Chemistry
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New answer posted
10 months agoContributor-Level 10
log (k? /k? ) = (Ea / 2.303R) [1/T? - 1/T? ]
log (3.555) = (Ea / (2.303R) [1/303 - 1/313]
1.268 * 8.314 * 303 * 313 = 10Ea
So, Ea = 100 kJ
New answer posted
10 months agoContributor-Level 10
According to IUPAC convention for naming of elements with atomic number more than 100, different digits are written in order and at the end ium is added. For digits following naming is used.
0 -nil
1-un
2-bi
3-tri
and so on.
New answer posted
10 months agoContributor-Level 10
(x/m) = k (P)¹/?
log (x/m) = log k + (1/n) log P
Slope = 1/n = 2 So n = ½
Intercept ⇒ log k = 0.477 So k = Antilog (0.477) = 3
So (x/m) = k (P)¹/? = 3 [0.04]² = 48 * 10?
New answer posted
10 months agoContributor-Level 10
Misch metal consists of Lanthanide metal (? 95%) and iron (? 5%) and traces of S, C, Ca and Al.
New answer posted
10 months agoContributor-Level 10
Na? SO? + H? SO? → SO? + Na? SO? + H? O
SO? + 2NaOH → Na? SO? + H? O
Na? SO? + SO? + H? O → 2NaHSO?
New answer posted
10 months agoContributor-Level 10
For concentration cell E? cell = 0
Anode: Cu (s) → Cu²? (aq)?
Cathode: Cu²? (aq)? → Cu (s)
Overall: Cu²? (aq)? → Cu²? (aq)?
As ΔG = −nFEcell
If ΔG = -ve than Ecell is positive
Ecell = E? cell − (0.059/2) log (C? /C? )
Ecell = − (0.059/2) log (C? /C? )
Ecell > 0 ⇒ C? > C?
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