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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 of
potassium?
2. How many formula units are present in 5.32 mol of baking soda
(NaHCO3 )?
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Practice Problem # 1
What is the amount (mol) of K atoms present in 19.5 g of potassium?
Collect and Organize
m of K = 19.5 g
M of K = 39.0983 g·mol−1
Analyze
To calculate the amount in a certain mass, divide by molar mass.
n(mol) =
Solve
n(mol) =
m(g)
M(g · mol −1 )
19.5 g
= 0.499 mol
39.0983 g · mol −1
Think about it
Sure, about half a mole.
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Practice Problem # 1
What is the amount (mol) of K atoms present in 19.5 g of potassium?
Collect and Organize
m of K = 19.5 g
M of K = 39.0983 g·mol−1
Analyze
To calculate the amount in a certain mass, divide by molar mass.
n(mol) =
Solve
n(mol) =
m(g)
M(g · mol −1 )
19.5 g
= 0.499 mol
39.0983 g · mol −1
Think about it
Sure, about half a mole.
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Practice Problem # 1
What is the amount (mol) of K atoms present in 19.5 g of potassium?
Collect and Organize
m of K = 19.5 g
M of K = 39.0983 g·mol−1
Analyze
To calculate the amount in a certain mass, divide by molar mass.
n(mol) =
Solve
n(mol) =
m(g)
M(g · mol −1 )
19.5 g
= 0.499 mol
39.0983 g · mol −1
Think about it
Sure, about half a mole.
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Practice Problem # 1
What is the amount (mol) of K atoms present in 19.5 g of potassium?
Collect and Organize
m of K = 19.5 g
M of K = 39.0983 g·mol−1
Analyze
To calculate the amount in a certain mass, divide by molar mass.
n(mol) =
Solve
n(mol) =
m(g)
M(g · mol −1 )
19.5 g
= 0.499 mol
39.0983 g · mol −1
Think about it
Sure, about half a mole.
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Practice Problem # 1
How many formula units are present in 5.32 mol of baking soda
(NaHCO3 )?
Collect and Organize
n of NaHCO3 = 5.32 mol
NA = 6.0221×1023 mol−1
Analyze
To calculate the number of entities in a certain amount, multiply by
Avogadro’s number.
N = n(mol) × NA
Solve
N = 5.32 mol × 6.0221 × 1023 mol −1 = 3.20 × 1024 formula units
Think about it
Yes, that’s 5 times Avogadro’s number.
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Practice Problem # 1
How many formula units are present in 5.32 mol of baking soda
(NaHCO3 )?
Collect and Organize
n of NaHCO3 = 5.32 mol
NA = 6.0221×1023 mol−1
Analyze
To calculate the number of entities in a certain amount, multiply by
Avogadro’s number.
N = n(mol) × NA
Solve
N = 5.32 mol × 6.0221 × 1023 mol −1 = 3.20 × 1024 formula units
Think about it
Yes, that’s 5 times Avogadro’s number.
9 / 31
Practice Problem # 1
How many formula units are present in 5.32 mol of baking soda
(NaHCO3 )?
Collect and Organize
n of NaHCO3 = 5.32 mol
NA = 6.0221×1023 mol−1
Analyze
To calculate the number of entities in a certain amount, multiply by
Avogadro’s number.
N = n(mol) × NA
Solve
N = 5.32 mol × 6.0221 × 1023 mol −1 = 3.20 × 1024 formula units
Think about it
Yes, that’s 5 times Avogadro’s number.
10 / 31
Practice Problem # 1
How many formula units are present in 5.32 mol of baking soda
(NaHCO3 )?
Collect and Organize
n of NaHCO3 = 5.32 mol
NA = 6.0221×1023 mol−1
Analyze
To calculate the number of entities in a certain amount, multiply by
Avogadro’s number.
N = n(mol) × NA
Solve
N = 5.32 mol × 6.0221 × 1023 mol −1 = 3.20 × 1024 formula units
Think about it
Yes, that’s 5 times Avogadro’s number.
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Practice Problem # 2
What is the amount (mol) of calcium carbonate CaCO3 present in
58.4 g of chalk (CaCO3 )?
0.583 mol
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Practice Problem # 2
What is the amount (mol) of calcium carbonate CaCO3 present in
58.4 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 H2 O.
M of carnotite = 902.176 g·mol−1
nU = 100.0 g carnotite ×
1 mol carnotite
2 mol U
×
902.176 g
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 H2 O.
M of carnotite = 902.176 g·mol−1
nU = 100.0 g carnotite ×
1 mol carnotite
2 mol U
×
902.176 g
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 H2 O.
M of carnotite = 902.176 g·mol−1
nU = 100.0 g carnotite ×
1 mol carnotite
2 mol U
×
902.176 g
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 H2 O.
M of carnotite = 902.176 g·mol−1
nU = 100.0 g carnotite ×
1 mol carnotite
2 mol U
×
902.176 g
1 mol carnotite
0.2217 mol U
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Big Bang Consequences
• H and He atoms in stars fuse
to form heavier elements.
• Subatomic particles fuse to
form H and He nuclei.
• Existence of subatomic
particles.
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Primordial Nucleosynthesis
1p
1
+ 10n −−→ 21D
2 21D −−→ 42He
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Primordial Nucleosynthesis
1p
1
+ 10n −−→ 21D
2 21D −−→ 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 c 2 )
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 is
1.1324×10−12 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 c 2 )
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 is
1.1324×10−12 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 c 2 )
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 is
1.1324×10−12 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 c 2 )
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 is
1.1324×10−12 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 c 2 )
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 is
1.1324×10−12 J
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Formation of heavier elements
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Stellar Nucleosynthesis
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Beyond iron-56
Neutron capture
56Fe
26
+ 3 10n −−→
59Fe
26
β Decay
59Fe
26
−−→
59Co
27
+ -10β
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Beyond iron-56
Neutron capture
56Fe
26
+ 3 10n −−→
59Fe
26
β Decay
59Fe
26
−−→
59Co
27
+ -10β
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Beyond iron-56
Neutron capture
56Fe
26
+ 3 10n −−→
59Fe
26
β Decay
59Fe
26
−−→
59Co
27
+ -10β
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Beyond iron-56
Neutron capture
56Fe
26
+ 3 10n −−→
59Fe
26
β Decay
59Fe
26
−−→
59Co
27
+ -10β
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