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New answer posted

7 months ago

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V
Vishal Baghel

Contributor-Level 10

A = [, [-1, 4]. |A| = 2 - 1 = 1.
Characteristic equation: λ² - tr (A)λ + |A| = 0 ⇒ λ² - 3λ + 1 = 0.
By Cayley-Hamilton, A² - 3A + I = 0. A? ¹ (A² - 3A + I) = A - 3I + A? ¹ = 0.
A? ¹ = 3I - A.
Comparing with A? ¹ = αI + βA, we get α=3, β=-1.
4 (α - β) = 4 (3 - (-1) = 16.

New answer posted

7 months ago

0 Follower 38 Views

V
Vishal Baghel

Contributor-Level 10

L.H.L = lim (x→0? ) (1 + |sin x|)³? /|sin x| = lim (h→0) (1 + sinh)³? /sinh = e³?
R.H.L = lim (x→0? ) e^ (cot 4x / cot 2x) = lim (x→0? ) e^ (tan 2x / tan 4x) = e¹/².
f (0) = b.
For continuity, e³? = e¹/² = b.
3a = 1/2 ⇒ a = 1/6. b = e¹/².
6a + b² = 6 (1/6) + (e¹/²)² = 1 + e

New answer posted

7 months ago

0 Follower 9 Views

R
Raj Pandey

Contributor-Level 9

Fe? ³ + e? → Fe? ² E° = 0.77V
Zn (s) → Zn? ² + 2e? ; E° = 0.76V
Cell reaction: 2Fe? ³ + Zn → 2Fe? ² + Zn? ² E°cell = 1.53V
Ecell = E°cell - (0.059/2)log ( [Zn? ²] [Fe? ²]²/ [Fe? ³]²)
1.5 = 1.53 - (0.06/2)log (1 * [Fe? ²]²/ [Fe? ³]²)
-0.03 = -0.03 log ( [Fe? ²]/ [Fe? ³])²
1 = log ( [Fe? ²]/ [Fe? ³])² => [Fe? ²]/ [Fe? ³] = 10
Let total iron = T. [Fe? ³] + [Fe? ²] = T. [Fe? ³] + 10 [Fe? ³] = T. 11 [Fe? ³] = T.
fraction of Fe? ³ = [Fe? ³]/T = 1/11 ≈ 0.09
This solution seems to differ from the image. Let's follow the image's steps.
log ( [Fe? ²]/ [Fe? ³])² = 1 => ( [Fe? ²]/ [Fe? ³])² = 10
[Fe? ²]/ [Fe?

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New answer posted

7 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

Normal to the required plane is perpendicular to the normals of the given planes.
n = n?  * n?  = (2i + j - k) * (i - j - k) = -2i + j - 3k.
Equation of the plane is -2 (x+1) + 1 (y-0) - 3 (z+2) = 0
-2x - 2 + y - 3z - 6 = 0
-2x + y - 3z - 8 = 0
2x - y + 3z + 8 = 0
Comparing with ax + by + cz + 8 = 0, we get a=2, b=-1, c=3.
a+b+c = 2-1+3 = 4.

New answer posted

7 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Cr? O? ²? + 6Fe²? + 14H? → 2Cr³? + 6Fe³? + 7H? O
n-factor for Cr? O? ²? = 6, for Fe²? = 1
M.E Cr? O? ²? = M.E Fe²?
M? V? n? = M? V? n?
0.02 * 15 * 6 = M? * 10 * 1
M? = (0.02 * 15 * 6)/10 = 0.18 M = 18 * 10? ² M

New answer posted

7 months ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

|a * b|² + |a . b|² = |a|²|b|²
8² + (a . b)² = 2² * 5²
64 + (a . b)² = 100
(a . b)² = 36
a . b = 6 (since angle seems acute from options, but could be -6).

New answer posted

7 months ago

0 Follower 24 Views

V
Vishal Baghel

Contributor-Level 10

a = i + j + 2k
b = -i + 2j + 3k
a + b = 3j + 5k
a . b = -1 + 2 + 6 = 7
a * b = |i,  j,  k; 1, 2; -1, 2, 3| = -i - 5j + 3k
(a - b) * b) = (a * b) - (b * b) = a * b
(a * (a - b) * b) = a * (a * b) = (a . b)a - (a . a)b = 7a - 6b
. The expression becomes (a + b) * (7a - 6b) * b)
= (a + b) * (7 (a * b)
= 7 [ (a * (a * b) + (b * (a * b) ]
= 7 [ (

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New answer posted

7 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

P (CO? ) = K? X (CO? )
X (CO? ) = P (CO? )/K? = 0.835 / (1.67 * 10³) = 0.5 * 10? ³
X (CO? ) = n (CO? )/ (n (CO? ) + n (H? O) ≈ n (CO? )/n (H? O) (since n (CO? ) << n (H? O)
n (H? O) in 0.9L = 900g/18gmol? ¹ = 50 mol
n (CO? ) = X (CO? ) * n (H? O) = 0.5 * 10? ³ * 50 = 25 * 10? ³ moles = 25 mmol

New answer posted

7 months ago

0 Follower 25 Views

R
Raj Pandey

Contributor-Level 9

[Reactions showing reduction of various functional groups]
(1) Nitrile reduced to primary amine with H? /Ni.
(2) Nitrile and aldehyde reduced to primary amine and alcohol with LAH.
(3) Nitro group reduced to primary amine with Na/Hg, C? H? OH.
(4) Nitrile reduced to aldehyde with SnCl? /HCl.
Products of reaction 1, 2 and 3 are 1° amines, so react with Hinsberg's reagent to form sulphonamide. Product of (4) is an aldehyde, which does not react.

New answer posted

7 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

 

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