Proszę o rozwiązanie zadania !! MAXYMALNIE BRAK 1 odpowiedzi !! 2- NADUZYCIE !! P.S TO NIE JEST KLAS.... Question from @XXDIZZYXx

December 2018 1 26 Report

Proszę o rozwiązanie zadania !! MAXYMALNIE BRAK 1 odpowiedzi !! 2- NADUZYCIE !!
P.S TO NIE JEST KLASOWKA ! (CZARNE PUNKTY - MOJE DANE)

robek150 ----------------------------------ZADANIE.1----------------------------------
$M_{NH_3}}=14 \frac{g}{mol}+3*1 \frac{g}{mol}=17 \frac{g}{mol}$
$m_{NH_3}}=n*M= \frac{1}{6,02*10^{23}}mol*17 \frac{g}{mol}=2,82392*10^{-23}g$

----------------------------------ZADANIE.2----------------------------------
$m_{HF}=2,35g$
$M_{HF}=1 \frac{g}{mol}+19 \frac{g}{mol}=20 \frac{g}{mol}$
$n_{HF}}= \frac{m}{M}= \frac{2,35g}{20 \frac{g}{mol} }=0,1175 mol$
$n_{HF_{at}}=2*n_{HF}}=2*0,1175mol=0,235mol$

$M_{CH_4}}=12 \frac{g}{mol}+4* 1\frac{g}{mol}=16 \frac{g}{mol}$
$n_{CH_4_{at}}=0,235mol$
$n_{CH_4}= \frac{n_{CH_4_{at}}}{5}= \frac{0,235mol}{5} =0,047mol$
$m_{CH_{4}}=nM=0,047mol*16 \frac{g}{mol}= 0,752g$

----------------------------------ZADANIE.3----------------------------------
$M_{N_XO_Y}=92,02 \frac{g}{mol}$
$\%N_2=30,45\%$
$\%O_2=69,55\%$

$\left \{ {{\%N= \frac{m_{N}}{m_{N}+m_{O}}} \atop {\%O= \frac{m_{O}}{m_{N}+m_{O}}}} \right.$
$\left \{ {{\%N= \frac{n_{N}M_{N}}{m_{N}+m_{O}}} \atop {\%O= \frac{n_{O}M_{O}}{m_{N}+m_{O}}}} \right.$
$\left \{ {{ \frac{\%N}{n_{N}M_{N}} = \frac{1}{m_{N}+m_{O}}} \atop {\frac{\%O}{n_{O}M_{O}} = \frac{1}{m_{N}+m_{O}}}} \right.$
$\frac{\%N}{n_{N}M_{N}}= \frac{\%O}{n_{O}M_{O}}$
$n_{O}= \frac{\%OM_{N}}{\%_{N} M_{O}} *n_{N}$
$n_{O}= \frac{(0,6955)(14 \frac{g}{mol} )}{(0,3045)( 16\frac{g}{mol} )} *n_{N}$
$n_{O}\approx2n_{N}$

$n_O*M_O+n_N*M_N=M_{N_XO_Y}$
$\left \{ {{2n_N*M_O+n_N*M_N=M_{N_XO_Y}} \atop {n_O*M_O+ \frac{1}{2} n_O*M_N=M_{N_XO_Y}}} \right.$
$\left \{ {{n_N= \frac{M_{N_XO_Y}}{2M_O+M_N}} \atop {n_O= \frac{M_{N_XO_Y}}{M_O+ \frac{1}{2} M_N}}} \right.$
$\left \{ {{n_N= \frac{92\frac{g}{mol} }{2*16 \frac{g}{mol} +14 \frac{g}{mol} }} \atop {n_O= \frac{92 \frac{g}{mol} }{16 \frac{g}{mol} + \frac{1}{2}* 14 \frac{g}{mol} }}} \right.$
$\left \{ {{n_N=2} \atop {n_O=4}} \right.$

$N_2O_4$

----------------------------------ZADANIE.4----------------------------------
$7,5\% \ \ \ \ \ \ \ \ 7\ (=22-15)$
$\\ . \ \ \ \ \ \ 15\% \\ 22\%\ \ \ \ \ \ \ \ 7,5\ (=15-7,5)$
$\frac{7}{7,5}= \frac{x}{10kg}$
$x= \frac{7}{7,5} *10kg= \frac{28}{3}kg \approx9,33kg$

----------------------------------ZADANIE.5----------------------------------
$m_{H_{2}O}=150g$
$V_{roztw_{1}}=100cm^3=0,1dm^3$
$C_n_{H_{2}SO_{3}}=2,59 \frac{mol}{dm^3}$
$\rho_{roztw_{1}}=1,155 \frac{g}{cm^3}=1155 \frac{g}{dm^3}$
$M_{H_{2}SO_{3}}=2 \frac{g}{mol}+32 \frac{g}{mol}+3*16 \frac{g}{mol}=82\frac{g}{mol}$

$m_{{roztw}_{1}}=\rho__{roztw_{1}}*V_{roztw_{1}}=1155 \frac{g}{dm^3}*0,1dm^3=115,5g$
$m_{H_{2}SO_{4}}=n_{H_{2}SO_{3}}*M_{H_{2}SO_{3}}=C_n_{H_{2}SO_{3}}*V_{roztw_{1}}*M_{H_{2}SO_{3}}$
$m_{H_{2}SO_{3}}=2,59 \frac{mol}{dm^3}*0,1dm^3*82 \frac{g}{mol}=21,238g$

$Cp= \frac{m_{H_2SO_4}+{m_{{roztw}_{1}}+m_{{H_2O}__2}}}$
$Cp= \frac{21,238g}{115,5g+150g}=7,999\%$

0 votes Thanks 1