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CHEMISTRY 10
21: The Mole and the Avogadro constant
Week 13
Slide 1 - Diapositive
W E L C O M E ! Today is the 4th of November, 2024.
Where will you put yourlearning materials: notebook, coursebook, pen and pencil, and fully-charged laptop?
Have you kept your phone in your locker?
Have you gone to the bathroombefore the class starts?
Have you checked any announcements or missing work on our Google Classroom?
Are you seated in your assigned chair?
timer
2:00
Slide 2 - Diapositive
definewhat a mole is.
calculatethe (a) amount of substance, (b) mass, (c) molar mass, (d) relative atomic mass or relative molecular / formula mass, and (e) number of particles using the value of the Avogadro constant.
Slide 3 - Diapositive
Which is easier to count: A or B ?
Slide 4 - Diapositive
Which is easier to count: A or B ?
Slide 5 - Sondage
Which is easier to count: A or B ?
Slide 6 - Diapositive
Which is easier to count: A or B ?
Slide 7 - Diapositive
Slide 8 - Diapositive
Mole (symbol: mol)
amount of substance that contains as many particles (atoms, molecules, ions formula units, or electrons) as precisely 12 g of carbon-12.
Slide 9 - Diapositive
How many atoms are there in each mass of element?
Slide 10 - Diapositive
Molar Mass (symbol: Mr)
the mass, in grams, of 1 mole of a substance
Slide 11 - Diapositive
How would you calculate the number of particles in a given mole?
Slide 12 - Diapositive
How would you calculate the number of particles in a given mole?
Slide 13 - Diapositive
Sample
2Mg (s) + O2 (g) → 2MgO (s)
Strategy:
2Mg = 2 x (6.02 x 10²³ atoms) = 1.20x10²⁴ atoms
1O2 = 1 x (6.02 x 10²³ molecules) = 6.02 x 10²³ molecules
2MgO = 2 x (6.02 x 10²³ formula units) = 1.20x10²⁴ formula units
2Mg²⁺ = 2 x (6.02 x 10²³ Mg²⁺ cations) = 1.20x10²⁴ Mg²⁺ cations
2O²⁻ = 2 x (6.02 x 10²³ O²⁻ anions) = 1.20x10²⁴ O²⁻ anions
Slide 14 - Diapositive
Guided Practice
Continue on page 10 of your learning module.
Study the sample question.
Do (a) to (e).
If you have any questions, please raise your hand.
timer
10:00
Slide 15 - Diapositive
Pres_Template
CHEMISTRY 10
21: The Mole and the Avogadro constant
Week 13
Slide 16 - Diapositive
Think about these:
1 mole of CO₂ = 6.02 x 10²³ molecules
1 mole of CO₂ = 44.0 g
44.0 g of CO₂ has 6.02 x 10²³ molecules
How many carbon dioxide molecules and moles are in 22.0 g of carbon dioxide?
Slide 17 - Diapositive
Slide 18 - Diapositive
Sample
A flask contains 0.750 mol of carbon dioxide gas, CO₂. What mass of carbon dioxide gas is in this sample?
Strategy:
CO₂ n = 0.750 mol
CO₂ Mᵣ = ? C = 12.0 g ; O₂ = 2 x 16.0 Total Mᵣ = 12.0 + 32.0 = 44.0 g / mol
CO₂ m = ?
CO₂ m = n x Mᵣ
CO₂ m = 0.750 mol x 44.0 g / mol
CO₂ m = 33.0 g
Answer: The mass of 0.75 mol of carbon dioxide is 33.0 g.
Slide 19 - Diapositive
Guided Practice
Continue on pages 13-14 of your learning module.
Study the sample question.
Do (Activity 4) to (Activity 5).
If you have any questions, please raise your hand.
timer
10:00
Slide 20 - Diapositive
Share your thought process.
Slide 21 - Diapositive
Slide 22 - Diapositive
I can define what a mole is.
Yes, I can confidently do it.
Yes, I can but need more practice.
No, I lack confidence so I need more practice or ask help from my teacher.
Slide 23 - Sondage
I can calculate the (a) amount of substance, (b) mass, (c) molar mass, (d) relative atomic mass or relative molecular / formula mass, and (e) number of particles using the value of the Avogadro constant.
Yes, I can confidently do it.
Yes, I can but need more practice.
No, I lack confidence so I need more practice or ask help from my teacher.