# 66 - General Chemistry Questions Answers

Joshi sir comment

Joshi sir comment

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

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?

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.

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Joshi sir comment

In first case reactions involved are

In second case reaction involved is${\mathrm{As}}_{4}{\mathrm{O}}_{6}+4{{\mathrm{I}}_{3}}^{-}\to {\mathrm{As}}_{4}{\mathrm{O}}_{10}+12{\mathrm{I}}^{-}$Now solve

inform if any issue

Balance

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

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

Asked By: BONEY HAVELIWALA
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
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
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
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
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Joshi sir comment

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