Chemistry
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New answer posted
10 months agoContributor-Level 10
KE_max = hc/λ - φ = 6.63e-19 - 4.41e-19 = 2.22e-19 = 222e-21 J.
New answer posted
10 months agoContributor-Level 10
E°_cell = E° (cathode)-E° (anode) = 0.16-0.52=-0.36V.
lnK = nFE°/RT = 1*0.36/0.025 = 14.4. Ans is x10? ¹, so 144.
New question posted
10 months agoNew answer posted
10 months agoContributor-Level 10
HI cleavage of ether. B is CH? =CHI, gives yellow PPT with AgNO? C is C? H? CH? OH.
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