213 - Chemistry Questions Answers

Q  what vol. of O2 OBTAINED AT ANODE AT STP IF Aq SOLn OF CuSO4 IS ELECTROLYSED BY PASSG 100A CURRENT FOR 60 MIN? ANS 20.89L

Asked By: SARIKA
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Joshi sir comment

Total charge flown in the circuit = 100*60*60 = 360000 coulomb

so no of faradays of electricity = 360000/96500 = 3600/965 = 3.73 F

so amount of O2 = 3.73*5.6 = 20.89 litre

Here 5.6 litre is eq. volume corresponding to 8 gm. of O2

Q ONE LITRE MIXTURE OF CO &CO2 IS PASSED THRU RED HOT CHARCOAL IN A TUBE THE NEW VOL . BECOMES 1.2L WHAT % OF CO IN MIXTURE INITIALLY?

Q AN Aq SOLn CONTAINING 100 g OF DISSOLVED MgSO4 IS FED TO A CRYSTALLIZER WHERE 80% SALT CRY. OUT AS MgSO4.6H2O CRYSTAL HOW MANY gm OF HEXAHYDRATE SALT CRY. ARE OBTAINED FROM CRYSTALLIZER=?

ELECTROCHEM

Fe3+  +  e- -----> Fe2+ = 0.77V

Fe --------.>  Fe3+  + 3e-     E = +0.04V

WHAT IS VALUE OF E* for Fe2+  +2e--------> Fe ? 

Asked By: SARIKA
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Joshi sir comment

CO2 + C (charcoal) ------->  2CO

let us consider that initially volume of CO2 is x litre and that of  CO is (1-x) litre

so after the above written process net CO in final = 2x+1-x = 1.2      given 

on solving x = 0.2 and 1-x = 0.8   now calculate %

 

MgSO4  -------------->   MgSO4 .6H2O

it means 120 gm (mw of compound) gives 228 gm MgSO4 . 6H2

so 80 gm will give 228*80/120     solve it

here 80 is used for 80% of 100 gm.

 

 

Fe+++  +  e-  ----->  Fe++                  E=  +0.77 V

Fe                ----->  Fe+++  + 3e-       E = +0.04 V

 

on adding we get

Fe                ----->  Fe++  + 2e-         

so E for the new reaction = [1F(0.77) + 3F(0.04)]/2F = 0.89/2 = 0.445

SIR , I WANT TO KNOW THAT IN THE EQUILM QUES WHY YOU TAKE MOL FRACT FOR CH3COCH3=1-1/6-1/6.WHY YOU SUB FROM 1.

Asked By: SARIKA
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Joshi sir comment

If we multiply mole fraction to pressure, we get the pressure exerted by the component so it is a good way to get answer. and 1 represents net sum of mole fractions of all the components. here first and second 1/6 stands for mole fraction of ethane and carbon monooxide. 

Q In NAcL type str of A+ &B- ,2A ATOMS ARE REMOVED FROM EDGES &2B ATOMS ARE REMOVED FROM CORNER. FORMULA=?  ANS =A14B15 . HOW?

Asked By: SARIKA
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Joshi sir comment

In NaCl like structure B covers all the corners and face centre and A covers all the edge centres and body centre

Net contribution of B = 8*1/8 + 6*1/2 = 4

Net contribution of A = 12*1/4 + 1*1 = 4

now 2 A ATOMS ARE REMOVED FROM EDGES so remaining contribution = 10*1/4 + 1*1 = 7/2

and 2 B ATOMS ARE REMOVED FROM CORNE so remainnig contribution = 6*1/8 + 6*1/2 = 15/4

so formula will be A7/2B15/4  or A14B15

Q CH3COCH3 (g) eqm CH3CH3 +CO

  initial pressure of CH3COCH is 300 mm when eqm is setup mole fraction of CO(g) = 1/6 hence total pressure will be =?

ANS 360mm of hg .SOL =?

Asked By: SARIKA
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Joshi sir comment

At equilibrium mole fraction of CO = 1/6, that of CH3CH= 1/6

so mole fraction of CH3COCH3  = 1- 1/6 - 1/6 = 2/3

let the total pressure at equilibrium = p

so pressure of CH3CH3 and CO will be p/6 and p/6

and that of CH3COCH= 2p/3 

now consider the following dissociation data

CH3COCH3 ----> CH3CH3 + CO

300 mm                        0                      0                            at starting

300-x  mm                    x                       x                            at  equilibrium 

here x = p/6  (1)  and 300-x = 2p/3   (2)                                         add these equations and get the value of p

sir, can please write atopic on amins for  reactions?

 

Asked By: SHUBHAM VED
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Joshi sir comment

Yes 

very soon

Sir can you please post a topic on ionic equilibrium ( especially for numerical solving).

Asked By: AMIT DAS
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Solution by Joshi sir

YES

very soon

Structure of Na2S4O6

Asked By: NITIN RAO
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Joshi sir comment

A decapeptide (Mol.wt 796) on complete hydrolysis gives glycine(Mol.wt 75) , alanine and phenylalanine.  Glycine contributes 47 % to the total weight  of the hydrolysed products . The number of glycine units present in the decapeptide is

Asked By: NITIN RAO
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Joshi sir comment

47 % of (796+162) = 958*47/100 = 450 approx

so no of glycene units = 450/75 = 6 approx

here 162 is taken for the weight of 9 water molecules required for breaking the decapeptides.

The change in standard gibbs free energy of given cell at NTP

Zn I Zn(2+) II OH- I Ag2O I Ag         Eo =1.104V

Asked By: NIKHIL VARSHNEY
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Joshi sir comment

Zn  ----->    Zn++   + 2e-

Ag2O  +  H2O + 2e-  ------>    2Ag  +  2OH-

so Gibbs free energy change = nFE = 2*96500*1.104

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