Qualification > Sciences
IGCSE CHEMISTRY DOUBTS
gaurav95:
--- Quote from: gaurav95 on June 05, 2010, 04:55:56 pm ---Some more doubts
4bi reason
7b again i want a reason.like what was the reaction and all the stuff thx
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--- Quote from: Ivo on June 05, 2010, 05:02:34 pm ---w04?
4) b) i) It would increase because silver is less reactive than tin, so now the difference in reactivity between zinc and silver is greater, therefore the voltage would increase.
Next time, please state which session and year, because I think you're asking me doubts from both s04 and w04, I'm getting confused!
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and I want reason for 7b u havent explained b4
the paper is attached w04
Ghost Of Highbury:
Coppe nitrate -> Copper + Nitrogen dioxide + Oxygen
Reddish-brown gas and brown solid will be seen.
Vin:
--- Quote from: anonymous7 on June 06, 2010, 05:56:48 am ---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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--- Quote from: red_911 on May 24, 2010, 10:45:34 am ---variant one
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Ok remember when concentrated NaCl is electrolyzed:
First the solution contains Na+ Cl- and H+ and OH- (from water). The positive inos go to the cathode, and negative ions to the anode.
At the CATHODE
the H+ ions accept electrons, since it is less reactive than Na (always elements which is more reactive 'likes' to stay in the ionic form)
2H+ + 2e- ---> H2
At the ANODE
the Cl- ions give up the electrons more readily than the OH- ions do.
2Cl- --> Cl2 + 2e-
what left in the solution is Na+ and OH- which combine to form NaOH
jus understand the concept ..
Vin:
--- Quote from: anonymous7 on June 06, 2010, 05:56:48 am ---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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Ok remember when diluteNaCl is electrolyzed:
First the solution contains Na+ Cl- and H+ and OH- (from water). The positive inos go to the cathode, and negative ions to the anode.
At the CATHODE
the H+ ions accept electrons, since it is less reactive than Na (always elements which is more reactive 'likes' to stay in the ionic form)
2H+ + 2e- ---> H2
At the ANODE
the OH- ions give up the electrons more readily as they are in abundance.
4OH- --> O2 + 2H2O + 4e-
what left in the solution is NaCl, which gets concentrated.
username:
winter 09
question 6 c
how do u do that???
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