Chemistry 3/4 2013 Thread

[Homer]

nah only messaging

[teletubbies_95]

haha. sos im not very technological. its a great idea but( for peeps with smart phones)
i have a 10 year old nokia block phone ( which i love <3 <3) with like one game on it

[saba.ay]

Use the relevant half-equations provided in the electrochemical series for the oxidation of H₂S to S by Cr₂O₇ ^2- to deduce the net ionic equation for the reaction.

Ok, a relatively easy question, but the wording is putting me off. Please help. Thanks in advance.

[Mr Keshy]

nah only messaging

Homer, get on our 2013 group Skype chat, we do occasionally talk about education

[Homer]

I barely use skype :/

[teletubbies_95]

okay in the electrochemical series (data booklet) uses :
H2S+ 2H+ +2e- ----->S(reduction)
therefore deducing the oxidation ( i kinda memorised this that oxidisaiton to an ion , therefore Cr 3+),
Cr2O7(-2) +14H+6e----->2Cr3+ + 7H20

Therefore , the overall equation would be
3H2S+14H+Cr2O7(2-) ---->3S+2Cr3+6H + 7H20

cancel the H+

therefore,
3H2S +8H+ Cr2O7(2-)---->+ 2Cr3+7H20+3S//
hope that helped!

[brightsky]

oxidation half-equation:
H2S --> S
H2S --> S + 2H (+)
H2S --> S + 2H (+) + 2e (-)

reduction half-equation:
Cr2O7 (2-) --> Cr (3+)
Cr2O7 (2-) --> 2Cr (3+)
Cr2O7 (2-) --> 2Cr (3+) + 7H2O
Cr2O7 (2-) + 14H (+)--> 2Cr (3+) + 7H2O

we have:
H2S --> S + 2H (+) + 2e (-)
Cr2O7 (2-) + 14H (+) + 6e (-) ---> 2Cr (3+) + 7H2O

multiply first equation by 3:
3H2S --> 3S + 6H (+) + 6e (-)
Cr2O7 (2-) + 14H (+) + 6e (-) ---> 2Cr (3+) + 7H2O
--------------------------------------------------------------------
3H2S + Cr2O7(2-) + 14H(+) --> 3S + 6H (+) + 2Cr (3+) + 7H2O
3H2S + Cr2O7(2-) + 8H(+) --> 3S + 2Cr (3+) + 7H2O

hopefully no errors

[Mr Keshy]

Should make like a whatsapp chem group! would be much easier to upload photos or questions

We could stick to this and use an image hosting site like Flickr. That's what I'm using now

[saba.ay]

Thanks Guys. Didn't know what Cr2O7 would have oxidized to.

[teletubbies_95]

1. Why is H2S04(LIQUID) needed instead of H2SO4 (AQUEOUS) as a catalyst when esters are formed?

2. According to the new study design, it says that there will be more details about the level of interpretation of IR and NMR spectra, what is an example of that ?

[teletubbies_95]

3. I know of three ways of calculating ppm: ug/g , mg/L and ug/mL . Are there any more ways ?

4. Sulfur dioxide, SO2, is an irritant gas that attacks the throat and lungs and can cause asthma and chronic bronchitis. Its main sources are the burning of coal and oil in power stations, industry and domestic heating. The current EPA limit to the concentration of sulfur dioxide in the air is 0.20 ppm if exposure is only for 1 hour, 0.08 ppm if exposure is over 24 hours, and 0.02 ppm if exposure occurs all year.
Express the limit for 1 hour in %(v/v).[/i][/u]

How do you work out v/v using the information???  it only gives ppm?

[polar]

3.

the first part needs to be 10^6 times smaller than the second (n parts per 1 million), for example (where )

[Shenz0r]

1. Why is H2S04(LIQUID) needed instead of H2SO4 (AQUEOUS) as a catalyst when esters are formed?

H₂SO₄ (l) is concentrated Sulfuric Acid. H₂SO₄ (aq) is not.

The more concentrated the sulfuric acid is, the greater the rate of reaction, therefore the faster the esterification. This is because there will be a greater amount of molecules present in the reaction mixture, leading to a greater number of collisions occurring between reactant molecules at a particular time frame.

(Mainly concerns Collision Theory, explored in U4)

[REBORN]

^ successful collisions

[teletubbies_95]

Thanks!!!!!