1. Mieszanina \(\displaystyle{ Hg_{2}I_{2}}\) i \(\displaystyle{ HgI_{2}}\) zawiera równe ilości wagowe rtęci i jodu. Oblicz zawartość procentową \(\displaystyle{ HgI_{2}}\) w tej mieszaninie.
2. Dwutlenek węgla zanieczyszczony azotem i tlenem poddano analizie i stwierdzono, że badana próbka zawiera \(\displaystyle{ 25 \%}\) węgla.
Ile procent stanowią zanieczyszczenia w tym gazie?
Proszę o pomoc.
zawartość procentowa pierwiastków
-
pesel
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zawartość procentowa pierwiastków
1.
\(\displaystyle{ \% HgI_{2}= \frac{m_{HgI_{2}}}{m_{HgI_{2}}+m_{Hg_{2}I_{2}}} \cdot 100 \%= \frac{1}{1+ \frac{m_{Hg_{2}I_{2}}}{m_{HgI_{2}}} } \cdot 100 \%}\)
\(\displaystyle{ m_{Hg}=m_{Hg_{2}I_{2}} \frac{2M_{Hg}}{M_{Hg_{2}I_{2}}} +m_{HgI_{2}} \frac{M_{Hg}}{M_{HgI_{2}}}}\)
\(\displaystyle{ m_{I}=m_{Hg_{2}I_{2}} \frac{2M_{I}}{M_{Hg_{2}I_{2}}} +m_{HgI_{2}} \frac{2M_{I}}{M_{HgI_{2}}}}\)
\(\displaystyle{ m_{Hg}=m_{I}}\)
\(\displaystyle{ m_{Hg_{2}I_{2}} \frac{2M_{Hg}}{M_{Hg_{2}I_{2}}} +m_{HgI_{2}} \frac{M_{Hg}}{M_{HgI_{2}}}=m_{Hg_{2}I_{2}} \frac{2M_{I}}{M_{Hg_{2}I_{2}}} +m_{HgI_{2}} \frac{2M_{I}}{M_{HgI_{2}}}}\)
\(\displaystyle{ m_{Hg_{2}I_{2}} \left ( \frac{2M_{Hg}}{M_{Hg_{2}I_{2}}} - \frac{2M_{I}}{M_{Hg_{2}I_{2}}} \right )=m_{HgI_{2}} \left ( \frac{2M_{I}}{M_{HgI_{2}}}- \frac{M_{Hg}}{M_{HgI_{2}}} \right )}\)
\(\displaystyle{ \frac{m_{Hg_{2}I_{2}}}{m_{HgI_{2}}} = \frac{M_{Hg_{2}I_{2}}}{2M_{HgI_{2}}} \cdot \frac{2M_{I}-M_{Hg}}{M_{Hg}-M_{I}}}\)
i wstawiamy do pierwszego równania.
2.
\(\displaystyle{ m_{C} =m_{CO_{2}} \frac{M_{C}}{M_{CO_{2}}}}\)
\(\displaystyle{ \%C= \frac{m_{C}}{m_{CO_{2}}+m_{z}} \cdot 100 \%=\frac{ m_{CO_{2}}\frac{M_{C}}{M_{CO_{2}}}}{m_{CO_{2}}+ m_{z}} \cdot 100 \%=\frac{ \frac{M_{C}}{M_{CO_{2}}}}{1+ \frac{m_{z}}{m_{CO_{2}}} } \cdot 100 \%=25 \%}\)
\(\displaystyle{ \frac{M_{C}}{M_{CO_{2}}}=0.25\cdot \left (1+ \frac{m_{z}}{m_{CO_{2}}} \right )}\)
\(\displaystyle{ \frac{M_{C}}{M_{CO_{2}}}-0.25= \frac{M_{C}-0.25M_{CO_{2}}}{M_{CO_{2}}} =0.25 \frac{m_{z}}{m_{CO_{2}}}}\)
\(\displaystyle{ \frac{4M_{C}-M_{CO_{2}}}{M_{CO_{2}}} =\frac{m_{z}}{m_{CO_{2}}} \to \frac{m_{CO_{2}}}{m_{z}}= \frac{M_{CO_{2}}}{4M_{C}-M_{CO_{2}}}}\)
i wstawiamy to do równania poniżej:
\(\displaystyle{ \%z= \frac{m_{z}}{m_{CO_{2}}+m_{z}} \cdot 100 \%= \frac{1}{1+ \frac{m_{CO_{2}}}{m_{z}} } \cdot 100 \%}\)
\(\displaystyle{ \% HgI_{2}= \frac{m_{HgI_{2}}}{m_{HgI_{2}}+m_{Hg_{2}I_{2}}} \cdot 100 \%= \frac{1}{1+ \frac{m_{Hg_{2}I_{2}}}{m_{HgI_{2}}} } \cdot 100 \%}\)
\(\displaystyle{ m_{Hg}=m_{Hg_{2}I_{2}} \frac{2M_{Hg}}{M_{Hg_{2}I_{2}}} +m_{HgI_{2}} \frac{M_{Hg}}{M_{HgI_{2}}}}\)
\(\displaystyle{ m_{I}=m_{Hg_{2}I_{2}} \frac{2M_{I}}{M_{Hg_{2}I_{2}}} +m_{HgI_{2}} \frac{2M_{I}}{M_{HgI_{2}}}}\)
\(\displaystyle{ m_{Hg}=m_{I}}\)
\(\displaystyle{ m_{Hg_{2}I_{2}} \frac{2M_{Hg}}{M_{Hg_{2}I_{2}}} +m_{HgI_{2}} \frac{M_{Hg}}{M_{HgI_{2}}}=m_{Hg_{2}I_{2}} \frac{2M_{I}}{M_{Hg_{2}I_{2}}} +m_{HgI_{2}} \frac{2M_{I}}{M_{HgI_{2}}}}\)
\(\displaystyle{ m_{Hg_{2}I_{2}} \left ( \frac{2M_{Hg}}{M_{Hg_{2}I_{2}}} - \frac{2M_{I}}{M_{Hg_{2}I_{2}}} \right )=m_{HgI_{2}} \left ( \frac{2M_{I}}{M_{HgI_{2}}}- \frac{M_{Hg}}{M_{HgI_{2}}} \right )}\)
\(\displaystyle{ \frac{m_{Hg_{2}I_{2}}}{m_{HgI_{2}}} = \frac{M_{Hg_{2}I_{2}}}{2M_{HgI_{2}}} \cdot \frac{2M_{I}-M_{Hg}}{M_{Hg}-M_{I}}}\)
i wstawiamy do pierwszego równania.
2.
\(\displaystyle{ m_{C} =m_{CO_{2}} \frac{M_{C}}{M_{CO_{2}}}}\)
\(\displaystyle{ \%C= \frac{m_{C}}{m_{CO_{2}}+m_{z}} \cdot 100 \%=\frac{ m_{CO_{2}}\frac{M_{C}}{M_{CO_{2}}}}{m_{CO_{2}}+ m_{z}} \cdot 100 \%=\frac{ \frac{M_{C}}{M_{CO_{2}}}}{1+ \frac{m_{z}}{m_{CO_{2}}} } \cdot 100 \%=25 \%}\)
\(\displaystyle{ \frac{M_{C}}{M_{CO_{2}}}=0.25\cdot \left (1+ \frac{m_{z}}{m_{CO_{2}}} \right )}\)
\(\displaystyle{ \frac{M_{C}}{M_{CO_{2}}}-0.25= \frac{M_{C}-0.25M_{CO_{2}}}{M_{CO_{2}}} =0.25 \frac{m_{z}}{m_{CO_{2}}}}\)
\(\displaystyle{ \frac{4M_{C}-M_{CO_{2}}}{M_{CO_{2}}} =\frac{m_{z}}{m_{CO_{2}}} \to \frac{m_{CO_{2}}}{m_{z}}= \frac{M_{CO_{2}}}{4M_{C}-M_{CO_{2}}}}\)
i wstawiamy to do równania poniżej:
\(\displaystyle{ \%z= \frac{m_{z}}{m_{CO_{2}}+m_{z}} \cdot 100 \%= \frac{1}{1+ \frac{m_{CO_{2}}}{m_{z}} } \cdot 100 \%}\)