CHEMISTRY COMMUNITY

Robert

the O with the double bond becomes a single bond as those electrons are pointed to the oxygen atom* see the other arrow curving to the right. When that occurs, the carbon that is connected to the oxygen is a carbocation and has a positive charge.

Robert

The equation when solving for H is essentially E x 5 /.0592 = log Q Solve for Q = (10^E*5 /.0592) And solve for H. Take the negative log and you should get -1.667 OR When you have log Q, plug in the other concentrations and you should get log( 1000*[H+]^8) and separate the inside by the addition of ...

Robert

I believe there is a rule that if there is a substituent on one of the double bonds, then that carbon is labeled as the first one.

Robert

That's how I determined the anode. Whichever reduction potentiometer was more negative/least positive!

Robert

The units should be in Kelvin because the formula k=e^-Ea/RT

The constant R is in J/mol*K

Robert

viewtopic.php?f=75&t=3418

This was answered here :)

Robert

The reaction rate is decreased by 1/4 because Cl2 is second order. (.5x)^2 = .25(x^2)

Robert

The heat given off by the copper is gained by the surroundings: water. qsystem= -qsurr

So use the formula q= mCdeltaT and solve for the final temp

You need to get the specific heat capacities of copper and water from a constant sheet.

Robert

Just to confirm, when you have sec or ter before an alkyl, do you disregard those italicized prefixes when alphabetizing. But when you have iso or neo, those are accounted for in the alphabetizing

Robert

Internal alkenes/alkynes are more stable than terminal ones because when the bond is internal and connected to more than one carbon-secondary, tertiary, quaternary--, the pi bonds are more stabilized by the surrounding carbons. Also, I think that having a terminal double/triple bond would allow more...

Robert

NO also has resonances; double and triple bonds

Robert

Ethane has more degrees of freedom than that of ethene. Ethene has a double bond so it stays in a rigid structure. Each degree of freedom contributes 1/2R to the heat capacity Cv,m

Robert

Question 7B on the Winter 2011 Final Exam states: "Give the full IUPAC name for this molecule." The molecule features a (Z) alkene with a carboxylic acid, chlorine, and alkyl group as its substituents. The molecule is labeled as cis-Z and the sustituents are H, C2H5 on the left and Cl, and...

Robert

It'd be cool if we could "like" or mark a post as "helpful."

Robert

On page 49, the 3 does not indicate where the double bond is; it is indicating the -iodo part of the hydrocarbon.

Double bonds have priority when numbering, so in 3-iodobutene, you number the carbons from the left to right.

Robert

Edit: in the hw problems and the practice finals, it seems both are accepted unless specified.

Robert

"any correct newman projection" refers to eclipsed, gauche, and anti. +2

to get the other 2 points, "THE correct" one is the anti conformation which is the most stable.

Robert

I googled energy minimization and several websites state: "Perform an energy minimization of the system, by iteratively adjusting atom coordinates. Iterations are terminated when one of the stopping criteria is satisfied. At that point the configuration will hopefully be in local potential ener...

Robert

The OH- and Br- are substituted. The first step involves the nucleophile OH- attaching to the CH3Br- to reach the transition state. The reaction shown to be bimolecular because the rate limiting step has an overall order of 2

Page 159 of the Intro to Organic Chemistry text.

Robert

OH^- is left in the rate law because it's concentration is so small (from the fact that [H3O+]=[OH-]= 10^-7) that it does have influence.

Robert

k= Ae^-Ea/RT

When an enzyme is added, it lowers the activation energy barrier Ea. Therefore, the rate constant will increase.

Robert

You're probably referring to the question: Calculate the time required for the concentration of N2O... If you notice the rate constant has the units 1/s, that tells you that it's a first order reaction. First order reactions have the units 1/s because… rate= k[A] Units: M/s= k* M k= 1/s So use the i...

Robert

k= Ae^-Ea/RT

Here the rate constant changes as a function of temperature. Ea is also a constant*

Robert

Thank you, you're right. One method is pre equilibrium and the other is steady state approx. Just recalled that from lecture...

Robert

Page 73 of the course reader will answer your question. It's because in a real life instance, the balanced reaction is unknown.

when you collect data, plot the points and obtain at straight line, then you can get the k value (proportionality constant)

Robert

Step 1 A fast bimolecular dimerization and its reverse: NO + NO --> N2O2 Rate of formation of N2O2 = k1[NO]2 N2O2 ---> NO + NO Rate of formation of N2O2 = k1'[N2O2] Step 2 A slow bimolecular reaction in which an O2 molecule collides with the dimer: O2 + N2O2 ---> NO2 + NO2 Rate of consumption of N2...

Robert

In dilute solutions where H₂O is the solvent, it can be omitted because the concentration does not noticeably change.

Robert

I just finished this problem too… Yes, the overall reaction is determined that way by adding all three reactions and canceling what's on both sides. The rate of reaction is determined by the slowest step, which in this case is the 2nd step (very slow). Intermediates cannot be in the rate law, so you...

Robert

All the central atoms are carbon surrounded by 4 molecules. The difference between these four species though is the size. Larger size, complex molecules have higher entropies than smaller size molecules because they have higher vibrational energies (energy levels are closer together when larger)

Robert

To determine the charge of the Mn, you look at the molecule overall and its oxidation numbers. So permanganate, MnO4- has an overall charge of -1. Oxygen molecules have an oxidation number of -2 normally and there are 4 of them so -2 x 4 = -8. So now you have to determine what value + -8 = -1 Solvin...

Robert

The E value of the reduction potential that is the more negative/least positive has a higher reducing power , or is more likely to be oxidized ; therefore, at the anode, Cd is oxidized to Cd2+. To calculate the standard E of the cell, you can either flip the sign of the reduction potential of the an...

Robert

You're right. The combustion of glucose is exothermic and has a negative delta H value. The heat released from the reaction is now used to heat the water in the child's body, so the value changes to positive value. the heat lost by the system = the heat gained by the surroundings q sys = -q surr In ...

Robert

Chapter 13 problem 41 What you have to do is write out the half reactions for the cathode and the anode. When you add the two equations together, you will have concentrations and partial pressures on the product side and the same for the reactants. Then you can write Q by taking the multiplication ...

Robert

That was very helpful! I appreciate your comprehensive explanation.

Robert

Use the data in Appendix 2B and the fact that, for the half- reaction F2(g) + 2 H+(aq) + 2 e– --> 2 HF(aq), E° = +3.03 V, to calculate the value of Ka for HF. How are the significant figures applied in this problem? The book uses natural log, so when solving lnK = (2)(-.16)/(.02569)= -12.45 …. the ...

Robert

Thank you!

Robert

(c)Pt(s)|Cl2(g, 250 Torr)|HCl(aq, 1.0 M)||HCl(aq, 0.85 M)|H2(g, 125 Torr)|Pt(s) I know how to solve the problem using the correct equations; I just have a question on the reaction quotients. The reaction quotients have both partial pressure and concentrations, which i have never seen before. Is this...

Robert

Actually, I think I figured it out. I set up O2 --> H+ and O2 ---> OH- and balanced the half reactions by using the procedures if in acidic solution and basic solution, respectively. Probably had to do with the fact that water is amphoteric and has acidic and basic properties?

Robert

Write the half-reactions and the balanced equation for the cell reaction for each of the following galvanic cells:

(d) Pt(s)|O2(g)|H+(aq)||OH-(aq)|O2(g)|Pt(s)

How do I set up the two half reactions for each one? This one seems different from all the others.

Robert

(a) Calculate K at 25°C for the reaction Br2(g) #4 2 Br(g) from the thermodynamic data provided in Appendix 2A. (b) What is the vapor pressure of liquid bromine? (c) What is the partial pressure of Br(g) above the liquid in a bottle of bromine at 25°C? I only have a question for part c. The solutio...

Robert

Thanks!

Robert

I would not think so because a quotient is just a ratio and if you convert the units to another type of units, it still maintains that same ratio. For ex, 1 kPa to 2 kPa is still the same ratio in atm--1/2.

Robert

When analyzing a heat curve graph (heat vs Temp), I noticed that the slopes matter when heating a solid, liquid, and vapor. So if you're trying to determine the molar specific heat, C which is equal to q /[(moles) X deltaT], then you would just look at the q/ delta T (moles is just a constant)…. The...

Robert

For Quiz one, there are a couple problems I have encountered that require certain values (specific heat of iron (Q2) and heat of formation of glucose(Q5)) that are not provided in the constant sheets. These, of course, will be given to us on the quiz, correct?

Thank you

Robert

Delta G is calculated by the (sum of G_f of the products - sum of G_f of reactants).

O2 was not included because that value is 0.

:)

Robert

but for the three reactions of the three isomers, the solution manual has set up equations for the following and found the del S to be 1) isomer 1--> isomer 2 del S= -4.33 J/mol-K 2) isomer 1-->isomer 3 del S = -7.11 3) isomer 2-->isomer 3 del S= -2.78 How do you find the Sm for each isomer. my atte...

Robert

The entropy values are given in Joules in the appendix, so remember to divide entropy by 1000 to keep everything in kJ. Also for the sake of simplicity when using the equations such as delta G= delta H - TdeltaS

Robert

I know that organic means containing carbon, C. So any liquid hydrocarbons, alcohols, etc. look at appendix 2A, A16

Robert

Precisely!

dS = nR ln (P1/P2) --the formula when pressure is involved. notice that P initial is on top..

Robert

Thank you!

Robert

Three isomeric alkenes have the formula C4H8 (see the following table). (a) Draw Lewis structures of these compounds. (b) Calculate dG°, dH°, and dS° for the three reactions that interconvert each pair of compounds. (c) Which isomer is the most stable? (d) Rank the isomers in order of decreasing Sm...

Robert

dS = R*ln (V2/V1)

If it is compress to 1/3 its initial volume, then V2= (1/3) V1

Plug in V2 into the equation and the V1 will cancel, so you have ln (1/3)

R=8.314 J/K*mol

change in molar entropy = 8.314 * ln (1/3) = -9.13 J/mol*K

Robert

The equation you have written is derived from integration, which takes all the small changes in volume that occur. Not just the initial and final states, I believe.

Robert

Thanks Iris!

Robert

They took the enthalpy of combustion values given in Appendix 2A and multiplied it by the stochiometric ratios particular to each combustion reaction to get kJ/ mol CO₂.

Robert

I would think so given from the formula dS = nR ln (V2/V1).

"n" and "R" would be given constants and analyzing the graph of a ln function, an increase in volume would make V2/V1 positive and as the final volume gets larger, the magnitude of "ln (V2/V1)" does as well.

Robert

Problem: Calculate the standard entropy of vaporization of water at 85°C, given that its standard entropy of vaporization at 100.°C is 109.0 J/K-mol and the molar heat capacities at constant pressure of liquid water and water vapor are 75.3 J/mol-k and 33.6 J/mol-k, respectively, in this range So t...

Robert

NH (g) + HCl(g) --> NH4Cl(s) N2(g) + 3 H2(g) ---> 2 NH3(g) N2(g) + 4 H2(g) + Cl2(g) ---> 2 NH4Cl(s) You want: H2 + Cl2 --> 2HCl Do not divide the second equation by 3, reverse it!Since you reverse the 1st equation the NH3 is on the product side and if you reverse the second equation the NH3 will be...

Robert

Sorry I explained the whole problem; I interpreted your question incorrectly :(

Robert

Under the enthalpy section of the textbook, there's a chart that tells you what molecular geometry corresponds to the value of Cvm.

Robert

Treat them as separate ions. So Zn²⁺ and 2 Cl^-

Robert

When you multiple the variables, you get the units: L-atm. however work is the units Joules because it's a measure of energy. They multiply by this constant provided in the constant sheet/book. which is 101.325 J = 1 L-atm

Robert

If you mean delta H rxn (overall heat of the reaction), you would multiply the coefficients of the first equation by 2. By doing so, you must also multiply the delta H of that reaction. When you add the two equations together, the NO2 cancels out on both sides. Also add the heats of the reactions to...

Robert

delta H = delta U + (change in moles) R*T

-3920 kJ = delta U + (-3) (8.314) (298K)

*Make sure to convert [(-3) (8.314) (298K)] to kJ because the units are in Joules.

-3920kJ + (3)(8.314)(298) / 1000 = delta U

-3912.567

3 sig figs from initial "-3920"

so -3.91 x 10^3 kJ

Robert

7.53 Hydrochloric acid oxidizes zinc metal in a reaction that produces hydrogen gas and chloride ions. A piece of zinc metal of mass 8.5 g is dropped into an apparatus containing 800.0 mL of 0.500 M HCl(aq). If the initial temperature of the hydrochloric acid solution is 25°C, what is the final tem...

Robert

It's amount of heat required to raise the temperature of the water from 22- 100 degrees celsius. The calculation is mass of the water (400g) x specific heat capacity of water (4.184) x Change in temperature (373-295K) which equals to 1.3 x 10^5.

Robert

At the stochiometric point because pK_in = pKa. pKa= pH @ stochiometric point in weak-acid strong base weak base- strong acid titrations

Robert

empirical formula is the smallest whole number ratio of atoms; therefore, the molecular formula cannot be smaller than the empirical. However, the molecular formula and empirical formula can be the same in some cases when the ratio of the molecular formula cannot be reduced.

Robert

Look valence electron configuration and look at the very last subshell (3s, 2p, etc.). If it's p 2 then there are 2 unpaired electrons. Why? The p subshell has 3 orbitals px, py, pz. Each can hold a max of 2 electrons (pauli exclusion principle). Then, you have 1 e- in px and 1e- in py (following Hu...

Robert

First, you would set up the dilution equation M 1 V 1 = M 2 V 2 because you're using concentrated acid and diluting it to make a solution with a certain molarity. To find the volume you would need, divide the M 1 by both sides so you have V 1 =… M 2 V 2 = the number of moles HCl that you want, which...

Robert

If you think about it on a step by step basis, from the beginning to the stochiometric pt., the base is neutralizing the acid by taking a proton and creating more conjugate base. At the stochiometric point, the all acetic acid has been neutralized into its conjugate base. You found the volume of nee...

Robert

When you have a weak acid and a strong base titration and you titrate with a strong base before reaching the stochiometric point, you use the equation with the weak acid giving off a proton to water, making hydronium and the conjugate base. That's because the base is still pulling protons off the we...

Robert

CH₃NH₃+ would act as an acid because if you know that it's the conjugate acid of the weak base CH3NH2. (pg 451)

conjugate acids of weak bases give off protons...

Robert

NaCl exists as ions in aqueous solutions, not as a solid; therefore it is written as NaCl (aq). Mg(OH)2 is a solid because according to the solubility rules, compounds with OH- are insoluble except if they have cations of Ca2+, Ba2+, Sr2+. We didn't learn solubility rules in class but if you've take...

Robert

Probably very weak because if you look at the acid H3PO4, which has 3 ionization constants, the one for HPO4- (Ka3) is 2.1 x 10^-13. I was just looking at the chart on page 455 for polyprotic acids

Robert

Metal oxides tend to be basic whereas nonmetal oxides turn acidic.

Example: nonmetal oxide co2 + h20 > h2co3 ( acidic)

Metal oxide CaO + h2o > Ca(OH)2

Oxides with metalloid are amphoteric.

Robert

Just look at the graph sideways. However you would never draw a graph like that because the x axis is the independent variable and the y axis is the dependent variable. The volume of titrant added is the independent variable (variable being manipulated) and the pH is "dependent" on that.

Robert

I would say NH3 is a Bronsted base because it accepts a proton. and ammonium chloride is separate in an aqueous solution.

To your second question: HCl is a bronsted acid (proton donor). Lewis acids accept electron pairs, which it does not in aqueous solutions

Robert

Well what if you have a salt in solution, CH3NH3+ and Cl-. The cation that acts as an acid donates one H and is not polypro tic (at least i think so). Also NH4+. It donates 1 proton, although having 4 multiple hydroden atoms.

Robert

You can find the polyprotic Ka values on page 455 of the textbook. Good idea to bookmark them because the problems refer to tables 11.1,2 etc a lot.

Robert

If you have a low pKa, that means that your Ka value is high. Give by the equation -log [Ka] = pKa -> 10^-pKa = Ka. A lower pKa means the Ka value is higher and a higher Ka value means the acid dissociates more readily because it has a larger concentration of Hydronium ions (H 3 O + ). If you have a...

Robert

The species HCl is a Lewis acid because it acts as a proton donor, the proton being h+

Robert

I did that problem earlier. The answer was that [H30+] does not equal [HA]. Since it's a weak acid, not all of the H+ is released from the acid (deprotonated) to make H30+. pKa of 10 means the Ka is very small, also reinforcing that the acid doesn't deprotonate readily like a strong acid.

Robert

When in aqueous solution, the ions exist as nh4+ and br-. Br- is the conjugate base of a strong acid therefore it is neutral according to the rules in the text and in course reader I believe, and has no effect on the solution. Nh4+ is the conjugate acid of a weak base nh3. The rule is: conjugate aci...

Robert

I think it involves the concept that small, highly charged cations can be acidic... They are surrounded by water molecules (hydration shell) and can pull the electrons of the oxygen on one H2o closely and kick out a hydrogen, resulting in an oh and h+. The equation is similar to 11.71(e) if you look...

Robert

The solution from the back of the textbook is in the correct order. Is says the h2SeO3 is the weakest acid.

Robert

Metalloids are amphoteric

Robert

Nonmetal oxides tend to be acidic when reacted with water and metal oxides tend to be basic when reacted with water. The reason is that oxides O2- and water form hydroxide ions

Robert

Its actually Cuprate, using the latin prefix.

Use Cuprate when the overall charge in the complex is negative. Use copper if it is neutral or positive.

Robert

The old convention was to alphabetize but put the anions before the neutral ligands. That's why OH was before NH3. I think the new convention is to simply alphabetize the actually symbols in the formula.

Robert

Do you mean converting concentration (mol/L) using the stochiometric ratios. And I believe you can use that on the C row, where you usually put -x, +x, etc.,

Robert

For your first question, you alphabetize. I2, nh3, then o2. Same rule applies to multiple anions.

Robert

Oxygen is linear and can only bring to the metal cation on one side/ it can coordinate to one bonding region and contribute a pair of electrons. Although it has two paid of electrons. For other molecules with complex geometries that can surround the action like a claw, those ligand s can occupy mult...

Robert

We won't be dealing with cubics, but if you do run into them in the future, you could graph it and find the zeros (on a graphing calc of course)

Robert

The one that is linked to the metal ion will be underlined.

Robert

I think you got mixed up with the h and hcl reaction which is 4 x 10^-18.

The K value for the reaction according to the chart says 160. It's the third reaction listed :)

Robert

The book says prefixes aren't included when you alphabetize. But for polydentates and Ligands with a prefix already in the name, such as ethylene diaamine, you use the prefixes bis, tris, and tetrakis. Oxalate is a bidentate; therefore, you would use those prefixes. And you don't write biaqua becaus...

Robert

You're right! It's just rearranged differently to emphasize the part thats giving the electrons, which is oxygen.

Robert

Add the ending -ate to to the cation if the overall charge is negative!

Robert

I think they're equivalent because if you increase the concentration, then the partial pressure increases as well.

Formula for pressure is

P= (n/v) RT

(n/v)= concentration

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