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Chemistry Paper 3 HELP!

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1)Write an ionic equation for the precipitation of the insoluble salt Silver (I) chloride.
2)Write an ionic equation for the reaction between zinc atoms and silver(I) ions.

Can someone do both and explain?
Thanks.
 
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1) could be
Ag+ (aq) + Cl- (aq) -> AgCl (s)
in a double displacement reaction
2) could be
Zn + Ag+ -> Ag + Zn 2+
The zinc displaces the silver.

i'm sure about the second one, my first one could be wrong
 
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im not sure with the first one.
but the second one is:

Zn + 2Ag+ ---> 2Ag + Zn2+

why 2Ag+
because:

zinc giv out 2 electrons to become Zn2+
and so, 2 silver will need to take the electrons.
that's why there is a "2" in front of Ag+
 
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i understood the second one, But why does zinc need to be 2+, why not just 1+???
 
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1- Ag(+) + Cl(-) --> AgCl
2- Zn + 2Ag(+) --> Zn(2+) + 2Ag
Am sure :D. Good luck tomorrow. It's gonna be east, trust me. :p
 

XPFMember

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narutogirl said:
hey can u help me out with this :/
http://www.xtremepapers.me/CIE/Cambridge IGCSE/0620 - Chemistry/0620_s09_qp_3.pdf
Q)5 b
i don't get it at all :(
Assalamoalaikum!!

check this out

xIshtar said:
I have seen many people ask about Question 5.b)

So, I thought I would explain it to everyone here :p

From the graph, 8cm^3 of the Phosphate reacts fully with 12cm^3 of T Nitrate.

So, identifying only the atoms we need:

T : Phosphate

Taking a ratio of volumes:

8 : 12

Multiply to find the LCM of both sides.

3T : 2 Phosphate

=> Equal volumes:

24 : 24.

This implies that the formula is T3(PO4)2

if this doesn't help or if u have doubt let me know :)
 
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Math_angel said:
narutogirl said:
hey can u help me out with this :/
http://www.xtremepapers.me/CIE/Cambridge IGCSE/0620 - Chemistry/0620_s09_qp_3.pdf
Q)5 b
i don't get it at all :(
Assalamoalaikum!!

check this out

xIshtar said:
I have seen many people ask about Question 5.b)

So, I thought I would explain it to everyone here :p

From the graph, 8cm^3 of the Phosphate reacts fully with 12cm^3 of T Nitrate.

So, identifying only the atoms we need:

T : Phosphate

Taking a ratio of volumes:

8 : 12

Multiply to find the LCM of both sides.

3T : 2 Phosphate

=> Equal volumes:

24 : 24.

This implies that the formula is T3(PO4)2

if this doesn't help or if u have doubt let me know :)

thanks but..in my graph the 8 cm^3 of phosphate stop reacting at 16 not 12 cm^3..so :S
n how is the is the LCM 3:2 i guess its the oppiste and then he/she cross multiplied
umm and from where is the equal volume, and y is it mentioned?

sorry :S
 

XPFMember

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''To 12.0 cm3 of an aqueous solution of the nitrate of metal T was added x cm3 of aqueous sodium phosphate''

well read the question carefully :D
16 what u r saying is the height of precipitate! we're adding diff. volumes of sodium phosphate to find out how much is required exactly...so that is 8cm3
 
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oh..lol..stupid me :p
i got it now..if i have only used my brain into thinking actually into just reading
umm but what did he/she mean by equal volume 24=24 ?
and thanks a lot ^__^
 

XPFMember

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just leave that part... :D

before that, till 3T : 2PO4 u understood right?

that means for every 3 T we need 2 PO4
valency of PO4 is 3-

so formula is gonna be smthng like T3(PO4)x and this x we found it as 2...bcoz we just came to conclusion that 3 T requires 2 PO4 !!
 
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yes i understood that thanx a lot for the help ^___^
would've taken me forever to understand that i have to read the question and not just look and say omg omg omg
lol...thnx again :D
 
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