Magnus Ehingers undervisning

Facit

Betygsgränser

Fullständig lösning krävs på alla frågor, om inte annat anges

1. 1. Metallbindning
   2. Vätebindning
   3. Jonbindning
   4. Kovalent bindning
2. 1. \(m = (39,1+35,5)\text{u} = 74,6\text{u}\)
   2. \(n = \frac {m}{M} = \frac {12,5\text{g}}{74,6\text{g/mol}} = 0,16756032\text{mol} \approx 0,168\text{mol}\)
3. \(c = \frac {n}{V} = \frac {0,175\text{mol}}{0,250\text{dm}^3} = 0,700\text{M}\)
4. 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.
5. \(c_1V_1 = c_2V_2\)
   \(V_2 = \frac {c_1V_1}{c_2} = \frac {0,150\text{mol/dm}^3 \cdot 0,010\text{dm}^3}{0,055\text{mol/dm}^3} =\)
   \(= 0,0272727\text{dm}^3 \approx 27\text{ml}\)
6. \(n_{\text{NaHCO}_3} = \frac {m_{\text{NaHCO}_3}}{M_{\text{NaHCO}_3}} =\)
   \(= \frac {4,0\text{g}}{(23,0 + 1,008 + 12,0 + 16,0 \cdot 3)\text{g/mol}} = 0,04761451\text{mol}\)

   
   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

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

7. Cl^–(aq) + Ag⁺(aq) → AgCl(s)

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

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

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

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

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

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

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

9. 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:
   \(n_{\text{CO}_2} = 8n_{\text{C}_8\text{H}_18} = 8 \cdot 492,3605271\text{mol} = 3938,884216\text{mol}\)

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

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