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4 months agoContributor-Level 9
CrCl? ·3NH? ·3H? O gives 3 moles of AgCl precipitate. This means all three Cl? are outside the coordination sphere.
The complex is [Cr (NH? )? (H? O)? ]Cl?
The 3 chloride ions satisfy only primary valency.
secondary valency satisfied by chloride ion = 0
New answer posted
4 months agoContributor-Level 9
PV = nRT
n = PV/RT = (1 atm * 4*10? L) / (0.083 LatmK? ¹mol? ¹ * 300 K) = 1.6 * 10? mol
Mass = n * Molar Mass = 1.6 * 10? mol * 16 g/mol = 25.6 * 10? g ≈ 26 * 10? g
New answer posted
4 months agoContributor-Level 9
Energy per second = 1000 J / 10 s = 100 J/s
Energy of one photon E = hc/λ = (6.626*10? ³? * 3*10? ) / (400*10? ) = 4.965 * 10? ¹? J
Number of electrons ejected = Total energy / Energy per photon = 100 / (4.965 * 10? ¹? ) = 20.14 * 10¹? ≈ 2 * 10²?
New answer posted
4 months agoContributor-Level 9
A + B? 2C
Initial: 1, 1
At eq: 1-x, 1+2x
K = [C]²/ ( [A] [B]) = (1+2x)²/ (1-x)² = 100
(1+2x)/ (1-x) = 10
1+2x = 10-10x => 12x = 9 => x = 3/4
[C] = 1+2x = 1+2 (3/4) = 1+1.5 = 2.5M = 25 * 10? ¹M
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4 months 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
4 months 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
4 months 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
4 months agoContributor-Level 9
C? H? + 13/2 O? → 4CO? + 5H? O
1 mole C? H? (58 g) produces 5 mole H? O (90 g)
∴ 90 g H? O obtained from 58 g C? H?
∴ 72g H? O obtained from (58/90) * 72g = 46.4 g
= 464 * 10? ¹g
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