CHEMISTRY COMMUNITY

Ishan Saha 1L

i believe that when using a porous disk we use a single line, and when using a salt bridge we use a double line.

Ishan Saha 1L

Hi, I believe that Test 2 question 8 is different for different discussion sections. Would you be able to send the problem?

Ishan Saha 1L

The problem should tell us whether or not we should use a porous disk or a salt bridge.

Ishan Saha 1L

Hi! Because it asks for the rate your body generates heat "in your surroundings" the formula has a -q. This because the system is releasing heat into the surroundings, or heat is flowing out of the system (with the system being the body).

Ishan Saha 1L

The way I understand it, Intermediates come into the reaction as a product, and are later consumed so they do not show up in the overall reaction equation. Catalysts are reactants that speeds up the reaction and go away but show up again in the product, so they also cancel out, and do not show up in...

Ishan Saha 1L

Yes entropy, enthalpy and Gibb's free energy are all examples of state functions in that they do not depend on the pathway, and instead only rely on the initial and final values. It is important to note that Cell potential (E) is NOT a state function, so if we need to add the values together, we nee...

Ishan Saha 1L

I don't think Dr. Lavelle went over that collision theory formula in lecture, so we shouldn't need to know that, but knowing the concepts would be beneficial for the final.

Ishan Saha 1L

You can determine whether or not the reaction is endothermic or exothermic by looking at the Activation energies. Because the activation of the forward reaction is greater than that of the reverse reaction, it can be determined that the reaction is endothermic. If you think of this graphically, the ...

Ishan Saha 1L

Every step of the overall reaction has its own unique activation energy so on the graph you will observe different Ea for each step.

Ishan Saha 1L

this question asks- "Under what conditions can these two mechanisms be distinguished by using kinetic data?" Could someone please explain what this means and what the answer might be? Thanks!

Ishan Saha 1L

For this problem, we are given a reaction profile. From this we can tell that there are three steps to the reaction. In part c we are told that a catalyst is added that accelerates the third step only. the question in part c asks, "What affect if any will the catalyst have on the rate of the ov...

Ishan Saha 1L

A good way to memorize some of them is by thinking of how some molecules are like others with a little change. For example you could think of a ketone as an aldehyde that has a double-bonded O in the middle rather than at the end, and a carboxylic acid is pretty much a ketone that has an OH (alcohol...

Ishan Saha 1L

Reversible expansion occurs when the pressure outside the system matches that inside the system, so that if there was a slight change in pressure cause the reaction to move a specific way. Irreversible expansion occurs when you are told that the system is working with a constant pressure or against ...

Ishan Saha 1L

The Cell Diagram should be Pt(s)|O3(g),O2(g)|OH-(aq)||O2(g),O3(g)|H+(aq)|Pt(s)

Ishan Saha 1L

We were supposed to use the half reactions on the back sheet provided to us. By comparing the two reactions, we can find which reaction will take place in the anode by looking at the one with the lower reduction potential. This means that this should be an oxidation reaction, so we can flip the equa...

Ishan Saha 1L

an inert metal conductor like Pt(s) needs to be added to the left side because I2(s) is a nonconductive nonmetal solid.

Ishan Saha 1L

If the catalyst shows up in the reactant side, then it must show up again in the products side of the equation at some step (not necessarily only the second) because catalysts cannot be totally consumed by the reaction.

Ishan Saha 1L

Hi! To answer solve this problem you had to first identify the two half reactions involved. They were (Cr2O7)^2- (aq) +14H+(aq) +6e- ==> 2(Cr)^3+ (aq) + 7(H2O)(l) with a standard cell potential of 1.33V and Cr^3+ +3e- ==> Cr(s) with a standard cell potential of -0.74V. THEN because cell potential is...

Ishan Saha 1L

Hi, this is just an example of Hess' Law where we add the reactions together to get the final overall balanced overall reaction and the overall reaction enthalpy. I do not think we will be expected to know the balanced equations of formations of compounds without any information given to us on the f...

Ishan Saha 1L

Hi, I2 and Br2 are actually in the reactants side. They accept 2e- and thus become 2I- ad 2Br-. We don't reverse these equations because this is naturally what happens. I think what you were talking about was whether or not we had to flip the equation for Al because that is in the product side. We d...

Ishan Saha 1L

Hi! Sorry if this is phrased in a confusing way- I was wondering why collision theory allows us to use the number of molecules colliding (the coefficient) to find the order of the reaction when we are finding reaction mechanisms, when typically for chemical reactions the order must be determined exp...

Ishan Saha 1L

for an element to have high oxidizing power, it would have to have a more positive value for reduction potential. This is because the higher the reduction potential, the better the element is at being reduced. An element that is being reduced is called the oxidizing agent because it is accepting ele...

Ishan Saha 1L

Hi, The question was "Which of the following statements is correct?" and both A and B (shown below) were accepted as correct answers on the past midterm. My question is, in the future, which answer is correct? Should we think that any gas always has more entropy than any liquid? If not, ho...

Ishan Saha 1L

Hey are you made of Curium and Tellurium??
'Cause dang you're CuTe!

Ishan Saha 1L

On this problem on the midterm, both A and B were considered correct answers. For future reference, what answer is correct? Should we think that any gas always has more entropy than any liquid? If not, how do we know if a molecule is complex enough to have more entropy as a liquid than a simple mole...

Ishan Saha 1L

This is possible because HCl is aqueous, and thus dissolves into H+ and Cl- ions. Since the other reactant/product is H2, the half reaction only involves the H+ ion from HCl.

Ishan Saha 1L

Are we allowed to change the sign (as in multiply by -1) of the standard cell potentials of the half reactions if we are looking for the standard cell potential of the reverse half reaction? I had thought that this was not possible because reversing a reduction half reaction makes it an oxidation re...

Ishan Saha 1L

Substances that more readily decompose are less stable, and thus decompose fast causing it to have a shorter half life.
Substances that do not readily decompose are more stable, and thus decompose relatively slowly, causing it to have a longer half life.

Ishan Saha 1L

the universe can be considered isolated because energy nor matter can leave or enter the universe.

Ishan Saha 1L

Hi! I also made this same mistake and treated it as though pressure was not constant. The thing is that it is never explicitly states that the system is reversible, or that the external pressure matches the internal pressure. Therefore you can assume that pressure remains constant throughout the pro...

Ishan Saha 1L

The point of telling us that a gas is ideal and monoatomic is so that we know we can use the constants C_v= 3/2R and C_p= 5/2R.This allows us to calculate work and entropy when only either volume or pressure remains constant, or the reaction is reversible.

Ishan Saha 1L

Hi, In this problem we actually use both of those methods to ultimately solve for k at each temperature. We cannot directly use the equation delta G(standard) = -RTlnK because we do not know G(standard) in 11.83. Thus we need to solve for it using the formula delta G(standard)= delta H(standard)- T*...

Ishan Saha 1L

Hi! When solving a problem where heat transfers from one object to another, like from copper to water for instance, how do we know which heat is negative and which is positive? I thought that in a situation where heat is being transferred from a piece of copper to the water, the copper would have ne...

Ishan Saha 1L

I was just wondering if there was any trend for bond enthalpy? Does it correspond to bond strength or ionization energy or anything like that?

Ishan Saha 1L

To my understanding, an irreversible reaction can only go in one direction, such that when the reactants react to become the product, the product cannot convert back into reactants. These reactions are represented with a unidirectional arrow, and can never be at equilibrium. On the contrary, reversi...

Ishan Saha 1L

[HCl]0 does represent the initial value. for 12.19, we do not need to know the actual value for [HCl]0 as this value cancels out because it is in both the numerator and denominator. Hope this helps!

Ishan Saha 1L

it depends on whether the molecule has resonance structures or not. If the pi bond is in a molecule that has resonance, then it is delocalized. If it is in a molecule that does NOT have resonance such that the pi bond is between only two atoms, then it is localized. Hope this Helps!

Ishan Saha 1L

Ka describes the rate at which the acid deprotonizes and gives an H+ atom to H20. The concentration of H30+ is described by pH. Kb is the same except for bases, where Kb describes the rate at which the base gains protons from H20, and pOH represents the concentration of OH- molecules.

Ishan Saha 1L

You can also use the 5% rule, which states that if the value of x is less than or equal to 5% of the initial, then you can assume the equilibrium amount to be approximately the same value.

Ishan Saha 1L

I also share the same question. I believe that the solution manual may be wrong because this molecule does not have resonance and thus the pi bond is localized, which means the presence of a double bond should decrease the Cl-P-Cl bond angles to make them less than 109.5 degrees. If this is wrong, I...

Ishan Saha 1L

A bond can be polar if there is a large difference in electronegativity between two atoms. This difference in electronegativity also expresses how big the dipole moment is. A molecule can have polar bonds and dipole moments between two atoms, but if the bonds in this molecule are symmetrical such th...

Ishan Saha 1L

A localized pi bond occurs when there is only 1 pi bond between two atoms and the molecule has no resonance structures.

Ishan Saha 1L

Another way you can find the limiting reactant is by converting the masses of each reactant to moles and than comparing that molar ratio to that which is described by the balanced equation. For example if the left side of an equation was 2Cl + 4H and once converting the mass of each reactant to mole...

Ishan Saha 1L

To add on to the previous response, a Lewis acid is an electron PAIR acceptor, and a lewis base is a electron PAIR donor.

Ishan Saha 1L

To add on to the previous response, C2O2HF3 is a stronger than C2O2HCl3, because the F atom is more electronegative than the Cl atom, which means that it will pull pull more electrons from the Oxygen, which thereby makes it easier for the oxygen to loose the H+ ion.

Ishan Saha 1L

1.60 actually has 3 sig figs, because all 0's after nonzero values after the decimal count as sig figs. I believe the correct answer would be 1.6, with 2 sig figs as was used in the original question.

Ishan Saha 1L

None of these molecules have dipole moments because the more electronegative atoms in each compound are surrounded symmetrically around the central atom. For example in PCl5, the Chlorine atoms surround the central atom equally in a trigonal bi-pyramidal fashion. Dipole movements would only occur if...

Ishan Saha 1L

If I understand what you are asking correctly, You look at the number of bonding regions (bonds and lone pairs) that a central atom in a molecule has. For Example, CH₄ has 4 bonding regions, and will thus be sp³ hybridization. Hope this helped!

Ishan Saha 1L

If I understood it correctly, I believe that the lone pairs will always be equatorial first, before going to the axial positions (so seeing lone pairs in the axial position is quite uncommon). This is because lone pairs have more repulsion (they don't want to have a smaller bond angle in relation to...

Ishan Saha 1L

Also to find the total hydrogen produced, you need to add the hydrogen produced in the first reaction to the hydrogen produced in the second reaction.

Ishan Saha 1L

HI! Perhaps you read the solution manual wrong? you are right in that Beryllium should have a greater value for electropositivity than Lithium. Checking my solutions manual for the same problem you are referring to, the answer agrees with that answer and is correct.

Ishan Saha 1L

Higher Frequency means higher energy, as can be told from the formula E=hv, where v is frequency. Because v and wavelength (λ) are inversely related in c=λ v, the larger the wavelength, the lower the frequency, and so the lower the energy.

Ishan Saha 1L

Hi! So because the uncertainty in velocity is 1.5 +/- 4m/s, the delta v is 0.8 m/s. To find delta p, you multiply delta v by the mass of an electron (9.11 x10^-31 kg). Then to find delta x, you do h/(4pi)/delta p which numerically is (6.626 x 10 ^-34)/(4 x pi)/(0.8 x (9.11 x 10-31)) and you get the ...

Ishan Saha 1L

Hi! I'm guessing that the second reaction uses H2O in excess? If so, then first you find the limiting reagent in the first reaction, and calculate the maximum amount of CO that can be produced while keeping the limiting reactant in mind. The product, CO, from the first reaction is used as a reactant...

Ishan Saha 1L

Hi! I was just wondering if the Lewis structure of SO2, sulfur dioxide, has two double bonds between S-O with a lone pair on Sulfur (making expanded octet), OR if it has one single bond and one double bond, with one lone pair on Sulfur (allowing it to stay within the octet rule). The first option al...

Ishan Saha 1L

4d has a greater energy than 5s when the 4d orbitals are empty. However when the 4d orbital starts to fill up with electrons, it has a lower energy than 5s because the electrons in the 4d orbital shield the electrons in the 5s orbital. Therefore when an element in the d-block is ionized, or loses el...

Ishan Saha 1L

Yes, the electron configuration for Zn1+ is [Ar]3d^10 4s^1, the same as that of Cu. This is because electrons are removed from the higher energy level, which in this case would be the 4s level. Further more, due to electron shielding that results from all the electrons in the 3d orbital, the 4s ener...

Ishan Saha 1L

Hi! I was just wondering what the electron configuration would be of a cation of Cr, as Cr is an exception to the usual notation? So if one electron was lost from Cr, which has the neutral notation of [Ar](3d^5)(4s^1), what would the notation become?

Ishan Saha 1L

Hi! I just wanted to clarify if when we are calculating for the energy emitted by an electron that is relaxing from n=5 to n=1, would the initial energy level be n=5 and the final level be n=1? Or would the initial be n=1 and the final be n=5?

Ishan Saha 1L

In addition to what everyone else has already said, it is important to note that DeBroglie's Eq is reserved for objects that have momentum. An objects momentum (p) is equal to its mass (m) times its velocity (v). This is why DeBroglie's equation has the notation λ=h/p, where λ is wavelength, h is Pl...

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