Question Number 32151 by Tinkutara last updated on 20/Mar/18 Answered by Rio Michael last updated on 16/Nov/19 $$\left(\mathrm{3}\right) \\ $$ Terms of Service Privacy Policy…
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Question Number 31961 by Tinkutara last updated on 17/Mar/18 Answered by rahul 19 last updated on 18/Mar/18 $$\left(\mathrm{3}\right)\:{Scandide}\:{contraction}\:\left({d}−{block}\:\right. \\ $$$$\left.{contraction}\right). \\ $$ Commented by Tinkutara…
Question Number 31960 by Tinkutara last updated on 17/Mar/18 Answered by rahul 19 last updated on 18/Mar/18 $$\left(\mathrm{2}\right)\:{Tc}. \\ $$ Commented by Tinkutara last updated…
Question Number 31959 by Tinkutara last updated on 17/Mar/18 Answered by rahul 19 last updated on 18/Mar/18 $$\left(\mathrm{3}\right)\:{Si},{C}\:. \\ $$$${Si}\:{has}\:{vacant}−{d}\:{orbitals}\:! \\ $$$${All}\:{other}\:{options}\:{are}\:{factual}. \\ $$ Commented…
Question Number 31899 by Tinkutara last updated on 16/Mar/18 Answered by rahul 19 last updated on 19/Mar/18 $$\left(\mathrm{3}\right)\:. \\ $$ Commented by Tinkutara last updated…
Question Number 31863 by Tinkutara last updated on 16/Mar/18 Commented by rahul 19 last updated on 16/Mar/18 $$\left(\mathrm{3}\right)\:? \\ $$ Commented by Tinkutara last updated…
Question Number 31193 by Tinkutara last updated on 03/Mar/18 Answered by ajfour last updated on 03/Mar/18 $${E}_{{n}} =−\frac{{e}^{\mathrm{2}} }{\mathrm{4}\pi\epsilon_{\mathrm{0}} \left(\mathrm{2}{r}\right)}+{m}_{{e}} {v}^{\mathrm{2}} \\ $$$$\frac{{m}_{{e}} {v}^{\mathrm{2}} }{{r}}=\frac{{e}^{\mathrm{2}}…
Question Number 30957 by Tinkutara last updated on 01/Mar/18 Answered by ajfour last updated on 01/Mar/18 $${mvr}=\frac{{nh}}{\mathrm{2}\pi} \\ $$$${T}=\frac{\mathrm{2}\pi{r}}{{v}}\:=\:\frac{\mathrm{2}\pi{r}}{\left(\frac{{nh}}{\mathrm{2}\pi{mr}}\right)}\:=\:\frac{\mathrm{4}\pi^{\mathrm{2}} {mr}^{\mathrm{2}} }{{nh}}\:. \\ $$ Commented by…
Question Number 30626 by Tinkutara last updated on 23/Feb/18 Commented by Cheyboy last updated on 23/Feb/18 $${All}\:{have}\:{thesame}\:{cuz}\:{all}\:{have}\:{thesame} \\ $$$${number}\:{of}\:{electrons} \\ $$ Commented by rahul 19…