200 - Chemistry Questions Answers

Q.WRITE CELL RECTION OF THE FOLLOWING- .{1} ZN | ZN2+ || cu2+ |cu    {2}    PT {H2} | HCL || CL2{PT}                                       

Asked By: PINKU UMAR
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Joshi sir comment

Zn                  ----------->   Zn++   +   2e-

Cu++ + 2e-   ----------->   Cu

 

 

H2  -------->  2H+  +   2e

Cl+ 2e-      -------->  2Cl-

If each orbital can hold a maximum of three electrons , then the number of elements in 9th period of periodic table

Asked By: VAIBHAV GUPTA
is this question helpfull: 8 3 read solutions ( 1 ) | submit your answer
Joshi sir comment

3/2 times of the previous number of elements

At 752 mm Hg of total Pressure and 301 K Temperature 150 ml H2 is produced and collected over the water, when current is passed through acidic water for 1 hour. How many Amp. current should have passed? ( at 301 K temperature, The pressure of water is 28 mm Hg )

A) 0.320 A

B) 310 A

C) 321 A

D) 0.305 A

please mention correct answer.

Asked By: BONEY HAVELIWALA
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Solution by Joshi sir

by formula PV/T = constant

(752-28)*150/301 = 760*V/273

solve for V, this will be the equivalent volume at STP

now convert this volume in gm by the formula 2V/22400

then finally m = zit, where 

z = 1/96500

solve it for i

one molar solution of sulphuric acid is equal to  2N. how???

sir  as we know for sul.acid n factor = 2

so  normality will be = 2xM  thus M = N/2 

but ans is 2N so plz give account of it .

Asked By: SARIKA
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Joshi sir comment

molarity = 1 

and N = nM = 2(1) = 2

hence it is clear 

Two flasks of equal volume connected by narrow tube(of negligible volume) are at 300K and contains 0.7 mole of H2(.35 mole in each flask) at 0.5 atm.One of the flask is then immersed in a bath kept at 400k while the other remains at 300K. Caculate the final pressure and the number of moles of H2 in each flask.

Asked By: ABHISHEK PAHI
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Joshi sir comment

by gas eq.  PV/nT = constant

so 0.5*2V/0.7*300 = PV/n1400 + PV/n2300        (1)

no. of moles are also constant

so 0.7 = n1 + n2

similarly by gas eq. 

PV = n1R400                      for first flask

PV = n2R300                      for second flask

now solve

Calculate molecular diameter of  He from its Van der Waal's constant  b=24ml/mole.

Asked By: ABHISHEK PAHI
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Joshi sir comment

b = 4Vwhere V is volume of 1 mole particles

so 24*10-6 cm3/mole = 4*[4πr3/3] cm3/mole 

now solve

 

 

The Total energy of one mole of an ideal monoatomic gas at 300K is____.

Asked By: ABHISHEK PAHI
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Joshi sir comment

= 3RT/2 

 

if there are Three Fe+2  For a Fe+3 in FexO, What is the value of x? 

 

Asked By: BONEY HAVELIWALA
is this question helpfull: 1 1 read solutions ( 5 ) | submit your answer
Joshi sir comment

3FeO.1Fe2O3 = Fe5O6 = Fe5/6O

 

 

 

 

 

Zinc but not copper is used for recovery  of Ag From the complex [Ag(CN2)]-  . Why?

 

Asked By: AAYUSH TANEJA
is this question helpfull: 4 3 read solutions ( 1 ) | submit your answer
Joshi sir comment

Zn is better reducing agent in comparison to Cu for the said complex so it is used

besides it answer by sarika is correct too

CH3-(CHBr)-CH2-CH3   +  Na  --------> ?

 

 

Asked By: AAYUSH TANEJA
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Joshi sir comment

in presence of ether it is Wurtz reaction and as obvious many products will be formed.

Major product will be CH3-CH2-CH(CH3)-CH(CH3)-CH2-CH3

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