217 - Chemistry Questions Answers

starch turns iodine blue 

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

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

so it will be -0.445 for the reverse, you are asking for reduction and i submitted the answer for oxidation so the duality created.

according to the given reaction we get 2L monooxide by 1L dioxide,

initially litres of CO₂  = x so it will form 2x litres of CO

and 1-x litres CO are present from the start so total litres of CO = 2x+1-x = 1+x

for any reaction obtained by other two reactions formula used will be nE = n'E' + n''E''

in second question 2x is used because x litre CO₂ will give 2x litre CO 

Very soon, we will submit

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

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 . 6H₂O 

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

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. 

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 A_7/2B_15/4  or A₁₄B₁₅

At equilibrium mole fraction of CO = 1/6, that of CH₃CH₃ = 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 CH₃COCH₃ = 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

Yes 

very soon