Messages posted by krishna.gopal

Compton shift delta_lamda = (h/mc)*(1-cos(theta)) where theta is deviation from intial direction. If the electron beams get deflected to backward direction theta = 180. Hence compton shift delta_lamda = 2h/mc.

As we go down the group size of atom increases and hence charge density on lone pair decreases. Hence basicity decreases down the group.

Yes benzaldehyde gives positive tollen's test.

2[Ag(NH₃)₂]⁺ + PhCHO + 3OH^- --> 2Ag + 4NH₃ + PhCOO^- + 2H₂O

logaritham or log (short form) is a function which is inverse of power function.

If x and y are two numbers related by y = a^x, where a is a positive number, then x = log_a(y). It is read as log of y to the base a is x.

Antilog is inverse of log or power function.

For an example 100 = 10².

So log₁₀(100) = 2.

It is used to convert multiplication and division into addition and subtraction and hence make calculation easier.

For example if x = 10.38*24.73/0.0312 then

log_a(x) = log_a(10.38)+log_a(24.73)-log_a(0.0312)

log_a(x) = 1.016+1.393-(-1.506) = 3.915

x = antilog(3.915) = 10^3.915 = 8227

A chemical reaction is a process in which one set of substances transfer into another by breakage and/or formation of chemical bonds.

Write Vanderwaal's equation. At a given pressure write P as a function of V. At critical point, dP/dV = 0 and d²P/dV² = 0. Solve vanderwall's equation and these two equations to get Pc, Vc and Tc.

delta_G = - nFE

In this reaction n=4 , F = 96500, delta_G = -827000 J. Solving we get E = 2.14 V

Daylight contains light of all colours. A piece of cloth looks red because it absorbs certain wavelengths and reflects red light. If we illuminate this cloth in night by a light which do not contain red light then this cloth will absorb all the light falling on it and it will appear dark.

Right approach by rohan. The answer is 1 for both the question.

If x is directly proportional to y then x1/x2 = y1/y2 =k. You are right.

As it is 0/0 form L hospital's rule can be used. But here i will solve it using series expansion.

tan(x) = x +x³/3+ higher terms

we can expand each term upto x³

[e^tan(x)-e^x]/[tanx-x] = [{1 + (x+x³/3) +x²/2 +x³/6}-{1+x+x²/2+x³/6}]/[{x+x³/3}-x] =[x³/3]/[x³/3] =1

It is a sigma coordinate bond.

In this process no chemical reaction takes place. Q = 0 as vessel is thermally insulated. So increase in internal energy = work done on system.

Work done in expansion = 0 (expansion against vaccum)

Let initial pressure and temp of gas are Pi, Vi.

After expansion P=Pi/2 and V = 2Vi (As Q=0,w=0 means dU = 0 or T is constant)

When gas is compressed from 2Vi to Vi its a adibatic compression and Pf = Pi*2^1.4

So w = PiVi*(2^1.4-1)/(1.4-1) = 4.1*PiVi = 4.1*nRT = 4.1*1.3*8.314*T (n =1.3 as 30% nitrogen are dissociated)

In this case increase in Internal energy =w = 44.3T (where T is initial Temperature of gas)

Distilation of 3 moles of acetone with sulphuric acid gives 1 mole of mesitylene and 3 moles of water.

A has 97.7/14 mole N and 2.3/1 mole H. So ration of N to H is 3:1 and compound is N3H. In the other case ration of N:H = 1:2 so the hydride is N2H4. Now let 1 mole of N3H reacts with xmoles of N2H4. SO the salt has (3+2x)*14 g N and (1+4x)*1 g H. As Hydrogen percent is 6.7 we get (1+4x)*100/(43+32x) = 6.7. SOlving this equation we get x = 1.

It is because of Joule Thomson effect. A real gas on free expansions needs to work against its intermolecular forces which leads to decrease in temperature (Same effect give cooling in AC and fridge). In this case temperature is low enough to solidify the CO2 which is called dry ice.

I do not know what do you mean by ma. But if ma according to you is force then force = kq^2/(4r^2) where q was the initial charge and k = 9*10^9

I do not know what do you mean by ma. But if ma according to you is force then force = kq^2/(4r^2) where q was the initial charge and k = 9*10^9

I am not sure of OBC but in general your rank should be around 10000

Right (6,5) is right solution but (-5,-6) can be other solution.

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