Wednesday 28 September 2011

Dimensional Analysis

  • how many dollars is € 35.00 ?          (1 $ 14)
            € 35 x  $ 14/  € 1 = $ 49
  • what is 100 km/h in miles/hour ?    (1.6 km = 1 mi)         
               100 km/h x 1 mi/1.6km = 62.5 mi/h
  • just like converting between currencies in chemistry it is usually necessary to convert between units
  • this process is called dimensional analysis

STEPS:

  • there are only 4 steps 
  1. Identify what units you want to end up with
  2. Find the conversion factors 
  3. Place units in the appropriate places
  4. Cancel units

EXAMPLES:

  • How many miles are equal to 120 km ?
                 120 km x 1 mi/1.6 km = 75 mi  
  • How many seconds are in 1.4 hours ?
              --> 1.4 h x 60 min/1 h x 60 s /1 min = 5040 s
              --> 1.4 h x 3600 s/1 h = 5040 s
  • What is 50 km/h in m/s ?
                 50 km/h x 1 h/3600 s x 1000 m/1 km = 13.88 m/s = 14 m/s
  • Convert 150 kJ/h into J/s :
                 150 kJ/h x 1 h/3600 s x 1000 J/1 kJ = 41.7 J/s
  • Convert 230 mg/s into g/min :
                 230 mg/s x 60 s/1 min x 1 g/1000 mg = 13.8 g/min

-Eva    


    Sunday 25 September 2011

    Sig Figs and Scientific Notation




    I. Significant Figures
         A. Non-zero numbers are significant
              a. Ex. 1, 2, 3, 4, ect.
         B. Place-keeping zeros are not significant
              a. Ex. 2,000 -> the zeros after the 2 are not significant
         C. Numbers after a decimal are significant
              a. Ex. 0.2310
         D. Zeros (that are not place-keeping zeros), that are found after a number are significant
              a. 0.0006405300
         E. Adding and Subtracting
              a. Round to the lowest significant digit
                   1. Ex. 12.33 + 22.1234 = 34.4534 -> 34.45 FOUR SIG FIGS
         F. Multiplying and Dividing
              a. Round to the lowest significant digit
                   1. Ex. 7.364 x 6.0 = 44.184 -> 44 TWO SIG FIGS
    II. Scientific Notation
         A. Used to express very large or very small numbers
         B. Construction
              a.  a \times 10^b
                   1. a = any real number
                   2. b = exponent
              b. Ex. 2.1 x 10³ = 2100
       


    - Nicole

    Tuesday 20 September 2011

    Classification of chemicals

    Classification of matter
    • understanding matter begin with how we name it
    • we can divide matter into 2 types: homogeneous substances and heterogeneous substances
    • homogeneous: consist of only one visible component (destilled water, oxygen, graphite)
    • heterogeneous: contain more than one visible component (chocolate chip cookie, granite)




    Pure substances
    •  there are 2 types of pure substances
    •  elements: substances that cannot be broken down into simpler substances  by chemical reactions (oxygen, iron, magnesium)
    •  compounds: substances that are made up of 2 or more element and can be changed into elements (or other compounds) by chemical reactions (water, sugar)

    Solution
    •  a solution is a homogeneous mixture of 2 or more substances
    •  solutions usually involve liquids but don´t have to (fog, steel)
    •  the component present in greater amount is the solvent
    •  water is the most common solvent
    •  the component present in smaller amount is the solute

    Mixtures 
    •  many mixtures are easy to identify (chocolate chip cookie), but others are easily confused as pure substances
    •  in heterogeneous mixtures the different parts are clearly visible (granite, sand, fog)
    •  in homogeneous mixtures the different parts are not visible (salt water, air, brass)

    Separating mixtures
    •  there are many methodes to separate mixtures, depending on the type of mixture
    1. by hand 
    2. filtration     --> physical       (heterogeneous
    3. destillation  -->   changes              mixtures only)
    4. crystallization
    5. chromatography

     -Eva

    Monday 19 September 2011

    Physical and Chemical Changes, Endothermic and Exothermic, and How to Balance an Equation!

    I. Matter Can Undergo 3 Types of Changes:
         1. Physical Changes
              A. Change of state: (s) (l) (g) (aq)

              B. No new substances are produced
         2. Chemical Changes
              A. New substances are formed
              B. The new substances also gain new properties
                   Ex. Conductivity, acidity, color
                   Ex. Burning wood, iron rusting, digesting food
         3. Nuclear Changes
              A. Involves the fusion or fission of a nucleus or an atom
                   a. Fusion: the combining of a nucleus or an atom
                   b. Fission: the splitting of a nucleus or an atom

    II. Energies Associated with each Phase Change
         1. Endothermic: to gain energy
              a. Sublimation, melting, and evaporation are endothermic
         2. Exothermic: to produce energy
              A. Freezing, condensation, and sublimation/deposition are exothermic

    III. Balancing Chemical Equations
         1. Matter cannot be destroyed nor created therefore all the matter in the reactants 
         should be equal to the matter in the products
         2. Make an element inventory
              A. List all the elements and how many of them you have on the reactant side and on
              the product side in a T-chart
         3. Write numbers in front of each element until there is the same amount of that element
         on both sides of the equation and on the T-chart
           Ex.
              
         B. Tips
              a. Balance elements in elemental form last
              b. Balance polyatomic ions as a group- only if they do no break apart!
              c. Balance oxygen and hydrogen last


    -Nicole

    Wednesday 14 September 2011

    Examples

    • a solution of aluminium chloride, carbon dioxide, and water can be prepared mixing pure aluminium carbonate with a solution of hydrochloric acid
                       Al2(CO3)3 + 6 HCl -> 2 AlCl3 + 3 CO2 + 3 H2O
    • write a balanced equation for the reaction of aqueous Barium Hydroxide neutralising a solution of Phosphonic Acid if solid barium phosphate and water are produced
                      3 BaOH2 (aq) + 2 H3PO4 (aq) -> Ba3(PO4)2 (s) + 6 HOH (l)

          
                                                                                                                           -Eva




    diatomic molecules
    polyatomic molecules

                                                                                              -Eva

    Balancing & Word Equations

    Phase symbols
    • are subscripts that indicate the phase of the chemical
    • Al(s), H2O(l), H2(g), AgNO3(aq)->dissolved in water

    Chemical equations from word equations
    •  in chemistry 11, a solution means something is dissolved in water,      therefore the phase is aqueous
    •  must include phase symbols and balance the equation
    •  diatomic molecules: H2, N2, O2, F2, Cl2, Br2, I2
    •  polyatomic molecules: P4, S8
    •  a solution of barium phosphate is mixed with aqueous sodium sulphate to yield solid barium sulphate and aqueous sodium phosphate
                    Ba3(PO4)2 (aq) + 3 Na2SO4 (aq) -> 3 BaSO4 (s) + 2 Na3PO4 (aq)






                                                                                                                               -Eva

    Monday 12 September 2011

    Getting Familiar With the Equipment


    -Nicole

    Top Ten Safety Tips for the Lab

    1. Always wait for instructions
    2. No eating/ drinking
    3. Wear safety equipment
    4. Tie long hair back
    5. No open-toed shoes
    6. No running/fooling around
    7. Be aware of your surroundings
    8. Do not directly inhale chemicals, waft instead
    9. Do not experiment and mix chemicals without instructed to do so
    10. If there is an accident, notify the teacher right away

    -Nicole