CHEMISTRY COMMUNITY

SantanaRodriguezDis1G

its the one that's the steepest

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no,If something is aqueous it is implied that there is water

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you want E to be as positive as it can because positive standard potential means spontaneity

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if delta v is equal to zero then work is equal to zero by the equation w=-pdeltav

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And since there are assigned homework problems involving unique rates i assume it can be asked on the test

SantanaRodriguezDis1G

I'm sure some of the review sessions will have practice problems in preparation for the final

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in Part A you are using the half life equation where in part b and c you use the formula: (1/[a]) = (1/[a]0) +kt and solve for t

SantanaRodriguezDis1G

the order shows how the reactants effect the rate

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The units of k can change the units for rate when you are using it in a calculation

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i am unsure if the order can be negative, good question

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cathode is reduction (gaining electrons) and anode is oxidation (losing electrons )

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cathode is reduction and anode is oxidation

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RED CAT is another way to remember that reduction is cathode

SantanaRodriguezDis1G

in class he mentioned log is included because that is what is used in the ph scale

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if you think it is an error it may be under Lavelle's solution manual mistakes on his website

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What did one ion say to the other?
I got my ion you

SantanaRodriguezDis1G

to start the cell diagram it helps to label each side as anode or cathode

SantanaRodriguezDis1G

you use the 2nd and 3rd equation to find delta G for three moles of NADH and then divide it by the first equation because the first equation represents NADH per mole of ATP

SantanaRodriguezDis1G

I was also wondering how you know to set up the equation in this manner

SantanaRodriguezDis1G

the total entropy change is equal to zero not just entropy of the system or surroundings

SantanaRodriguezDis1G

on the test we should be given the value of Cp and you only use Cp if it explicitly says constants pressure.

SantanaRodriguezDis1G

How do you know whether to use the work equation -p deltaV or -nRTln(v2/V1)?

SantanaRodriguezDis1G

Why does the solution to chapter 9 question 57 use h20 (aq) when the problem gives it as h20 (l) and i don't believe h20 aq exists?

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You can also change the energy of a system by heating or cooling it( if its a closed system) and by doing work on the system (compressing a piston)

SantanaRodriguezDis1G

it doesn't matter which side has the negative. either way it would turn out to be the same answer

SantanaRodriguezDis1G

in question 8.21 what does the term negligible mean in regards to the concept of the question ?

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for example a pot on the oven with no lid is open, a pot on the oven with a lid is closed, and a really good thermal cup wouldn't allow for the transfer of energy or matter so it would be isolated

SantanaRodriguezDis1G

by looking at the enthalpies you can see that the melting should happen twice as fast as vaporization which is one indication as to why the answer is c

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The negative indicates that the system is losing energy

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Why in the solutions manual does it say that the reaction enthalpy is 358.8 kj. mol^-1 versus in the question it is simply written as 358.8 kj ?

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it is the enthalpy change when a bond is broken. where enthalpy is the state function that allows us to keep track of energy changes at constant pressure

SantanaRodriguezDis1G

Can someone also please explain where the 12.5 J/mol c comes from ?

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in question 8.31 part b what does the equation Cvm = (3/2)R represent ?

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ionic radius are different because cations are smaller than their parents while anions are larger than their parents so you cant look at ionic radius as you do atomic radius. I hope this helps

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Thank you for the help

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Should we know the Henderson hassleback formula for the final ?

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for bronsted most of a strong acid is depronated in solution and for a weak acid only a small fraction are

SantanaRodriguezDis1G

some elements have to follow the octet rule, and you need to make sure that the formal charge adds up to the total charge

SantanaRodriguezDis1G

What are the bond angles for something that is t-shaped?

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Can someone please explain to me why part a of number 11(sulfur tetrachloride) in chapter 4 is trigonal bipyramidal instead of seesaw?

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Why is the lone pair an equatorial instead of axial in a seesaw shape?

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How do lone pairs effect the calculation of hybridization?

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In the solution manual it is written with d at the end.

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A radical is electrons with unpaired spins. I assume you must first evaluate all the elements of the compound. For example on the midterm it would not make sense to put the one electron on oxygen or hydrogen because their octets are already full, leaving the electron to be put on carbon.

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In the review session one of the examples was [HC2O4]^-1
I do not understand why the h atom attaches to an O atom instead of the C. Could someone please explain why ?

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Why is S in the middle of the lewis structure?

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you need to consider the formal charge to make the molecule more stable

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Phosphorus is a bigger size and has an empty d-orbital

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also how would you get the electron configuration for Th(thorium) would it include an f-orbital or just (Rn) 7s^26d^2 ?

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You would start by using the Rydberg equation: E=Z^2hR(1/n^2lower- 1/n^2 upper)

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if you just write 2p2 it does not specify what the ml quantum number would be. So you would need to include (for example: 2px1, 2py1)

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It depends on the energy of the incoming photon

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I was also told that you should know that the Balmer series corresponds with visible light and that the Lyman series corresponds to ultraviolet

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I believe the conversion will be given, but the conversion is 1ev=1.602x10^-19J and in problem 23 you would have to first convert it from Kev to ev.

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You would start by using the 0.50g of NaCl given and divide it by the molar mass of NaCl (58.44g mol^-1) and then multiply that by the ratio of moles of NaCl to Cl ions which is 1:1. And then you would add that to the concentration in KCl by starting with the given (0.30g KCl) and dividing that by t...

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Yes, because by the law of conservation of mass, matter can not be created nor destroyed

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