Chemistry 3/4 2013 Thread

[Edward21]

I learnt a different 3: stable when heated. :S

That sounds very plausible

[Stick]

Well that's what the textbook says. If it isn't stable when heated, the chemical composition of the precipitate may change.

[Limista]

I learnt a different 3: stable when heated. :S

How is that different to what Thushan said? My understanding of this is:

When heated, the substance is inert in air or stable and unreactive in the atmosphere.

Well that's what the textbook says. If it isn't stable when heated, the chemical composition of the precipitate may change.

...Which would in turn change the molar/formula mass of the substance only if it is not a primary standard. However this does not mean that the precipitate is not suitable for use in standard solution. So yeah...

[Stick]

Fair enough. What's a primary standard?

[Limista]

Fair enough. What's a primary standard?

Can be used to make a more accurate standard solution, because it has certain properties (e.g. high molar mass to decrease likelihood or chance of percentage error, exact known chemical formula etc...)

[Stick]

And what's a standard solution? I don't recall learning any of this.

[Limista]

And what's a standard solution? I don't recall learning any of this.

You probably haven't learnt it yet because it's part of gravimetric analysis (I think! or was it volumetric?.. LOL) in the 3/4 course. Have you started studying the 3/4 course? If not, that's probably why you aren't aware of it yet. That's cool - just flip to it in your 3/4 textbook! I was just looking ahead in the 3/4 course myself a couple of days back. 

[KevinooBz]

And what's a standard solution? I don't recall learning any of this.

standard solution is made from primary standard from memory. It just reduces the potential error. Imagine trying to add 1g of a powder. When doing titration you can't really control how exact your adding of primary standard is and your solution will almost change colour instantly so you can't really calculate anything accurately. If you diluted it with 50ml of water, you can be much more accurate when slowly adding the standard solution. My teacher briefly went through this when we were performing a titration experiment.

[Stick]

You probably haven't learnt it yet because it's part of gravimetric analysis (I think! or was it volumetric?.. LOL) in the 3/4 course. Have you started studying the 3/4 course? If not, that's probably why you aren't aware of it yet. That's cool - just flip to it in your 3/4 textbook! I was just looking ahead in the 3/4 course myself a couple of days back. 

I've done gravimetric analysis, but I haven't looked at volumetric analysis just yet. I have a gravimetric analysis SAC in the first week back so I'm thinking it's something that's covered in the next chapter. I hope so!

[Reckoner]

^You're correct Stick don't fret, they're part of volumetric analysis.

[Stick]

Alright, thanks everyone. Also, when do we learn to use units like ppm, w/w etc? Was I supposed to have already learnt it?

[Math Nerd]

The combustion of octane in excess air can represented as: C8H18(g)+12.5 O2(g) -> 8CO2(g)+9H2)(l)
To produce 176g of CO2 in this reaction:

x mole of C8H18 needs to react with x mole of O2

*So for ive worked out n(CO2)=176/44=4mol, im not sure what to do next.

[Stick]

Use the mole ratio in the given equation. It says 1 mole of octane and 12.5 moles of oxygen will produce 8 moles of carbon dioxide.

[Math Nerd]

Yay got it

[paulsterio]

And what's a standard solution? I don't recall learning any of this.

Solution made from a primary standard with a KNOWN concentration.

It's just definitions, you'll get to it.