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Final Exam Review Session 1

Final Exam Review Session 1. General Chemistry Mr. Mata. Metric Conversions. 672 g x ________ mg (1000 mg = 1 g) g 672 g x 1000 mg = 672,000 mg 1 g. Metric Conversions. 4850 cm x ________ m (1 m = 100 cm)

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Final Exam Review Session 1

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  1. Final Exam Review Session 1 General Chemistry Mr. Mata

  2. Metric Conversions 672 g x ________ mg (1000 mg = 1 g) g 672 g x 1000 mg = 672,000 mg 1 g

  3. Metric Conversions 4850 cm x ________ m (1 m = 100 cm) cm 4850 cm x 1 m = 48.5 m 100 cm

  4. Metric Conversions 27.7 kg x ________ g (1000 g = 1 kg) kg 27.7 kg x 1000 g = 27,700 g 1 kg

  5. Metric Conversions 18.6 mL x ________ L (1 L = 1000 mL) mL 18.6 mL x 1 L = 0.0186 L 1000 mL

  6. Density Calculations (Density = mass/volume) What is the density of a 22 gram piece of metal with a volume of 650 cm3? Density = 22 gram = 0.034 gram/cm3 650 cm3

  7. Density Calculations (Density = mass/volume) Find the density of a cube with a length of 7 cm and a mass of 522 grams. (volume = length x width x height) Volume = 7 cm x 7 cm x 7 cm = 343 cm3 Density = 522 grams = 1.52 gram/cm3 343 cm3

  8. Density Calculations (Density = mass/volume) Calculate the density of an object with a volume of 16 mL and a mass of 48 grams. Density = 48 grams = 3.0 gram/mL 16 mL

  9. Grams -> moles How many moles are there in 120 grams NaOH? 120 grams NaOH x 1 mol NaOH = 3.0 mol NaOH 40 g NaOH Molar mass NaOH = 23 + 16 + 1 = 40 grams

  10. Moles -> grams How many grams are there in 18.7 moles NaNO3? 18.7 mol NaNO3 x 85 g NaNO3 = 1589.5 g NaNO3 1 mol NaNO3 Molar mass NaNO3 = 23 + 14 + 48 = 85 grams

  11. Moles -> atoms How many atoms are in 178 moles of Ca(NO3)2? (Hint: 1 mole = 6.02 x 1023 atoms) 178 mol Ca(NO3)2 x 6.02 x 1023 atoms Ca(NO3)2 = 1 mol Ca(NO3)2 = 1.1 x 10 26 atoms Ca(NO3)2

  12. Electron Configuration F 9 F 9 = 1s22s22p5

  13. Electron Configuration Cl 17 Cl 17 = 1s22s22p63s23p5

  14. Electron Configuration Fe 26 Fe 26 = 1s22s22p63s23p64s23d6

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