Chemistry
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New answer posted
3 months agoContributor-Level 10
E°cell = E°cathode - E°anode = (+1.510) - (1.229) = +0.287 V. As E°cell is positive the cell will work.
New answer posted
3 months agoContributor-Level 10
O? (15) will have configuration σ1s²σ1s²σ2s²σ2s²σ2p? ² (π2p? ²=π2p? ²) (π*2p? ¹). This ion is paramagnetic.
New answer posted
3 months agoContributor-Level 10
Gadolinium has outer configuration of [Xe]4f?5d¹6s².
Its third ionization energy is low due to highly exchange energy and hence stability of the half-filled f subshell.
New answer posted
3 months ago
Contributor-Level 10
Kjeldahl's method does not work for heterocyclic and azo compounds.
New answer posted
3 months ago
Contributor-Level 10
Cyclohepta-1, 3, 5-triene is not aromatic as one carbon is saturated (sp³).
New answer posted
3 months agoContributor-Level 10
[Ag (H? O)? ]? [Ag (CN)? ]? is called diaquasilver (I)dicyanidoargentate (I).
New answer posted
3 months agoContributor-Level 10
pH = pK? + log ( [CH? COONa] / [CH? COOH])
pH = 4.57 + log (0.1 / 0.01) = 5.57
New answer posted
3 months agoContributor-Level 10
MgH? is an ionic or saline hydride, GeH? is an electron precise hydride with 8 electrons around Ge, B? H? is an electron deficient hydride and HF is electron rich.
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