Search found 26 matches
- Fri Mar 17, 2017 5:37 pm
- Forum: *Cycloalkanes
- Topic: 2015 Final Q9A
- Replies: 1
- Views: 742
Re: 2015 Final Q9A
The main carbon chain is a propane because there is an -OH functional group attached to the second carbon, so the ending will be propanol. The 3 carbons are numbered from right to left, so the cyclohexyl is attached to the first carbon, and the methyl is attached to the second carbon. The cyclohexyl...
- Tue Mar 14, 2017 8:27 pm
- Forum: *Alcohols
- Topic: Functional Groups [ENDORSED]
- Replies: 2
- Views: 741
Re: Functional Groups [ENDORSED]
I believe my TA told us that we will only have to deal with one functional group as a substituent!
- Sun Mar 12, 2017 10:16 pm
- Forum: *Cycloalkenes
- Topic: Multiple double or triple bonds
- Replies: 2
- Views: 690
Re: Multiple double or triple bonds
Hi Chelsey!! I don't think you have to put a prefix indicating the amount of double or triple bonds that are present in a cyclo-alkene/alkyne; the numbering in front of the name should be enough to indicate the position and amount of double or triple bonds involved in the molecule. Actually, a pref...
- Mon Mar 06, 2017 6:51 pm
- Forum: *Alkanes
- Topic: Iso and Neo Prefixes
- Replies: 5
- Views: 765
Re: Iso and Neo Prefixes
The prefixes 'iso-' and 'neo-' are used for common naming instead of IUPAC naming; 'iso-' being used when there are 4 or less carbons, and 'neo-' when there are five or more. Common naming doesn't follow the rule of numbering the longest carbon chain, but instead counts all of the carbons. Using com...
Re: Naming
Lone pairs should be included in hybridization, because they are included in the number of regions of electron density.
However, I think that you don't need to mention them when naming the shape, although it does alter the structure of the molecules.
However, I think that you don't need to mention them when naming the shape, although it does alter the structure of the molecules.
- Sat Feb 25, 2017 12:53 am
- Forum: *Electrophiles
- Topic: Nucleophiles and Electrophiles
- Replies: 2
- Views: 422
Re: Nucleophiles and Electrophiles
Nucleophiles are negatively charged and donate electrons, while electrophiles are positively charged and are attracted to electrons.
- Sun Feb 19, 2017 12:17 am
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Kinetics Quiz
- Replies: 1
- Views: 396
Re: Kinetics Quiz
Dr. Lavelle mentioned in a post that the quiz will only cover up to pg. 73, so there shouldn't be any problems about the material after that!
- Sun Feb 12, 2017 8:00 pm
- Forum: Balancing Redox Reactions
- Topic: Cell Diagrams
- Replies: 1
- Views: 322
Re: Cell Diagrams
I think you'll always include H+ and OH- in the cell diagrams because they contribute to the overall charge, but since H2O has no charge it is not included.
- Sun Feb 05, 2017 9:19 pm
- Forum: Phase Changes & Related Calculations
- Topic: Balancing Equations/When To Use Integers and When Not To?
- Replies: 3
- Views: 470
Re: Balancing Equations/When To Use Integers and When Not To?
Usually, the only time we have really needed to use fractions for balancing equations is if we are calculating the standard enthalpy of formation; that is just so the manipulated equations will look like the final. Besides that, we will balance equations with integers.
- Wed Jan 25, 2017 6:29 pm
- Forum: Phase Changes & Related Calculations
- Topic: Quiz 1
- Replies: 11
- Views: 1191
Re: Quiz 1
The quiz will cover chapters 8, 9, and part of 11.
- Wed Jan 25, 2017 9:26 am
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: 8.67
- Replies: 1
- Views: 292
Re: 8.67
The problem I think you're referring to is #73a, but you are looking at the answer to #73b? The easiest way to approach these problems is to draw out the molecular structures, then you can compare the differences between the reactants and the products. For this problem, there will be 3 carbon-carbon...
- Fri Jan 20, 2017 4:45 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Chapter 8, Problem 61
- Replies: 5
- Views: 820
Re: Chapter 8, Problem 61
You only have to reverse the equation so the molecules will match the final equation. For example, since you were left with HBr on the reactants side of the equation, you would reverse the equation so that HBr will be on the product side.
- Tue Jan 17, 2017 10:01 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Problem 8.57 (Hess's Law)
- Replies: 2
- Views: 421
Re: Problem 8.57 (Hess's Law)
Also, the reason why the solutions have some of the coefficients as fractions is so that the equations, when added together, match the appearance of the final equation that is given.
- Sun Jan 15, 2017 2:28 pm
- Forum: Student Social/Study Group
- Topic: Tips/ Tricks for those New to Chemistry Community
- Replies: 4
- Views: 959
Re: Tips/ Tricks for those New to Chemistry Community
If you scroll a little bit down when you're asking/answering a question, there is a box that says "attachments", and you can upload a picture there or drag the file into the message box! Hope this helps
- Wed Nov 30, 2016 12:49 am
- Forum: General Science Questions
- Topic: Sig Figs
- Replies: 3
- Views: 786
Re: Sig Figs
Usually for my calculations I include about 4-5 places after the decimal for each step, then for the final answer round to however many significant figures the problem calls for.
- Wed Nov 30, 2016 12:46 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 11.117
- Replies: 2
- Views: 409
Re: 11.117
In the problem, you're given two values: 0.245 mol SO 3 and 0.240 mol SO 2 . The volume is 1.00 L, so your concentration will be of the same values. When you create your ICE table, the reaction initially starts off with none of the product (SO 2 ), and ends up with 0.240, you can conclude that the c...
- Sun Nov 27, 2016 4:31 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Problem 11. 87
- Replies: 1
- Views: 350
Re: Problem 11. 87
You know the reaction is endothermic because between the first and second image, it appears that several of the molecules have broken apart. This means that heat must have been added to the container, thus breaking the chemical bonds of the molecules. And when heat is added, that reaction is endothe...
- Sun Nov 27, 2016 10:27 am
- Forum: Naming
- Topic: Naming Compounds
- Replies: 1
- Views: 312
Re: Naming Compounds
The only compounds that should end in -ate are the ones that have a negative charge. So if a compound looks like [xx]2-, it will have the suffix -ate.
Hope this helps!
Hope this helps!
- Wed Nov 16, 2016 1:56 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Naming Ligands
- Replies: 5
- Views: 932
Re: Naming Ligands
They're exactly the same compounds, just the ones ending in -ido are from the new IUPAC naming.
- Sun Nov 13, 2016 1:04 am
- Forum: Naming
- Topic: Trans and Cis
- Replies: 3
- Views: 540
Re: Trans and Cis
Trans and cis molecules are isomers, meaning that they have the same molecular formula, and the basic difference between these molecules is the placement of its atoms. Also, trans molecules are non-polar while cis molecules are polar.
- Fri Nov 04, 2016 7:55 pm
- Forum: Hybridization
- Topic: Hybridization for expanded octets
- Replies: 3
- Views: 1080
Re: Hybridization for expanded octets
I think it should actually be written as dsp3, but other than that your answer is correct.
- Wed Oct 26, 2016 5:14 pm
- Forum: Lewis Structures
- Topic: CH3 HW questions
- Replies: 3
- Views: 528
Re: CH3 HW questions
So for #57c, by having three double bonds, the formal charge for the molecule is 0. If there were two double bonds, the overall charge on the molecule would be +2. Having more double bonds just reduces the formal charge of the molecule, since you want to be as close to 0 as possible. For #99, the re...
- Fri Oct 21, 2016 7:22 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: Question 2.19 in textbook
- Replies: 1
- Views: 322
Re: Question 2.19 in textbook
You only need to take the d into consideration when determining the number of orbitals, which in this case, is 5.
- Fri Oct 14, 2016 11:08 pm
- Forum: Significant Figures
- Topic: Addition and subtraction of significant figures with scientific notation?
- Replies: 4
- Views: 1010
Re: Addition and subtraction of significant figures with scientific notation?
When using significant figures during addition and subtraction, you have to round to the decimal place of the least accurate number.
- Sat Oct 08, 2016 6:04 pm
- Forum: Empirical & Molecular Formulas
- Topic: 2015 Quiz 1 Preparation Problem 5 from Workbook [ENDORSED]
- Replies: 1
- Views: 377
Re: 2015 Quiz 1 Preparation Problem 5 from Workbook [ENDORSED]
You did all of the work correctly, but just multiplied by the wrong factor to get the molecular formula. You said you ended up with C=1, H=2.38, and O=1, but you should round 2.38 up to 2.4 so when you multiply everything by 5, you should get C5H12O5. If you check the molar mass, it equals 152.15
- Sat Oct 01, 2016 4:33 pm
- Forum: Significant Figures
- Topic: Is there a sig fig resource in the textbook? [ENDORSED]
- Replies: 2
- Views: 528
Re: Is there a sig fig resource in the textbook? [ENDORSED]
There are two areas in the textbook that have some useful information on sig figs. In the front of the book pg. F10 has the definition and some examples, and the appendix in the back (A5) has some more detail on significant figures!