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Edwin Ng 1G

It doesn't matter, but your two choices for naming are 1,6-dichlorocyclohexene or 2,3-dichlorocyclohexene. It wouldn't be 1,2-dichlorocyclohexene because the double bond wouldn't be between the first and second carbon. The first one is correct because 1<2.

Edwin Ng 1G

Check the back of the book!

Edwin Ng 1G

If there is a C=C double bond, the pi bond can break and form a bond with the electrophile. This is not the case with a C-C single bond because it doesn't have a pi bond.

Edwin Ng 1G

When you integrate (1/[A]²) it becomes (-1/[A]), and so the two negatives cancel

Edwin Ng 1G

You need to match the half-reactions with the total reaction given (HF -> H + F). Because the first equation with E=3.03 was flipped, that one becomes negative.

Edwin Ng 1G

Most likely the week after you take the quiz. If not, then the second week for sure.

Edwin Ng 1G

There isn't a reason why. I think it's just more convenient.

One mole of a compound is usually more than one gram of a compound, so if they were both expressed in joules, the molar heat capacity can become a large number.

Edwin Ng 1G

When you take away energy/heat from the 800K reservoir, entropy decreases. Less energy means that the molecules are not moving as fast, which lead to less disorder

Edwin Ng 1G

Most likely the numbers you used for the calculation are different than what the book used, in that yours might have been more exact. On a test, if your final answer is slightly off compared to the answer key but the all the work done is correct, you will most likely get full credit.

Edwin Ng 1G

Yep, a combustion reaction is written like this

(Substance) + O₂ -> H₂O + O₂

When written like this, the reaction is exothermic (the substance burns and releases heat) and so delta H is negative

Edwin Ng 1G

You can probably memorize the strong acids/bases since there aren't a lot.

Here's a website that lists the 7 strong acids and 8 strong bases.

http://mccord.cm.utexas.edu/courses/summer2005/ch302/strong.html

Edwin Ng 1G

In general.. Acidic oxides are formed when the oxygen is bound to a nonmetal . Basic oxides are formed when the oxygen is bound to a metal . Amphoteric oxides are formed when the oxygen is bound to a metalloid . For more info see this link: http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive...

Edwin Ng 1G

The atomic radius of the F atom is smaller than the Cl atom and so the bond in HF is stronger, making it harder to dissociate.

For some of the halogens, HI is a stronger acid than HBr, which is stronger than HCl, which is stronger than HF.

Edwin Ng 1G

Even when I use the new initial values of 4,0 and 2, my calculator spits out x=0 instead of 2.6*10^-33. On a side note, will we be dealing with these very small numbers on quizzes and exams?

Edwin Ng 1G

Do you have an example? You could be mixing up OH₂ (aqua) with OH^- (hydroxido)

Edwin Ng 1G

You can calculate the net charge of -1 by calculating the formal charge on each atom and adding them all up (step 4 in example 3.6).

Alternatively, you know that KSCN is a neutral compound, and since K has a charge of +1, SCN is -1.

Edwin Ng 1G

Well, here are two bent molecules with different bond angles, SCL 2 (103°) and H 2 O (104.45°) https://en.wikipedia.org/wiki/Sulfur_dichloride"onclick="window.open(this.href);return false; https://en.wikipedia.org/wiki/Properties_of_water"onclick="window.open(this.href);return false; So there isn't ...

Edwin Ng 1G

Boron only has 3 valence electrons for it to share with other atoms so it usually forms 3 bonds, which is also enough for a formal charge of 0. Fluorine usually forms an octet.

Edwin Ng 1G

My TA said that if the last digit is off by 1 then it should be okay. If the solution manual says 4.3x10^4, 4.4 would be okay, but 4.5 would not be correct. Anyone else have an idea?

Edwin Ng 1G

Well I just plugged in numbers and checked if it works. A bit more insight... the problem states that the wavelength is in the ultraviolet spectrum, and for hydrogen, that corresponds to the Lyman series! And for this, n1 = 1. .88848=((1/n1^2)-(1/n2^2)) You also know n2 > n1 (otherwise the value on ...

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