Class 12th
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New answer posted
a year agoContributor-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?
New answer posted
a year agoContributor-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
a year agoContributor-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
a year agoContributor-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
a year agoContributor-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 [ (
New answer posted
a year agoContributor-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
a year ago
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
a year agoContributor-Level 9
Cl?C=CCl? is not used for dry cleaning of clothes due to safety purpose.
New answer posted
a year ago
Contributor-Level 9
(a) Rate vs Time for zero order is a horizontal line (Rate = const.)
(b) t?/? vs Concentration for zero order is a straight line through the origin (t?/? ∝ [C?])
(c) Concentration vs Time for first order is an exponential decay (C? = C?e?)
(d) Concentration vs Time for a reaction is a decreasing curve, not a horizontal line.
(e) Rate vs Concentration for first order is a straight line through the origin (Rate = k[Conc.]¹)
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