Search found 20 matches

by Christina Quigg 4G
Wed Mar 09, 2016 8:31 pm
Forum: *Alkenes
Topic: 2011 Final Exam Q7B
Replies: 1
Views: 348

Re: 2011 Final Exam Q7B

It comes the double bond. The carbon attached to the carboxylic acid is considered to be 1. This is because functional groups get priority over double bonds in naming. Because of this, the double bond starts on carbon 2. Also, -enoic acid is the suffix for carboxylic acid. I hope this helps!
by Christina Quigg 4G
Fri Mar 04, 2016 1:23 pm
Forum: *Electrophilic Addition
Topic: Quiz 3 prep 3 #5 and #8 (p.50)
Replies: 1
Views: 331

Re: Quiz 3 prep 3 #5 and #8 (p.50)

I'm not sure about #8 but here is #5! IMG_0462.JPG In the first step, the double bond breaks as well and the bond between the H and the Br. The H is delta+ so it is attracted the delta- double bond. This causes the double bond to break. The H than goes to the one of the Cs from where the double bond...
by Christina Quigg 4G
Thu Feb 25, 2016 7:28 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Catalyst
Replies: 1
Views: 315

Re: Catalyst

In a multi-step reaction, a catalyst is something that is used up earlier in the reaction and reappears later. For example, catalysts usually appear on the reactants side of the equation first and then reappear in the products side in a later step. Example: A+E->G A+B->C C->B+D D+E->F F->G In this e...
by Christina Quigg 4G
Thu Feb 18, 2016 7:36 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: pre-equilibrium approach in course reader
Replies: 1
Views: 289

Re: pre-equilibrium approach in course reader

The 1/2(d[NO2]/dt) comes from the concept of 1/a(d[a]/dt). There are two moles of NO2 so d[NO2]/dt is multiplied by 1/2.
by Christina Quigg 4G
Wed Feb 10, 2016 2:53 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: winter 2011 question 5a
Replies: 3
Views: 502

Re: winter 2011 question 5a

You are balancing the equation . When you do this, everything except for glucose has a coefficient of 6.
by Christina Quigg 4G
Thu Feb 04, 2016 1:47 pm
Forum: Balancing Redox Reactions
Topic: 14.5b balance reaction
Replies: 1
Views: 334

Re: 14.5b balance reaction

The half reaction, Br_{2}^{-}\rightarrow 2BrO_{3}^{-} , is basic so in order to balance it, OH^{-} is used. It works the same way as the acidic redox reactions where you add H_{2}O to the side without O. Here is the problem worked out: Br_{2}^{-}\rightarrow 2BrO_{3}^{-} Br_{2}^{-}+6H_{2}O\rightarrow...
by Christina Quigg 4G
Thu Jan 28, 2016 9:07 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Volume change on entropy
Replies: 1
Views: 383

Re: Volume change on entropy

Entropy change can cause an increase or a decrease in volume. When the volume decreases, there is a smaller area for the energy to be so the "randomness" or entropy decreases. When the volume increases, there is a larger area for the energy to be so the "randomness" or the entrop...
by Christina Quigg 4G
Thu Jan 21, 2016 9:10 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Internal Energy
Replies: 2
Views: 483

Re: Internal Energy

Internal energy is considered a state function because the only thing that matter is the current value, not how this value was achieved. Having the same level of internal energy can be achieved different ways (such as having different heat and work values) so this is why it is a state function.
by Christina Quigg 4G
Thu Jan 14, 2016 9:32 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Enthalpy
Replies: 2
Views: 559

Re: Enthalpy

I think the difference between kJ and kJ/mol has to do with what the question wants you to find. If a substance is created and there are multiple mols involved, then it is kJ/mol. This is because kJ/mol has to do with quantity of kJs per mol of molecules. If only one mol is formed, then it would be ...
by Christina Quigg 4G
Wed Jan 06, 2016 10:12 pm
Forum: Phase Changes & Related Calculations
Topic: Explaining kJ/mol
Replies: 3
Views: 766

Re: Explaining kJ/mol

I think what Dr. Lavelle was talking was paying attention to what the question is asking to figure out the units. The question we were doing was about the formation of a substance. Enthalpies of formations are expressed as kJ/mol. If the question was asking about the enthalpy of forming 1 mol, then ...
by Christina Quigg 4G
Tue Dec 01, 2015 6:58 am
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: Determining whether a compound is an acid or a base
Replies: 1
Views: 514

Re: Determining whether a compound is an acid or a base

Think about it like you would for titrations. For KBr, the chemical equation is KOH+HBr makes KBr. KOH is a strong base and HBr is a strong acid. As we learned in titrations, strong acid and strong base together makes a neutral solution. For KF, it is KOH+HF produces KF. Like before KOH is a strong ...
by Christina Quigg 4G
Wed Nov 25, 2015 10:39 pm
Forum: Calculating the pH of Salt Solutions
Topic: Salts as Acids and Bases
Replies: 1
Views: 370

Re: Salts as Acids and Bases

From my understanding, anions of strong acids, such as Cl-, produce weak bases. This comes from the idea of having a conjugate base to an acid. Since Cl- can produce a strong acid like HCl, it will be a weak base. Because it is a weak base, it has no real effect on the pH and therefore is seen as ne...
by Christina Quigg 4G
Wed Nov 18, 2015 9:30 am
Forum: Naming
Topic: Naming
Replies: 1
Views: 319

Re: Naming

I believe that bis- and tris- are used when the molecule it applies to already has a Greek prefix or/and it is a ligand that is polydentate. For example, (en)_{2} would be bisethylenediamine because it is both a a polydentate ligand and the prefix di- is used in it. For writing out the molec...
by Christina Quigg 4G
Tue Nov 10, 2015 3:29 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Electron configurations for ions
Replies: 4
Views: 690

Re: Electron configurations for ions

For the d block electrons, with the exception of chromium and copper, the 4s electrons will be lost first. This is because the 4s orbital is the outer orbital, even though it is filled up before the 3d orbital. I hope this helps.
by Christina Quigg 4G
Tue Nov 03, 2015 2:31 pm
Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
Topic: Order of filling Molecular Orbitals
Replies: 1
Views: 586

Re: Order of filling Molecular Orbitals

The \pi bonding orbitals needs to be filled completely before the sigma 2pz bond can be filled. So with three electrons in the p orbital, it would be \left ( \pi 2px \right )^{2} and \left ( \pi 2py \right )^{1} . Like you said, it would be because the pi bonds would have slightly lo...
by Christina Quigg 4G
Wed Oct 28, 2015 4:22 pm
Forum: Octet Exceptions
Topic: Bonds in ClO2
Replies: 2
Views: 732

Re: Bonds in ClO2

There doesn't need to be any double bonds since the formal charge of the equation is 0. Both the Cl and and the O have a 0 charge. I believe when the charge is 0, the molecule is the most stable. I hope this helps.
by Christina Quigg 4G
Fri Oct 23, 2015 10:26 am
Forum: Electronegativity
Topic: HW 3.77(b)
Replies: 2
Views: 502

Re: HW 3.77(b)

The answer would be tex]CF_{4}[/tex] since F is the most electronegative element there is. So therefore, the electronegativity difference would be higher since nothing can be more electronegative than F. I hope this helps.
by Christina Quigg 4G
Sat Oct 17, 2015 1:01 pm
Forum: Lewis Structures
Topic: Determining between double bonds or single bonds
Replies: 1
Views: 350

Re: Determining between double bonds or single bonds

I think the reason why the double bond is between the O and the N is because each atom wants an octet. Because F has 7 valence electrons, it can easily get its octet by forming a single bond with N. O, on the other hand, has only 6 valence electrons. The only way both N and O can get an octet is by ...
by Christina Quigg 4G
Thu Oct 08, 2015 11:57 pm
Forum: DeBroglie Equation
Topic: Question #4 on Page 19 of Workbook
Replies: 2
Views: 446

Re: Question #4 on Page 19 of Workbook

I'm confused if you are talking about #4 on page 18 or #5 on page 19 so I will solve both. 4. c=\lambda \nu 3X10^8=580X10-9v v=5.2X10^14Hz Fairly straight forward. You use the equation c=\lambda \nu to find frequency. 5. For this problem, I used de Broglie to find the wavelength and I got the right ...
by Christina Quigg 4G
Sat Oct 03, 2015 7:33 am
Forum: Properties of Light
Topic: Photelectric effect when electron has 0 KE?
Replies: 3
Views: 583

Re: Photelectric effect when electron has 0 KE?

I believe that the electron would come off of the metal and be attracted the detector. With no KE, as long as the photon has reached the threshold energy, this would happen. The energy from the photon just needs to be greater than or equal to the energy needed to remove the electron for the photoele...

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