CHEMISTRY COMMUNITY

Jonathan Omens 1K

A catalyst can be present in the rate law if it is a reactant in the rate determining step. Because it is a reactant in that step, it is included in the rate law.

Jonathan Omens 1K

I agree with Henry's answer except instead of adding OH- equal to double the amount of H2O initially added, find how many H one side needs to make it balanced and and add twice that amount of H2O to that side and that amount of OH- to the other side. Also don't forget to balance all the other reacta...

Jonathan Omens 1K

They are not related, they just happen to be written the same.

Jonathan Omens 1K

t_1/2 = 0.693/k
to get to 1/64 the original power, just do it 6 times because 2^6=64

Jonathan Omens 1K

Use experimental data to calculate the order of reactants in a reaction. Divide one experiment by another as shown in class on March 1st. The order shows how many of that molecule interact in the rate determining step.

Jonathan Omens 1K

The two are equivalent, but typically when referring to the unique rate you would be referring to the equation that describes the rate with respect to one reactant or product while the rate law describes it in terms of all reactants. Eg. Unique rate = -\frac{1}{a}\frac{d[A]}{dt} for a as a reactant ...

Jonathan Omens 1K

A concentration cell is a galvanic cell with both sides containing the same solutions at different molarities. The E^o is always zero because it is calculated using the standard reduction potentials of both sides which are the same because the solutions contain the same components.

Jonathan Omens 1K

Everything that is added to the solutions that participates in the redox reaction is put into the cell diagram except for water and the acid/base that is added to the solution to allow the reaction to happen. This means that the electrode, and oxidized/reduced materials are added, as well as other m...

Jonathan Omens 1K

E o is always 0 for a concentration cell because E o refers to the potential of a solution at 1M concentration. For the concentration cell, E o would be calculated using 1M solutions of the solutions on each side, and since in a concentration cell, both solutions have the same components, the differ...

Jonathan Omens 1K

Expanding on Aili's answer, the standard potential for the anode should be negative already because it is oxidized so it will be subtracted when you add them.

Jonathan Omens 1K

An inert conductor is used when the ions on the cathode side will not form a solid on their own. This should be given I don't think we need to know when to use it and when not to.

Jonathan Omens 1K

Comparing the half reactions is just one method for calculating the overall cell potential. The standard hydrogen electrode is an electrode that is used to measure the potentials of substances and to standardize them. A galvanic cell is set up and potential is measured between the cell of interest a...

Jonathan Omens 1K

The molar Gibbs free energy refers to the amount of Gibbs free energy per mole of a substance.
Standard Gibbs free energy is the total amount of energy, not just the amount of energy in each mole of the substance.

Jonathan Omens 1K

K is just a constant in the equation that relates entropy and enthalpy in a system.
If you want to know more I suggest you look at the section in the thermodynamics tab called Van't Hoff Equation.

Jonathan Omens 1K

Gibbs free energy is a measure of energy contained within a system. More specifically, this energy is the energy in the system that can be accessed to do work. Enthalpy is the measure of energy due to temperature. Entropy is a measure of the number of quantum states that a system can occupy. Heat is...

Jonathan Omens 1K

Each SO₂F₂ can be in 1 of 6 configurations. Since we are calculating the molar entropy or entropy per mole, use the Boltzmann equation with n=6.022x10²³
So, $S=k_{b}lnW=1.38\times 10^{-23}ln6^{6.022\times 10^{23}}=14.9J\cdot K^{-1}$

Jonathan Omens 1K

5/2 R should be used when pressure is constant and 3/2 R when volume is constant. They are derived from ideal gas equations not discussed in class.

Jonathan Omens 1K

You won't have to integrate to solve any problems. They were just used to derive others from class.
As for knowing which equations to use when certain values are constant, you just have to remember which to use in which scenario.

Jonathan Omens 1K

The way that you would find out the enthalpy of a molecule would be by using the enthalpies of the other molecules in the reaction which should be given. You could also add up the bond enthalpies if those are given.

Jonathan Omens 1K

Jonathan Omens 1K

In this class, use atm when reporting pressures.

Jonathan Omens 1K

You should memorize the general pattern of rises and plateaus but not the specific numbers.

Jonathan Omens 1K

Adding on to that, because enthalpy is a state function, you can calculate it by adding the enthalpies of multiple reactions to find the enthalpy of the overall reaction.

Jonathan Omens 1K

You should know all three methods because you will need to choose one according to what you are given in each problem.

Jonathan Omens 1K

If you are given a K value, an acid with a K_a < $1\times 10^{-3}$ or a base with K_b less than that value will be a weak acid/base respectively.
(This is also a general rule but works in the majority of situations)

Jonathan Omens 1K

The dissociation of F₂ is more stable. The K value is larger meaning there is a higher ratio of products to reactants. The dissociated molecule is the product so there should be more products made in the F₂ dissociation than the Cl₂ dissociation.

Jonathan Omens 1K

Did you make sure to use concentrations instead of moles of each chemical? (convert using moles / liters)
Also, in the K_c value don't forget to put the proper exponents on the terms that have coefficients in the balanced equation.

Jonathan Omens 1K

When the two equations are combined in full, the result will be: 2BrCl + H 2 + Cl 2 <--> Br 2 + 2HCl + Cl 2 To write the full equilibrium constant equation for this, we would get: K = \frac{[Br_{2}][2HCl][Cl_{2}]}{[2BrCl][H_{2}][Cl_{2}]} This is the same as both K values multiplied together. The Cl ...

Jonathan Omens 1K

In this problem, both reactions were combined and their K values were multiplied together. Is this the case for all problems of this type or are there other factors to consider when combining the two K factors?

Jonathan Omens 1K

You are correct in saying that we use Kp for gases, but if necessary, you can convert using the ideal gas law: PV=nRT or P=(Concentration)RT

Jonathan Omens 1K

The phosphoric acid will donate a H⁺ to each of 3 ammonia molecules. This will result in PO₄^3- and 3NH₄⁺. They come together to form the salt (NH₄)₃PO_4.

Jonathan Omens 1K

If the molecule dissociates into an ion that doesn't affect the pH such as Na⁺, it can be left out of the net ionic equation.

Jonathan Omens 1K

First calculate the amount of moles of Ba(OH) 2 using the molar mass. Then remember that it completely dissociates into that number of moles of Ba and twice the number of moles of OH. Divide by .1L to get the molarity of Ba and OH. Then calculate the pOH using the molarity of OH and subtract it from...

Jonathan Omens 1K

Yes, whenever an atom is bonded to another atom, its regions of electron density are hybridized into sp, sp², etc. orbitals.

Jonathan Omens 1K

There are certain characteristics to acids and bases that are good to remember. For example, a compound consisting of one hydrogen and one halogen will always be a strong acid. Molecules with group 1 or 2 metals and an OH group will be basic. Also, any compound with an N atom that has a lone pair wi...

Jonathan Omens 1K

Polydentate ligand lone pairs are in the same orientation as atoms in the VSEPR model. For example, if there are 6 lone pairs from ligands bonding to the transition metal, they will be in an octahedral shape. The lone pairs should be the correct distance apart to form these shapes.

Jonathan Omens 1K

If you draw the Lewis structure, there should be 28 e - (21 from the 3 fluorine, 7 from bromine). The bromine is bonded to each of the 3 fluorine atoms by a single bond. That leaves 4 electrons which are put around the bromine atom as 2 lone pairs. This gives a total of 5 regions of electron density...

Jonathan Omens 1K

The 1 and 2 after the H and C represent the 1 in 1s or 2 in 2sp shell respectively. This is to distinguish between other shells such as 2s, 3s, 4s, etc. or a 3sp hybrid orbital. The hydrogen has 1s because a H atom has a valence electron in the 1s shell. The C has 2sp because the orbitals are a hybr...

Jonathan Omens 1K

In a resonance structure, the hybridization is found the same way as a non resonant structure. That is, by looking at the regions of electron density around the atom. If a lone pair in one resonance structure is a bonding pair in another resonance structure, the lone pair is considered to be in the ...

Jonathan Omens 1K

I believe that we do not need to know these concepts for class.

Jonathan Omens 1K

Because the difference in their electronegativities is only about .76, the bond has mostly covalent character. Only bonds with a difference in electronegativity of 2 or greater are said to have mostly ionic character.

Jonathan Omens 1K

In this problem, the POCl 3 molecule has a tetrahedral shape. In class we learned that tetrahedral molecules always have bond angles of 109.5 o and this is the correct answer in the solutions manual. However, the bond angles are not exactly 109.5 because (I assume) of the double bond that the oxygen...

Jonathan Omens 1K

The unit for dipole moments is the debye which is 3.336x10^-30 coulomb meter.

charge (coulombs) x distance (meters) = debye (coulomb meter)

Jonathan Omens 1K

What determines if a substance is a solid, liquid, or gas is not necessarily bonds but interactions between the atoms/molecules. These can be hydrogen bonds, Van der Waals' interactions, etc. If these interactions are stronger, then the atoms/molecules will be held more tightly together so it will t...

Jonathan Omens 1K

It has to do with the difference in electromagnetically between the two elements. Ag and F have very different electromagnetically so their bond has a high ionic character. This means that it will be soluble in water. As you go down the group, Cl, Br... the electronegativities get closer to that of ...

Jonathan Omens 1K

The polarizability gets bigger as the ion gets larger. The radius of O^2- is less that the radius of N^3-. Thus, it has a smaller polarizability. Also, N^3- is less electronegative which also contributes.

Jonathan Omens 1K

A) The brackets are just to distinguish between the Cl and NH4. B) The Molecule is K 3 P. The three K + ions surrounding the P 3- ion show that there are 3 potassium ions per phosphorous ion in this salt. The arrangement of K + ions around the phosphorous is arbitrary. C) Again here the brackets are...

Jonathan Omens 1K

The bond order is simply the number of bonds between two atoms.

For a single bond, the bond order is 1.
For a double bond, the bond order is 2.
etc.

Jonathan Omens 1K

This rule only comes into effect for elements above argon (Z=18). You are correct that 3d is written before 4s, etc. when writing electron configurations However the 4s shell will fill before the 3d shell. This can be seen for the elements Potassium and Calcium (19 & 20). K: [Ar] 4s 1 , Ca: [Ar]...

Jonathan Omens 1K

The 4s orbital is always written after the 3d orbitals because it has higher energy. Remember higher n means higher energy. However, The 4s orbital will fill before the 3d orbital which is why potassium and calcium have electrons in the 4s orbital but not the 3d orbital. Keep in mind the exceptions ...

Jonathan Omens 1K

It is generally best to choose the atom with the lowest ionization energy as the central atom. Start with this atom and arrange the others symmetrically around it.

Jonathan Omens 1K

The answer is C.
Since the absorbed wavelengths are detected, the spectrometer can produce a spectrum with absorbed lines, hence the name absorption line spectrum. Because it is measuring absorbed lines, the experiment is called absorption spectroscopy.

Jonathan Omens 1K

All particles have particle and wave-like properties.
Electrons do have a wavelength and since they are small enough, the wavelength can usually be measured.
Larger objects such as a baseball have wavelengths that cannot be measured.

Jonathan Omens 1K

We do not need to know radial nodes.
Angular nodal surfaces and nodal planes are the same so the question is asking for how many nodal planes in each.
Calculating them would involve the schrodinger equation which we are not using in class so just memorize them.

Jonathan Omens 1K

Use the equation to find the energy of the incoming radiation: E_{photon}=E_{K}+\phi . The velocity of the ejected electron is given and the mass of the electron is in the list of constants, so you can calculate E_{K} (kinetic energy of electron) using \frac{1}{2}mv^2 . In part (b) you calculated th...

Jonathan Omens 1K

The intensity of a source of light is the number of protons that come from the source. This is different from the energy of the light wave, which is dependent on the frequency of the wave, hence the equation: E=hv.

Jonathan Omens 1K

keV is a kilo-electron Volt, or 1000 electron volts. 1 Joule is equal to $6.24*10^{18}$ electron volts. So 1 Joule = $6.24*10^{15}$ keV.

Jonathan Omens 1K

Use the most accurate molar mass given to you. If the periodic table you are using gives the molar mass of carbon as 12.011 g/mol, then use that. Because the problem you gave an example of starts with 3.45g carbon (3 sig figs), the final result should be rounded to 3 significant figures. In other wo...

Jonathan Omens 1K

Look at how many atoms of each color there are and write the formula based off of the key it gives you.
Eg. F1 (7th edition). Dark grey = C, White = H, Red = O
10 Dark grey atoms
16 white atoms
1 red atom
Molecular formula - C10H16O
The order in which you put the atoms doesn't really matter.

Jonathan Omens 1K

I agree with Matthew. Because the question says that the lab can manipulate individual atoms, it is safe to assume that each circle on the scale represents a single atom. Therefore, the answer will be 3 atoms of astatine rather than 3 moles of astatine.

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