Chemistry: Post your doubts here!

[Muhammad Bin Anis]

the least negative one goes fwd as it is more "feasible" than the more negative one.

Another way of remembering :

(the more reactive the metal is , the more stable it is in the ionic form)

the one that has the lower electrode potential value is more reactive - so it prefers to remain in the ionic form.


Therefore it will not reduce from +2 or +3 (or any positive value) to zero.

 

[Amy Bloom]

ok so here i have a chemistry p5 question 2 ....... so dont skip this question i really need the answer!!!!
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_5.pdf
so in Q2 part (f) they asked to draw construction lines
so can you tell me what these construction lines are and how to make them on the graph ????

I just quoted a graph from Google. Construction lines are lines which i have denoted by thick dotted lines. These enables you to read the variable (y-axis value if u wish) at a particular point (the x axis value). Hope this helped & Good luck.
P.S the exception here is that in Paper 5 you'll have to draw these lines on a Grid (Graph paper). Here, its a blank space on the background.

 

[NouranAyman]

I have a question!! urgent pleaseee.... in paper one i see questions asking stuff related to the length of the bond, can anybody explain!! PLEASEEE

 

[Muhammad Bin Anis]

I have a question!! urgent pleaseee.... in paper one i see questions asking stuff related to the length of the bond, can anybody explain!! PLEASEEE

They say it like this:

the shorter the bond length the stronger the bond is.... (and vice versa...)

 

[DANGERBP]

A solution contains 50.0 g dm–3 of the chloride of an alkaline earth metal; To 25.0 cm3
of this solution was added an excess of aqueous silver nitrate; 3.77 g of silver chloride,
AgCl, was precipitated. Calculate the value of the relative atomic mass of the metal and
suggest its identity.
(XCI2 + 2AgNO3 → X(NO3)2 + 2AgCI)
number of moles AgCI )1(
143 5.
77.3
= 0.0263 ⇒ 0.0131 mol XCI2 (1)
Mr(XCI2) = 25/1000 × 50 (1) = 95.2 –71 = 24.2 (1)
0.0131 (1)
why do we need the 0.0131 .. if they found the mass number by subtractin ??


HELPPPP !!!!!

 

[Adhiraj Shinde]

where on the forum can i find the compiled organic synthesis sheet for AS and A2 chemisry?

 

[sidbloom]

hello i need some help
i am in a bit of a hard situation here
i live in jeddah
my chemistry teacher is not good and i dont know any good tution over here or the ones that have there name going for them is too expensive.
my AS result was not really good....(bellow C)
my parents are now asking me to move to Pakistan to some private academy and study for AL there
i am really confused what to do and will i even get any admission there in the first place and how do i know that the teachers there are good enough......

 

[DANGERBP]

A solution contains 50.0 g dm–3 of the chloride of an alkaline earth metal; To 25.0 cm3
of this solution was added an excess of aqueous silver nitrate; 3.77 g of silver chloride,
AgCl, was precipitated. Calculate the value of the relative atomic mass of the metal and
suggest its identity.
(XCI2 + 2AgNO3 → X(NO3)2 + 2AgCI)
number of moles AgCI )1(
143 5.
77.3
= 0.0263 ⇒ 0.0131 mol XCI2 (1)
Mr(XCI2) = 25/1000 × 50 (1) = 95.2 –71 = 24.2 (1)

0.0131 (1)
why do we need the 0.0131 .. if they found the mass number by subtractin ??


HELPPPP !!!!!

again ! help ...

 

[InnocentAngel]

Hey friends i need the marking scheme for a certain paper the teacher gave us to work .I just have this written at the bottom ' 9256/5 S 99 .
Could someone tell me which where to get the ms. And yeah, is it a cie paper or another examination board ?
Thanks in advance.

 

[Dhaval Shah]

A solution contains 50.0 g dm–3 of the chloride of an alkaline earth metal; To 25.0 cm3
of this solution was added an excess of aqueous silver nitrate; 3.77 g of silver chloride,
AgCl, was precipitated. Calculate the value of the relative atomic mass of the metal and
suggest its identity.
(XCI2 + 2AgNO3 → X(NO3)2 + 2AgCI)
number of moles AgCI )1(
143 5.
77.3
= 0.0263 ⇒ 0.0131 mol XCI2 (1)
Mr(XCI2) = 25/1000 × 50 (1) = 95.2 –71 = 24.2 (1)
0.0131 (1)
why do we need the 0.0131 .. if they found the mass number by subtractin ??


HELPPPP !!!!!

Lol have patience! The part where you said "25/1000 × 50" you have to divide that by 0.0131 to get 95.4, which you later subtract 71 from to get the Mr of the earth metal. You have to do that because by doing 25/1000 × 50 you are finding the mass, in grams, so to find the Mr you would then divide by no. of moles right? I hope you get it...

 

[DANGERBP]

Lol have patience! The part where you said "25/1000 × 50" you have to divide that by 0.0131 to get 95.4, which you later subtract 71 from to get the Mr of the earth metal. You have to do that because by doing 25/1000 × 50 you are finding the mass, in grams, so to find the Mr you would then divide by no. of moles right? I hope you get it...

im no patient =p
ugh sht now i forgot how i got the 0.0131 -.-'

 

[snowbrood]

can anyone tell me where are chemistry a level atp question paper i mean past papers... which paper is that 51,21 or 11?

 

[Dhaval Shah]

im no patient =p
ugh sht now i forgot how i got the 0.0131 -.-'

Lol you see you need to relax, maybe then you will remember! 0.0131 is the moles of XCI2, which we found by first calculating moles of AgCI, which is (3.77(the mass))/(143.5(Mr)) = 0.0263 mol. Then isn't it using molar ratio we get the ratio XCI2 : AgCI, which is 1:2 from the formula, so moles of XCI2 is 0.0263/2 = 0.0131. Simple

 

[DANGERBP]

Lol you see you need to relax, maybe then you will remember! 0.0131 is the moles of XCI2, which we found by first calculating moles of AgCI, which is (3.77(the mass))/(143.5(Mr)) = 0.0263 mol. Then isn't it using molar ratio we get the ratio XCI2 : AgCI, which is 1:2 from the formula, so moles of XCI2 is 0.0263/2 = 0.0131. Simple

cant control that ..
ugh finally .. !
thaaaaanks man !!

 

[NouranAyman]

Guysss Please since exam is soo close!!! Pleaseee can anyone upload god chemistry AS notes to revize from and I need May/june 2012 papers also if u can provide a practical note then it will be really helpfull pleaseeeeeeeeeeeeeeeeeee!!!!!!!!!!! helppppp!!!

 

[VelaneDeBeaute]

NouranAyman This was all I had!

 

[VelaneDeBeaute]

And this as well!

 

[thats_me]

Guys can i have the link to 90's & the latest papers?

 

[NouranAyman]

And this as well!

Thank you sooooo muchhhhhhh

 

[aatifdesai]

Re: All Chemistry help here!! Stuck somewhere? Ask here!


jaldi btao....i calculated bt not confident wid ma anser

when i calculated the pH, it was 5.372234608 which rounds off to 5.37.