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Aanya
Questions (14)
Two particles A and B has the same mass and are being kept on a rough horizontal table "d" distance apart.B is kept still and A
3 answers
561 views
I really dont know how to do this one...Can someone please explain? Many thanks!!!
a travelling microscope is focused on a mark
2 answers
501 views
If sec a=cos a + sin a
how do we prove that cos 2a= [tan(pi/2 - a) ]^2 ?
2 answers
370 views
How do we evaluate
lim x-->0 [ cos(2x^3)-1]/[(sin2x^6] I tried the substitution 2@=2x^3 but that leaves a 'x^2' in the
3 answers
456 views
What is the reason for considering N2 has a boiling point lower than O2,because of its low molecular mass.
Just curious about
2 answers
454 views
Why N2(g) has a lower boiling point?
3 answers
453 views
What is the reason for the higher ability of Fe3+ , than Fe2+, for being hydrolyzed?
Does this has something to do with Fe3+
1 answer
665 views
What would be more acidic? SO2 or SO3?
Does this have to do something with the higher oxidation state? And why is that? Or can we
2 answers
2,744 views
If 1) a b 0 , 2) 0 a b , 3)b 0 a =0,
Where 1) ,2),3) represents the three rows of a (3*3) matrix which is equal to 0) show that
1 answer
475 views
What compounds(or anions or cations) will make FeCl3 colourless, when it is added to the initial compound(anion or cation)?
4 answers
647 views
How do we find the sum of n terms of the following?
Tr= 5^r/[(r+1)(r+2)] I can see there's an arithmetic progression and a
2 answers
491 views
How do we find the value of a without using L'hospitals rule?
lim x-->0 [(1+ax) - (1+x)^(1/2)]/(x^2) } =(1/8)
1 answer
404 views
Question : ka1 and ka2 values of H2X is given 1*(10)^-5 M and 1*(10)^-9 M.50cm^3 of 0.2 M KOH and 50 cm^3 of H2X of unknown
7 answers
1,068 views
How do we show that the value of the expression is not less than (b^2 - a^2)?
I don't have any idea about proceeding on.Do we
2 answers
361 views
Answers (26)
E°cell= E°cathode-E°anode E°cell=-0.44-(-0.74) = 0.30V Ecell= E°cell - 0.0591/n log [Cr3+]^2/[Fe2+]^3 Ecell= 0.30- 0.0591/6 log[0.1]^2/[0.1]^3 Ecell= 0.30 - 0.0591/6 log 10 Ecell= 0.30 - 0.0591= 0.2409V
But they have given it like that in the question, that A reaches the previous point.
*1/2m(w^2)=aRd
Radius of the beaker?
Thanks a lot!!
The same here :-(
Than O2?
1 picometer= 1*(10)^-12 m
Is that what you meant?
*H2CO will create?
Yes that is what I meant! :-)
Thank you! I was worried about ths question stating Fe3+ becoming colourless
Sn^(2+)?
So what are the changes which should be made,if we titrate the final solution with NaOH,not with HCl? Change of indicators or molarities etc.
First find the moles of H2X with the help of the NaOH moles.Now you get the moles of H2X in 25ml.Multiply it by 4 to find the moles in 100ml. Find the concentration of NaOH with the given data in the titration of KHP. Find the molar mass of H2X by dividing
Emily there are two reactions of H2X with NaOH ,as of what I've learned.. H2X+NaOH ---> NaHX + H2O when there's excess NaOH, H2X+ 2NaOH ---> Na2X +2H2O
Some corrections should be added! x^2==>2*(10)^-6 M x==>1.41*(10)^-3 M y^2= 9*1.41*(10)^-8 M y^2=12.69*(10)^-8 M y ==>3.56*(10)^-4 M [H+]=> 3.56*(10)^-4 M
DrBob222 don't we have to consider the hydrolysis of HX- here? H2O(l)+ HX-(aq)H2X(aq)+OH-(aq)-->(1) equ. (0.1-x) x x (moldm-3) Kh= [H2X][OH-]/[HX-] Kh=kw/ka(HX-) But we are not given the kW value and we can take it as 14 only if the temperature is 25°C.
Thank you very much! Now it sounds more reasonable! And I have another question regarding this. If we titrate the final solution with NaOH of same concentration as HClmentioned in the last part what are the changes which should be done in this experiment?
Thank you very much! Now it sounds more reasonable! And I have another question regarding this. If we titrate the final solution with NaOH of same concentration what are the changes which should be done in this experiment?
Thank you very much for your concern on this question in the first place. And I forgot to include that It is given that the temperature is 25°C. Again I'm also having doubts.I will wait for your answer and in the mean time I'll try this again.
Thank you very much DrBob222.I totally forgot about the Henderson equation! I think you cleared out all my doubts.Thank you very much for the help..
There's a typo in the in the fourth line from the bottom.It should be corrected as, 4x +20 +2x +10 = * 120 * is to indicate the place I've corrected
Let the shortest side be x(cm).So the longest side equals to (x+10)cm and the other side equals to (x+x+10)/2 So perimeter of the triangle= x + (x+10)+[(x+x+10)/2] =60 2x+10+[(2x+10)/2] =60 Now let's multiply both sides by 2 to get, 2(2x+10) + (2x+10) =
Thank you very much Emily! And if we titrate 25cm^3 of the final solution with 0.1 M HCl,using Me.O as the indicator,is there any reaction taking place? I think no as they are both acids(KOH is reacted completely) and the concentrations of OH- we get from
x^2 +2ax+b is the expression #PsyDAG