Magnus Ehingers undervisning

Allt du behöver för A i Biologi, Kemi, Bioteknik, Gymnasiearbete m.m.

Kemi 1

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Prov 2012-02-14 i Kemiska beräkningar - Facit

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Facit

Betygsgränser

Max: 18,0
Medel: 12,94
E: 6,0
D: 9,5
C: 12,0
B: 15,0
A: 17,5

Fullständig lösning krävs på alla frågor, om inte annat anges

  1. f
  2. c
    1. m = (39,1 + 35,3)u = 74,6u
    2. \(n = \frac {m}{M} = \frac {12,5\text{g}}{74,6\text{g/mol}} = 0,167560322mol \approx 0,168\text{mol}\)
    3. \(N = n \cdot N_{\text{A}} = 2 \cdot 0,167560322\text{mol} \cdot 6,022 \cdot 10^{23}/\text{mol} =\)
      \(= 2,018096515 \cdot 10^{23} \approx 2,02 \cdot 10^{23}\)
    1. \(c = \frac {m_{\text{C}_{12}\text{H}_{22}\text{O}_{11}}}{m_{\text{tot}}} = \frac {25\text{g}}{(25+60)\text{g}} = 0,294118 \approx 29 \%\)
    2. \(n_{\text{C}_{12}\text{H}_{22}\text{O}_{11}} = \frac {m_{\text{C}_{12}\text{H}_{22}\text{O}_{11}}}{M_{\text{C}_{12}\text{H}_{22}\text{O}_{11}}} = \frac {25\text{g}}{(12,0 \cdot 2 + 1,008 \cdot 22 + 16,0 \cdot 11)\text{g/mol}} =\)
      \(=0,07306182\text{mol}\)

      \(c_{\text{C}_{12}\text{H}_{22}\text{O}_{11}} = \frac {n_{\text{C}_{12}\text{H}_{22}\text{O}_{11}}}{V} = \frac {0,07306182\text{mol}}{0,060\text{dm}^3} =\)
      \(= 1,21769694\text{mol/dm}^3 \approx 1,2\text{M}\)
  3. \(n_{\text{NiSO}_4} = \frac {m_{\text{NiSO}_4}}{M_{\text{NiSO}_4}} = \frac {0,590\text{g}}{(58,7+32,1+16,0\cdot 4)\text{g/mol}} =\)
    \(= 0,00381137\text{mol}\)

    \(n_{\text{H}_2\text{O}} = \frac {m_{\text{H}_2\text{O}}}{M_{\text{H}_2\text{O}}} = \frac {(1,00-0,590)\text{g}}{(1,008 \cdot 2 + 16,0)\text{g/mol}} = 0,02275755\text{mol}\)

    \(\frac {n_{\text{H}_2\text{O}}}{n_{\text{NiSO}_4}} = \frac {0,02275755\text{mol}}{0,00381137\text{mol}} = 5,970963883 \approx 6\)

    Alltså måste den rätta formeln vara NiSO4·6H2O
    1. 2C2H3Cl(s) + 5O2(g) → 4CO2(g) + 2H2O(g) + 2HCl(g)
      1. \(m_{\text{C}_2\text{H}_3\text{Cl}} = 0,80\text{kg} = 800 \text{g}\)
      2. \(n_{\text{C}_2\text{H}_3\text{Cl}} = \frac {m_{\text{C}_2\text{H}_3\text{Cl}}}{M_{\text{C}_2\text{H}_3\text{Cl}}} = \frac {800 \text{g}}{(12,0 \cdot 2 + 1,008 \cdot 3 + 35,5)\text{g/mol}} =\)
        \(= 12,7950867\text{mol}\)
      3. \(n_{\text{CO}_2} = 2n_{\text{C}_2\text{H}_3\text{Cl}} = 2 \cdot 12,8950867\text{mol} = 25,5901734\text{mol}\)
      4. \(m_{\text{CO}_2} = n_{\text{CO}_2} \cdot M_{\text{CO}_2} =\)
        \(= 25,5901734\text{mol} \cdot (12,0+16,0 \cdot 2)\text{g/mol} =\)
        \(= 1125,9673\text{g} \approx 1,1\text{kg}\)
  4. Vi kan anta att vi har 1000 g vatten. Då kan vi beräkna substansmängden vatten:
    \[n_{\text{H}_2\text{O}} = \frac {m_{\text{H}_2\text{O}}}{M_{\text{H}_2\text{O}}} = \frac {1000\text{g}}{(1,008 \cdot 2 + 16,0)\text{g/mol}} = 55,5062167\text{mol}\]
    Vattnets densitet är 1,0 g/dm3. Då motsvaras 1000 g vatten av 1,0 dm3. Då kan vi beräkna koncentrationen:
    \[c_{\text{H}_2\text{O}} = \frac {n_{\text{H}_2\text{O}}}{V} = \frac {55,5062167\text{mol}}{1,0\text{dm}^3}=55,5062167\text{mol/dm}^3 \approx 56\text{M}\]
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