Chemistry
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New answer posted
9 months agoNew answer posted
9 months agoContributor-Level 10
M = (a³ * d * N_A) / Z = (3.608 * 10? )³ * 8.92 * 6.022 * 10²³) / 4
M = (46.96 * 10? ²? * 8.92 * 6.022 * 10²³) / 4 = 63 g/mole
the closest answer is choice (1)
P = (nRT) / V = (2 * 0.0831 * 300) / 10 = 4.986 bar
New answer posted
9 months agoContributor-Level 10
Oxidation state in neutral medium changes from +7 to +4 for MnO? to MnO?
New answer posted
9 months agoContributor-Level 10
Ecell = 1.05 - (0.059 / 2) log ( [Ni²? ] / [Ag? ]²)
= 1.05 - (0.059 / 2) log ( [10? ³] / [10? ³]²)
= 1.05 - (0.059 * 3) / 2 = 1.05 - 0.0885 = 0.9615 volt
There is a misprint in the question. The E? cell is incorrectly given as 10.5 V. This should have been 1.05 volt.
New answer posted
9 months agoContributor-Level 10
Statement I is true but statement II is incorrect as 3°-alcohols are the most reactive and give immediate turbidity with Luca's reagent.
New answer posted
9 months agoContributor-Level 10
Hematite: Fe? O?
Magnetite: Fe? O?
Calamine: ZnCO?
Kaolinite: [Al? (OH)? Si? O? ]
New answer posted
9 months agoContributor-Level 10
kc = [O? ]² / [O? ]³
3 * 10? = [O? ]² / [4 * 10? ²]³
[O? ]² = 192 * 10? ⇒ [O? ] = 4.38 * 10? ³²
New answer posted
9 months agoContributor-Level 10
k = (2.303 / 5) log (0.1 / 0.001) = (2.303 * 2) / 5 = 4.606 / 5 = 0.921 min? ¹
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