Chemistry
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New answer posted
9 months agoContributor-Level 10
E_cell = E? _cell - (0.059/2)log { [Cu²? ]/ [Ag? ]²}
E? = E? _cell - (0.059/2)log [0.1/ (0.01)²] = 0.3095
E? _cell = 0.3095 + 0.0885 V
E? _cell = 0.398
Again, E? = E? _cell - (0.059/2)log [10? ²/ (10? ³)²]
E? = 0.398 - (0.059/2) x 4 V
E? = 0.28 V
E? = 28 x 10? ² V
New answer posted
9 months agoContributor-Level 10
1. The number of chiral carbons in open-chain aldohexose (such as glucose) is four, therefore, the number of stereoisomers? 2? 16.
New answer posted
9 months agoNew answer posted
9 months agoContributor-Level 10
A → 2B
t = 0: 1 mole 0
after 100 min: (1 - 0.1) mol 0.2 mol
K = (1/t)ln [a/ (a-n)]
K = (2.303/100)log (1/0.9) min? ¹
0.693/t? /? = (2.303/100) [log10 - log9] = (2.303/100) x 0.046
t? /? = 69.3 / (2.303 x 0.046) = 654.15 min
Ans. = 654 (the nearest integer)
New answer posted
9 months agoContributor-Level 10
C (graphite) → C (diamond)
ΔH? = Δ? H° (graphite) - Δ? H° (Diamond) = -832.8 - [-834.8] = 2kJ/mole
New answer posted
9 months agoContributor-Level 10
ppm of O? = (wt. of O? ) / (wt. of H? O) * 10?
= (10.3 mg) / (1.03 * 10? mg) * 10?
= 10ppm
New answer posted
9 months agoContributor-Level 10
i = total no. of particles after dissociation/association / total no. of particle before dissociation/association
HA? H? + A? (Dissociation)
0.5a
2HA → (HA)? (Association)
0.5a/2
i = (a + 0.25a)/a = 125 x 10? ²
Ans. = 125
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