IGCSE CHEMISTRY DOUBTS

[gaurav95]

Some more doubts
4bi reason

7b again i want a reason.like what was the reaction and all the stuff thx

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!

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]

Question:

Products at the electrodes. What's the difference for a dilute aqueous electrolyte and concentrated aqueous electrolyte? Take sodium chloride for example.

variant one

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]

Question:

Products at the electrodes. What's the difference for a dilute aqueous electrolyte and concentrated aqueous electrolyte? Take sodium chloride for example.

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???

[princess12]

winter 09
question 6 c

how do u do that???

number of moles of FeSO4 used = 12.16/152 = 0.08
number of moles of Fe2O3 formed = 0.04
mass of one mole of Fe2O3 = 160 g
mass of iron(III) oxide formed = 0.04 × 160 = 6.4 g
number of moles of gases formed = 0.08
volume of sulfur trioxide formed = 0.08 × 24 = 1.92dm3

[princess12]

did u get it furqan

[username]

yh
i think so
thx vry much!!!

[NidZ- Hero]

1.23 g of barium carbonate produces 0.12dm3 [of co2 when treated wid excess hydrochloric acid
calculate the % purity of barium carbonate
thanz if any 1 culd help
 

[WARRIOR]

1.23 g of barium carbonate produces 0.12dm3 [of co2 when treated wid excess hydrochloric acid
calculate the % purity of barium carbonate
thanz if any 1 culd help
 

80%?

[gaurav95]

Coppe nitrate -> Copper + Nitrogen dioxide + Oxygen

Reddish-brown gas and brown solid will be seen.

Thx a lot........

[WARRIOR]

Thx a lot........

its wrong

copper oxide + nitrogen dioxide + oxygeb

blak solid will be seen + brown fumes

[Vin]

its wrong

copper oxide + nitrogen dioxide + oxygeb

blak solid will be seen + brown fumes

yes

2 Cu(NO3)2 -----> 2 CuO + 4 NO2 + O2

[Ghost Of Highbury]

Sry, my bad.

[J.Darren]

Eek, 19 hours left till D :O