[OSK KIMIA 2019]

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masbambang
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Posts: 23
Joined: Thu Oct 04, 2018 7:01 pm
Bidang OSN: Kimia SMA
Tahun OSN: 2018
Deskripsi: mantap gan :D
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Wed Feb 27, 2019 1:09 pm

[DISCLAIMER]
semua jawaban berikut ini bukanlah jawaban resmi dan hanya merupakan interpretasi penulis
  1. PILIHAN GANDA
    1. rumus molekul = $\ce{ CN3H5 }$
      $\ce{ \begin{align*} \%N &= \frac{n_N*A_r N}{M_r} *100\% \\ &= \frac{3*14}{59}*100\% = 71.19\%\end{align*} }$
      (D)
    2. .
      $\ce{ Ag2S ${(s)}$ -> 2Ag+ ${(aq)}$ + S^{2-} ${(aq)}$ }$
      $\ce{ \begin{align*} n_{Ag_2S} &= \frac{m_{Ag_2S}}{M_r Ag2S} \\ n_{Ag} &= 2nAg2S \\ &= \frac{2m_{Ag_2S}}{M_r Ag2S} \\ m_{Ag} &= n_{Ag}*A_r Ag \\ &= \frac{2m_{Ag_2S}*A_r Ag}{M_r Ag2S} \\ \%Ag &= \frac{m_{Ag}}{m_{sampel}}*100\% \\ &= \frac{2m_{Ag_2S}*A_r Ag}{M_r Ag2S*m_{sampel}}*100\% \\ &= \frac{2*0.124 g*108 g mol^{-1}}{248 g mol^{-1}*1.5 g}*100\% \\ &= 7.2 \%\end{align*} }$
      (C)
    3. $\ce{ KAl2(AlSi3O_{10})(OH)2 -> 3Si + ... }$
      $\ce{ \begin{align*} n_{KAl_2(AlSi_3O_{10})(OH)_2} &= \frac{1}{3}n_{Si} \\ m_{KAl_2(AlSi_3O_{10})(OH)_2} &= n_{KAl_2(AlSi_3O_{10})(OH)_2}*M_r KAl2(AlSi3O_{10})(OH)2 \\ &= \frac{1}{3}*\frac{m_{Si}}{A_r Si}*M_r KAl2(AlSi3O_{10})(OH)2 \\ &= \frac{1}{3}*\frac{0.42 g}{28 g mol^{-1}}*398 g mol^{-1} \\ &= 1.99 g\end{align*} }$
      (C)
    4. $\ce{ H2SO4 ${(aq)}$ + 2NaOH ${(aq)}$ -> Na2SO4 ${(aq)}$ + 2H2O ${(l)}$ }$
      $\ce{ \begin{align*} n_{H_2SO_4} &=\frac{1}{2}n_{NaOH} \\ [H2SO4] &= \frac{n_{H_2SO_4}}{V_{H_2SO_4}} \\ &= \frac{[NaOH]*V_{NaOH}}{2V_{H_2SO_4}} \\ &= \frac{1.47 M*23.7 {mL}}{2*1.5 {mL}} \\ &= 11.6 M\end{align*} }$
      (D)
    5. .
      $\ce{ \begin{align*} n_{HCl} &= 4n_{Cl_2} \\ &= \frac{4m_{Cl_2}}{M_r Cl2} \\ m_{HCl} &= n_{HCl}*M_r HCl \\ m_{sampel} &= m_{HCl}*\frac{100\%}{36\%} \\ &= \frac{4m_{Cl_2}*M_r HCl}{M_r Cl2}*\frac{100}{36} \\ &= \frac{4*2.5 g*36.5 g mol^{-1}*100}{71 g mol^{-1}*36} \\ &= 14.3 g\end{align*} }$
      (C)
    6. volume STP
      $\ce{ \begin{align*} n_{H_2} &= 2n_{SnO_2} \\ V_{H_2} &= 22.4 L mol^{-1}*n_{H_2} \\ &= \frac{22.4 L mol^{-1}*2*2 g}{150.7 g mol^{-1}} \\ &= 0.595 L = 595 {mL}\end{align*} }$
      (E)
    7. .
      $\ce{ \begin{align*} PV &= \frac{m}{M_r}*RT \\ M_r &= \frac{mRT}{PV} \\ &= \frac{0.238 g*0.0821 L atm mol^{-1} K^{-1}*(273 + 14) K}{\frac{600 {mmHg}}{760 {mmHg} atm^{-1}}*0.1 L} \\ &= 71 g mol^{-1} -> Cl2\end{align*} }$
      (C)
    8. (B), sudah jelas
    9. (C), sudah jelas
    10. (D), inert (sukar bereaksi) $\ce{ -> }$ tidak mudah terbakar
    11. orde total $\ce{ -> }$ 3
      $\ce{ $\frac{v_2}{v_1}$ = 2^3 = 8 }$(E)
    12. .
      $\ce{ \begin{align*} -\frac{1}{3}\frac{\Delta[O2]}{\Delta{t}} &= \frac{1}{2}\frac{\Delta[O3]}{\Delta{t}} \\ \frac{\Delta[O2]}{\Delta{t}} &= -\frac{3}{2}*2*10^{-7} \\ &= -3*10^{-7}\end{align*} }$
      (D)
    13. $\ce{ [Benz-] = [HBenz] }$
      $\ce{ \begin{align*} [OH-] &= \sqrt{\frac{K_w}{K_a}[Benz-]} \\ K_a &= \frac{K_w*[Benz-]}{[OH-]^2} \\ {pH} &= -log\left(\sqrt{K_a[HBenz]}\right) \\ &= -log\left(\frac{[Benz]}{[OH-]}\sqrt{K_w}\right) \\ &= -log\left(\frac{0.2}{10^{8.65 - 14}}\sqrt{10^{-14}}\right) \\ &= 2.35\end{align*} }$
      (C)
    14. (D), sudah jelas #cmiiw
    15. (E), $\ce{ K_b > K_a }$
    16. .
      $\ce{ \begin{align*} [I-] &= \frac{K_{sp} AgI}{[Ag+]} \\ &= \frac{8.3*10^{-17}}{0.1} \\ &= 8.3*10^{-16} M\end{align*} }$
      (A)
    17. $\ce{ 2Co ${(s)}$ + 3NaOCl ${(aq)}$ + 3H2O ${(l)}$ -> 2Co(OH)3 ${(s)}$ + 3NaCl ${(aq)}$ }$
      (A)
    18. (C), sudah jelas
    19. .
      $\ce{ \begin{align*} E^o_{(Pu^{3+}, \space Pu^{4+}\space{//} Cl_2, \space Cl^-)} &= E^o_{(Cl_2,\space Cl^-)} - E^o_{(Pu^{4+}, \space Pu^{3+})} \\ E^o_{(Pu^{4+}, \space Pu^{3+})} &= E^o_{(Cl_2,\space Cl^-)} - E^o_{(Pu^{3+}, \space Pu^{4+}\space{//} Cl_2, \space Cl^-)} \\ &= 1.36 - 0.35 V = 1.01 V\end{align*} }$
      (B)
    20. .
      $\ce{ \begin{align*} m &= \frac{I*t*A_r}{n e^-*F} \\ A_r &= \frac{m*n e^-*F}{I*t} \\ &= \frac{7.11 g*2*96500 C mol^{-1}}{3 A*2*3600 s} \\ &= 63.5 g mol^{-1} -> Cu\end{align*} }$
      (D)
    21. pH = 11 $\ce{ -> }$ pOH = 3, $\ce{ [OH-] = 10^{-3} M }$
      $\ce{ \begin{align*} K_{sp} &= [Mg^{2+}][OH-]^2 \\ [Mg^{2+}] &= \frac{K_{sp}}{[OH-]^2} \\ &= \frac{8*10^{-12}}{10^{-6}} \\ &= 8*10^{-6} M\end{align*} }$
      (E)
    22. (B), sudah jelas
    23. (D), sudah jelas
    24. (B), elektron terdorong menuju gugus aril klorida dan triklorometil, dan yang paling kuat terkumpul di gugus triklorometil
    25. (E), alkil benzena apa pun yang bereaksi dengan larutan $\ce{ KMnO4 }$ pasti menghasilkan asam benzena alkanoat
    26. (A), sudah jelas
    27. (B)
      IMG_20190227_123943.jpg
      IMG_20190227_123943.jpg (28.57 KiB) Viewed 1377 times
    28. (C)
    29. (E), reaksi adisi
    30. (B), terbentuk 1,2-dibromo-1,2,2-trikloroetana, khiral di atom C no 1
  2. ESAI
    1. Pembahasan esai no. 1 dapat dilihat di sini
    2. Pembahasan esai no. 2 dapat dilihat di sini
    3. Pembahasan esai no. 3 dapat dilihat di sini
    4. Pembahasan esai no. 4 dapat dilihat di sini
    5. Pembahasan esai no. 5 dapat dilihat di sini
Last edited by masbambang on Wed Feb 27, 2019 7:02 pm, edited 14 times in total.
masbambang
Member
Member
Posts: 23
Joined: Thu Oct 04, 2018 7:01 pm
Bidang OSN: Kimia SMA
Tahun OSN: 2018
Deskripsi: mantap gan :D
Has thanked: 4 times
Been thanked: 3 times

Wed Feb 27, 2019 1:09 pm

lelah mengetik mhank
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