CHEMISTRY COMMUNITY

Danah Albaaj 1I

This is what I picture the dilute saying whenever a precipitate forms in a concentration cell.

Danah Albaaj 1I

With an isothermal process, the heat is equal to the opposite of the work value, remember the equation ∆U=q+w. The first law of thermodynamics, relating to the law of conservation of matter and energy explains why when a process is isothermal ∆U=0, q=-w. As long as you are capable of calculating the...

Danah Albaaj 1I

Well, because both open and closed systems are capable of exchanging energy with their surroundings this means either one could have an adiabatic process. Heat is a form of energy, as long as the system has the capability to exchange energy, it will also have the possibility of being an adiabatic pr...

Danah Albaaj 1I

If a system is isothermal we can be positive there is no change in the system's internal energy because of the equation U=3/2nRT. This does not mean that the system could have no change in heat, because if the system is adiabatic as well there would be no change in heat.

Danah Albaaj 1I

The irreversible equation is not isothermal because the -nRT within the equation for reversible, isothermal expansion is in a way replacing the Pex from the irreversible equation. The isothermal is used to make -nRT a constant during the integration of the volume of the reversible expansion (pg 265 ...

Danah Albaaj 1I

You just have to look at the molecules within each step, and check to see if any are repeated within the product of one step and the reactant of another. This would make them an intermediate of the reaction. If you still struggle with determining the intermediates and the overall reaction, you could...

Danah Albaaj 1I

The reaction would be first order because the reaction rate is increased in direct proportion to each reactant's concentration.

Danah Albaaj 1I

No, because we are able to set up similar equations for zero and second order reactions using our integrated rate laws. 15.39 is a perfect example on how one such problem would look when being asked to do so.

Danah Albaaj 1I

No, because as he has mentioned many times. We won't really be be involving ourselves with reactions that are more than second order due to their complexity.

Danah Albaaj 1I

No, the answer would just be -1 because you still have to follow the formula and dividing the fractions will cancel out the value of a and leave you with -1.

Danah Albaaj 1I

You add H+ to reactions in acidic solutions and OH- to reactions in basic solutions. For H+ you add the ion to the oxidizing agent, and for the OH- ion you add it to the reducing agent. From there you make sure they are on opposite sides of the half-reactions so that they will cancel out when adding...

Danah Albaaj 1I

When solving for ∆G˚ you should be getting your answers in J/mol because the Coulombs in Faraday's constant and voltage will cancel out. Voltage is J/Coulomb while Faraday's constant is in Coulomb/mol, multiplying them together will leave you with the J/mol.

Danah Albaaj 1I

I was also quite confused by this question. In the solutions manual, they used the values of permanganate and the iron for finding half-reactions in appendix 2B but I wasn't sure on why.

Danah Albaaj 1I

It's really confusing to be asked which is the oxidizing agent and which is the reducing agent. It's the opposite of the reaction the molecule would be in, so O3 is in the reduction reaction which means it is the oxidizing agent.

Danah Albaaj 1I

After doing multiple problems with entropy calculations with other classmates, UAs and TAs, as well as checking in the textbook and on here I am still confused on when to use Cv or Cp for ∆S= -nClnT2/T1. The problems all involved a diatomic gas at a changing volume and temperature, which meant I was...

Danah Albaaj 1I

In Chapter 8, we are given the equation ∆U=3/2nR∆T to relate the change in internal energy to the change in temperature. If a system is isothermal, this means that ∆U=0 for this system. Using the first law of thermodynamics, we can determine the q=-w, rather than 0 .

Danah Albaaj 1I

The bond dissociation energies are from a table in Chapter 3. The F-F bond is lower in energy than the other bonds because of its short bond length which in turn causes the electron pairs between the two F atoms to have stronger electron repulsion. This repulsion causes the dissociation energy of th...

Danah Albaaj 1I

Using Q is for when the reaction is not at equilibrium. So then you would use both the result from when the reaction at equilibrium found from ∆G˚= -RTlnK and the value of RTlnQ. Professor Lavelle went over the equation at the end of lecture on Friday before the Van't Hoff equation if you would like...

Danah Albaaj 1I

The volume of the balloon decreased, which means there was expansion work done. The definition of expansion work is from the change in the volume of the system, which is how we get the equation w = -P∆V

Danah Albaaj 1I

Exothermic processes lead to an increase in entropy because the heat is being released into the surroundings which causes the overall entropy to increase. In this situation, the change in entropy of the surroundings will be greater than that of the system (exothermic process), which is why it is imp...

Danah Albaaj 1I

The first equation is at a constant temperature, while the second equation is with changing temperatures. Because of this difference, you should not expect to see the same trend in the entropy. In the second equation, the value from dividing the final temperature by the initial temperature will alwa...

Danah Albaaj 1I

It has to do with the kinetic energy of the ideal gas. I recommend reading section 8.7 where it goes into great detail on how the numbers come into play through the use of Boltzmann's constant and the Ideal Gas Law alongside the equipartition theorem. Professor Lavelle also briefly went over it whil...

Danah Albaaj 1I

I believe it depends on whether the question is asking you for the entropy per mole of the substance or not.

Danah Albaaj 1I

I'm sure the standard states of molecules not seen as often as those similar to H2 and O2 will be given to us. Otherwise, it wouldn't hurt to be able to recognize the common elements' standard states.

Danah Albaaj 1I

The values of each standard enthalpy of combustion they provided can just be plugged in to find the reaction enthalpy. So your equation to find the enthalpy will look like this deltaH = [1 mol(-1560 kJ/mol)] - [1 mol(-1300 kJ/mol) + 2 mol(-286 kJ/mol)] The final answer will then be -312 kJ. If you'r...

Danah Albaaj 1I

Can someone help me with the final steps of this problem? I found the heat for water using q=mCdeltaT, but I'm not sure what I'm supposed to do with the given information regarding the photon's wavelength. I know I have to use Avogadro's constant, but each combination I try is not the answer of 2.49...

Danah Albaaj 1I

Also a follow-up, can someone explain why the answer to part c is 12.5 kJ/mol instead of 12.5 J/mol?
Thank you!

Danah Albaaj 1I

In homework question 8.103 they ask for the molar kinetic energy in joules/mole of Krypton gas at two temperatures with a deltaT of 1 degree celsius and using the answers I'm supposed to find the specific heat capacity. My TA covered this concept a little bit in discussion today, but I was wondering...

Danah Albaaj 1I

A closed system has a fixed amount of matter within it, but it can exchange energy with its surroundings. The coolant in a refrigerator coil is an example of a closed system because the type of matter doesn't change (coolant) and there is an energy exchange (temperature of refrigerator). Isolated sy...

Danah Albaaj 1I

A closed system has a fixed amount of matter within it, but it can exchange energy with its surroundings. Isolated systems can exchange neither energy or matter with its surroundings, essentially having no contact with its surroundings.

Danah Albaaj 1I

The problem provides you with the volume in centimeters cubed and the pressure in atm. In this case it is best if you convert the volume to liters in order to perform the conversions for joules. If the pressure had already been in pascals, then having the volume be in liters would have been unnecess...

Danah Albaaj 1I

For this question, you use the given pKb to calculate the pKa by subtracting the pKb from 14. From there you are able to calculate the molecule's Ka using the equation: Ka=10^-pKa The resulting value should then help you determine the strength of each acid and rank them accordingly. The whole proces...

Danah Albaaj 1I

I feel as if it would be wise to know which elements are exceptions the trends in the periodic table, especially if they are within the first 5 rows and could apply to any of the main topics like oxygen does. I would recommend checking Dr. Lavelle's lecture outlines and see if in any of them mention...

Danah Albaaj 1I

We use Kw because it is has a calculated experimental constant for autoprotolysis at 25* C, which allows us to determine how the formation of hydronium and hydroxide within an equilibrium can be used to determine the pH and pOH based on their concentrations. We use Kw to find the pH and pOH because ...

Danah Albaaj 1I

You would want to consider how trichlorophenol has atoms with a higher electronegativity on it, which correlates with its strength as an acid. This is because the bonds of atoms with higher electronegativity typically are weaker therefore increasing the transfer of H+

Danah Albaaj 1I

The two carbons are bonded together. Then you place the 2 hydrogens on of one of the carbons. This is a bent shape. For the other carbon you have a hydrogen by itself and the oxygen bonded to both the nitrogen and carbon. The C-C-N bond will be one diagonal line, and make the shape of that part of t...

Danah Albaaj 1I

I'm not entirely sure how to approach this problem. It states a 25.0-g sample of ammonium carbamate, NH4(NH2CO2), was placed in an evacuated 0.250-L flask and kept at 25 C. At equilibrium, 17.4 mg of CO2 was present. What is the value of Kc for the decomposition of ammonium carbamate into ammonia an...

Danah Albaaj 1I

I wouldn't say that the energy of a hybridized orbital is the average of its contributing orbitals. Instead, I like to think about how much each orbital contributed to the hybridized orbital's energy. For example, a sp^3 orbital has 25% s-character and 75% p-character. It would make sense that the h...

Danah Albaaj 1I

This problem asks me to draw a Lewis structure of benzyne (C6H4) including the hybridization of each carbon atom, then use my knowledge of bonding to explain why benzyne might be a reactive molecule. I know that the triple bond within the molecule would play a significant role, I'm just not exactly ...

Danah Albaaj 1I

Another way to think of it is trigonal pyramidal is a tetrahedral shape that has lost a bond and replaced it with a lone pair. You can see it when you look at the VSEPR notation for both shapes, tetrahedral is AX4 while trigonal pyramidal is AX3E.

Danah Albaaj 1I

I agree with what was stated above, and in regards to where you should put a lone pair on the central atom. You want to put it the farthest you possibly can from any bonds or other lone pairs while keeping some symmetry with the rest of the bonds and their angles.

Danah Albaaj 1I

The formal charge of a double bonded oxygen atom is 0, while the formal charge of a double bonded chlorine atom is +1. The formal charge of 0 is more stable and has a lower energy, which is what we want.

Danah Albaaj 1I

Hi! So for the Lewis structure, you have to connect the central carbon to its own two hydrogen as well as the NH2 and COOH. So basically just put one carbon in the center, then attach a nitrogen on one side, and another carbon on the other. From there bond the remaining parts of each compound to eac...

Danah Albaaj 1I

The textbook tells us that one of the elements is oxygen, and I believe it goes more in depth about how we can find these exceptions.

Danah Albaaj 1I

I believe the most efficient way would be to calculate the formal charge of the atom. As it says in our notes, the most stable bond is when the formal charge is equal to 0. Finding the bonds necessary for the formal charge to equal 0 should only take a few steps, especially if we are able to anticip...

Danah Albaaj 1I

Oxygen has a lower ionization energy than nitrogen and fluorine because of the repulsion between its electrons. Nitrogen only has 3 electrons in the 2p-orbital which means each electron is unpaired. Fluorine has 5 electrons in the 2p-orbital meaning only one electron is unpaired, which makes it more...

Danah Albaaj 1I

On the test I took yesterday, the values were given to me. However, I agree with what others have been saying and recommend knowing their differences because wavelength relates to the key concepts of this chapter.

Danah Albaaj 1I

The homework due any week, is 4 questions relating to what has been discussed in lecture. This week you only have to turn in 4 more problems from Chapter 1, unless you decided to read ahead and do some of the Chapter 2 problems.

Danah Albaaj 1I

You can also find the masses of both particles on the class website! It's in a worksheet named constants and equations, just in case you wanna have a separate page to look at so you're not constantly flipping the pages of your book :)

Danah Albaaj 1I

Question #25 states: Sodium vapor lamps, used for public lighting, emit yellow light of wavelength 589 nm. How much energy is emitted by (a) an excited sodium atom when it generates a photon; (b) 5.00 mg of sodium atoms emitting light at this wavelength; (c) 1.00 mol of sodium atoms emitting light a...

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