Qualification > Sciences
IGCSE CHEMISTRY DOUBTS
Vin:
okay!!! so, we just divide 0.2/2 to get into 1:1 ratio, got it. If yes, then cool +rep. If you say no and start another explanation then imma -rep you >:( .. CHILL just joking ;) Its just that we were never taught such stuff(even if I were I don't remember). THANK YOU SO MUCH IVO, YOU'VE BEEN GREAT HELP! :D
Ivo:
--- Quote from: ~VIN1094~ on June 05, 2010, 11:05:51 pm ---okay!!! so, we just divide 0.2/2 to get into 1:1 ratio, got it. If yes, then cool +rep. If you say no and start another explanation then imma -rep you >:( .. CHILL just joking ;) Its just that we were never taught such stuff(even if I were I don't remember). THANK YOU SO MUCH IVO, YOU'VE BEEN GREAT HELP! :D
--- End quote ---
Yep, I guess the top tip for working out limiting reagents and which is in excess is first to convert the substances to moles. Then for comparison, change to 1:1 ratio.
You got it!
Vin:
--- Quote from: Ivo on June 05, 2010, 10:53:13 pm ---Also, can someone please explain the colour changes and what happens for the potassium manganate and potassium iodide, I have no idea what they mean. Thanks in advance.
--- End quote ---
Potassium manganate (VII) is an oxidising agent, purple compound. The oxidation state of manganese is +VII. BUt it is much more stable in oxi. state +II. So it is strongly driven to gain electrons and reduce its oxi. state to +II.
So it takes electrons from other substances, in the presence of a little acid. It itself is reduced causing a colour change..
(reduction)
MnO4- ----------> Mn2+
manganate(VII)ion manganese ion(II)
purple colourless
This means KMnO4 can be used to test for a reducing agent.
Vin:
--- Quote from: Ivo on June 05, 2010, 11:14:41 pm ---Yep, I guess the top tip for working out limiting reagents and which is in excess is first to convert the substances to moles. Then for comparison, change to 1:1 ratio.
You got it!
--- End quote ---
And buddy you got some rep ;) Thanks again.. :)
About Potassium Iodide, a reducing agent.
H2O2 + 2KI + H2SO4 -----> I2 K2SO4 + 2H2O
H2O2 here is loses O2, therefore reduced. KI helps this to happen also it itself is oxidised to iodine which brings a colour change::
(oxidation)
2I-1 ------------ > I2
colourless brown
anonymous7:
Question:
Products at the electrodes. What's the difference for a dilute aqueous electrolyte and concentrated aqueous electrolyte? Take sodium chloride for example.
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