CHEMISTRY COMMUNITY

CameronJohari1J

The book uses 1.5R for the Cv of O2 in step 2, but since O2 is a linear molcule shouldn't the Cv be 2.5R?

CameronJohari1J

If we ignore the unbalanced oxygen atoms, conservation of mass is not upheld which seems illogical.

CameronJohari1J

I am in doubt that the back of the book as well as lavelle's solution manual answer are correct for this problem. Immediately upon seeing the equation provided for the reaction: CH4(g) + H2O(g)\rightarrow CO2(g) + 3H2(g) I realized that the equation is NOT balanced as...

CameronJohari1J

Why does H+ in an aqeouous solution have an enthalpy of formation of zero? Isn't H2 (gas) the most stable form of hydrogen thus it is 0 enthalpy of formation for hydrogen?

CameronJohari1J

I understand that all the reactants and products need to be in their standard states and that only 1 mole of substance needs to be burned. Since the table for these data values is given at 25C, would H2O be in liquid state for products? Thank you.

CameronJohari1J

I understand that all the reactants and products need to be in their standard states and that only 1 mole of substance needs to be burned. Since the table for these data values is given at 25C, would H2O be in liquid state for products? Thank you.

CameronJohari1J

The steeper the slope of a heat curve, the smaller the heat capacity is? can someone provide also provide a mathematical example to prove/disprove my statement as well.

CameronJohari1J

The steeper the slope of a heat curve, the smaller the heat capacity is? can someone provide also provide a mathematical example to prove/disprove my statement as well.

CameronJohari1J

Why is helium melting not an endothermic process? Isn't the liquid molecules of any element more energetic than that of its solid state.

CameronJohari1J

The change in internal energy for an irreversible expansion is not always zero. See part b of example 8.1 in the book if you need an example. Delta U will be zero if an expansion is isothermal

CameronJohari1J

How is it possible for a bomb calorimeter to be constant volume if the water's density is not constant at all temperatures? Unless I am misunderstanding and the constant volume is for the total volume of the calorimeter assuming that it isn't filled to the brim with water. Any clarification would be...

CameronJohari1J

Do enzymes always speed up the reaction by lowering the activation energy or can they also act as inhibitors and slow down the reaction?

CameronJohari1J

How are we supposed to identify that this expansion is irreversible or reversible? Are there any hints or clues in the problem that imply one type of expansion over the other? Thank you in advance.

CameronJohari1J

Because copper has the lowest specific heat capacity, it is the most efficient metal listed for boiling water. Conceptually speaking, copper requires the least amount of joules to raise its temperature per gram. It is important to look at specific heat capacity and not molar heat capacity because al...

CameronJohari1J

Quick note, enzymes are not always catalysts and can be inhibitors. Inhibitors work the opposite to catalysts as they increase the activation energy for the reaction.

CameronJohari1J

I was wondering what factors would change the value of k for a reaction?

CameronJohari1J

In general, it is useful to keep k units for time in terms of seconds. You can always convert to other units of time if the problem specifically asks for it, aka minutes or hours.

CameronJohari1J

Pt is so commonly used because it is an inert metal and does not participate in redox reactions under normal conditions.

CameronJohari1J

A real world example of a zero order reaction is N2O decomposing into nitrogen and oxygen with hot platinum wire as a catalyst. Once all the platinum is covered by N2O, the concentration no longer matters until new space is available.

CameronJohari1J

The units of the rate of reaction,K, is solely dependent on the order of the reaction. A zero order reaction will have the units M/s, a first order reaction will have the units of 1/s, a second order reaction will have the units of 1/Ms.

CameronJohari1J

The solutions manual has a 2 subscript for the nickel hydroxide in the final step but shouldn't it be a 3 subscript so that hydroxides cancel out?

CameronJohari1J

I do not understand the structure or importance of a salt bridge. Could someone explain to me physcially what it is, which type(s) of cells require it, and why it is important? Thank you in advance.

CameronJohari1J

I am confused how the textbook knew how to use the constant for constant pressure not constant pressure for example 9.6, step 3, pg329.

CameronJohari1J

To my understanding, the heat capacity of a system or molecule cannot be negative, as it represents the amount of energy required to increase or decrease the temperature by one degree. Energy is inherently a positive value unless with regards to a system.

CameronJohari1J

The E of the cathode is -.76V and the E of the anode is unknown. The E of the cell overall is .16V. Set E of cell to E cathode minus E anode. Solving for E of cathode results in -.92V for M(s) to M4+

CameronJohari1J

Use the second and third reaction to get the total amount of energy released in the regeneration of ATP. Make sure to multiply he second equation by 3 to get the correct factor. Then divide the free energy by 30.5 (energy required in the first reaction) to get the number of moles of ATP that would b...

CameronJohari1J

This is rare instance where the solutions manual is incorrect. Please refer to the link on his website Solution Manual Errors for this problem.

CameronJohari1J

The pH isn't relevant to this problem. Multiply both the second and third equation by 3 to represent 3 moles of NADH being used. Add the free energy sums of the two equations and divide by 30.5kJ which is the energy reuquired for the formation of ATP to calculate the moles of ATP produced.

CameronJohari1J

He sent out an email earlier today saying that he will tell us more information about the midterm in this upcoming week.

CameronJohari1J

Positional residual entropy is basically the absolute entropy and the equation was derived statically. The formula is S=Kb*lnW. Regular entropy to my understanding is usually delta S of the system.

CameronJohari1J

I think you are the using the wrong gas constant. Try using 8.314 J/molK for the gas constant.

CameronJohari1J

The situation occurs when the system experiences isothermal change, thus delta U is zero. Subsequently, q=-w means that heat loss is equal work done on the system and visa versa is also true.

CameronJohari1J

9.13 is one of the problems listed in the errors in solution manual. Definitely check out the link here: https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/Solution_Manual_Errors_6Ed.pdf

CameronJohari1J

9.13 is one of the problems, listed in errors in solution manual. Definitely, check that link out here: https://lavelle.chem.ucla.edu/wp-conten ... rs_6Ed.pdf[/url]

CameronJohari1J

All combustion reactions can be treated as if they are in a bomb calorimeter as long as all the heat and matter are accounted for.

CameronJohari1J

Conceptually, the difference is that heat is transferred between surroundings and system while delta heat is used to describe the change of heat in the system.

CameronJohari1J

Because no heat or matter is allowed to go between the system and surrounding for an isolated system and such occurs in a calorimeter and its surroundings.

CameronJohari1J

NO2 has greater enthalpy than NO because there are more bonds that require more energy thus greater enthalpy.

CameronJohari1J

If I am not mistaken d is the distance that the surface moved and gives us the correct units for volume

CameronJohari1J

The differences between the two properties is that extensive properties are dependent on mass while intensive properties are not. An example for each respectively would be mass for extensive and density for intensive

CameronJohari1J

Delta H is change in enthalpy while delta U is change in internal energy. The units for each is joules.

CameronJohari1J

Buffers are chapter 13 material which is not included in the acid base.

CameronJohari1J

The particle in a box is a simplification of finding an electron within a certain region. Both require use of Schrodinger's equation. It is useful primarily in quantum theory.

CameronJohari1J

To find pOH, the formula is -log(OH molarity).
I would write the chemical equation just to be safe.
Bases do react with water:
NH3 + H2O <--> NH4+ + OH-

CameronJohari1J

An additional amphiprotic substance is hydrogen carbonate ion HCO3-
It can either donate its hydrogen to become CO3 (2-) or accept another hydrogen to become H2CO3

CameronJohari1J

The following are strong acids:
HCl
HNO3
HBr
HI
H2SO4
HClO4
All other acids are considered weak.

The following are strong bases:
All group 1 metal hydroxides
LiOH
NaOH
KOH
RboH
CsOH
Some of the heavier group 2 metal hydroxides
Ca(OH)2
Sr(OH)2
Ba(OH)2
All other bases are considered weak.

CameronJohari1J

Look at the charge of the ligands and then the overall charge of the coordination complex. The sum of the ligand charges plus that of the transition metal equals the overall charge. From this equation solve for the transition metal charge which is the oxidation state.

CameronJohari1J

The k value of the reverse reaction is inverse of k.
If k is the equilibrium value for A+B<-->C K=C/(A*B)
then C<-->A+B 1/K=(A*B)/C

CameronJohari1J

For this example, as long as all the units are the same the equilibrium constant will be the same.

CameronJohari1J

According to the chemistry email, the test will go up to HW 11.63

CameronJohari1J

Think of the reaction based energy change, endothermic or exothermic. If the reaction is endothermic then energy can be treated as a species on the right side of the chemical equation. If the reaction is exothermic then energy can be treated as a species on the left side of the chemical equation. If...

CameronJohari1J

When delta G is negative, the reaction is spontaneous in the forward direction. Subsequently when it is positive, the reaction is non-spontaneous in the forward direction and requires additional energy to progress. When delta H is positive and delta S is negative, the reaction is always non-spontane...

CameronJohari1J

Also when an inert gas is added to a system in equilibrium in a rigid container aka constant volume, the concentration of the respective products and reactants does not change. However, the total pressure of the system will increase by adding the inert gas.

CameronJohari1J

Even if we do need to convert between different units of pressure, it would ultimately just be dimensional analysis.

CameronJohari1J

The dashed line and the wedge/solid triangle are merely drawn to the portray that the molecule is three dimensional. The dashed line usually shows that the bonded pair is going into the plane of the page and the wedge is going out of the plane of the page. Use them when the molecule cannot be drawn ...

CameronJohari1J

The shape of a molecule is independent of the dipole moments and based off the numbers of bonded atoms and lone pairs of electrons. ClO2+ has one central atom, two bonded pairs, and one lone pair of electrons (AX2E1). The lone pair-bonding pair has greater repulsion than the bonding pair-bonding pai...

CameronJohari1J

The structure is circular with the sulfur and nitrogen atoms alternating. One of the sulfur atoms will have a double bond for each of the adjacent nitrogen atoms and a lone pair. Another sulfur will have a single bond, a double bond, and two lone pairs. The final sulfur has two single bonds and two ...

CameronJohari1J

The general trend for electron affinity is that it increases across periods and decreases down groups with a few exceptions in the upper corner of the p-block. Group 15 elements generally have lower electron affinities than that of group 14 because the electron is entering an already half full p orb...

CameronJohari1J

For a comprehensive review you can do all the assigned problems for chapter 2 and review chapter 1 problems from section 1.6 onwards. According to the website we will only be tested on information from October 27th and before. The test will be whenever you have your scheduled discussion in the upcom...

CameronJohari1J

Just a follow up, you can use the coulomb potential energy equation to justify the fact that strong bonds have a shorter internuclear distance than weak bonds. The equation states that electrostatic energy is proportional to charge times charge divided by distance. If the charges are similar for the...

CameronJohari1J

The energy level 6d is actually lower than that the 5f orbital for thorium, so the d orbital would get filled first because that is more energetically favorable. Also, I think I remember Dr. Lavelle mentioning in class that f block elements are rare and radioactive and that we'll probably not focus ...

CameronJohari1J

From my understanding the electron configuration of silver would be 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1 4d10. This occurs because the d (n-1) orbital is in a lower energy level than the s (n) orbital, so the d orbital is filled instead of the s orbital. I am pretty sure this exception will be expla...

CameronJohari1J

Quick reminder that once you convert ounces to grams, then grams must then be converted to kg, as SI units must be used to solve debroglie equation problems. The other SI unit in the problem is meters per second for velocity.

CameronJohari1J

Another main difference between the Balmer and Lyman Series is their respective regions of the electromagnetic spectrum. The Balmer series generally occurs in the visible light region whereas the Lyman series occurs in the ultraviolet spectrum.

CameronJohari1J

To go from the empirical formula to the molecular formula, the molar mass of the molecular formula must be known. Divide the molecular formula molar mass by the empirical formula molar mass to get a natural number. Then multiply all the coefficients in the empirical formula by that number to get the...

CameronJohari1J

If you are still confused, the link "Everything you want to know about Sig Fig" on Dr. Lavelle's website is extremely helpful. Another resource would be the appendix section 1C for the 6th edition of the textbook which covers sig figs in depth.

Search found 66 matches