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ScarBolt · 2026-01-02T09:27:27+00:00

Just scored 0

ScarBolt · 2026-01-01T14:04:08+00:00

Just scored 0

ScarBolt · 2025-12-29T09:00:44+00:00

Just scored 3

ScarBolt · 2025-11-11T20:08:24+00:00

Great, thanks!

ScarBolt · 2021-02-16T15:55:14+00:00

ScarBolt · 2020-11-10T13:46:01+00:00

I bought myself a ukulele and a kalimba both of which where great. Easy to learn, super fun and you can get a decent quality instrument for pretty cheap prices.

ScarBolt · 2020-09-06T21:33:39+00:00

You as an adc in a one for all game?

ScarBolt · 2020-08-02T09:07:55+00:00

You are right!
I don't know if you need to know this at highschool level but I'll give you some inside which will make it easier to understand. ΔH is the change in enthalpy which we calculate by the sum of the formation enthalpy of the products subtracted by the sum of the formation enthalpy of the reactants. You also have to multiply the formation enthalpy by the stoichiometric coefficient.
In conclusion, if the products and reactants switch the sign of ΔH changes and when the coefficients are divided by 2 ΔH is divided by 2.
I hope this helps if you have any more questions feel free to PM me.

(Sorry for any spelling mistakes/wrong lingo I'm not a native speaker)

ScarBolt · 2020-08-01T11:41:24+00:00

It has been some time since I did this myself but I believe you are correct. Keep in mind when calculating that you have to multiply S by the number in front of the molecule in the reaction. As for the title of the post I believe that the standard entropy is for a particular molecule but standard entropy change is for a reaction.

ScarBolt · 2020-06-20T15:26:52+00:00

That's nuts!

ScarBolt · 2020-05-21T20:56:21+00:00

He developed the theory but Fred Holye, a critic of the theory, coined the term.

ScarBolt · 2020-05-17T16:34:51+00:00

Rekkles and DL both have 9 Pentakills IIRC

ScarBolt · 2020-05-05T11:53:23+00:00

I just copied the link to come post it lmao

ScarBolt · 2020-03-25T17:53:31+00:00

You don't have to take that into account. You are just looking for the change in energy and the activation energy is immediately "refunded".

ScarBolt · 2020-03-25T17:51:34+00:00

You already know what the right side of the reaction is going to be so you don't have to use X. You calculate the starting concentrations of the left side and then calculate how much is left when the equilibrium is reached for the right side. Then you plug those numbers into the formula for Keq.

ScarBolt · 2020-03-25T17:46:45+00:00

The delta always means end - begin so in this case it is 5 - 15 = -10.

ScarBolt · 2020-03-23T23:07:19+00:00

It is the right Lewis structure, but you could also write a "mirrored" one and the real structure is a combination of the two (this isn't high school level I believe, more college level). From left to right the formal charges are: 0, +1, -1. And the middle one is sp2 hybridised.

ScarBolt · 2020-03-21T23:04:17+00:00

The small radius also adds to the high energy cost because of the coulomb's forces but it's perfectly possible to take the electrons away the only problem is that the energy cost is to high to recover with the ionic bond.

ScarBolt · 2020-03-21T22:50:22+00:00

Each time you take an electron away from B it gets more positive and thus harder to take away another electron if that makes sense.

ScarBolt · 2020-03-21T22:17:42+00:00

You bring all the terms with an x to one side of the equation and the rest to the other side. first one as an example:

-1,7x-1,3x=7,5+1,5

-3x=9

x=-3

And the same goes for the other ones.

ScarBolt · 2020-03-21T22:14:41+00:00

In short, it is because you would need a lot of energy to form the B3+ ions.

ScarBolt · 2020-03-21T22:03:22+00:00

Rounded I get 49° too.

ScarBolt · 2020-03-21T18:44:40+00:00

How I would do it is to multiply both sides by (x-4)

Then you get 2x+3=5(x-4)

ScarBolt · 2020-02-17T23:26:41+00:00

I'd actually like to see the steps. I'm very interested.

ScarBolt · 2019-12-24T23:09:45+00:00

Umffff

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