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Not Pure Substance= Mixture

T C = (T F – 32)  5. 9. T F = (T C  9 ) + 32. 5. Not Pure Substance= Mixture. Slide 1. Alpha,  = particles with +2 charge Beta,  = high speed e - Gamma,  = high energy light (no charge). Dr. Ali Bumajdad. m. n =. M.m. Law of constant composition =1.

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Not Pure Substance= Mixture

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  1. TC = (TF – 32)  5 9 TF = (TC  9 ) + 32 5 Not Pure Substance= Mixture Slide 1 Alpha,  = particles with +2 charge Beta,  = high speed e- Gamma,  = high energy light (no charge) Dr. Ali Bumajdad

  2. m n = M.m. Law of constant composition =1 Law of Multiple Proportion =2 Isotopes have same number of protons Empirical Formula = I need no. of moles • Find the moles elements • (for %mass assume you have 100 g) • 2) Divide by the smallest mole value 2 masses = may be limiting reactant Slide 1 Fe3O4 M.w. = 231.55 g mol Al M.w. = 27 g/mol Fe M.w. = 55.85 g/mol Dr. Ali Bumajdad

  3. M i V i = M f V f TC = TK – 273.15 TF = (TC  9 ) + 32 5 Dilution = one substance + water Oxidation: Loss of electrons 113.7634725 Slide 1 Law of Multiple Proportion = 2 compound from the same elements Dr. Ali Bumajdad

  4. A X Element Symbol Z Mass Number Atomic Number Alpha,  = particles with +2 charge Beta,  = high speed e- Gamma,  = high energy light (no charge) Slide 1 Slide 1 Dr. Ali Bumajdad

  5. N n = NA M.m.M.F. M.F. = E.F. × M.m.E.F. m n = M.m. No. of Objects = I need n and NA Molecular Formula = I need no. of moles and Molar mass • Find the moles elements • (for %mass assume you have 100 g) • 2) Divide by the smallest mole value C 3.45 H 3.44 O 3.44 3) Spectator ions disappear in the net ionic equation Dr. Ali Bumajdad

  6. nH M aicd V acid = nOHM base V base Slide 1 2 masses = may be limiting reactant Slide 1 O2 M.w. = 32.0 g mol Al M.w. = 27.0 g/mol Al2O3 M.w. = 55.85 g/mol Slide 1 Dr. Ali Bumajdad

  7. 2.016 8 = 32 X Isotopes have same number of protons but different number of neutrons Law of constant composition =1 2 masses = may be limiting reactant C2H4 M.w. = 26.036 g/mol O2 M.w. = 32.0 g/mol CO2 M.w. = 44.0 g/mol Slide 1 Start by balancing those elements that appear in only one reactant and one product. Dr. Ali Bumajdad

  8. N n = m of 1 mol in g NA m of 1 atom in g= NA M.m.M.F. M.F. = E.F. × M.m.E.F. n M = VLsolution Slide 1 NPCl2 M.w. = 115.87 g/mol 2 moles 3 moles 4 moles No 5 moles C12H22O11 M.w. = 342 g/mol Dr. Ali Bumajdad

  9. Spectator ions disappear in the net ionic equation Slide 1 Slide 1 Law of Multiple Proportion = 2 compound from the same elements Dr. Ali Bumajdad

  10. Slide 1 Slide 1 Slide 1 Slide 1 Slide 1 Dr. Ali Bumajdad

  11. M.m.M.F. M.F. = E.F. × M.m.E.F. Slide 1 Slide 1 Molecular Formula = I need no. of moles and Molar mass • Find the moles elements • (for %mass assume you have 100 g) • 2) Divide by the smallest mole value 3) Dr. Ali Bumajdad

  12. N n = NA m n = CH2F2 M.w. = 52.0 g /mol M.m. No. of Objects = I need n and NA • Find the moles elements • (for %mass assume you have 100 g) • 2) Divide by the smallest mole value NO2 M.w. = 52.0 g /mol H2O M.w. = 18.0 g/mol Slide 1 Dr. Ali Bumajdad

  13. nH M aicd V acid = nOHM base V base n M = VLsolution Dr. Ali Bumajdad

  14. Kuc02.kuniv.edu.kw/~bumajdad/

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