Messages posted by Kulvinder Singh

Simply raise tickets and stop wasting time on Chill zone.

P1V1=n1RT1

P2V1=n2RT2

=> 900/500 = N1/N2

2A -> A2

a
a-x x/2

a-x/2 = N2

After the reaction, moles of O2 left = 0.03 - 3/2 * 0.01 = 0.015 and CO produced = 0.02 moles

Total P = 101.325kPa. Now, you can get the pressure of CO and O2 at 383K. Then use, P/T constant to get new P of each species.

Cheers
Kulvinder Singh

HgI2 will react with KI to form K2HgI4 thus decreasing the number of solute molecules causing an increase in vapor pressure.

Cheers
Kulvinder Singh

x% by volume means x gm of solute in 100 cc of solution (i.e. 100d gm of solution)

Molarity = (x/Mo)/0.1

Molality = (x/Mo)* 1000/(100d-x)

Mole fraction of solute = (x/Mo)/(x/Mo + (100d-x)/1

Mole fraction of solvent = 1- mole fraction of solute

Cheers
Kulvinder Singh

I think there is a misprinting in that question. Total volume should be 500 ml.

Cheers
Kulvinder Singh

You can only ask questions from VMC material. For other questions, go to www.100percentile.com

Cheers
Kulvinder Singh

Statement - 1 is incorrect because R0 is the resistance at 0 degrees. So, to calculate alpha, use :

100 = R0 (1 + alpha x 27) - (i)

150 = R0 (1 + alpha x 227) - (ii)

Cheers
Kulvinder Singh

The answer is (3) only. We will get this corrected ASAP.

Cheers
Kulvinder Singh

With cis isomer, there will be two enantiomers and for trans isomer, there will be two enantiomers. So, total stereoisomers = 4.

Cheers
Kulvinder Singh

Solution is correct.

Cheers
Kulvinder Singh

Yes. You are correct. Answer is (1).

Cheers
Kulvinder Singh

I dont know why there is so much confusion on dipole moment. Since, it is a vector quantity and N is more and less electronegative in NH3 and NF3 respectively, dipole moment will be in upward and downward directions respectively.

Cheers
Kulvinder Singh

Mean free path is inversely proportional to P.

Cheers
Kulvinder Singh

BrF5 : sp3d2 : Square pyramidal

XeOF4 : sp3d2 : Square pyramidal

Cheers
Kulvinder Singh

It is just about passion for a subject. You just need to spend more time on Chemistry. Chemical bonding is a complicated as well as interesting subject. Nothing to worry about. Make a determination and you will find it very easy. Any competitive exam is all about 3Ps.

Passion,
Practice,
Performance.

In short, only a passoinate practice can make u perform.

Cheers
Kulvinder Singh

Please tell me your doubt. It is a solved example.

Cheers
Kulvinder Singh

Please tell me your doubt. It is a solved example.

Cheers
Kulvinder Singh

It is a solved example. Please tell me where are you having doubt.

Cheers
Kulvinder Singh

Can you just give me a link or type in the question so that i can check what went wrong ?

Cheers
Kulvinder Singh

I think it is more than sufficient if you could complete VMC modules or your institute modules 3-4 times. Studying more books doesnt mean more chances of success.

Cheers
Kulvinder Singh

Third one is similar to as in Illustration.

Cheers
Kulvinder Singh

Yes. First start mugging up Inorganic chem from NCERT. As you study more and more concepts, you will come to know that Inorganic is not only about mugging. There are concepts associated with up.

Best of Luck.

Cheers
Kulvinder Singh

It is just that one of the 2 factor will dominate. More stronger factor will decide the way density should behave.

Cheers
Kulvinder Singh

The concept of excitation of electrons from low energy level to higher energy level (e.g. 4s to 3d) is favorable only in d and f block elements. Thats why we never use this concept in s and p orbitals.

Cheers
Kulvinder Singh

If you consider the reactant and intermediate complex, it is Activation energy in forward direction and If you consider the product and intermediate complex, it is Activation energy in backward direction

Cheers
Kulvinder Singh

Take it as open beaker.

Cheers
Kulvinder Singh

For 1 mole of O2, heat liberated = 220 KJ and for 1 mol of H2O, heat absorbed = 132 KJ.

x * 220 = y * 132 (net hear should be zero to keep T constant).

=> x/y = 0.6

Cheers
Kulvinder Singh

When it is reacted with NaOH to reduce the acid content to half, [salt] = [acid] => pH = pKa (Acidic buffer)

Now, use : pH = 0.5 (pKa - logC) = 4.5 => pKa = 8

Cheers
Kulvinder Singh

Delta (S) system = Delta H/T = -401.7*1/368 J/K

Delta (S) surrounding = q(surr)/T = -q(sys)/T = -Delta(H sys)/T = 401.7/273

Delta (S) total = Delta (S) sys + Delta (S) surr

Cheers
Kulvinder Singh

Please post the reason and the confusion rather than leaving us guessing.

Cheers
Kulvinder Singh

qrev/T is the entropy change in a reversible isothermal process. It is not equal to entropy.

Cheers
Kulvinder Singh

Standard enthalpy of formation of H+ ion = 0

Cheers
Kulvinder Singh

mmoles of OH- should be 0.1 after the neutralisation.

[OH-] = 0.1/20 = 1/200 M => pOH = 2.3 => pH = 11.7

Cheers
Kulvinder Singh

pH = 7 + 0.5 (10-log(0.1)) = 11.5

Cheers
Kulvinder Singh

An isothermal process has to be in Thermal equilibrium with surroundings. So, T of the System and surrounding were taken as equal.

Cheers
Kulvinder Singh

It is a salt of weak acid and weak base and pKa for CH3COOH = pKb for NH4OH

Cheers
Kulvinder Singh

(1)Which of the following expression does not represent Charles law?
  (a)Vt=a+bt (b)Vt=V0(273.15+t/0C) (c)Vt=(V0/273.15k)T (d)Vt=V0 t [0C=celcius]

  (2)The kelvin tempreture of an ideal gas is
  (a)proportional to the average potential energy of the molecules
  (b)propostional to the average speed of the molecules
  (c)propostional to the average kinetic energy of the molecules
  (d)inversely propostional to the partial pressure

  Ans : KE avg = 3/2RT => (C)

  (3)One way to decrease the average speed of the molecules of an ideal gas is to
  (a)add more gas in the container at constant tempreture
  (b)expand the gas into vaccum
  (c)compress the gas adiabatically
  (d)expand the gas adiabatically against a constant pressure

  Ans : To decrease the avg speed, decrease the temperature of gas => (D)

  (4)A sample of Cl2 gas having a volume of 0.75L at 20(degree celcius) and 725mm Hg has a mass equal to
  (a)1600g (b)10.6g (c)2.11g (d)5.3g

  Ans : PV=nRT => 725/760 * 0.75 = nx0.0821 x 293, mass = n x 71 gm

  (5)The molar kinetic energy of molecules in 2mol 0f a gas at STP would be
  (a)3/2*8.314*273 J mol^-1 (b)1/2*8.314*273
  (c)3*8.314*273 J mol^-1 (d)(3/2)(8.314*273/6.023*10^23)J

See, its molar kinetic energy. So, answer has to be (A)

Cheers
Kulvinder Singh

M is the molecular mass of the gas in kg/mol

You should also put in some effort from your side by studying topics sincerely.

Cheers
Kulvinder Singh