CHEMISTRY COMMUNITY

Deap Bhandal L1 S1J

Activation energy can be negative if the rate decreases as temperature rises (not common). However, I do not believe we have had a problem with this case, so I personally would double check that answer if I got it.

Deap Bhandal L1 S1J

I attached the Lewis structure pictures and Sahaj Gill is right.

Deap Bhandal L1 S1J

I believe it could be a fair concept to test on. As for problems, 15.30 explicitly asks about pseudo first order, but 15.29 and 15.37 contain related concepts.

Deap Bhandal L1 S1J

The specific heat capacity is the amount of energy it takes to raise a certain unit of mass a certain unit of temperature. Usually this unit of mass is grams and the unit of temperature is Celsius or Kelvin. For example 1 calorie is the amount of energy it takes to raise the temperature by one degre...

Deap Bhandal L1 S1J

The oxidation number of elements will usually add up to the overall charge of the ion. For example, in MnO4 with a charge of 1-, the oxidation numbers of the atoms must add up to -1. Oxygen will be -2 in most cases, and since there are 4 oxygen atoms, Mn would have an oxidation number of +7. +7 +4(-...

Deap Bhandal L1 S1J

As electrons leave the anode to the cathode, the anode part becomes positively charged while the cathode part becomes negatively charged. The porous barrier allows for ions usually sulfate to move from the cathode to the anode part, thus keeping the parts relatively neutral so the reaction can proce...

Deap Bhandal L1 S1J

Temperature cannot change the activation energy, but it does allow the collision frequency of the molecules to increase, meaning the reaction has an easier time crossing the activation energy barrier with an increase in temperature.

Deap Bhandal L1 S1J

Yes, for example, the k for a 10th order reaction would be M^(-9) * s^(-1).

Deap Bhandal L1 S1J

I do not believe we have.

Deap Bhandal L1 S1J

I believe the problem should specify since they are different.

Deap Bhandal L1 S1J

Since the test is from 15.1 to 15.6, I'm assuming the last point will be:
Know how to determine the order of a reaction, its rate law, and its rate constant for reactions with more than one reactant.

Deap Bhandal L1 S1J

A unique rate is the rate of a substances consumption or formation divided by the stoichiometric coefficient of that substance. This number will be the same for every substance in the reaction.

Deap Bhandal L1 S1J

That negative sign changes to plus since pH is the -log[H+]. When they changed log[H+] to pH they had to multiply it by -1. Hope this makes sense!

Deap Bhandal L1 S1J

I believe you would use the equation where it the ion goes to neutral. So like In3+ to In instead of In3+ to In2+.

Deap Bhandal L1 S1J

After multiplying the first reaction by 3 the Au on the left side of half reaction 2 will cancel out one of these on half reaction 1. If you are wondering why the electrons are on different sides for the reactions this is simply because Au is losing and gaining electrons. These are unique reactions ...

Deap Bhandal L1 S1J

Also I am assuming standard temperature and pressure.

Deap Bhandal L1 S1J

Items with a charge are usually aqueous, the metals are solid, while the diatomic molecules and O3 are gaseous. Basically, their location on the periodic table dictates their natural state if they are neutral.

Deap Bhandal L1 S1J

Actually you did most of the calculation right. One thing though. When you did it, your units should be L*atm*K^(-1). One liter times atm equals 101.33 J so when you multiply your answer: 0.0379 by 101.33 you get around 3.84 just like the solution. Hope this helps!

Deap Bhandal L1 S1J

I believe the more orientations a molecule has the more its entropy is. If a molecule did not have cis and trans versions it would have a lower entropy than a similar molecule that did.

Deap Bhandal L1 S1J

I believe they are talking about if a bunch of SO2F2 molecules take a disordered arrangement as a whole. In a crystal structure, all the molecules are aligned perfectly so that they all look alike 3 dimensionally. This question asks if they are not all aligned the same way (like the fluorine atom on...

Deap Bhandal L1 S1J

In 9.55 delta H (formation) equals delta H (reaction) since reactants for each equation you produce are in their standard states. These reactions have a total delta H of zero so subtracting the reactants like you did in 9.57 is not necessary.

Deap Bhandal L1 S1J

That molar entropy value you see is the actual entropy of that substance figuring in microstates and degeneracy and all that stuff. This can never be zero like enthalpy since all substances even ones in their natural states have entropy since the temperature never reaches 0 K. The standard entropy o...

Deap Bhandal L1 S1J

Basically for a delta H of a reaction, I would use the coefficients, whatever they may be. Even if all the numbers are doubled the delta h for the reaction would just get doubled and it would be the delta H for that reaction. If you are using J/mole divide the delta h reaction by the main product mo...

Deap Bhandal L1 S1J

I believe SO2 entropy change is positive and since the equation for delta g is enthalpy - T(delta S), as T increases the value of delta g becomes more negative making it more stable.

Deap Bhandal L1 S1J

Basically a negative delta G means its stable while a positive delta G means it is not. A positive delta G means that the formation reaction favors the reverse direction because of spontaneity so the product will degrade to its reactants most likely.

Deap Bhandal L1 S1J

I can explain part of your question. Basically work is being done in a reversible process but not for an irreversible process. When work is being done, q(surr) exists therefore entropy change of the surroundings does too. This will usually equal the entropy change of the system making total entropy ...

Deap Bhandal L1 S1J

For a reversible path, work is done, so there is a q(surr) this means entropy change for the surroundings is not 0. In fact it is the exact opposite of the entropy change in the system so they cancel out and make total entropy change equal to 0. for the irreversible process no work is being done, q(...

Deap Bhandal L1 S1J

There is actually a fourth equation too which derives delta g from a standard state cell potential: delta G= (-)moles of electrons*faraday's constant*cell potential. Depending on what information is given will tell you which equation to use. The accuracy of them depends on how accurate and precise t...

Deap Bhandal L1 S1J

Yes, from what I've learned so far, you have a good idea of what entropy is. Entropy is the degree of randomness or disorder rather than disorder or randomness itself. So the more microstates, the more the potential for a higher entropy.

Deap Bhandal L1 S1J

An example would be the delta H of a reaction is in kJ. However delta H formation could be written in kJ/mole of the product if we divide the original number by the molar coefficient of the product. The kJ/mole is specific for a substance and if you know the amount of moles of the substance you can ...

Deap Bhandal L1 S1J

Thanks guys!

Deap Bhandal L1 S1J

The engine requires gasoline and oxygen which are matter. It also dissipates heat. Since both matter and energy can enter/leave the system, it is considered open.

Deap Bhandal L1 S1J

I believe the piston had a string with mass attached to it on the outside of the chamber. Not fully sure, but I would recommend looking at the picture in your notes again.

Deap Bhandal L1 S1J

I know that if heat is released, q is negative. If the question asks for how much heat was released to we still put the negative sign on the number of Joules or calories?

Deap Bhandal L1 S1J

Actually delta u equals q + w, so delta u = delta h +pv. I think you are confusing it with the equation delta h = delta u -pv which is a derivation of the first one.

Deap Bhandal L1 S1J

Usually in the equation, they are using delta T and since both Celsius and Kelvin have the same unit interval, final temp minus initial temp would be the same.

Deap Bhandal L1 S1J

Normally the calorimeter is not included in the equations and is assumed that it does not loose heat, but it could be in certain cases. However each of the heat constants have units attached to them. If you are using q=mCdeltaT, make sure the c you choose has mass, energy, and a unit of temperature ...

Deap Bhandal L1 S1J

I believe you are correct. At a constant pressure, Delta H = Delta U - PV and when both pressure and volume are constant, Delta H = Delta U.

Deap Bhandal L1 S1J

Basically, from what I know, a state function usually has final state - initial state (path has no impact) while a non state function will require the whole path or value in addition to its initial and final. So like friction, a non state function, includes the total distance traveled by a particle....

Deap Bhandal L1 S1J

Yes, reversing a reaction changes the sign of delta H.

Deap Bhandal L1 S1J

Personally, I believe your TA would accept any 7 homework problems from the listed HW problems for chapter 8 on the syllabus. So far I know for sure he has talked about the measurement of heat which is 8.5 and the enthalpy of physical change which is 8.11.

Deap Bhandal L1 S1J

Overall, a standard enthalpy of formation is the change in enthalpy in the formation of 1 mole of the substance from the elements that make up the substance. For gases, we assume it follows the ideal gas law equation at 1 bar. For solutions, we assume 1 M and 1 bar.

Deap Bhandal L1 S1J

Yes, K and C have the same degree interval and to convert C to K you would add 273.15. In any equation I would recommend using K since that is SI. However in this case since the temp rises by 7.32 C; that means delta T is 7.32. If you converted to K and calculated delta T it would be the same since ...

Deap Bhandal L1 S1J

Partial pressure of each gas is a factor when determining Kp. A higher or lower partial pressure of a gas participating in the reaction can either shift the reaction to the reactants or products.

Deap Bhandal L1 S1J

When changing the pressure through volume, concentration is inversely affected since concentration is n over V. Adding an inert gas changes the pressure but has no effect on volume therefore no effect on concentration.

Deap Bhandal L1 S1J

pKb is not a constant. It is the negative log(base 10) of the base dissociation constant(Kb). Also I hypobromite is more basic than Morphine.

Deap Bhandal L1 S1J

I find it really funny when someone posts the answer while you’re typing it.

Deap Bhandal L1 S1J

https://lavelle.chem.ucla.edu/wp-conten ... ns_14A.pdf

It’s in his 14A website with the link: Final Exam Review Sessions.

Deap Bhandal L1 S1J

I believe so only that the transition metal does not follow the octet rule.

Deap Bhandal L1 S1J

The conjugate base of a strong acid placed in water is a week base. The conjugate acid of a strong base is a weak acid. I believe this should be true for both lewis and bronsted. Lewis definitions just include a wider array of acids and bases than bronsted.

Deap Bhandal L1 S1J

Lone pairs definitely affect the polarity of a molecule. Unless they cancel out, the dipole will point to the atom with the most lone pairs.

Deap Bhandal L1 S1J

Sometimes when using an ice table and plugging x back in, we must solve a quadratic formula. If K is small enough can we say x is negligible? If so, how small can K be so that x is negligible?

Deap Bhandal L1 S1J

I believe it depends if the metal is alone, has a charge, or is part of a larger compound. Metals with no charge are written just as their name appears on the periodic table. Metals with a charge are the same however the charge is written in parentheses. Metals part of compounds must have charges an...

Deap Bhandal L1 S1J

If a chemical reaction has both aqueous and gaseous substances, how would you determine K if given the concentrations? Would you use Kp or Kc? Also if using Kc, are gases included? Dr. Lavelle said only solid and liquids were excluded in the calculation.

Deap Bhandal L1 S1J

Does someone know the range of how many ligands a transition metal can receive? I know the common one is 6 but how higher and lower can they go?

Deap Bhandal L1 S1J

Well first I would draw the lewis structure, then count the regions of electron density for C and O since hybridization can explain their bonds. That number (3) corresponds with the sp2 hybrid orbital. The bond angle is determined by vsepr. The structure is a trigonal planar which results in 120 deg...

Deap Bhandal L1 S1J

In fact, the number of regions of electron density (number of lone pairs (count two as a pair) and bonds (single, double, triple all count as one)) will always be able to tell the hybridization configuration.

Deap Bhandal L1 S1J

It does not always have to be six, however that is a very common configuration for a coordination complex.

Deap Bhandal L1 S1J

Usually the least electronegative atom is the central atom.

Deap Bhandal L1 S1J

Radicals are in chapter 3, so a question regarding them on the midterm is possible. Radicals occur usually when there is an electron that is not part of a lone pair. An easy way to tell is if the compound has an odd number of outer electrons (the ones that go on the Lewis structure). To be sure you ...

Deap Bhandal L1 S1J

I’m assuming you mean in Lewis structures. Usually I just try to space them out. If there is one line pair, I just place it on the opposite side of the bond.

Deap Bhandal L1 S1J

A d orbital with a lower n value will always go before an s orbital of a higher n value. For example, Titanium would be 3d^2 4s^2

Deap Bhandal L1 S1J

In the p-orbital, for instance, does the l value from first to last go from -l to +l or +l to -l?

Deap Bhandal L1 S1J

When filling up orbitals does it matter if you start with spin downs or spin ups?

Deap Bhandal L1 S1J

Basically, the (n-1) is used for the d orbital since it is always one level lower in each configuration.

Deap Bhandal L1 S1J

Actually he drew the Lewis dot structures for salts like KF. The only difference was that he used square brackets to differentiate the individual charges of the ions. So, I believe you could do the same for ammonium sulfate (there would be two ammonium ions since it has a plus one charge while sulfa...

Deap Bhandal L1 S1J

Even though Xe is a noble gas why does it form molecules? And how would one do the Lewis dot structures for these molecules?

Deap Bhandal L1 S1J

I think this picture that was shown in the lecture illustrates a light wave pretty well.

Deap Bhandal L1 S1J

I really got a better understanding of this from Dr. Lavelle’s diagram. Normally for a large object we can use light beams to detect its position and velocity. However the same process will not work for an electron since the photons hitting the electron will cause the electron to take an unpredictab...

Deap Bhandal L1 S1J

For atomic size, it increases as you go down the columns but decreases as you go from left to right on the rows. Ionization energy follows the exact opposite trend. Metallic characteristics increase going left and down but decrease when going right and up.

Deap Bhandal L1 S1J

Another way to think of more photons is a higher amplitude. The wavelength determines whether the beam of light can eject an electron. Amplitude and wavelength can change independently from one another.

Deap Bhandal L1 S1J

In this case it's very similar. Velocity is just speed with direction. Sometimes a negative sign will denote a leftward direction and a positive will denote a rightward direction, but nowhere is this set in stone. As long as the reference point is kept constant throughout the experiment, there shoul...

Deap Bhandal L1 S1J

This equation is just a combination of the equations: E=hv and c= λv. When asked for the wavelength and given the energy or asked for the energy and given the wavelength, this equation is just quicker than using the two equations above.

Deap Bhandal L1 S1J

Thanks guys.

Deap Bhandal L1 S1J

Sometimes when I solve for the empirical formula I get moles like 3.1 or 2.9. Looking at the answers, I found out that rounds to 3. After what decimal digit (3.15 or 3.2) should I multiply all the moles by a constant to get better whole numbers?

Deap Bhandal L1 S1J

If the question asks for an amount in grams, then, using your example, O2 would have twice the amount of grams than O. If asking for moles, then it does not matter. However, in most questions oxygen means oxygen gas and that is O2 which has a molar mass of around 32g per mole.

Deap Bhandal L1 S1J

Hi Michelle,

Leading zeros are not significant digits so 0.020 would only have 2 sig figs. The 2 and the 0 after it are significant digits.

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