Magnus Ehingers undervisning

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

Kemi 1

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Prov 2011-03-01 i Kemiska beräkningar - Facit

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Facit

Betygsgränser

Max: 26,0
Medel:  12,81
G: 8,0
VG: 16,0
MVG: 21,0

Fullständig lösning krävs på alla frågor, om inte annat anges

    1. Metallbindning
    2. Vätebindning
    3. Jonbindning
    4. Kovalent bindning
    1. \(m = (39,1+35,5)\text{u} = 74,6\text{u}\)
    2. \(n = \frac {m}{M} = \frac {12,5\text{g}}{74,6\text{g/mol}} = 0,16756032\text{mol} \approx 0,168\text{mol}\)
  1. \(c = \frac {n}{V} = \frac {0,175\text{mol}}{0,250\text{dm}^3} = 0,700\text{M}\)
  2. Flaska B. I denna är totalkoncentrationen 0,1M, men eftersom det går två \({\sf \text{NO}_3^-}\) på varje Ca2+, blir [\({\sf \text{NO}_3^-}\)] = 0,2M.
  3. \(c_1V_1 = c_2V_2\)
    \(V_2 = \frac {c_1V_1}{c_2} = \frac {0,150\text{mol/dm}^3 \cdot 0,010\text{dm}^3}{0,055\text{mol/dm}^3} =\)
    \(= 0,0272727\text{dm}^3 \approx 27\text{ml}\)
  4. \(n_{\text{NaHCO}_3} = \frac {m_{\text{NaHCO}_3}}{M_{\text{NaHCO}_3}} =\)
    \(= \frac {4,0\text{g}}{(23,0 + 1,008 + 12,0 + 16,0 \cdot 3)\text{g/mol}} = 0,04761451\text{mol}\)


    Reaktionsformeln ger att:

    \(n_{\text{CO}_2\text{(g)}} = \frac {1}{2} n_{\text{NaHCO}_3} = \frac {0,04761451\text{mol}}{2} = 0,02380726\text{mol}\)


    Allmänna gaslagen \(pV = nRT\) ger att

    \(V = \frac {nRT}{p} = \frac {0,02380726\text{mol} \cdot 8,314 \frac{\text{Nm}}{\text{mol} \cdot \text{K}} \cdot (273,15+225)\text{K}}{101300\frac{\text{N}}{\text{m}^2}} =\)
    \(= 0,0009733523\text{m}^3 \approx 0,973\text{dm}^3\)

  5. Cl(aq) + Ag+(aq) → AgCl(s)

    \(n_{\text{AgCl}} = \frac {m_{\text{AgCl}}}{M_{\text{AgCl}}} = \frac {0,253\text{g}}{(107,9 + 35,5)\text{g/mol}} = 0,001764296\text{mol}\)

    \(n_{\text{Cl}^-} = n_{\text{AgCl}} = 0,001764296\text{mol}\)


    \(c_{\text{Cl}^-} = \frac {n_{\text{Cl}^-}}{V} = \frac {0,001764296\text{mol}}{0,250\text{dm}^3} =\)
    \(= 0,00705718\text{mol/dm}^3 \approx 7,06\text{mM}\)

  6. \(m_{\text{Ca i mjölk}} = 0,00123 \cdot 5000\text{kg} = 6,15\text{kg} = 6150\text{g}\)


    \(n_{\text{Ca}} = \frac {m_{\text{Ca}}}{M_{\text{Ca}}} = \frac {6150\text{g}}{40,1\text{g/mol}} = 153,366584\text{mol}\)


    \(n_{\text{CaCO}_3} = n_{\text{Ca}}\)


    \(m_{\text{CaCO}_3} = n_{\text{CaCO}_3} \cdot M_{\text{CaCO}_3} =\)
    \(= 153,366584\text{mol} \cdot (40,1 + 12,0 + 16,0 \cdot 3)\text{g/mol} =\)
    \(=15351,995\text{g} \approx 15,3\text{g}\)

  7. 2C8H18 + 25O2 → 16CO2(g) + 18H2O(g)

    \(n_{\text{C}_8\text{H}_18} = \frac {m_{\text{C}_8\text{H}_18}}{M_{\text{C}_8\text{H}_18}} =\)
    \(\frac {56000\text{g}}{(12 \cdot 8 + 1,008 \cdot 18)\text{g/mol}} = 492,3605271\text{mol}\)


    Reaktionsformeln ger att:
    \(n_{\text{CO}_2} = 8n_{\text{C}_8\text{H}_18} = 8 \cdot 492,3605271\text{mol} = 3938,884216\text{mol}\)


    \(m_{\text{CO}_2} = n_{\text{CO}_2} \cdot M_{\text{CO}_2} =\)
    \(m_{\text{CO}_2} = n_{\text{CO}_2} \cdot M_{\text{CO}_2} = 3938,884216\text{mol} \cdot (12,0+16,0\cdot 2)\text{g/mol} =\)
    \(= 173310,9052\text{g}\)


    \(m_{\text{CO}_2\text{, 12 mån}} = m_{\text{CO}_2} \cdot 12 = 173310,9052\text{g} \cdot 12 =\)
    \(=2079730,86288\text{g} \approx 2,08\text{ton}\)

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