Please enable JavaScript.

Coggle requires JavaScript to display documents.

Semester 2 Coggle (Thermal Chemistry (q = mC∆T (q = energy (J), m = mass…

- - - - 1 Calorie = 4.184 Joules
      - Specific heat of water
  - - - Also, during a phase change the latent heat must also be considered
  - - - Amount of energy required to change one mole of a substance from liquid to gas state and vice versa
        
        Q = m∆vap
        
        Ice = 2240 J/g
    - - Amount of energy required to change one mole of a substance from liquid to solid state and vice versa
        
        Q = m∆fus
        
        Ice = 333 J/g
  - - - Can be done through a calorimeter, where heat energy is used to heat up water
        
        To set, use -[mC∆T] = mC∆T
        
        (q = -q)
  - - - The change in amount of heat energy in a process
        
        Exothermic
        
        A process in which energy is expelled
        
        Endothermic
        
        A process in which energy is taken in
    - - Melting - Freezing
      - Vaporization - Condensation
      - Sublimation - Deposition
    - - The measurement of the ability to do work
    - - Temperature at which no amount of pressure can turn gas into liquid
    - - All three phases at once
- - - - Conjugate base
        
        Base from an acid that loses an electron
      - Conjugate acid
        
        Acid from a base that gains an electron
      - Amphoretic
        
        Capable of both donating and receiving electrons, such as H2O
  - - - Juices
      - Sour taste
      - Reacts with metals
      - aqueous or liquids
    - - Slippery
      - Cleaning suppies
      - Solids
      - Conducts electricity
  - - - H and OH to pH and pOH
        
        -log(H+) or -log(OH-)
      - pH to pOH and vice versa
        
        14 = pH + pOH
      - pH and pOH to H and OH
        
        10^-(pH) or 10^-(pOH)
      - H to OH and vice versa
        
        1.0*10^-14 = [H][OH]
      - H and OH to Ka and Kb
        
        ICE table
      - Ka, pKa, Kb, pKb calculations
        
        These calculations are equivalent to the other calculations
  - - - Each 1-14, 1 meaning 1.0*10^-1 mols of H+ ions present
      - pOH measures the amount of OH- in solution
  - - - Are product favored
    - - Are reactant favored
  - - - Since it is conjugate, the solution becomes the other of it
        
        Example
        
        Na4NO2
        
        Na+
        
        NaOH is strong
        
        NO2-
        
        HNO2
        
        C base of acid is base
        
        Solution is basic
    - - The bigger K value wins
- - - - Dissolved substance in solution
    - - Dissolver
  - - - Uniform
    - - Not uniform
  - - - Shows the ratio between solute and overall
      - solute/solvent *100
  - - - Still more can be dissolved
    - - More than the max, by heating before mixing
    - - Maximum solute dissolved by amount of solvent
- - - - R = k[A]^x[B]^y
        
        Only gases and aqeous
        
        Concentration of liquids and solids are constant
        
        If single step:
        
        Use the stoich ratio as exponent
        
        Most of the time, needs data to solve
        
        To solve, find trials in which one reactant molarity changes and the rate change as well
        
        Find the change amount of both and set up equation, ∆reactant^x = ∆rate. x = ?
        
        Once every reactant is done, choose a trial and solve for k.
        
        If not possible to find a trial in which only one product changes:
        
        Set up, Rate/Rate = k[a]^x[b]^y/k[a]^x[b]^y
        
        solve for x
    - - The order is the exponent, so the second order would be squared
      - Overall order is the added order
  - - - Divided by the coefficent
        
        Is negative because it is being lost
  - - - More particles present to collide
    - - Provides alternate energy pathways
    - - More exposure
    - - Higher temperature will result in more collsions
- - - - In order to use up
    - - In order to use up the extra reactants present
  - - - To solve:
        
        ICE
        
        Change
        
        -X for reactants, +X for products.
        
        Use the stoich ratios for x
        
        Equilbrium
        
        Solve by plugging in, use quadratic formula to solve
        
        Initial
        
        The original amount reactants present
        
        5 percent rule
        
        If the equilibrium constant divided by the molarity present impacts the results by less than five percent, then added and subtracted variables of x may be removed
      - Equilibrium constant
        
        Can be compared with the equilibrium quotient in order to figure out how the reaction will change
        
        Q<K
        
        The reaction will produce more products till equibrium
        
        Q=K
        
        At equibrium
        
        Q>K
        
        The reaction will produce more reactants till equibrium
        
        Only effected by temperature
        
        The constant that is unitless
        
        If k>1, then reaction will produce more products
        
        If k<1, then reaction will produce more reactants
        
        Catalysts do not effect the direction, just the rate by letting a separate pathway
    - - A and B are products
      - C and D are reactants
      - Raised to the power of the stoich ratio
    - - Concentrations are constant
- - - - r = 101.3 atm/mol * K
    - - r = 62.4 L mmHg/mol * K
    - - r = 8.314 KPa atm/mol * K
    - - Add 273 to Celsius amount
  - - - Gases are constantly in motion, therefore possess energy of motion
      - No force of attraction between gases
        
        Actually gases do experience inter molecular attraction
      - Gases are elastic: Don't lose energy after bouncing off wall
        
        Actually gases lose energy
      - Average kinetic energy depends on temperature not element
      - Gases consists of far apart tiny molecules
        
        Actually gases have volume
      - Pressure is just measurement of how often molecules bounce off surface
    - - High temperature
      - large polar molecules
      - High pressure
    - - Low temperature
      - small non-polar molecules
      - Low pressure
    - - Charle's Law
        
        V1/T1 = V2/T2
        
        Direct
      - Gay Lussac's Law
        
        P1/T1 = P2/T2
        
        Direct
      - Boyle's Law
        
        P1V1 = P2V2
        
        Inverse relationship
        
        Combined Gas Law
        
        P1V1/T1 = P2V2/T2
        
        Ideal Gas Law
        
        1 more item...
      - Avagadro's Law
        
        V1/n1 = V2/n2
        
        Direct
      - Dalton's law of Partial Pressure
        
        Ptotal = P1 + P2 +P3...
      - D = m/V
        
        So, D = MP/RT
- - - - Ionic Formulas are ALWAYS in empirical form
        
        How to find empirical formula:
        
        Divide by small
        
        Multiply till whole
        
        Mass to mole
        
        Percent to mass
    - - Combustion reactants will always include O2, Co2, and H2O. There may also be Nitrogen present as well
        
        To solve for unknown substance, the mass of it must be known
        
        To solve Combustion Analysis:
        
        Then, subtract the total amount of none Oxygen mass from Total mass to get mass of Oxygen
        
        Finally, use the method for finding empirical formula
        
        First isolate the amount of none Oxygen elements
        
        Since the masses for H2O and Co2 remains constant, use:
        
        H2O: x * 2.02/18.02
        
        Co2: x * 12.01/44.01
        
        Note: Nitrogen usually does not have any Oxygen, so no ration should be utilized
        
        Burned carbon ends up as Co2 and burned Hydrogen ends up as H2O
  - - - How to find molecular formula:
        
        Divide given mass by mass of empirical formula
  - - - Divide mass of element by total mass