Question Number 22837 by Tinkutara last updated on 22/Oct/17 $$\mathrm{The}\:\mathrm{effective}\:\mathrm{nuclear}\:\mathrm{charge}\:\mathrm{of}\:\mathrm{N}\:\mathrm{on}\:\mathrm{its} \\ $$$$\mathrm{last}\:\mathrm{electron}\:\mathrm{is}:− \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 22813 by Tinkutara last updated on 22/Oct/17 $$\mathrm{The}\:\mathrm{nucleus}\:\mathrm{Fe}^{\mathrm{57}} \:\mathrm{emits}\:\mathrm{a}\:\gamma-\mathrm{ray}\:\mathrm{of}\:\mathrm{energy} \\ $$$$\mathrm{14}.\mathrm{4}\:\mathrm{keV}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{the}\:\mathrm{nucleus}\:\mathrm{is} \\ $$$$\mathrm{56}.\mathrm{935}\:\mathrm{u},\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{recoil}\:\mathrm{energy}\:\mathrm{of} \\ $$$$\mathrm{nucleus}. \\ $$ Answered by ajfour last updated on…
Question Number 88339 by Rio Michael last updated on 10/Apr/20 $$\:\int\left(\:\sqrt{\mathrm{tan}\:{x}\:}\:+\:\sqrt{\mathrm{cot}\:{x}}\:\right){dx}\:=\:? \\ $$ Commented by jagoll last updated on 10/Apr/20 $$\sqrt{\mathrm{tan}\:\mathrm{x}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}\:\mathrm{x}}}\:=\:\frac{\mathrm{tan}\:\mathrm{x}+\mathrm{1}}{\:\sqrt{\mathrm{tan}\:\mathrm{x}}} \\ $$$$\sqrt{\mathrm{tan}\:\mathrm{x}}\:=\:\mathrm{t}\:,\:\mathrm{tan}\:\mathrm{x}\:=\:\mathrm{t}^{\mathrm{2}} \\ $$$$\mathrm{2sec}\:^{\mathrm{2}}…
Question Number 22797 by Tinkutara last updated on 22/Oct/17 $$\mathrm{He}^{+} \:\mathrm{in}\:\mathrm{ground}\:\mathrm{state}\:\mathrm{is}\:\mathrm{hit}\:\mathrm{by}\:\mathrm{photon} \\ $$$$\mathrm{thus}\:\mathrm{exciting}\:\mathrm{electron}\:\mathrm{to}\:\mathrm{second}\:\mathrm{excited} \\ $$$$\mathrm{state}.\:\mathrm{The}\:\mathrm{energy}\:\mathrm{absorbed}\:\mathrm{in}\:\mathrm{this} \\ $$$$\mathrm{process}\:\mathrm{is}\:\left(\mathrm{approx}\right) \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}.\mathrm{18}\:×\:\mathrm{10}^{−\mathrm{14}} \:\mathrm{J} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{7}.\mathrm{75}\:×\:\mathrm{10}^{−\mathrm{18}} \:\mathrm{J} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{4}.\mathrm{5}\:×\:\mathrm{10}^{−\mathrm{18}}…
Question Number 22782 by Tinkutara last updated on 22/Oct/17 $$\mathrm{Successive}\:\mathrm{ionisation}\:\mathrm{energies}\:\mathrm{of} \\ $$$$\mathrm{element}\:\mathrm{is}\:\mathrm{738},\:\mathrm{1450},\:\mathrm{10540},\:\mathrm{13630} \\ $$$$\mathrm{kJ}/\mathrm{mole}\:\mathrm{respectively}.\:\mathrm{The}\:\mathrm{element}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Na} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Mg} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Al} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{K} \\ $$ Commented…
Question Number 22744 by uttam1024 last updated on 22/Oct/17 $$\mathrm{How}\:\mathrm{can}\:\mathrm{i}\:\mathrm{copy}\:\mathrm{and}\:\mathrm{paste}\:\mathrm{from}\:\mathrm{this}\: \\ $$$$\mathrm{keyboard}\:\mathrm{to}\:\mathrm{other}\:\mathrm{apps}\:\mathrm{in}\:\mathrm{my}\:\mathrm{mobile}?\:\mathrm{i}\:\mathrm{can}\:\mathrm{copy}\:\mathrm{but}\: \\ $$$$\mathrm{can}\:\mathrm{not}\:\mathrm{getting}\:\mathrm{paste}\:\mathrm{in}\:\mathrm{other}\:\mathrm{apps}.\: \\ $$$$\mathrm{what}\:\mathrm{to}\:\mathrm{do}? \\ $$$$ \\ $$ Commented by $@ty@m last updated…
Question Number 22743 by Tinkutara last updated on 22/Oct/17 $$\mathrm{The}\:\mathrm{Enthalpy}\:\mathrm{of}\:\mathrm{neutralization}\:\mathrm{of} \\ $$$$\mathrm{acetic}\:\mathrm{acid}\:\mathrm{and}\:\mathrm{sodium}\:\mathrm{hydroxide}\:\mathrm{is} \\ $$$$−\mathrm{55}.\mathrm{4}\:\mathrm{kJ}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{enthalpy}\:\mathrm{of} \\ $$$$\mathrm{ionisation}\:\mathrm{of}\:\mathrm{acetic}\:\mathrm{acid}? \\ $$ Answered by Sahib singh last updated on…
Question Number 88263 by M±th+et£s last updated on 09/Apr/20 $${prove}\:{that}\: \\ $$$$\mid\frac{{e}^{{z}} −{e}^{−{z}} }{\mathrm{2}}\mid^{\mathrm{2}} +{cos}^{\mathrm{2}} {y}={sinh}^{\mathrm{2}} {x}\:\:\:\:\:{when}\:{z}={x}+{iy} \\ $$$$ \\ $$ Commented by M±th+et£s last…
Question Number 22703 by Sahib singh last updated on 22/Oct/17 Commented by Sahib singh last updated on 22/Oct/17 Answered by ajfour last updated on 22/Oct/17…
Question Number 88236 by Rio Michael last updated on 09/Apr/20 $$\:\mathrm{Evaluate}\:\:\int\sqrt[{\mathrm{3}}]{\frac{\mathrm{27}}{{x}^{\mathrm{3}} −\mathrm{6}}}\:{dx}\: \\ $$ Answered by MJS last updated on 09/Apr/20 $$\int\sqrt[{\mathrm{3}}]{\frac{\mathrm{27}}{{x}^{\mathrm{3}} −\mathrm{6}}}{dx}=\mathrm{3}\int\frac{{dx}}{\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} −\mathrm{6}}}= \\…