Vraag 7

Voorbeelden van een juist antwoord zijn:(\frac{4}{8}\times3{,}42+\frac{4}{8}\times2{,}42-\frac{2}{8}\times8{,}24-\frac{8}{8}\times0{,}923)\cdot10^5=-6\cdot10^3(\mathrm{~J~mol}{ }^{-1}\ \mathrm{HCl})(\frac{4}{8}\times3{,}42+\frac{4}{8}\times2{,}42-\frac{2}{8}\times8{,}24-\frac{8}{8}\times0{,}923)\cdot10^5=-6\cdot10^3(\mathrm{~J~mol}{ }^{-1}\mathrm{HCl})(\frac{4}{8}\times3{,}42+\frac{4}{8}\times2{,}42-\frac{2}{8}\times8{,}24-\frac{8}{8}\times0{,}923)\cdot10^5=-6\cdot10^3(\mathrm{~J~mol}{ }^{-1}\mathrm{HCl})$(\frac{4}{8} \times 3{,}42+\frac{4}{8} \times 2{,}42-\frac{2}{8} \times 8{,}24-\frac{8}{8} \times 0{,}923) \cdot 10^{5}=-6 \cdot 10^{3}(\mathrm{~J} \mathrm{~mol}{ }^{-1} \mathrm{HCl})of

$-E_{\text {begin }}+E_{\text {eind }}=-\left[\frac{4}{8} \times(-3{,}42 \cdot 10^{5})+\frac{4}{8} \times(-2{,}42 \cdot 10^{5})\right]

+\left[\frac28\times(-8{,}24\cdot10^5)+\frac88\times(-0{,}923\cdot10^5)\right]=-0{,}06\cdot105(\mathrm{~J~mol}{ }^{-1}\ \mathrm{HCl})+\left[\frac28\times(-8{,}24\cdot10^5)+\frac88\times(-0{,}923\cdot10^5)\right]=-0{,}06\cdot10^3(\mathrm{~J~mol}{ }^{-1}\ \mathrm{HCl})+\left[\frac28\times(-8{,}24\cdot10^5)+\frac88\times(-0{,}923\cdot10^5)\right]=-0{,}6\cdot10^3(\mathrm{~J~mol}{ }^{-1}\ \mathrm{HCl})+\left[\frac28\times(-8{,}24\cdot10^5)+\frac88\times(-0{,}923\cdot10^5)\right]=-06\cdot10^3(\mathrm{~J~mol}{ }^{-1}\ \mathrm{HCl})+\left[\frac28\times(-8{,}24\cdot10^5)+\frac88\times(-0{,}923\cdot10^5)\right]=-6\cdot10^3(\mathrm{~J~mol}{ }^{-1}\ \mathrm{HCl})+\left[\frac28\times(-8{,}24\cdot10^5)+\frac88\times(-0{,}923\cdot10^5)\right]=+\left[\frac28\times(-8{,}24\cdot10^5)+\frac88\times(-0{,}923\cdot10^5)\right]=-6\cdot10^3(\mathrm{~J~mol}{ }^{-1}\ \mathrm{HCl})+\left[\frac28\times(-8{,}24\cdot10^5)+\frac88\times(-0{,}923\cdot10^5)\right]=+\left[\frac28\times(-8{,}24\cdot10^5)+\frac88\times(-0{,}923\cdot10^5)\right]+\left[\frac28\times(-8{,}24\cdot10^5)+\frac88\times(-0{,}92\cdot10^5)\right]+\left[\frac28\times(-8{,}24\cdot10^5)+\frac88\times(-0{,}9\cdot10^5)\right]+\left[\frac28\times(-8{,}24\cdot10^5)+\frac88\times(-0{,}\cdot10^5)\right]+\left[\frac28\times(-8{,}24\cdot10^5)+\frac88\times(-0\cdot10^5)\right]+\left[\frac28\times(-8{,}24\cdot10^5)+\frac88\times(-2{,}42\cdot10^5)\right]+\left[\frac28\times(-8{,}24\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]+\left[\frac28\times(-8{,}2\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]+\left[\frac28\times(-8{,}\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]+\left[\frac28\times(-8\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]+\left[\frac28\times(-3{,}42\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]+\left[\frac{4}{8}\times(-3{,}42\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]+E\left[\frac{4}{8}\times(-3{,}42\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]+E_{}\left[\frac{4}{8}\times(-3{,}42\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]+E_{\text{e}}\left[\frac{4}{8}\times(-3{,}42\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]+E_{\text{ei}}\left[\frac{4}{8}\times(-3{,}42\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]+E_{\text{ein}}\left[\frac{4}{8}\times(-3{,}42\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]+E_{\text{eind}}\left[\frac{4}{8}\times(-3{,}42\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]+E_{\text{eind }}\left[\frac{4}{8}\times(-3{,}42\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]+E_{\text{eind }}=\left[\frac{4}{8}\times(-3{,}42\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]+E_{\text{eind }}=-\left[\frac{4}{8}\times(-3{,}42\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]+E_{\text{eind }}=-\left[\frac{4}{8}\times(-3{,}42\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]+E_{\text{eind }}=-\left[\frac{4}{8}\times(-3{,}42\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]+E_{\text{eind }}=-\left[\frac{4}{8}\times(-3{,}42\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]-+E_{\text{eind }}=-\left[\frac{4}{8}\times(-3{,}42\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]-E_{\text{be}}+E_{\text{eind }}=-\left[\frac{4}{8}\times(-3{,}42\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]-E_{\text{beg}}+E_{\text{eind }}=-\left[\frac{4}{8}\times(-3{,}42\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]-E_{\text{begi}}+E_{\text{eind }}=-\left[\frac{4}{8}\times(-3{,}42\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]-E_{\text{begin}}+E_{\text{eind }}=-\left[\frac{4}{8}\times(-3{,}42\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]-E_{\text{begin }}+E_{\text{eind }}=-\left[\frac{4}{8}\times(-3{,}42\cdot10^5)+\frac{4}{8}\times(-2{,}42\cdot10^5)\right]+\left\lbrack\right\rbrack+

of

$-E_{\text {begin }}+E_{\text {eind }}=-\left[4 \times(-3{,}42 \cdot 10^{5})+4 \times(-2{,}42 \cdot 10^{5})\right]$+\left[2 \times(-8{,}24 \cdot 10^{5})+8 \times(-0{,}923 \cdot 10^{5})\right]=-0{,}504 \cdot 10^{5}(\mathrm{~J}per 8 mol\mathrm{HCl})\mathrm{HCl}))$)

Opmerking

De volgende berekening goed rekenen:\frac{4}{8}\times3{,}42+\frac{4}{8}\times2{,}42-\frac{2}{8}\times8{,}24-\frac{8}{8}\times0{,}923=-6\cdot10^3(\mathrm{~J~mol}^{-1}\ \mathrm{HCl})\frac{4}{8}\times3{,}42+\frac{4}{8}\times2{,}42-\frac{2}{8}\times8{,}24-\frac{8}{8}\times0{,}923=-6\cdot10^3(\mathrm{~J~mol}^{-1}\mathrm{HCl})\frac{4}{8}\times3{,}42+\frac{4}{8}\times2{,}42-\frac{2}{8}\times8{,}24-\frac{8}{8}\times0{,}923=-6\cdot10^3(\mathrm{~J~mol}^{-1}\mathrm{HCl})$\frac{4}{8} \times 3{,}42+\frac{4}{8} \times 2{,}42-\frac{2}{8} \times 8{,}24-\frac{8}{8} \times 0{,}923=-6 \cdot 10^{3}(\mathrm{~J} \mathrm{~mol}^{-1} \mathrm{HCl})