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Hallo flyte Lily beregn koncentrationen af azid ioner i opløsningen ekvator Kjemiker spiller

$\begin{array}{lllll} a)&n_{NaCl}=\frac{m}{M}=\frac{50\; g}{58.44\; \frac{g}{mol}}=0.86\; mol\\\\ &c_{NaCl}=\frac{n}{V}=\frac{0.86\; mol}{0.5\; L}=1.72\; M\\\\&\left [Cl^- \right ]=1.72\; M\\\\\\\\ b)&n_{CaCl_2}=V\cdot c=\left (100\; mL \right )\cdot \left ( 2\; \frac{mol}{1000\; mL} \right )=0.2\; mol\\\\ &m_{CaCl_2}=n\cdot M=(0.2\; mol)\cdot \left ( 110.98\; \frac{g}{mol} \right )=22.20\; g\\\\ &\textup{opl\o selighed i vand ved 20}\degree\textup{ er }\frac{74.5\; g}{100\; mL}=\frac{18.63\; g}{25\; mL}\qquad \textup{opl\o singen er derfor m\ae ttet}\\\\ &\left [ CaCl_2 \right ]=\frac{\frac{m}{M}}{V}=\frac{\frac{18.63\; g}{110.98\; \frac{g}{mol}}}{0.025\; L}=6.7\; M\\\\&\left [ Cl^- \right ]=2\cdot \left ( 6.7\; M \right )=13.4\; M \end{array}$

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

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

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

$\begin{array}{llll} \textup{pufferligning:}&pH=pK_s+\log\left ( \frac{n_b}{n_s} \right )\\\\ \textup{titreringspuffer}&pH=pK_s+\log\left ( \frac{V\cdot c}{V_{slut}\cdot c-V\cdot c} \right )\\\\ &pH=pK_{s2}+\log\left ( \frac{V}{V_{slut}-V} \right )\qquad 0<V<V_{slut}\\\\ V\textup{ er tilsat volumen}\\ \textup{titrator (NaOH)} \end{array}$