CHEMISTRY COMMUNITY

Emma Strassner 1J

In this case it is included because it is part of the reverse reaction from the first step that is being used to create the rate law you need. You would included it even though it is an intermediate.

Emma Strassner 1J

I agree, you need to use the Arrhenius equation for this problem. Do it for the catalyzed and uncatalyzed reactions by dividing them, and you'll notice that A cancels out for both. From there just plug in the needed values and solve (remember to watch the units, as R is in J and the given values are...

Emma Strassner 1J

Like people said, if you know the general rules about elements like O and H, you can usually figure out the other oxidation numbers from there.

Emma Strassner 1J

Can someone explain how when we're given multiple steps whether to tell the difference between what's an intermediate and what is a catalyst? I feel like a catalyst would be present in the fast step because that's what makes it fast and we would see it also in the product but I don't know if this i...

Emma Strassner 1J

It is rare, but theoretically yes. All of the reactants would have to be zero order for this to occur.

Emma Strassner 1J

Hi, I agree with what you're thinking. The rate laws come from the balanced reactions and the stoichiometric coefficients, but just remember that intermediates are not included in the rate law itself.

Emma Strassner 1J

Mari Williams 1K wrote:Is the order of the reaction referring to the total sum of the order of the reactants?

The overall order of the reaction is the sum of the orders of each of the reactants in your rate law.

Emma Strassner 1J

The zero order graphs are linear with the given data, [A] vs. Time. First order graphs come when ln[A] is plotted vs. time and gives a straight line. Second order graphs come when 1/[A] is plotted vs. time and gives a straight line.

Emma Strassner 1J

Knowing the laws and half-life equations is good but memorizing how they are derived is not as important, just having an understanding of how they were derived is good.

Emma Strassner 1J

Thermodynamics is primarily concerned with stability of a reaction, but kinetics is concerned with the rate of the reaction.

Emma Strassner 1J

The log equation is used when you are at 298 K (25 C). Otherwise, use the ln equation because it has the variable for temperature in it.

Emma Strassner 1J

The anode must go on the left and the cathode must go on the right so that it is consistent when you see how the electrons are flowing.

Emma Strassner 1J

I'm seeing both 273.15 K and 298 K on this thread for standard temperature. I thought that it was 298 K (25 degrees celsius), but could someone please clarify which one it is? The temperature at STP is 273.15 K (0 C) and the temperature at SATP is 298 K (25 C). Standard conditions for delta G naugh...

Emma Strassner 1J

Okay great, thanks!

Emma Strassner 1J

Calculate the standard potential, E∘, for this reaction from its ΔG∘ value.

X(s)+Y2+(aq)⟶X2+(aq)+Y(s)ΔG∘=63.0 kJ

Which equation would we use for this problem?

Emma Strassner 1J

That makes more sense, thank you!

Emma Strassner 1J

Write a balanced overall reaction from these unbalanced half-reactions.

In⟶In3+

Cd2+⟶Cd

I tried this problem by adding electrons but didn't get the right answer, how would you add the electrons to find the overall reaction?

Emma Strassner 1J

The anode is always the oxidized side because that is where the electrons are leaving for the reduced side.

Emma Strassner 1J

E is not a state function because you need to know the path taken. You cannot just figure it out from the final state to the initial, you need to know the process.

Emma Strassner 1J

You would also need to know the two temperatures and the equilibrium constants themselves (depending on the version you are using). Knowing delta H would also be useful.

Emma Strassner 1J

Okay, I think it was the units messing me up, thanks!

Emma Strassner 1J

For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 bar for all species. For the reaction N2(g)+3H2(g)↽−−⇀2NH3(g) the standard change in Gibbs free energy is ΔG∘=−69.0 kJ/mol. What is ΔG for this reaction at 298 K when the partial pressures are PN2=0.400 bar, PH2=0.350 ...

Emma Strassner 1J

That makes sense, thank you guys!

Emma Strassner 1J

Hi, I am having some trouble finding delta H for this problem, which equation would I use for that?

Emma Strassner 1J

Like people said above, the best way is to approach it as a Hess-like situation. You need to manipulate the reactions so that you can cancel things out to get the final result. Just remember that any changes you make there should be reflected in the delta H and delta S values before you plug them in...

Emma Strassner 1J

Okay, just wanted to make sure, thank you!

Emma Strassner 1J

A constant‑volume calorimeter was calibrated by carrying out a reaction known to release 1.32 kJ of heat in 0.600 L of solution in the calorimeter (q=−1.32 kJ) , resulting in a temperature rise of 2.18 C . In a subsequent experiment, 300.0 mL of 0.40 M HClO2(aq) and 300.0 mL of 0.40 M NaOH(aq) were ...

Emma Strassner 1J

Okay that makes sense, thanks!

Emma Strassner 1J

A sample of an ideal gas in a cylinder of volume 3.84 L at 298 K and 2.10 atm expands to 8.65 L by two different pathways. Path A is an isothermal, reversible expansion. Path B has two steps. In the first step, the gas is cooled at constant volume to 1.07 atm . In the second step, the gas is heated ...

Emma Strassner 1J

Compression is the easiest example for me to remember when it comes to a positive work value because the work is going into the system.

Emma Strassner 1J

At constant volume, the heat of combustion of a particular compound, compound A, is −3931.0 kJ/mol. When 1.157 g of compound A (molar mass =120.61 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 6.587 ∘C. What is the heat capacity (calorime...

Emma Strassner 1J

Endothermic reactions will always be positive because they require heat to occur, and exothermic reactions will always be negative because they release heat and therefore lose energy.

Emma Strassner 1J

If the surroundings are getting warmer, the reaction is exothermic because it is releasing heat into the system. If they are getting colder, it is endothermic because heat is being absorbed from the system.

Emma Strassner 1J

You can arrange the reactions any way you want, just remember to make the necessary changes to the delta H values.

Emma Strassner 1J

If K it is less than 10^-3 you can assume, but to be safe it is better to approximate if K is less than 10^-4.

Emma Strassner 1J

You would use the inverse when you are dealing with the reverse reaction.

Emma Strassner 1J

That makes sense, thanks!

Emma Strassner 1J

The Kb for an amine is 9.365×10−5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.570 ? Assume that all OH− came from the reaction of B with H2O. Hi, I found the concentration of [OH-] but I cannot get the right percentage, how do I find the correct percentage p...

Emma Strassner 1J

That makes sense, thank you!

Emma Strassner 1J

The Ka of a monoprotic weak acid is 0.00362. What is the percent ionization of a 0.142 M solution of this acid?

Hi, is this one of the instances where you can disregard the x in 0.142-x (from the ICE table) when working with the Ka equation?

Emma Strassner 1J

Memorizing the strong acids and bases is the best bet because then you can easily identify which ones are weak acids and bases during an exam.

Emma Strassner 1J

Hi, ICE tables are applicable when working with both concentration (Kc) and pressure (Kp).

Emma Strassner 1J

Hi, you would use Kelvin and make sure to convert from other units like Celsius.

Emma Strassner 1J

Hi, you would use 0.700 M for [NO] because you are trying to find the concentrations when equilibrium is reestablished after making 0.700 the amount, so you wouldn't want to use the first value given.

Emma Strassner 1J

Kaleb Tuliau 3E wrote:So total pressure is all of our equilibrium pressures combined?

Yes, the total pressure is all of the equilibrium pressures (found in the last row of the ICE table) combined!

Emma Strassner 1J

Yes, Q and K are both the same process of using the ratio of products to reactants. The only difference is that Q is when the reaction is not at equilibrium.

Emma Strassner 1J

Yes, if you are given the initial concentration for a reverse reaction, the products would not equal 0.

Emma Strassner 1J

Yes, these terms all mean the same thing in describing that formation of the products is favored, and I agree that thinking of it in the other terms besides "shifting" is easiest for me to remember.

Emma Strassner 1J

That makes sense, thank you!

Emma Strassner 1J

At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3. H2(g)+I2(g)↽−−⇀2HI(g) Kc=53.3 At this temperature, 0.300 mol H2 and 0.300 mol I2 were placed in a 1.00 L container to react. What concentration of HI is present at equilibrium? Hi, how would you solve the ICE table fo...

Emma Strassner 1J

Hi, in general for these calculations you should convert from mL to L because Liters is the unit used in the formula itself.

Emma Strassner 1J

In each of the following salts, either the cation or the anion is a weak acid or a weak base in water. Write the chemical equation for the proton transfer reaction of this cation or anion with water: (a) NaC6H5O; (b) KClO; (c) C5H5NHCl; (d) NH4Br. Hi, I looked at the answers in the textbook but I am...

Emma Strassner 1J

That helps tons, thank you!

Emma Strassner 1J

A monoprotic weak acid, HA , is ionized according to the reaction HA(aq) + H2O(l) ↽−−⇀ A−(aq) + H3O + (aq) pKa=3.78 where A− is the conjugate base to HA. For this weak monoprotic acid, the predominant species present at pH 6.49 is: neutral. charged. unknown. Hi, how would I go about solving this pro...

Emma Strassner 1J

The easiest way to know is just to remember that Bronsted is dealing with protons, while Lewis is dealing with electrons. It helps me to think back to Lewis structures, because we focused on the electrons during that topic, so that relates to the focus on electrons here.

Emma Strassner 1J

That makes sense, thank you!

Emma Strassner 1J

Hi, I agree that you just use a space. You only attach the things that are inside the brackets into one word, and you put the stuff outside the brackets into separate words that are either before or after the compound, based on where they are in relation to the brackets.

Emma Strassner 1J

Hi, coordination number is the number of ligands attached to the transition metal. It is easiest to look at the number of ligands inside the brackets.

Emma Strassner 1J

Which of the following ligands can be polydentate? If the ligand can be polydentate, give the maximum number of places on the ligand that can bind simultaneously to a single metal center: (a) HN(CH2CH2NH2)2; (b) CO32−; (c) H2O; (d) oxalate. Hi, what is the easiest way to figure out which of these ar...

Emma Strassner 1J

Hi, naming the compounds like that follows alphabetical order without the prefixes included. You only add in the prefixes after you have determined the order of the ligand names themselves.

Emma Strassner 1J

Counting the regions of electron density an atom has is the quickest way to figure it out. You just add the number of atoms connected to the atom to the number of lone pairs attached and then that gives you the values (2 regions is sp, 3 is sp2, etc.).

Emma Strassner 1J

Hi! I'm assuming that yes, you should know the general bond angles for atoms that have no lone pairs to affect the values. That means just generally knowing that linear is 180, trigonal planar is 120, tetrahedral is 109.5, etc. because those are set and from there you can make some assumptions about...

Emma Strassner 1J

Remembering molecular geometry is really just easiest through memorization, and I use flashcards to keep all of the details straight. I also do what Faith said in terms of remembering what the roots of the words mean, and it is also easy to think about the differences between a planar shape and a py...

Emma Strassner 1J

The 120 degree angle comes from the distance between the three Cl equatorial atoms around the P. They are all on the same plane, and the plane (360 degrees) is divided into three equal sections of 120.

Emma Strassner 1J

The easiest way to tell is just to remember that trigonal planar has three regions of electron density that are all on the same plane. Trigonal pyramidal adds a lone pair to create four regions, forming the "pyramid" shape that gives it its name.

Emma Strassner 1J

Hi, we do count regions of electron density because they affect the shape of the molecule by repelling other regions.

Emma Strassner 1J

Hi, I agree with the concept that the dipole moments won't cancel out. If the atoms are the same electronegativity, they will cancel, but otherwise the difference between the two means they are polar.

Emma Strassner 1J

All resonance structures have sigma and pi bonds, and when drawing the structures you just change where they go.

Emma Strassner 1J

That makes sense, thank you!

Emma Strassner 1J

I also think it becomes possible for that one to have lone pairs on the central atom because there could be one on the top of the central atom and one on the bottom, so they would repel and cancel each other out in terms of altering the linear structure.

Emma Strassner 1J

Predict whether each of the following molecules is likely to be polar or nonpolar:
(c) CHCl3
(trichloromethane, also known as chloroform, a common organic solvent and once used as an anesthetic).

Why exactly is this molecule polar?

Emma Strassner 1J

VSEPR notation can tell you the shape because it tells you the bonded atoms and the pairs. From there you can figure out the bond angles if they're slightly different from expected.

Emma Strassner 1J

Okay cool, I was just going to wait through the second pass and hope enough people drop so that we can find a spot :( hopefully some spots open up!

Emma Strassner 1J

Can we still enroll in chem 14b if the class is full and closed? Or do we have to choose another class?

Emma Strassner 1J

That makes sense, thank you!

Emma Strassner 1J

Is the only way to determine the oxidation number of something by memorizing the rules? Or is there a process we could follow?

Emma Strassner 1J

The oxygen atoms with a single bond have a formal charge of -1, while the other has a formal charge of zero because of the double bond.

Emma Strassner 1J

That makes sense, thank you so much!

Emma Strassner 1J

Hi, how would you find the formal charge for C03^2-? For the sapling platform does it go on the central atom?

Emma Strassner 1J

If you're talking about the negative on the molecular formula itself, you use the brackets to put it outside the drawing of the Lewis structure. Otherwise, you are talking about the negative that goes with the charge of each atom in the molecule individually, so it goes with the atoms themselves.

Emma Strassner 1J

I agree with what has been said about each atom having its own formal charge. It all depends on the individual bonds and lone pairs that accompany each atom, which determines the charge.

Emma Strassner 1J

Hi, the drawings themselves are not very particular about the lengths of the bonds. You can draw them the same and it'll be fine, just remember that they are shorter in reality even though it is not needed in the drawing.

Emma Strassner 1J

Hi, it goes in the middle because it has the most open spaces to accept new bonds and electrons, so it is the most useful there.

Emma Strassner 1J

The radius decreases across the period because the effective nuclear charge increases as you go across. This causes greater attraction to electrons that pulls them in closer to the nucleus, resulting in a smaller radius.

Emma Strassner 1J

Atomic radius increases as you go down a group because more shells are being added as n increases. It decreases across a period because the effective nuclear charge increases, which pulls the electrons in closer to the nucleus of the atom.

Emma Strassner 1J

I agree, electrons in the outer shells are there because they were excluded from inner shells due to the Pauli Exclusion Principle. Because they are physically not able to be closer, they feel less of the effects of the nucleus which lessens the energy required to remove it from the atom.

Emma Strassner 1J

Thank you, I didn't understand that you had to add the exponents together!

Emma Strassner 1J

Hi, how exactly do we figure out how to go from the excited state of an atom to the ground state?

Emma Strassner 1J

That makes sense, thank you!

Emma Strassner 1J

The electron affinity of thulium has been measured by a technique known as laser photodetachment electron spectroscopy. In this technique, a gaseous beam of the anions of an element is bombarded with photons from a laser. Electrons from the anion are then ejected and their energies are detected. The...

Emma Strassner 1J

Okay, that makes sense, thank you!

Emma Strassner 1J

The E. coli bacterium is about 2.3 μm long. Suppose you want to study it using photons of that wavelength or electrons having that de Broglie wavelength. What is the energy Eelectron of the electron?

How do I find the energy of the electron for this problem?

Emma Strassner 1J

Hi, I agree with what is said above. Use the amount that is specified on the formula sheet and try to avoid rounding many decimal places until the end of the problem to get the closest decimal possible. It will make it easier to match your answer to whatever is closest on the multiple choice!

Emma Strassner 1J

Hi! I think you have to convert the wavelengths from cm to meters before you calculate the energy, or else you end up with a different answer. If you convert the wavelength first, you should get an energy of 2.13*10-20. Hope this helps. Hi, I agree with this. Converting from cm to meters first is n...

Emma Strassner 1J

Hi, I agree with using the equation v=R[(1/n1^2)-(1/n2^2)]. By plugging in 1 for n1, and 6 for n2, you are ensuring that you do not end up with a negative answer, which you would get if you switched those values. Hope that helps!

Emma Strassner 1J

Okay, that makes sense, thank you!

Emma Strassner 1J

Calculate the wavelength, in nanometers, of the spectral line produced when an electron in a hydrogen atom undergoes the transition from the energy level n=4 to the level n=2. Hi, for this question it is asking for the wavelength, but the constant on the formula sheet is in Hz. Do we have to solve i...

Emma Strassner 1J

Hi! I was missing the division by Avogadro's number, so that makes much more sense! Thank you!

Emma Strassner 1J

Light hits a sodium metal surface and the velocity of the ejected electron is 6.61 x 105 m.s-1. The work function for sodium is 150.6 kJ.mol-1. How much energy is required to remove an electron from one sodium atom? Hi, I know this is the problem where the value for the work function comes into play...

Emma Strassner 1J

I agree, using a vacuum is necessary in this case because without a vacuum, collisions with the air molecules cause them to lose energy before reaching the detector. In order for the detector to obtain the correct readings, the vacuum is needed to reduce electron collisions and keep the maximum ener...

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