Chemistry Unit 3 Equations

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Chemistry Unit 3 Equations
- Specific Heat Capacity
  - How much energy is needed to increase the temperature of 500 g of lead from 20ºC to 45ºC? The specific heat capacity of lead is 128 J/kg ºC.
    - mass of lead = 500/1000 = 0.5kg temperature change = 45-20 = 25c energy needed = 0.5 x 128 x 25 = 1600j or 1.6kj
  - E =m x c x t
    - m = mass
    - c = specific heat capacity
    - t = temperature change
- Bond Energy Calculations
  - You will be given bond energies for example h-h = 436kj per mole.
    - Energychange = sum of bonds broken - sum of bonds made
- Titration Equation
  - M1 x V1 divided by N1 = M2 x V2 divided by N2
    - E.G. 25.00cm of an unknown strength HCL reacted with 38.90cm of 0.100 mole per cm solution of Na OH calculate the concentration of HCL?
      - HCL + NaOH = NaCL +H2O
        So where m1 is unknown v1 = 25.00cm n1 = 1 m2 = 0.100 moles per cm v2 =38.9cm and n2 = 1
        M1 = M2 x V2 x N1 divided by N2 x V1
        M1 = 0.100 x 38.90 x 1 divided by 1 x 25.00
How much energy is needed to increase the temperature of 500 g of lead from 20ºC to 45ºC? The specific heat capacity of lead is 128 J/kg ºC.
- mass of lead = 500/1000 = 0.5kg temperature change = 45-20 = 25c energy needed = 0.5 x 128 x 25 = 1600j or 1.6kj

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