I’ll be using this for my PC doubts permanently.

Please help in these whenever you’re free.

For Q7,

ΔH**solution** = ΔH **hydration** + ΔH **lattice energy**

From This You Get ΔH.

ΔS is given.

T = 273+25 = 298 K

Now use the formula ΔG = ΔH - TΔS.

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For 5: calculate th change in pressure i.e.

P-X+X+X/2=p+x/2=133 which gives X=38mm

Now ∆x/∆t=38/20 mm/sec.

For molarity use pv=nrt

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Got it thank you

Was making mistake founnd it

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Question 45 solution -

K(NaCl+H2O) = 0.2346/7600=3.08×10^(−5)

KNaCl = K(NaCl+H2O) − K(H2O) =(3.08−2.55)×10^(−5)=5.3×10^(−6)

C(NaCl) = K(NaCl) × 1000/n = 5.3×10^(−6)×1000/126.5 = 4.180×10^(−5)

C = n/V => V = n/C = 8.547/(4.19×10^(-5)) = 2×10^(5)

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Haan. O and H ka change nahi hota sirf C ka hi hota hh isliye only C ka change nikalte

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Answer 3 hai na?

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Yup

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Mujhse nikal nhi rhe OS ek baar btaoge kaise? Jaise Formaldehyde me mere hisab se 4, methanol me 6…galat h… Pr ho nhi rhe

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disproportion reaction h…toh n factor = n1.n2 \over n1+n2 se 1 aagya… ab equivalent weight = 30 \over 1 aagya h

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Maine kal parso bheja tha ruko vapis bhejta hu

Isme O.S. hh carbon ke functional groups me

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formaldehyde mein Carbon ka zero hoga x + 2(1) - 2 = 0

x + 1(4) - 2 = 0

x = -2

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Shit!! I m again out of mind… I was considering - 1 For H

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|For the couple at 25°C Ag(s) | AgCl(s)| || | Ag(s) where ksp(AgCl) = 10–10, = 0.06, = 0.64.

If initially in a 10 litre container, 1 mole KCl and 2 mole AgNO3 are taken along with sufficient amount of Ag and AgCl in it then find the moles of AgCl changed after long time. |
---|

Correct Answer: 0.10

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this qn is weird, man…

initially they have given cell potentials and stuff, but then after that they havent said anything about what the electrodes are etc. so im guessing its just a normal Ksp problem…

So, [Ag+][Cl-] = Ksp

From that, we get the final amount of [Ag+] = 0.1, and [Cl-] \approx 0

So, 1 mole of AgCl is precipitated… so 1 should be the answer?

sorry, i know i didnt get the answer… just putting it here tho

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Thank u so much for sharing the main Idea

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