chem p4 help plzzz;''(

[angel eyez]

GUYZ PLZ SEE N/10/41 QS# 4 B...
HERE V HAVE TO FIND THE EMPIRICAL FORMULA...
I CANNOT UNDERSTAND WAT TO DO AFTER TAKING OUT THE MOLES...

ANYONE?? :Search:

 

[tuckletoe]

After that take ratio.
Divide it by the smaller number of moles. The answer would be 1:1.33
so we have to give it like in whole number. Need to multiply it with the number which give that ratio like this, eg 4:5 o 6:5
So see which number would do that.
Multiply it by 2. any success? naaa
By 3? Any success yes. Its increased to 3:3.99 which is close to 3:4

 

[bionology]

www.chemguide.co.uk

I'll advise you all not to rush for the past paper practise . Just do June/Nov 10 all variants and know every tit bit of your chem book (Ayub is the best)
Can anyone please suggest me a website to practice the skeletal formulas which have start coming?

 

[angel eyez]

thanx...
m trying n/10 pp for the first tym...

well i heard that ayub khan book is really gud but i dont have it

even i want to noe a site for skeletal formula...anyone?? :Search:

i'll let u noe as soon as i get to noe any...

LISTEN ANYONE WHO HAS PRACTICED N/10/43 QS# 5 B
WHY R V SUBTRACTING CATHODE FROM ANODE..?? ISN'T IT CATHODE MINUS ANODE...??? :%)

 

[bionology]

it's just higher E* - lower E*
don't get into the confusion of cathode and anode, just look into the E* of the cells and apply the formula i told you

 

[beacon_of_light]

it's just higher E* - lower E*
don't get into the confusion of cathode and anode, just look into the E* of the cells and apply the formula i told you

that formula might work but the actual formula is
E*(specie reduced) - E*(specie oxidised) ...

 

[bionology]

yeah but it's easier to remember and it's quicker!
who's gonna look into what's oxidised and what's reduced when you have a lot to do

 

[beacon_of_light]

Look when negative voltages come ... you might make some mistake...
but again go with the formula you feel easy with...

 

[beacon_of_light]

Plus when you have been given the overall emf of the cell and the voltage of one cell... only this formula will work when finding the voltage of the other cell... you can try that question in one of the old past papers..

 

[angel eyez]

it's just higher E* - lower E*
don't get into the confusion of cathode and anode, just look into the E* of the cells and apply the formula i told you

that formula might work but the actual formula is
E*(specie reduced) - E*(specie oxidised) ...

but in the qs m asking wen m using the formula E(reduced)-E(oxidised) m getting the wrong ans..(something in negative)
now??

 

[beacon_of_light]

it's just higher E* - lower E*
don't get into the confusion of cathode and anode, just look into the E* of the cells and apply the formula i told you

that formula might work but the actual formula is
E*(specie reduced) - E*(specie oxidised) ...

but in the qs m asking wen m using the formula E(reduced)-E(oxidised) m getting the wrong ans..(something in negative)
now??

soory didn't get ya... Look when you have two equations... both of them must show reduction potentials... then calculate...well you can tell me the questions...will explain you in a much better way ...

 

[bionology]

look, use this formula of high E*-low E*
for lower it's -0.82. -(-0.82) makes it plus and overall, it's (+)


for that reduced one, i don't think so it's needed. when an unknown metal is given, normally the polarity of second electrode is given, so u can still use the formula of (high E* - low E*)

 

[aisha231]

Bionology aap over nai ho

 

[angel eyez]

N/10/41 QS# 4 B...
well this is the qs m talking abt...

 

[beacon_of_light]

Look this is how you'll solve this question...
If the oxide contains 9.3% oxygen so it will contain 100% - 9.3% = 90.7% of lead...
Now the same mole ratios... divide both by their respective atomic masses ...you get ... Pb : O = 0.4382 : 0.581 divide the ratios by the smaller value .i.e 0.4382 and you get Pb : O = 1: 1.3 .. now you'll have to multiply this ratio by a number that will convert 1.3 into an integer value...multiply by 3... so u get 3:3.9 = 3:4 ...
hope that helps

 

[tuckletoe]

Skeletal formula, check them out.
http://www.friendlyscience.com/index.ph ... &Itemid=28
http://www.bookrags.com/wiki/Skeletal_formula
http://www.ivy-rose.co.uk/Chemistry/Org ... ecules.php

 

[tuckletoe]

it's just higher E* - lower E*
don't get into the confusion of cathode and anode, just look into the E* of the cells and apply the formula i told you

that formula might work but the actual formula is
E*(specie reduced) - E*(specie oxidised) ...

but in the qs m asking wen m using the formula E(reduced)-E(oxidised) m getting the wrong ans..(something in negative)
now??

Listen. When you actually subtract.
You don't change signs of either species.
Maybe you are changing signs. Please don't do that.
Jo sign jaise data booklet main hai us ko exactly waise equate karo and solve it.

 

[angel eyez]

Look this is how you'll solve this question...
If the oxide contains 9.3% oxygen so it will contain 100% - 9.3% = 90.7% of lead...
Now the same mole ratios... divide both by their respective atomic masses ...you get ... Pb : O = 0.4382 : 0.581 divide the ratios by the smaller value .i.e 0.4382 and you get Pb : O = 1: 1.3 .. now you'll have to multiply this ratio by a number that will convert 1.3 into an integer value...multiply by 3... so u get 3:3.9 = 3:4 ...
hope that helps

thnx but m really sorry i typed the wrong question...
the qs is N/10/43 QS# 5 B ...this q is related to the cathode anode thing v were discussing....

sorry once again :sorry:

 

[beacon_of_light]

OMG!
Okey okey np .. lol
well for b)i) H2) loses electrons to give Oxygen so is oxidised ... E* = +1.23V ... H2O gains electrons to form H2 and is reduced so E* = -0.83V ... so E* = E(reduced)-E(oxidised) = 1.23-(-0.83) = 2.06V
Here you can not use the OH- equation since this will result in a negative E* value.... reaction isn't feasible then!

 

[beacon_of_light]

Do the Cl2 one in the same way ...