Class 12th
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New answer posted
a year agoContributor-Level 9
Solubility of CdSO? is water ; S = 8 * 10? M
Using Ksp = S²
Ksp = 64 * 10? M
Now,
CdSO? Cd²? + SO? ²?
in H? SO? (0.01M)
Ksp = [Cd²? ] [SO? ²? ]
64 * 10? = S? (S? + 0.01)
S? << 0.01
So, S? = 64 * 10? M.
New answer posted
a year agoContributor-Level 9
Moles of benzoic acid = 6.1 / 121 = 0.05
Theoretical moles of m- bromobenzoic acid = 0.05
Observed moles of m- bromobenzoic acid = 7.8 / 200 = 0.039
% yield = (0.039 / 0.05) * 100 = 78%
New answer posted
a year agoContributor-Level 9
Half life, t? /? = 1min
Let, time of 99.9% completion of reaction be 't' min
Let the reaction is of first order
K = (2.303/t) log? ( [R]? / [R])
[R] = 0.001 [R]?
t = (2.303 * 3 min) / 0.693
t = 9.99 min
the nearest integer is 10.
New answer posted
a year agoContributor-Level 9
λ° (BaCl? ) = 280 Scm²mol? ¹
λ° (H? SO? ) = 860 Scm²mol? ¹
λ° (HCl) = 426 Scm²mol? ¹
λ° (BaSO? ) = λ° (BaCl? ) + λ° (H? SO? ) - 2λ° (HCl)
= 280 + 860 - 2 * 426
= 288 Scm² mol? ¹
New answer posted
a year agoContributor-Level 9
In Tollen's test for aldehyde, aldehyde is oxidized to carboxylic acid salt as:
R – CHO + H? O →R – COO? + 3H? + 2e?
So; 2e? are transferred per aldhyde group.
New answer posted
a year agoContributor-Level 9
Partial hydrolysis of XeF? gives XeO? F?
XeF? + 2H? O → XeO? F? + 4HF
Compound A is XeF? , so number of lone pair on Xe is 1.
New answer posted
a year agoContributor-Level 10
The reduction of aluminum oxide (Al? O? ) is performed through an electrolytic process in its molten state. This method is necessary because Al? O? is a highly ionic and stable compound.
New answer posted
a year agoContributor-Level 9
m=2 molal
ΔT? = 100.52 – 100 = 0.52°C
Using, ΔT? = iK? m
0.52 = I * 0.52 * 2
i = 0.5
Now using, α = (1-i) / (1-1/n)
Where, n=2 (dimerisation)
α = (1-0.5) / (1-0.5) = 1
So, percentage association = 100%
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