STUFF WE GOTTA KNOW

[/0]

Oh ok, when a secondary cell is discharging,

Code: [Select]

       ___(V)__
       |          |
       |          |
      \_/       \_/

Just say the right electrode is negative. It will be the anode. Then the left will be positive and cathode.

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       ___+Battery-__
       |            |
       |            |
      \_/         \_/

Now when charging, the right electrode is still negative, but now it's the cathode.
The left electrode is still positive but it's the anode.

[dekoyl]

I think the fusion in the sun only appeared in the Kilbaha exam and Kilbaha is sometimes a bit.. "eh".

[Mao]

Oh, also fun fact of the day - did you know nuclear fission is endothermic in the sun between elements from hydrogen to iron

Why u say between those elements its endothermic, theyre the only elements the sun can consist of, nothing heavier exists within a sun.

False. In a dying star, lots of things can happen. (Stars are universal element factories, they convert hydrogen to helium to lithium and keep going until Iron, in their healthy lifespan. Just before they completely die, they spit out the heavier more exotic elements such as lead, uranium, etc, basically everything in the periodic table).

[StringFever]

Efficiency-style questions:

(Let's suppose a percent, 75% OK?)
If the number required is larger than the answer you worked out (because you've only got 75% product from a particular process and you need to work out the amount of reactant, for example) then you multiply it by 100/75
But if the number is smaller than what you worked out (for example, because the process is 75% efficient), then multiply it by 75/100!

[hyperblade01]

Lol at the efficieny-style question

Thanks /0 for the diagrams, the original wording kinda confused me (probably 'cuz of my downtime or something)


Question about the star thing, why does it stop at Iron? irrelevant to the upcoming exam I know, but it sounds interesting

[StringFever]

Lol at the efficieny-style question

Thanks /0 for the diagrams, the original wording kinda confused me (probably 'cuz of my downtime or something)


Question about the star thing, why does it stop at Iron? irrelevant to the upcoming exam I know, but it sounds interesting

I don't remember

And yeah, positive reinforcement for the efficiency questions

[/0]

Iron is one of the most stable elements because it has a high binding energy, which means it is hard to pull it apart:
http://en.wikipedia.org/wiki/File:Binding_energy_curve_-_common_isotopes.svg

[dekoyl]

Efficiency-style questions:

(Let's suppose a percent, 75% OK?)
If the number required is larger than the answer you worked out (because you've only got 75% product from a particular process and you need to work out the amount of reactant, for example) then you multiply it by 100/75
But if the number is smaller than what you worked out (for example, because the process is 75% efficient), then multiply it by 75/100!

Sorry StringFever, could you (or someone else) explain that again?
I remember a VCAA paper that had 60% efficiency for some hydrogen cell. And, instead of multiplying by 0.6, you had to divide by 0.6. Why is this so?

It took me a while to get and I just wanna go over it again =\

[Over9000]

Oh, also fun fact of the day - did you know nuclear fission is endothermic in the sun between elements from hydrogen to iron

Why u say between those elements its endothermic, theyre the only elements the sun can consist of, nothing heavier exists within a sun.

False. In a dying star, lots of things can happen. (Stars are universal element factories, they convert hydrogen to helium to lithium and keep going until Iron, in their healthy lifespan. Just before they completely die, they spit out the heavier more exotic elements such as lead, uranium, etc, basically everything in the periodic table).

Thats weird I thought that it goes up to iron only when its dying and while its healthy it simply converted hydrogen to helium (I better go bak to the drawing board).

I thought once the sun got up to iron, the star was choked under the hihg weight of iron and then proceeded to die (but maybe that is the phase before becoming a red giant).

[StringFever]

Efficiency-style questions:

(Let's suppose a percent, 75% OK?)
If the number required is larger than the answer you worked out (because you've only got 75% product from a particular process and you need to work out the amount of reactant, for example) then you multiply it by 100/75
But if the number is smaller than what you worked out (for example, because the process is 75% efficient), then multiply it by 75/100!

Sorry StringFever, could you (or someone else) explain that again?
I remember a VCAA paper that had 60% efficiency for some hydrogen cell. And, instead of multiplying by 0.6, you had to divide by 0.6. Why is this so?

It took me a while to get and I just wanna go over it again =\

This is assuming that you had to find the hydrogen used in the cell, right?

Because if we think about it, the cell was 60% efficient right - therefore, only 60% of the hydrogen was used to produced energy.
For us, we've only got data about the energy produced from the cell, and that's 60% of the hydrogen used.
Since the actual amount of hydrogen is greater, we need to multiply by (100/60) which is the same as dividing by 0.6!

[/0]

Hey u know for redox reactions how for acidic medium you add then , does it work the opposite way for alkaline medium? I've done a few examples adding first and then but is that how you're meant to do it?

[kimtywong]

yeah, that 100/60 got me confused initially. Cos i remember doing 60/100 in year 11, all the way through that year. then this year, we had to do it the other way round. so yeah.

is it ALWAYS 100/x, x being efficiency percentile?

[StringFever]

Hey u know for redox reactions how for acidic medium you add then , does it work the opposite way for alkaline medium? I've done a few examples adding first and then but is that how you're meant to do it?

I find the easiest way to do alkaline-based half-equations is to balance it as if it were in an acidic solution (so with H+ ions), and then after that, balance both sides with the same number of OH- ions. In doing so, the H+ and OH- will form water, and all you need to do is clean it up and you have your basic half-equation.

[NE2000]

Hey u know for redox reactions how for acidic medium you add then , does it work the opposite way for alkaline medium? I've done a few examples adding first and then but is that how you're meant to do it?

I find the easiest way to do alkaline-based half-equations is to balance it as if it were in an acidic solution (so with H+ ions), and then after that, balance both sides with the same number of OH- ions. In doing so, the H+ and OH- will form water, and all you need to do is clean it up and you have your basic half-equation.

This = the chemguide way

[d0minicz]

Hey u know for redox reactions how for acidic medium you add then , does it work the opposite way for alkaline medium? I've done a few examples adding first and then but is that how you're meant to do it?

I find the easiest way to do alkaline-based half-equations is to balance it as if it were in an acidic solution (so with H+ ions), and then after that, balance both sides with the same number of OH- ions. In doing so, the H+ and OH- will form water, and all you need to do is clean it up and you have your basic half-equation.

can you give an example please?