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Overview

Practice Problems

Making Elements

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Practice Problem # 1

1. What is the amount (mol) of K atoms present in 19.5 g ofpotassium?

2. How many formula units are present in 5.32 mol of baking soda(NaHCO3)?

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Practice Problem # 1What is the amount (mol) of K atoms present in 19.5 g of potassium?Collect and Organizem of K = 19.5 gM of K = 39.0983 g·mol−1AnalyzeTo calculate the amount in a certain mass, divide by molar mass.

n(mol) =m(g)

M(g ·mol−1)

Solven(mol) =

19.5 g39.0983 g ·mol−1

= 0.499 mol

Think about itSure, about half a mole.

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Practice Problem # 1What is the amount (mol) of K atoms present in 19.5 g of potassium?Collect and Organizem of K = 19.5 gM of K = 39.0983 g·mol−1AnalyzeTo calculate the amount in a certain mass, divide by molar mass.

n(mol) =m(g)

M(g ·mol−1)

Solven(mol) =

19.5 g39.0983 g ·mol−1

= 0.499 mol

Think about itSure, about half a mole.

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Practice Problem # 1What is the amount (mol) of K atoms present in 19.5 g of potassium?Collect and Organizem of K = 19.5 gM of K = 39.0983 g·mol−1AnalyzeTo calculate the amount in a certain mass, divide by molar mass.

n(mol) =m(g)

M(g ·mol−1)

Solven(mol) =

19.5 g39.0983 g ·mol−1

= 0.499 mol

Think about itSure, about half a mole.

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Practice Problem # 1What is the amount (mol) of K atoms present in 19.5 g of potassium?Collect and Organizem of K = 19.5 gM of K = 39.0983 g·mol−1AnalyzeTo calculate the amount in a certain mass, divide by molar mass.

n(mol) =m(g)

M(g ·mol−1)

Solven(mol) =

19.5 g39.0983 g ·mol−1

= 0.499 mol

Think about itSure, about half a mole.

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Practice Problem # 1How many formula units are present in 5.32 mol of baking soda(NaHCO3)?Collect and Organizen of NaHCO3 = 5.32 molNA = 6.0221×1023 mol−1AnalyzeTo calculate the number of entities in a certain amount, multiply byAvogadro’s number.

N = n(mol)× NA

Solve

N = 5.32 mol × 6.0221× 1023 mol−1 = 3.20× 1024 formula units

Think about itYes, that’s 5 times Avogadro’s number.

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Practice Problem # 1How many formula units are present in 5.32 mol of baking soda(NaHCO3)?Collect and Organizen of NaHCO3 = 5.32 molNA = 6.0221×1023 mol−1AnalyzeTo calculate the number of entities in a certain amount, multiply byAvogadro’s number.

N = n(mol)× NA

Solve

N = 5.32 mol × 6.0221× 1023 mol−1 = 3.20× 1024 formula units

Think about itYes, that’s 5 times Avogadro’s number.

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Practice Problem # 1How many formula units are present in 5.32 mol of baking soda(NaHCO3)?Collect and Organizen of NaHCO3 = 5.32 molNA = 6.0221×1023 mol−1AnalyzeTo calculate the number of entities in a certain amount, multiply byAvogadro’s number.

N = n(mol)× NA

Solve

N = 5.32 mol × 6.0221× 1023 mol−1 = 3.20× 1024 formula units

Think about itYes, that’s 5 times Avogadro’s number.

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Practice Problem # 1How many formula units are present in 5.32 mol of baking soda(NaHCO3)?Collect and Organizen of NaHCO3 = 5.32 molNA = 6.0221×1023 mol−1AnalyzeTo calculate the number of entities in a certain amount, multiply byAvogadro’s number.

N = n(mol)× NA

Solve

N = 5.32 mol × 6.0221× 1023 mol−1 = 3.20× 1024 formula units

Think about itYes, that’s 5 times Avogadro’s number.

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Practice Problem # 2

What is the amount (mol) of calcium carbonate CaCO3 present in58.4 g of chalk (CaCO3)?

0.583 mol

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Practice Problem # 2

What is the amount (mol) of calcium carbonate CaCO3 present in58.4 g of chalk (CaCO3)?

0.583 mol

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Practice Problem # 3

The uranium used in nuclear fuel exists in nature in several minerals.Calculate the amount of uranium (mol) found in 100.0 g of carnotite,K2(UO2)2(VO4)2 · 3 H2O.M of carnotite = 902.176 g·mol−1

nU = 100.0 g carnotite ×1 mol carnotite

902.176 g× 2 mol U

1 mol carnotite

0.2217 mol U

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Practice Problem # 3

The uranium used in nuclear fuel exists in nature in several minerals.Calculate the amount of uranium (mol) found in 100.0 g of carnotite,K2(UO2)2(VO4)2 · 3 H2O.M of carnotite = 902.176 g·mol−1

nU = 100.0 g carnotite ×1 mol carnotite

902.176 g× 2 mol U

1 mol carnotite

0.2217 mol U

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Practice Problem # 3

The uranium used in nuclear fuel exists in nature in several minerals.Calculate the amount of uranium (mol) found in 100.0 g of carnotite,K2(UO2)2(VO4)2 · 3 H2O.M of carnotite = 902.176 g·mol−1

nU = 100.0 g carnotite ×1 mol carnotite

902.176 g× 2 mol U

1 mol carnotite

0.2217 mol U

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Practice Problem # 3

The uranium used in nuclear fuel exists in nature in several minerals.Calculate the amount of uranium (mol) found in 100.0 g of carnotite,K2(UO2)2(VO4)2 · 3 H2O.M of carnotite = 902.176 g·mol−1

nU = 100.0 g carnotite ×1 mol carnotite

902.176 g× 2 mol U

1 mol carnotite

0.2217 mol U

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Big Bang Consequences

• H and He atoms in stars fuseto form heavier elements.

• Subatomic particles fuse toform H and He nuclei.

• Existence of subatomicparticles.

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Primordial Nucleosynthesis

11p +

10n −−→ 21D

221D −−→ 42He

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Primordial Nucleosynthesis

11p +

10n −−→ 21D

221D −−→ 42He

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Mass Defect and Binding Energy

Mass of an α particle = 6.644466×10−27 kg

Total mass of nucleons = 6.69510×10−27 kg

The difference 0.05055×10−27 kg is equivalent in energy (E = m c2)to 4.534×10−12 J

or 1.134×10−12 J per nucleon

Berylium-8 does not form because its binding energy per nucleon is1.1324×10−12 J

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Mass Defect and Binding Energy

Mass of an α particle = 6.644466×10−27 kg

Total mass of nucleons = 6.69510×10−27 kg

The difference 0.05055×10−27 kg is equivalent in energy (E = m c2)to 4.534×10−12 J

or 1.134×10−12 J per nucleon

Berylium-8 does not form because its binding energy per nucleon is1.1324×10−12 J

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Mass Defect and Binding Energy

Mass of an α particle = 6.644466×10−27 kg

Total mass of nucleons = 6.69510×10−27 kg

The difference 0.05055×10−27 kg is equivalent in energy (E = m c2)to 4.534×10−12 J

or 1.134×10−12 J per nucleon

Berylium-8 does not form because its binding energy per nucleon is1.1324×10−12 J

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Mass Defect and Binding Energy

Mass of an α particle = 6.644466×10−27 kg

Total mass of nucleons = 6.69510×10−27 kg

The difference 0.05055×10−27 kg is equivalent in energy (E = m c2)to 4.534×10−12 J

or 1.134×10−12 J per nucleon

Berylium-8 does not form because its binding energy per nucleon is1.1324×10−12 J

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Mass Defect and Binding Energy

Mass of an α particle = 6.644466×10−27 kg

Total mass of nucleons = 6.69510×10−27 kg

The difference 0.05055×10−27 kg is equivalent in energy (E = m c2)to 4.534×10−12 J

or 1.134×10−12 J per nucleon

Berylium-8 does not form because its binding energy per nucleon is1.1324×10−12 J

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Formation of heavier elements

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Stellar Nucleosynthesis

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Beyond iron-56

Neutron capture

5626Fe + 3

10n −−→ 5926Fe

β Decay

5926Fe −−→

5927Co +

0-1β

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Beyond iron-56

Neutron capture

5626Fe + 3

10n −−→ 5926Fe

β Decay

5926Fe −−→

5927Co +

0-1β

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Beyond iron-56

Neutron capture

5626Fe + 3

10n −−→ 5926Fe

β Decay

5926Fe −−→

5927Co +

0-1β

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Beyond iron-56

Neutron capture

5626Fe + 3

10n −−→ 5926Fe

β Decay

5926Fe −−→

5927Co +

0-1β

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Practice ProblemsMaking Elements