# 63 - General Chemistry Questions Answers

Asked By: LUFFY
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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

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

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

A sample of fuming HNO3 is labelled as 110% . assume that the labelling of fuming is similar the labelling of oleum . What is the % of free N2O5 in the sample??

plz annex explanatn also ans is 60%

Asked By: SARIKA
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Joshi sir comment

x % HNO--------------------- 100 + (18/108)*x % of fuming HNO3

so 18x/108 = 10

or x = 60 %

which of the following has the maximum value for van der waals constant b - H2O (g) , AlCl3 (g), Cl2 (g)

Asked By: SWATI KAPOOR
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Joshi sir comment

b depends on size so on the basis of size order will be AlCl3>Cl2>H2O

This answer is given by Sarika and is correct answer

the type of interaction present between NO3- ion and I2 molecule is ion-dipole interactions or ion-induced dipole interactions.Explains the difference between the two also.

Asked By: SWATI KAPOOR
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Joshi sir comment

Iis non polar molecule so it will be ion-induced dipole interactions

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