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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