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Additional Practice problem Solutions Additional Practice problem Solutions

Additional Practice problem Solutions - PowerPoint Presentation

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Uploaded On 2024-01-03

Additional Practice problem Solutions - PPT Presentation

From chemistry class 330 Additional Practice Problems How many moles are equal to 903 x 10 24 atoms of mercury How many atoms are in 100 mole of sucrose C 12 H 22 O 11 How many atoms of O are in 365 moles of C ID: 1039241

practice mol mole atoms mol practice atoms mole c12h22o11 250 additional answers mass problems1 sucrose 1023 grams compounds sodium

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1. Additional Practice problem SolutionsFrom chemistry class 3/30

2. Additional Practice Problems How many moles are equal to 9.03 x 1024 atoms of mercury?How many atoms are in 1.00 mole of sucrose, C12H22O11?How many atoms of O are in 3.65 moles of C12H22O11?2

3. Answers to Practice Problems1. 9.03 x 1024 atoms x 1 mole 1 6.02 x 1023 atoms = 15.0 moles2. 1.0 mole x 6.02 x 1023 mlcs x 45 atoms 1 1 mole 1 mlc = 2.71 x 1025 atoms3. 3.65 mol x 6.02 x 1023 mlcs x 11 atoms of O 1 1 mole 1 mlc C12H22O11 = 2.42 x 1025 atoms O3

4. Additional Practice Problems1. What is the mass, in grams, of 1.72 mol CaCl2?2. Determine the mass of one mole of each of the following compounds and give their names. (Fill in the chart)4

5. Answers to Practice Problems1. 1.72 mol x 110.98 g 1 1 mol = 190.89 g5

6. Answers to Practice ProblemsCompoundMass in One Mole Name a. CO2 44.01 g carbon dioxide b. SO3 80.07 g sulfur trioxidec. NaOH 40.01 g sodium hydroxide d. Al2(SO4)3 342.17 g aluminum sulfate e. Ba(NO3)2 261.35 g barium nitrate 6

7. Additional Practice ProblemsCalculate the mass in grams for 0.250 mol of each of the following compounds:sucrose (C12H22O11)sodium chloridepotassium permanganate7

8. Answers to Practice ProblemsCalculate the mass in grams for 0.250 mol of each of the following compounds:sucrose (C12H22O11) 0.250 mol x 342 g = 85.50 g C12H22O11 1 1 mol sodium chloride 0.250 mol x 58.5 g = 14.63 g NaCl 1 1 mol potassium permanganate 0.250 mol x 158 g = 39.50 g KMnO4 1 1 mol 8