67 - General Chemistry Questions Answers

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Asked By: SHELLEY BROWNING
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Asked By: MITUL
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Joshi sir comment

 Joshi sir comment

Constant b is associated to volume correction so its value be less for molecule of smaller volume.

Asked By: MITUL
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Joshi sir comment

Does stabilization of a charge mean willingness of other atoms in the molecule to share the charge or the extent of sharing of charge ? If its the 1st then how does that stabilize the charge?

Asked By: LUFFY
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Joshi sir comment

Sharing is the case applicable for the atoms of same electronegativity, thus by sharing electrons (charges) atoms make their orbital configurations stable according to the rules defined for stability.

Both are separate questions....How to know what reaction is going to occur in each?
Asked By: LUFFY
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Joshi sir comment

In first case reactions involved areReduction of Fe+++,    Fe+++Fe++ Titration step,               6Fe+++K2Cr2O76Fe++++2Cr+++ By using titration reaction, solve the question   

 In second case reaction involved isAs4O6+4I3-As4O10+12I-Now solve

inform if any issue

 

 

 

 

Balance 

KMnO4+H2SO4+H2S -> K2SO4+MnSO4+S+H2O

                                              

Asked By: LUFFY
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Joshi sir comment

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

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 

 

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