Home

prosím potvrďte detekce Rozbít se beregn koncentrationen af azid ioner i opløsningen podstatné jméno umlčet Normální

$\begin{array}{rcllllll} \textup{reaktionsforhold} &HCl:NaOH&=&1:1\\\\ \textup{titerligning}&n_{HCl}&=&n_{NaOH}\\\\ &V_{HCl}\cdot c_{HCl}&=&V_{NaOH}\cdot c_{NaOH}\\\\ &\left ( 10\; mL \right )\cdot c_{HCl}&=&(25\; mL)\cdot (0.3\; M)\\\\ &c_{HCl}&=&\frac{25\; mL}{10\; mL}\cdot (0.3\; M)\\\\ &c_{HCl}&=&2.5\cdot \left ( 0.3\; M \right )\\\\ B)&c_{HCl}&=&0.75\; M \end{array}$

$\begin{array}{llllll} a)\\& \begin{array}{llllll} \left [ Fe(NO_3)_3\cdot9H_2O \right ]\;_{\textit{(s)}}\;\longrightarrow \;Fe^{3+}\;_{\textit{(aq)}}\:+\;3\;N{O_3}^-\;_{\textit{(aq)}}\;+\;9\,H_2O\;_{\textit{(l)}} \end{array}\\\\\\ b)\\& \begin{array}{llllll} M( Fe(NO_3)_3\cdot 9H_2O)=\left (241.86+9\cdot 18.015 \right )\;\frac{g}{mol}=403.995\;\frac{g}{mol}\\\\ n( Fe(NO_3)_3\cdot 9H_2O)=n(Fe(NO_3)_3)=\frac{2.85\;g}{403.995\;\frac{g}{mol}}=0.007055\;mol\\\\ \left [ Fe(NO_3)_3 \right ]=\frac{0.007055\;mol}{0.250\;L}=0.028218\;M \end{array}\\\\\\ c)\\& \begin{array}{llllll} \left [ Fe^{3+} \right ]=\left [ Fe(NO_3)_3 \right ]&=&0.028218\;M\\\\ \left [ N{O_3}^- \right ]=3\cdot \left [ Fe(NO_3)_3 \right ]&=&0.084655\;M \end{array}\\\\\\ d)\\& \begin{array}{llr} \left [ Fe^{3+} \right ]_{\textup{fort}}=\frac{1}{3}\cdot \left (0.028218\;M \right )&=&9.406\;mM\\\\ \left [ N{O_3}^- \right ]_{\textup{fort}}=\frac{1}{3}\cdot \left ( 0.084655\;M \right )&=&28.218\;mM \end{array} \end{array}$

$\begin{array}{llllll}&& \left [ Ca^{2+} \right ]\cdot \left [ S{O_4}^{2-} \right ]=4.93\cdot 10^{-5}\;M^2\\ \textup{kaldes den mol\ae re}\\\textup{opl\o selighed af }CaSO_4\\\textup{\textbf{s} haves:}\\&& s\cdot s=4.93\cdot 10^{-5}\;M^2\\\\&&s=\left (4.93\cdot 10^{-5}\;M^2 \right )^{\frac{1}{2}}=0.007021\;M\\ \textup{De aktuelle koncentra-}\\ \textup{tioner af }Ca^{2+}\textup{ og }S{O_4}^{2-}\textup{:} \\&&\left [ Ca^+ \right ]=\left [ S{O_4}^{2-} \right ]=0.007021\;M \end{array}$