The amount of uranium dissolved in the ocean.

Question

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Answer 1

Question

Chapter 30, Problem 12P

(a)

To determine

The amount of uranium dissolved in the ocean.

(a)

Expert Solution

Answer to Problem 12P

The amount of uranium dissolved in the ocean is

$4×015 g$ .

Explanation of Solution

Given Info:

The density is $3×10−3 g/m3$ .

The deapth is $4×103 m$ .

The radius is $6.38×106 m$ .

Formula to calculate the amount of uranium dissolved in the ocean is,

$mU=cVc(23A)hav=c(23(4πRE2))hav=8chavπRE23$

$mU$ is the amount of uranium dissolved
$c$ is the density
$V$ is the volume
$A$ is the area
$hav$ is the depth
$RE$ is the radius of Earth

Substitute $3×10−3 g/m3$ for $c$ , $4×103 m$ for $hav$ and $6.38×106 m$ for $RE$ to find $mU$ .

$mU=8(3×10−3 g/m3)(4×103 m)(3.14 rad)(6.38×106 m)23=4×1015 g$

Thus, the amount of uranium dissolved in the ocean is $4×015 g$ .

Conclusion:

The amount of uranium dissolved in the ocean is $4×015 g$ .

(b)

To determine

The time the uranium lasts.

(b)

Expert Solution

Answer to Problem 12P

The time the uranium lasts is

$5×103 year$ .

Explanation of Solution

Given Info:

The Avogadro number is $6.02×1023 atoms/mol$ .

The percentage of isotope present is $0.70%$

The energy per event $200 MeV/atoms$

The rate of energy consumption is $1.5×1013 J/s$

The mass per mol is $235 g/mol$

Formula to calculate the amount of uranium isotope is,

$m235U=ΔpmU$

$m235U$ is the mass of isotope
$Δp$ is the percentage of isotope

Formula to calculate the number of atoms is,

$N=NAMmolm235U=NAMmolΔpmU$

$NA$ is the Avogadro number
$Mmol$ is the mass per mol
$N$ is the number of atoms

Formula to calculate the energy released is,

$E=NE0=NAMmolΔpmUE0$

$E$ is the energy released
$E0$ is the energy released per atom

Formula to calculate the time is,

$t=EP=NAMmolΔpmUE0P=NAPMmolΔpmUE0$

$t$ is the energy released
$P$ is the energy released per atom

Substitute $6.02×1023 atoms/mol$ for $NA$ , $0.70%$ for $Δp$ , $200 MeV/atoms$ for $E$ , $1.5×1013 J/s$ for $P$ , $4×1015 g$ for $mU$ and $235 g/mol$ for $Mmol$ to find $t$ .

$t=(6.02×1023 atoms/mol)(1.5×1013 J/s)(235 g/mol)(0.70%)(4×1015 g)(200 MeV/atoms)=5×103 year$

Thus, the time the uranium lasts is $5×103 year$ .

Conclusion:

The time the uranium lasts is $5×103 year$ .

(c)

To determine

The source of uranium and the possibility of renewal of the energy source.

(c)

Expert Solution

Explanation of Solution

The uranium comes from natural resources as rocks and minerals.

The uranium comes from dissolving rock and minerals. Rivers carry such solutes into the oceans, however, the Earth’s supply of uranium is not renewable. However, if breeder reactors are used, the current ocean supply can last about a half- million years.

Conclusion:

The sources of uranium are natural sources as rocks and minerals which are not renewable.

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Answer 2

Question

Chapter 30, Problem 12P

(a)

To determine

The amount of uranium dissolved in the ocean.

(a)

Expert Solution

Answer to Problem 12P

The amount of uranium dissolved in the ocean is

$4×015 g$ .

Explanation of Solution

Given Info:

The density is $3×10−3 g/m3$ .

The deapth is $4×103 m$ .

The radius is $6.38×106 m$ .

Formula to calculate the amount of uranium dissolved in the ocean is,

$mU=cVc(23A)hav=c(23(4πRE2))hav=8chavπRE23$

$mU$ is the amount of uranium dissolved
$c$ is the density
$V$ is the volume
$A$ is the area
$hav$ is the depth
$RE$ is the radius of Earth

Substitute $3×10−3 g/m3$ for $c$ , $4×103 m$ for $hav$ and $6.38×106 m$ for $RE$ to find $mU$ .

$mU=8(3×10−3 g/m3)(4×103 m)(3.14 rad)(6.38×106 m)23=4×1015 g$

Thus, the amount of uranium dissolved in the ocean is $4×015 g$ .

Conclusion:

The amount of uranium dissolved in the ocean is $4×015 g$ .

(b)

To determine

The time the uranium lasts.

(b)

Expert Solution

Answer to Problem 12P

The time the uranium lasts is

$5×103 year$ .

Explanation of Solution

Given Info:

The Avogadro number is $6.02×1023 atoms/mol$ .

The percentage of isotope present is $0.70%$

The energy per event $200 MeV/atoms$

The rate of energy consumption is $1.5×1013 J/s$

The mass per mol is $235 g/mol$

Formula to calculate the amount of uranium isotope is,

$m235U=ΔpmU$

$m235U$ is the mass of isotope
$Δp$ is the percentage of isotope

Formula to calculate the number of atoms is,

$N=NAMmolm235U=NAMmolΔpmU$

$NA$ is the Avogadro number
$Mmol$ is the mass per mol
$N$ is the number of atoms

Formula to calculate the energy released is,

$E=NE0=NAMmolΔpmUE0$

$E$ is the energy released
$E0$ is the energy released per atom

Formula to calculate the time is,

$t=EP=NAMmolΔpmUE0P=NAPMmolΔpmUE0$

$t$ is the energy released
$P$ is the energy released per atom

Substitute $6.02×1023 atoms/mol$ for $NA$ , $0.70%$ for $Δp$ , $200 MeV/atoms$ for $E$ , $1.5×1013 J/s$ for $P$ , $4×1015 g$ for $mU$ and $235 g/mol$ for $Mmol$ to find $t$ .

$t=(6.02×1023 atoms/mol)(1.5×1013 J/s)(235 g/mol)(0.70%)(4×1015 g)(200 MeV/atoms)=5×103 year$

Thus, the time the uranium lasts is $5×103 year$ .

Conclusion:

The time the uranium lasts is $5×103 year$ .

(c)

To determine

The source of uranium and the possibility of renewal of the energy source.

(c)

Expert Solution

Explanation of Solution

The uranium comes from natural resources as rocks and minerals.

The uranium comes from dissolving rock and minerals. Rivers carry such solutes into the oceans, however, the Earth’s supply of uranium is not renewable. However, if breeder reactors are used, the current ocean supply can last about a half- million years.

Conclusion:

The sources of uranium are natural sources as rocks and minerals which are not renewable.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 30, Problem 12P

(a)

To determine

The amount of uranium dissolved in the ocean.

(a)

Expert Solution

Answer to Problem 12P

The amount of uranium dissolved in the ocean is

$4×015 g$ .

Explanation of Solution

Given Info:

The density is $3×10−3 g/m3$ .

The deapth is $4×103 m$ .

The radius is $6.38×106 m$ .

Formula to calculate the amount of uranium dissolved in the ocean is,

$mU=cVc(23A)hav=c(23(4πRE2))hav=8chavπRE23$

$mU$ is the amount of uranium dissolved
$c$ is the density
$V$ is the volume
$A$ is the area
$hav$ is the depth
$RE$ is the radius of Earth

Substitute $3×10−3 g/m3$ for $c$ , $4×103 m$ for $hav$ and $6.38×106 m$ for $RE$ to find $mU$ .

$mU=8(3×10−3 g/m3)(4×103 m)(3.14 rad)(6.38×106 m)23=4×1015 g$

Thus, the amount of uranium dissolved in the ocean is $4×015 g$ .

Conclusion:

The amount of uranium dissolved in the ocean is $4×015 g$ .

(b)

To determine

The time the uranium lasts.

(b)

Expert Solution

Answer to Problem 12P

The time the uranium lasts is

$5×103 year$ .

Explanation of Solution

Given Info:

The Avogadro number is $6.02×1023 atoms/mol$ .

The percentage of isotope present is $0.70%$

The energy per event $200 MeV/atoms$

The rate of energy consumption is $1.5×1013 J/s$

The mass per mol is $235 g/mol$

Formula to calculate the amount of uranium isotope is,

$m235U=ΔpmU$

$m235U$ is the mass of isotope
$Δp$ is the percentage of isotope

Formula to calculate the number of atoms is,

$N=NAMmolm235U=NAMmolΔpmU$

$NA$ is the Avogadro number
$Mmol$ is the mass per mol
$N$ is the number of atoms

Formula to calculate the energy released is,

$E=NE0=NAMmolΔpmUE0$

$E$ is the energy released
$E0$ is the energy released per atom

Formula to calculate the time is,

$t=EP=NAMmolΔpmUE0P=NAPMmolΔpmUE0$

$t$ is the energy released
$P$ is the energy released per atom

Substitute $6.02×1023 atoms/mol$ for $NA$ , $0.70%$ for $Δp$ , $200 MeV/atoms$ for $E$ , $1.5×1013 J/s$ for $P$ , $4×1015 g$ for $mU$ and $235 g/mol$ for $Mmol$ to find $t$ .

$t=(6.02×1023 atoms/mol)(1.5×1013 J/s)(235 g/mol)(0.70%)(4×1015 g)(200 MeV/atoms)=5×103 year$

Thus, the time the uranium lasts is $5×103 year$ .

Conclusion:

The time the uranium lasts is $5×103 year$ .

(c)

To determine

The source of uranium and the possibility of renewal of the energy source.

(c)

Expert Solution

Explanation of Solution

The uranium comes from natural resources as rocks and minerals.

The uranium comes from dissolving rock and minerals. Rivers carry such solutes into the oceans, however, the Earth’s supply of uranium is not renewable. However, if breeder reactors are used, the current ocean supply can last about a half- million years.

Conclusion:

The sources of uranium are natural sources as rocks and minerals which are not renewable.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 30, Problem 12P

(a)

To determine

The amount of uranium dissolved in the ocean.

(a)

Expert Solution

Answer to Problem 12P

The amount of uranium dissolved in the ocean is

$4×015 g$ .

Explanation of Solution

Given Info:

The density is $3×10−3 g/m3$ .

The deapth is $4×103 m$ .

The radius is $6.38×106 m$ .

Formula to calculate the amount of uranium dissolved in the ocean is,

$mU=cVc(23A)hav=c(23(4πRE2))hav=8chavπRE23$

$mU$ is the amount of uranium dissolved
$c$ is the density
$V$ is the volume
$A$ is the area
$hav$ is the depth
$RE$ is the radius of Earth

Substitute $3×10−3 g/m3$ for $c$ , $4×103 m$ for $hav$ and $6.38×106 m$ for $RE$ to find $mU$ .

$mU=8(3×10−3 g/m3)(4×103 m)(3.14 rad)(6.38×106 m)23=4×1015 g$

Thus, the amount of uranium dissolved in the ocean is $4×015 g$ .

Conclusion:

The amount of uranium dissolved in the ocean is $4×015 g$ .

(b)

To determine

The time the uranium lasts.

(b)

Expert Solution

Answer to Problem 12P

The time the uranium lasts is

$5×103 year$ .

Explanation of Solution

Given Info:

The Avogadro number is $6.02×1023 atoms/mol$ .

The percentage of isotope present is $0.70%$

The energy per event $200 MeV/atoms$

The rate of energy consumption is $1.5×1013 J/s$

The mass per mol is $235 g/mol$

Formula to calculate the amount of uranium isotope is,

$m235U=ΔpmU$

$m235U$ is the mass of isotope
$Δp$ is the percentage of isotope

Formula to calculate the number of atoms is,

$N=NAMmolm235U=NAMmolΔpmU$

$NA$ is the Avogadro number
$Mmol$ is the mass per mol
$N$ is the number of atoms

Formula to calculate the energy released is,

$E=NE0=NAMmolΔpmUE0$

$E$ is the energy released
$E0$ is the energy released per atom

Formula to calculate the time is,

$t=EP=NAMmolΔpmUE0P=NAPMmolΔpmUE0$

$t$ is the energy released
$P$ is the energy released per atom

Substitute $6.02×1023 atoms/mol$ for $NA$ , $0.70%$ for $Δp$ , $200 MeV/atoms$ for $E$ , $1.5×1013 J/s$ for $P$ , $4×1015 g$ for $mU$ and $235 g/mol$ for $Mmol$ to find $t$ .

$t=(6.02×1023 atoms/mol)(1.5×1013 J/s)(235 g/mol)(0.70%)(4×1015 g)(200 MeV/atoms)=5×103 year$

Thus, the time the uranium lasts is $5×103 year$ .

Conclusion:

The time the uranium lasts is $5×103 year$ .

(c)

To determine

The source of uranium and the possibility of renewal of the energy source.

(c)

Expert Solution

Explanation of Solution

The uranium comes from natural resources as rocks and minerals.

The uranium comes from dissolving rock and minerals. Rivers carry such solutes into the oceans, however, the Earth’s supply of uranium is not renewable. However, if breeder reactors are used, the current ocean supply can last about a half- million years.

Conclusion:

The sources of uranium are natural sources as rocks and minerals which are not renewable.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 30, Problem 12P

(a)

To determine

The amount of uranium dissolved in the ocean.

(a)

Expert Solution

Answer to Problem 12P

The amount of uranium dissolved in the ocean is

$4×015 g$ .

Explanation of Solution

Given Info:

The density is $3×10−3 g/m3$ .

The deapth is $4×103 m$ .

The radius is $6.38×106 m$ .

Formula to calculate the amount of uranium dissolved in the ocean is,

$mU=cVc(23A)hav=c(23(4πRE2))hav=8chavπRE23$

$mU$ is the amount of uranium dissolved
$c$ is the density
$V$ is the volume
$A$ is the area
$hav$ is the depth
$RE$ is the radius of Earth

Substitute $3×10−3 g/m3$ for $c$ , $4×103 m$ for $hav$ and $6.38×106 m$ for $RE$ to find $mU$ .

$mU=8(3×10−3 g/m3)(4×103 m)(3.14 rad)(6.38×106 m)23=4×1015 g$

Thus, the amount of uranium dissolved in the ocean is $4×015 g$ .

Conclusion:

The amount of uranium dissolved in the ocean is $4×015 g$ .

(b)

To determine

The time the uranium lasts.

(b)

Expert Solution

Answer to Problem 12P

The time the uranium lasts is

$5×103 year$ .

Explanation of Solution

Given Info:

The Avogadro number is $6.02×1023 atoms/mol$ .

The percentage of isotope present is $0.70%$

The energy per event $200 MeV/atoms$

The rate of energy consumption is $1.5×1013 J/s$

The mass per mol is $235 g/mol$

Formula to calculate the amount of uranium isotope is,

$m235U=ΔpmU$

$m235U$ is the mass of isotope
$Δp$ is the percentage of isotope

Formula to calculate the number of atoms is,

$N=NAMmolm235U=NAMmolΔpmU$

$NA$ is the Avogadro number
$Mmol$ is the mass per mol
$N$ is the number of atoms

Formula to calculate the energy released is,

$E=NE0=NAMmolΔpmUE0$

$E$ is the energy released
$E0$ is the energy released per atom

Formula to calculate the time is,

$t=EP=NAMmolΔpmUE0P=NAPMmolΔpmUE0$

$t$ is the energy released
$P$ is the energy released per atom

Substitute $6.02×1023 atoms/mol$ for $NA$ , $0.70%$ for $Δp$ , $200 MeV/atoms$ for $E$ , $1.5×1013 J/s$ for $P$ , $4×1015 g$ for $mU$ and $235 g/mol$ for $Mmol$ to find $t$ .

$t=(6.02×1023 atoms/mol)(1.5×1013 J/s)(235 g/mol)(0.70%)(4×1015 g)(200 MeV/atoms)=5×103 year$

Thus, the time the uranium lasts is $5×103 year$ .

Conclusion:

The time the uranium lasts is $5×103 year$ .

(c)

To determine

The source of uranium and the possibility of renewal of the energy source.

(c)

Expert Solution

Explanation of Solution

The uranium comes from natural resources as rocks and minerals.

The uranium comes from dissolving rock and minerals. Rivers carry such solutes into the oceans, however, the Earth’s supply of uranium is not renewable. However, if breeder reactors are used, the current ocean supply can last about a half- million years.

Conclusion:

The sources of uranium are natural sources as rocks and minerals which are not renewable.

Answer 6

Chapter 30, Problem 12P

(a)

To determine

The amount of uranium dissolved in the ocean.

(a)

Expert Solution

Answer to Problem 12P

The amount of uranium dissolved in the ocean is

$4×015 g$ .

Explanation of Solution

Given Info:

The density is $3×10−3 g/m3$ .

The deapth is $4×103 m$ .

The radius is $6.38×106 m$ .

Formula to calculate the amount of uranium dissolved in the ocean is,

$mU=cVc(23A)hav=c(23(4πRE2))hav=8chavπRE23$

$mU$ is the amount of uranium dissolved
$c$ is the density
$V$ is the volume
$A$ is the area
$hav$ is the depth
$RE$ is the radius of Earth

Substitute $3×10−3 g/m3$ for $c$ , $4×103 m$ for $hav$ and $6.38×106 m$ for $RE$ to find $mU$ .

$mU=8(3×10−3 g/m3)(4×103 m)(3.14 rad)(6.38×106 m)23=4×1015 g$

Thus, the amount of uranium dissolved in the ocean is $4×015 g$ .

Conclusion:

The amount of uranium dissolved in the ocean is $4×015 g$ .

Answer 7

Chapter 30, Problem 12P

(a)

To determine

The amount of uranium dissolved in the ocean.

Answer 8

(a)

Expert Solution

Answer to Problem 12P

The amount of uranium dissolved in the ocean is

$4×015 g$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given Info:

The density is $3×10−3 g/m3$ .

The deapth is $4×103 m$ .

The radius is $6.38×106 m$ .

Formula to calculate the amount of uranium dissolved in the ocean is,

$mU=cVc(23A)hav=c(23(4πRE2))hav=8chavπRE23$

$mU$ is the amount of uranium dissolved
$c$ is the density
$V$ is the volume
$A$ is the area
$hav$ is the depth
$RE$ is the radius of Earth

Substitute $3×10−3 g/m3$ for $c$ , $4×103 m$ for $hav$ and $6.38×106 m$ for $RE$ to find $mU$ .

$mU=8(3×10−3 g/m3)(4×103 m)(3.14 rad)(6.38×106 m)23=4×1015 g$

Thus, the amount of uranium dissolved in the ocean is $4×015 g$ .

Conclusion:

The amount of uranium dissolved in the ocean is $4×015 g$ .

Answer 13

(b)

To determine

The time the uranium lasts.

(b)

Expert Solution

Answer to Problem 12P

The time the uranium lasts is

$5×103 year$ .

Explanation of Solution

Given Info:

The Avogadro number is $6.02×1023 atoms/mol$ .

The percentage of isotope present is $0.70%$

The energy per event $200 MeV/atoms$

The rate of energy consumption is $1.5×1013 J/s$

The mass per mol is $235 g/mol$

Formula to calculate the amount of uranium isotope is,

$m235U=ΔpmU$

$m235U$ is the mass of isotope
$Δp$ is the percentage of isotope

Formula to calculate the number of atoms is,

$N=NAMmolm235U=NAMmolΔpmU$

$NA$ is the Avogadro number
$Mmol$ is the mass per mol
$N$ is the number of atoms

Formula to calculate the energy released is,

$E=NE0=NAMmolΔpmUE0$

$E$ is the energy released
$E0$ is the energy released per atom

Formula to calculate the time is,

$t=EP=NAMmolΔpmUE0P=NAPMmolΔpmUE0$

$t$ is the energy released
$P$ is the energy released per atom

Substitute $6.02×1023 atoms/mol$ for $NA$ , $0.70%$ for $Δp$ , $200 MeV/atoms$ for $E$ , $1.5×1013 J/s$ for $P$ , $4×1015 g$ for $mU$ and $235 g/mol$ for $Mmol$ to find $t$ .

$t=(6.02×1023 atoms/mol)(1.5×1013 J/s)(235 g/mol)(0.70%)(4×1015 g)(200 MeV/atoms)=5×103 year$

Thus, the time the uranium lasts is $5×103 year$ .

Conclusion:

The time the uranium lasts is $5×103 year$ .

Answer 14

(b)

To determine

The time the uranium lasts.

Answer 15

(b)

Expert Solution

Answer to Problem 12P

The time the uranium lasts is

$5×103 year$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given Info:

The Avogadro number is $6.02×1023 atoms/mol$ .

The percentage of isotope present is $0.70%$

The energy per event $200 MeV/atoms$

The rate of energy consumption is $1.5×1013 J/s$

The mass per mol is $235 g/mol$

Formula to calculate the amount of uranium isotope is,

$m235U=ΔpmU$

$m235U$ is the mass of isotope
$Δp$ is the percentage of isotope

Formula to calculate the number of atoms is,

$N=NAMmolm235U=NAMmolΔpmU$

$NA$ is the Avogadro number
$Mmol$ is the mass per mol
$N$ is the number of atoms

Formula to calculate the energy released is,

$E=NE0=NAMmolΔpmUE0$

$E$ is the energy released
$E0$ is the energy released per atom

Formula to calculate the time is,

$t=EP=NAMmolΔpmUE0P=NAPMmolΔpmUE0$

$t$ is the energy released
$P$ is the energy released per atom

Substitute $6.02×1023 atoms/mol$ for $NA$ , $0.70%$ for $Δp$ , $200 MeV/atoms$ for $E$ , $1.5×1013 J/s$ for $P$ , $4×1015 g$ for $mU$ and $235 g/mol$ for $Mmol$ to find $t$ .

$t=(6.02×1023 atoms/mol)(1.5×1013 J/s)(235 g/mol)(0.70%)(4×1015 g)(200 MeV/atoms)=5×103 year$

Thus, the time the uranium lasts is $5×103 year$ .

Conclusion:

The time the uranium lasts is $5×103 year$ .

Answer 20

(c)

To determine

The source of uranium and the possibility of renewal of the energy source.

(c)

Expert Solution

Explanation of Solution

The uranium comes from natural resources as rocks and minerals.

The uranium comes from dissolving rock and minerals. Rivers carry such solutes into the oceans, however, the Earth’s supply of uranium is not renewable. However, if breeder reactors are used, the current ocean supply can last about a half- million years.

Conclusion:

The sources of uranium are natural sources as rocks and minerals which are not renewable.

Answer 21

(c)

To determine

The source of uranium and the possibility of renewal of the energy source.

Answer 22

(c)

Expert Solution

Explanation of Solution

The uranium comes from natural resources as rocks and minerals.

The uranium comes from dissolving rock and minerals. Rivers carry such solutes into the oceans, however, the Earth’s supply of uranium is not renewable. However, if breeder reactors are used, the current ocean supply can last about a half- million years.

Conclusion:

The sources of uranium are natural sources as rocks and minerals which are not renewable.