homework Problem 3.65
As a general statement, oxygen is most stable when it has 2 bonded pairs and 2 lone pairs. For these structures, they break the octet rule because Xe can have more than 8 valence electrons. Giving oxygen a double bond in order to use all the electrons is the most stable way to draw these structures.
- Wed May 23, 2018 8:33 pm
- Forum: Resonance Structures
- Topic: 3.117
- Replies: 2
- Views: 476
Re: 3.117
The question is: "Structural isomers are molecules that have the same formula but in which the atoms are connected in a different order. Two isomers of disulfur difluoride, S2F2, are known. In each the two S atoms are bonded to each other. In one isomer each of the S atoms is bonded to an F ato...
- Wed May 16, 2018 9:40 am
- Forum: Formal Charge and Oxidation Numbers
- Topic: Reducing formal charge number
- Replies: 3
- Views: 844
Re: Reducing formal charge number
Yes, especially if you need to make a resonance structure you should check the formal charge on each atom to see which resonance structure is most stable. If there are more than one, then you would draw all of them because they are equally likely to occur. Methods to reduce formal charge would be to...
- Wed May 16, 2018 9:34 am
- Forum: Resonance Structures
- Topic: CN- Resonance structures [ENDORSED]
- Replies: 2
- Views: 1378
Re: CN- Resonance structures [ENDORSED]
The octet rule states the each element must have 8 surrounding electrons. One lone pair is 2 electrons and one bond is 2 electrons, so C would only have 4 electrons. So C would need more bonds or lone pairs. C typically prefers to have 4 bonds more than 4 lone pairs so it is more likely that C would...
- Tue May 15, 2018 10:39 pm
- Forum: Ionic & Covalent Bonds
- Topic: Name to Bond [ENDORSED]
- Replies: 1
- Views: 308
Name to Bond [ENDORSED]
What does it mean if a molecular name has parentheses? For example, what does thallium(III) chloride mean?
Thanks!
Thanks!
- Wed May 09, 2018 10:58 am
- Forum: Trends in The Periodic Table
- Topic: 2.67 part b
- Replies: 2
- Views: 315
Re: 2.67 part b
Electron affinity is the energy released when an electron is gained. This means that atoms that want to gain an electron will have a higher electron affinity. In order to fill Carbon's outer shell, it would have to gain 4 electrons. It would appear that Nitrogen only has to gain 3 electrons, so it w...
- Wed May 09, 2018 10:53 am
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: 2.59
- Replies: 2
- Views: 419
Re: 2.59
Ionic radius is not the same as being isoelectric. All of these ions have the same number of electrons, which makes them isoelectronic. However, ionic radius decreases as you go from left to right on the periodic table. Since P3- is the furthest left, it has the largest ionic radius and so forth.
- Wed May 09, 2018 10:35 am
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: 1.23 Homework
- Replies: 2
- Views: 387
Re: 1.23 Homework
So now that you have E= 2.2513 x 10^-14 J you can use the equations E=hv and c=vλ
You combine the equations to:
λ= hc/ E
λ= (6.626x10^-34 Js) (2.997x10^8m s-1)/ (2.2513x10^-14J)
λ=8.8237x10^-12 m
You combine the equations to:
λ= hc/ E
λ= (6.626x10^-34 Js) (2.997x10^8m s-1)/ (2.2513x10^-14J)
λ=8.8237x10^-12 m
- Tue May 01, 2018 11:14 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: Magnetic Quantum Number
- Replies: 2
- Views: 428
Re: Magnetic Quantum Number
In order to find this value you would have to be given another value. For example, if you were given that l=2 you would have
ml=−l,(−l+1),(−l+2),…,−2,−1,0,1,2,…(l–1),(l–2),+l
ml= -2, (-2+1), (-2+2), (2-1), 2
ml= -2, -1, 0, 1, 2
ml=−l,(−l+1),(−l+2),…,−2,−1,0,1,2,…(l–1),(l–2),+l
ml= -2, (-2+1), (-2+2), (2-1), 2
ml= -2, -1, 0, 1, 2
- Tue May 01, 2018 8:04 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: 2.43
- Replies: 7
- Views: 5452
2.43
Can some explain to me why silver is [Kr] 4d10 5s1 instead of [Kr] 4d9 5s2?
Thanks!
Thanks!
- Tue May 01, 2018 7:44 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Zeff
- Replies: 1
- Views: 277
Zeff
Can someone please explain what Zeff is. It's in problem 2.37 in the textbook.
Thanks for the help!
Thanks for the help!
- Thu Apr 26, 2018 8:51 am
- Forum: Photoelectric Effect
- Topic: Energy spectrum
- Replies: 4
- Views: 415
Re: Energy spectrum
The energy levels were found experimentally so they are not perfect like we would imagine if we just came up with the idea. Because Hydrogen only has one electron, it would make sense that the first shell has the highest energy, and every subsequent shell after that has less.
- Wed Apr 25, 2018 9:40 pm
- Forum: Photoelectric Effect
- Topic: Module Questions 33 and 34 [ENDORSED]
- Replies: 3
- Views: 2616
Re: Module Questions 33 and 34 [ENDORSED]
Can you explain the part about subtracting the ""leftover" energy (which will equal the energy of the ejected electron) by subtracting the work function (required energy to eject electron) from the energy of the photon that the material was hit with". Numbers might help. Thanks!
- Wed Apr 25, 2018 9:16 pm
- Forum: Photoelectric Effect
- Topic: Module Question 29
- Replies: 5
- Views: 486
Re: Module Question 29
Does anyone know why we don't add the kinetic energy to the energy to remove e-?
I thought that to find the energy needed to eject the e- we had to add the work function to the kinetic energy.
I thought that to find the energy needed to eject the e- we had to add the work function to the kinetic energy.
- Tue Apr 17, 2018 10:32 pm
- Forum: Properties of Light
- Topic: The speed of light
- Replies: 4
- Views: 453
Re: The speed of light
Yes, the speed of light is a constant that for the equations we use will always be 3.00x10^8 ms-1.
- Tue Apr 17, 2018 10:25 pm
- Forum: Properties of Light
- Topic: Series Question
- Replies: 3
- Views: 478
Series Question
Can someone explain to me exactly what a series is. I know that they are called Lyman, Balmer, etc but I don't understand how they are categorized into different series. Thanks!
- Tue Apr 17, 2018 10:23 pm
- Forum: Properties of Light
- Topic: HW 1.15
- Replies: 3
- Views: 472
Re: HW 1.15
You know that you need the equation v=R{(1/n1^2)-(1/n2^2)} to find the n values. But first you need to find the value of v. v=c/λ =3.00 x 10^8 ms-1/ 102.6 x10^-9 m = 2.922 x 10^15 s- You also know that n1=1 because of the Lyman series so the unknown you are looking for is n2. Rearrange the equation ...
- Wed Apr 11, 2018 9:02 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: question on fundamentals G
- Replies: 5
- Views: 845
Re: question on fundamentals G
First, you need to find the molarity of Na2CO3 M=n/V= (2.11gNaCO3mol/105.951g)/.25L = .08 molL-1 Then you can solve each equation for a volume a) V= 2.15x10-3 mol Na+ x 1 mol Na2C03 / .08 molNa2CO3 L-1 x 2mol Na+ = 1.35x 10-2L b) V= 4.98x10-3 mol CO3 x 1 mol Na2C03 / .08 molNa2CO3 L-1 x 2mol CO3 = 6...
- Wed Apr 11, 2018 8:44 pm
- Forum: Limiting Reactant Calculations
- Topic: Question 8 on UA's Practice Test
- Replies: 3
- Views: 402
Re: Question 8 on UA's Practice Test
Now that you know that you have .21 moles HCl, you want to get ?gAlCl3
.21moles HCl (1 mole AlCl3/ 1 mole HCl) (133.34 g AlCl3/mole AlCl3)= 28 g AlCl3
.21moles HCl (1 mole AlCl3/ 1 mole HCl) (133.34 g AlCl3/mole AlCl3)= 28 g AlCl3
- Wed Apr 11, 2018 8:37 pm
- Forum: Balancing Chemical Reactions
- Topic: Fastest way to balance chemical equation
- Replies: 9
- Views: 16796
Re: Fastest way to balance chemical equation
I'm not sure if this is the fastest way, but if you are stuck the easiest way to start is by making a system of equations. For example if you have the reaction HF + SiO2 --> SiF4 + H20 then you can set that up as aHF + bSiO2 --> cSiF4 + dH20 For each element go through so H: 1a= 2d F: 1a= 4c Si: 1b=...
- Sun Apr 08, 2018 9:00 pm
- Forum: Balancing Chemical Reactions
- Topic: H3 Image [ENDORSED]
- Replies: 2
- Views: 375
Re: H3 Image [ENDORSED]
the dark diamond is Si the dark circle is O and the open circle is H The first box (the reactants): 1 dark diamond is connected to 4 open circles (SiH4). This is repeated 2 times (2SiH4). 1 dark circle is connected to 2 open circles (H2O). This is repeated 4 times (4H2O). The second box (the product...
- Sun Apr 08, 2018 8:42 pm
- Forum: Balancing Chemical Reactions
- Topic: Catalysts [ENDORSED]
- Replies: 4
- Views: 348
Re: Catalysts [ENDORSED]
A catalyst just changes the speed of a reaction but does not change the amount of reactant or the amount of product that is produced.
- Sat Apr 07, 2018 1:00 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: Video 4 - post assessment [ENDORSED]
- Replies: 2
- Views: 297
Re: Video 4 - post assessment [ENDORSED]
A solution is prepared by dissolving 55.1 g of KCl in approximately 75 mL of water and then adding water to a final volume of 125 mL. What is the molarity of KCl in this solution? molar mass KCl= 74.55 gmol-1 Find the moles in 55.1g KCl 55.1g/74.55gmol-1 = .739 mol Find the Molarity of the solution ...