It is given that, radon is present as 1.0 g per 7.0 metric ton of a pitch ball. The number of radon atoms that can be isolated from given mass of pitch-blende is to be calculated. The number of radon atoms remain in 2025 is to be calculated if 15 mg radon is manufactured in 1925 . Concept introduction: Amount of radon left is calculated using the formula, A E = A 0 0.5 t / t 1 / 2 To determine: The number of radon atoms that can be isolated from 1.75 × 10 8 g pitch-blende; the number of radon atoms remain in 2025 , if 15 mg radon is manufactured in 1925 .

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Answer 1

Question

Chapter 20, Problem 116IP

Interpretation Introduction

Interpretation: It is given that, radon is present as $1.0 g$ per $7.0 metric ton$ of a pitch ball. The number of radon atoms that can be isolated from given mass of pitch-blende is to be calculated. The number of radon atoms remain in $2025$ is to be calculated if $15 mg$ radon is manufactured in $1925$ .

Concept introduction: Amount of radon left is calculated using the formula,

$AE=A00.5t/t1/2$

To determine: The number of radon atoms that can be isolated from $1.75×108 g$ pitch-blende; the number of radon atoms remain in $2025$ , if $15 mg$ radon is manufactured in $1925$ .

Expert Solution & Answer

Answer to Problem 116IP

Answer

Number of radon $(Ra)$ atoms that can be isolated from $1.75×108 g$ pitch-blende is $6.66×1022 atoms_$ . Number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ .

Explanation of Solution

Mass of radon $(Ra)$ is $25 g_$ .

Given

Mass of pitch-blende is $1.75×108 g$ .

The conversion of gram $(g)$ into kilogram $(kg)$ is done as,

$1 g=10−3 kg$

Hence,

The conversion of $1.75×108 g$ into kilogram is,

$1.75×108 g=(1.75×108×10−3) kg=1.75×105 kg$

Since, radon is present as $1.0 g$ per $7.0 metric ton$ of a pitch ball, mass of radon is calculated as,

$Mass of Ra=Mass of pitch-blende×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)$

Substitute the value of mass of pitch-blende in the above equation.

$Mass of Ra=Mass of pitch-blende×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)=1.75×105 kg×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)=25 g_$

Number of radon $(Ra)$ atoms that can be isolated from $1.75×108 g$ pitch-blende is $6.66×1022 atoms_$ .

Mass of radon is $25 g$ .

Atomic mass of radon is $226 g$ .

Formula

The number of radon atoms is calculated as,

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra$

Substitute the values of mass and atomic mass of radon in the above equation.

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra=25 g×1 mol Ra226.0 g×6.022×1023 atoms 1 mol Ra=6.66×1022 atoms_$

Number of radon $(Ra)$ atoms in $15 mg$ is $3.99×1019 atoms_$ .

Given

Mass of radon atoms is $15 mg$ .

Half life of radon is $1.60×103 years$ .

Atomic mass of radon is $226 g$ .

The conversion of milligram $(mg)$ into gram $(g)$ is done as,

$1 mg=10−3 g$

Hence,

The conversion of $15 mg$ into gram is,

$15 mg=(15×10−3) g=15×10−3 g$

Formula

The number of radon atoms is calculated as,

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra$

Substitute the values of mass and atomic mass of radon in the above equation.

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra=15×10−3 g×1 mol Ra226.0 g×6.022×1023 atoms 1 mol Ra=3.99×1019 atoms_$

Number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ .

Half life of radon is $1.60×103 years$ .

Number of radon $(Ra)$ atoms in $15 mg$ is $3.99×1019 atoms$ .

Total time from $1925$ to $2025$ is,

$2025−1925=100 years$

Formula

Amount of radon left is calculated using the formula,

$AE=A00.5t/t1/2$

Where,

$AE$ is the amount of radon left.
$A0$ is the original amount of radon.
$t1/2$ is the half-life.
$t$ is the total time.

Substitute the values of $t1/2,t$ and $A0$ in the above equation.

$AE=A00.5t/t1/2=3.99×1019 Ra atoms×(0.5)100 years1600 years=3.82×1019 Ra atoms_$

Conclusion

Conclusion

The calculated value of number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ . Number of radon $(Ra)$ atoms in $25 g$ is $6.66×1022 atoms_$

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Identify and provide an explanation of the differences between homogeneous and heterogeneous sampling in the context of sampling methods.

Г C-RSA CHROMATOPAC CH=1 DATA 1: @CHRM1.C00 ATTEN=10 SPEED= 10.0 0.0 b.092 0.797 1.088 1.813 C-RSA CHROMATOPAC CH=1 Report No. =13 ** CALCULATION REPORT ** DATA=1: @CHRM1.000 11/03/05 08:09:52 CH PKNO TIME 1 2 0.797 3 1.088 4 1.813 AREA 1508566 4625442 2180060 HEIGHT 207739 701206 V 287554 V MK IDNO CONC NAME 18.1447 55.6339 26.2213 TOTAL 8314067 1196500 100 C-R8A CHROMATOPAC CH=1 DATA 1: @CHRM1.C00 ATTEN=10 SPEED= 10.0 0. 0 087 337. 0.841 1.150 C-R8A CHROMATOPAC CH=1 Report No. =14 DATA=1: @CHRM1.000 11/03/05 08:12:40 ** CALCULATION REPORT ** CH PKNO TIME AREA 1 3 0.841 1099933 41.15 4039778 HEIGHT MK IDNO 170372 649997¯¯¯ CONC NAME 21.4007 78.5993 TOTAL 5139711 820369 100 3 C-R8A CHROMATOPAC CH=1 DATA 1: @CHRM1.C00 ATTEN=10 SPEED= 10.0 0.100 0:652 5.856 3 1.165 C-RSA CHROMATOPAC CH-1 Report No. =15 DATA=1: @CHRM1.000 11/03/05 08:15:26 ** CALCULATION REPORT ** CH PKNO TIME AREA HEIGHT MK IDNO CONC NAME 1 3 3 0.856 4 1.165 TOTAL 1253386 4838738 175481 708024 V 20.5739 79.4261 6092124…

Draw the product of the reaction shown below. Ignore small byproducts that would evaporate please.

Answer 2

Question

Chapter 20, Problem 116IP

Interpretation Introduction

Interpretation: It is given that, radon is present as $1.0 g$ per $7.0 metric ton$ of a pitch ball. The number of radon atoms that can be isolated from given mass of pitch-blende is to be calculated. The number of radon atoms remain in $2025$ is to be calculated if $15 mg$ radon is manufactured in $1925$ .

Concept introduction: Amount of radon left is calculated using the formula,

$AE=A00.5t/t1/2$

To determine: The number of radon atoms that can be isolated from $1.75×108 g$ pitch-blende; the number of radon atoms remain in $2025$ , if $15 mg$ radon is manufactured in $1925$ .

Expert Solution & Answer

Answer to Problem 116IP

Answer

Number of radon $(Ra)$ atoms that can be isolated from $1.75×108 g$ pitch-blende is $6.66×1022 atoms_$ . Number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ .

Explanation of Solution

Mass of radon $(Ra)$ is $25 g_$ .

Given

Mass of pitch-blende is $1.75×108 g$ .

The conversion of gram $(g)$ into kilogram $(kg)$ is done as,

$1 g=10−3 kg$

Hence,

The conversion of $1.75×108 g$ into kilogram is,

$1.75×108 g=(1.75×108×10−3) kg=1.75×105 kg$

Since, radon is present as $1.0 g$ per $7.0 metric ton$ of a pitch ball, mass of radon is calculated as,

$Mass of Ra=Mass of pitch-blende×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)$

Substitute the value of mass of pitch-blende in the above equation.

$Mass of Ra=Mass of pitch-blende×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)=1.75×105 kg×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)=25 g_$

Number of radon $(Ra)$ atoms that can be isolated from $1.75×108 g$ pitch-blende is $6.66×1022 atoms_$ .

Mass of radon is $25 g$ .

Atomic mass of radon is $226 g$ .

Formula

The number of radon atoms is calculated as,

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra$

Substitute the values of mass and atomic mass of radon in the above equation.

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra=25 g×1 mol Ra226.0 g×6.022×1023 atoms 1 mol Ra=6.66×1022 atoms_$

Number of radon $(Ra)$ atoms in $15 mg$ is $3.99×1019 atoms_$ .

Given

Mass of radon atoms is $15 mg$ .

Half life of radon is $1.60×103 years$ .

Atomic mass of radon is $226 g$ .

The conversion of milligram $(mg)$ into gram $(g)$ is done as,

$1 mg=10−3 g$

Hence,

The conversion of $15 mg$ into gram is,

$15 mg=(15×10−3) g=15×10−3 g$

Formula

The number of radon atoms is calculated as,

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra$

Substitute the values of mass and atomic mass of radon in the above equation.

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra=15×10−3 g×1 mol Ra226.0 g×6.022×1023 atoms 1 mol Ra=3.99×1019 atoms_$

Number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ .

Half life of radon is $1.60×103 years$ .

Number of radon $(Ra)$ atoms in $15 mg$ is $3.99×1019 atoms$ .

Total time from $1925$ to $2025$ is,

$2025−1925=100 years$

Formula

Amount of radon left is calculated using the formula,

$AE=A00.5t/t1/2$

Where,

$AE$ is the amount of radon left.
$A0$ is the original amount of radon.
$t1/2$ is the half-life.
$t$ is the total time.

Substitute the values of $t1/2,t$ and $A0$ in the above equation.

$AE=A00.5t/t1/2=3.99×1019 Ra atoms×(0.5)100 years1600 years=3.82×1019 Ra atoms_$

Conclusion

Conclusion

The calculated value of number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ . Number of radon $(Ra)$ atoms in $25 g$ is $6.66×1022 atoms_$

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Answer 3

Question

Chapter 20, Problem 116IP

Interpretation Introduction

Interpretation: It is given that, radon is present as $1.0 g$ per $7.0 metric ton$ of a pitch ball. The number of radon atoms that can be isolated from given mass of pitch-blende is to be calculated. The number of radon atoms remain in $2025$ is to be calculated if $15 mg$ radon is manufactured in $1925$ .

Concept introduction: Amount of radon left is calculated using the formula,

$AE=A00.5t/t1/2$

To determine: The number of radon atoms that can be isolated from $1.75×108 g$ pitch-blende; the number of radon atoms remain in $2025$ , if $15 mg$ radon is manufactured in $1925$ .

Expert Solution & Answer

Answer to Problem 116IP

Answer

Number of radon $(Ra)$ atoms that can be isolated from $1.75×108 g$ pitch-blende is $6.66×1022 atoms_$ . Number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ .

Explanation of Solution

Mass of radon $(Ra)$ is $25 g_$ .

Given

Mass of pitch-blende is $1.75×108 g$ .

The conversion of gram $(g)$ into kilogram $(kg)$ is done as,

$1 g=10−3 kg$

Hence,

The conversion of $1.75×108 g$ into kilogram is,

$1.75×108 g=(1.75×108×10−3) kg=1.75×105 kg$

Since, radon is present as $1.0 g$ per $7.0 metric ton$ of a pitch ball, mass of radon is calculated as,

$Mass of Ra=Mass of pitch-blende×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)$

Substitute the value of mass of pitch-blende in the above equation.

$Mass of Ra=Mass of pitch-blende×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)=1.75×105 kg×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)=25 g_$

Number of radon $(Ra)$ atoms that can be isolated from $1.75×108 g$ pitch-blende is $6.66×1022 atoms_$ .

Mass of radon is $25 g$ .

Atomic mass of radon is $226 g$ .

Formula

The number of radon atoms is calculated as,

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra$

Substitute the values of mass and atomic mass of radon in the above equation.

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra=25 g×1 mol Ra226.0 g×6.022×1023 atoms 1 mol Ra=6.66×1022 atoms_$

Number of radon $(Ra)$ atoms in $15 mg$ is $3.99×1019 atoms_$ .

Given

Mass of radon atoms is $15 mg$ .

Half life of radon is $1.60×103 years$ .

Atomic mass of radon is $226 g$ .

The conversion of milligram $(mg)$ into gram $(g)$ is done as,

$1 mg=10−3 g$

Hence,

The conversion of $15 mg$ into gram is,

$15 mg=(15×10−3) g=15×10−3 g$

Formula

The number of radon atoms is calculated as,

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra$

Substitute the values of mass and atomic mass of radon in the above equation.

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra=15×10−3 g×1 mol Ra226.0 g×6.022×1023 atoms 1 mol Ra=3.99×1019 atoms_$

Number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ .

Half life of radon is $1.60×103 years$ .

Number of radon $(Ra)$ atoms in $15 mg$ is $3.99×1019 atoms$ .

Total time from $1925$ to $2025$ is,

$2025−1925=100 years$

Formula

Amount of radon left is calculated using the formula,

$AE=A00.5t/t1/2$

Where,

$AE$ is the amount of radon left.
$A0$ is the original amount of radon.
$t1/2$ is the half-life.
$t$ is the total time.

Substitute the values of $t1/2,t$ and $A0$ in the above equation.

$AE=A00.5t/t1/2=3.99×1019 Ra atoms×(0.5)100 years1600 years=3.82×1019 Ra atoms_$

Conclusion

Conclusion

The calculated value of number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ . Number of radon $(Ra)$ atoms in $25 g$ is $6.66×1022 atoms_$

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Answer 4

Question

Chapter 20, Problem 116IP

Interpretation Introduction

Interpretation: It is given that, radon is present as $1.0 g$ per $7.0 metric ton$ of a pitch ball. The number of radon atoms that can be isolated from given mass of pitch-blende is to be calculated. The number of radon atoms remain in $2025$ is to be calculated if $15 mg$ radon is manufactured in $1925$ .

Concept introduction: Amount of radon left is calculated using the formula,

$AE=A00.5t/t1/2$

To determine: The number of radon atoms that can be isolated from $1.75×108 g$ pitch-blende; the number of radon atoms remain in $2025$ , if $15 mg$ radon is manufactured in $1925$ .

Expert Solution & Answer

Answer to Problem 116IP

Answer

Number of radon $(Ra)$ atoms that can be isolated from $1.75×108 g$ pitch-blende is $6.66×1022 atoms_$ . Number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ .

Explanation of Solution

Mass of radon $(Ra)$ is $25 g_$ .

Given

Mass of pitch-blende is $1.75×108 g$ .

The conversion of gram $(g)$ into kilogram $(kg)$ is done as,

$1 g=10−3 kg$

Hence,

The conversion of $1.75×108 g$ into kilogram is,

$1.75×108 g=(1.75×108×10−3) kg=1.75×105 kg$

Since, radon is present as $1.0 g$ per $7.0 metric ton$ of a pitch ball, mass of radon is calculated as,

$Mass of Ra=Mass of pitch-blende×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)$

Substitute the value of mass of pitch-blende in the above equation.

$Mass of Ra=Mass of pitch-blende×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)=1.75×105 kg×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)=25 g_$

Number of radon $(Ra)$ atoms that can be isolated from $1.75×108 g$ pitch-blende is $6.66×1022 atoms_$ .

Mass of radon is $25 g$ .

Atomic mass of radon is $226 g$ .

Formula

The number of radon atoms is calculated as,

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra$

Substitute the values of mass and atomic mass of radon in the above equation.

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra=25 g×1 mol Ra226.0 g×6.022×1023 atoms 1 mol Ra=6.66×1022 atoms_$

Number of radon $(Ra)$ atoms in $15 mg$ is $3.99×1019 atoms_$ .

Given

Mass of radon atoms is $15 mg$ .

Half life of radon is $1.60×103 years$ .

Atomic mass of radon is $226 g$ .

The conversion of milligram $(mg)$ into gram $(g)$ is done as,

$1 mg=10−3 g$

Hence,

The conversion of $15 mg$ into gram is,

$15 mg=(15×10−3) g=15×10−3 g$

Formula

The number of radon atoms is calculated as,

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra$

Substitute the values of mass and atomic mass of radon in the above equation.

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra=15×10−3 g×1 mol Ra226.0 g×6.022×1023 atoms 1 mol Ra=3.99×1019 atoms_$

Number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ .

Half life of radon is $1.60×103 years$ .

Number of radon $(Ra)$ atoms in $15 mg$ is $3.99×1019 atoms$ .

Total time from $1925$ to $2025$ is,

$2025−1925=100 years$

Formula

Amount of radon left is calculated using the formula,

$AE=A00.5t/t1/2$

Where,

$AE$ is the amount of radon left.
$A0$ is the original amount of radon.
$t1/2$ is the half-life.
$t$ is the total time.

Substitute the values of $t1/2,t$ and $A0$ in the above equation.

$AE=A00.5t/t1/2=3.99×1019 Ra atoms×(0.5)100 years1600 years=3.82×1019 Ra atoms_$

Conclusion

Conclusion

The calculated value of number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ . Number of radon $(Ra)$ atoms in $25 g$ is $6.66×1022 atoms_$

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Answer 5

Chapter 20, Problem 116IP

Interpretation Introduction

Interpretation: It is given that, radon is present as $1.0 g$ per $7.0 metric ton$ of a pitch ball. The number of radon atoms that can be isolated from given mass of pitch-blende is to be calculated. The number of radon atoms remain in $2025$ is to be calculated if $15 mg$ radon is manufactured in $1925$ .

Concept introduction: Amount of radon left is calculated using the formula,

$AE=A00.5t/t1/2$

To determine: The number of radon atoms that can be isolated from $1.75×108 g$ pitch-blende; the number of radon atoms remain in $2025$ , if $15 mg$ radon is manufactured in $1925$ .

Expert Solution & Answer

Answer to Problem 116IP

Answer

Number of radon $(Ra)$ atoms that can be isolated from $1.75×108 g$ pitch-blende is $6.66×1022 atoms_$ . Number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ .

Explanation of Solution

Mass of radon $(Ra)$ is $25 g_$ .

Given

Mass of pitch-blende is $1.75×108 g$ .

The conversion of gram $(g)$ into kilogram $(kg)$ is done as,

$1 g=10−3 kg$

Hence,

The conversion of $1.75×108 g$ into kilogram is,

$1.75×108 g=(1.75×108×10−3) kg=1.75×105 kg$

Since, radon is present as $1.0 g$ per $7.0 metric ton$ of a pitch ball, mass of radon is calculated as,

$Mass of Ra=Mass of pitch-blende×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)$

Substitute the value of mass of pitch-blende in the above equation.

$Mass of Ra=Mass of pitch-blende×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)=1.75×105 kg×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)=25 g_$

Number of radon $(Ra)$ atoms that can be isolated from $1.75×108 g$ pitch-blende is $6.66×1022 atoms_$ .

Mass of radon is $25 g$ .

Atomic mass of radon is $226 g$ .

Formula

The number of radon atoms is calculated as,

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra$

Substitute the values of mass and atomic mass of radon in the above equation.

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra=25 g×1 mol Ra226.0 g×6.022×1023 atoms 1 mol Ra=6.66×1022 atoms_$

Number of radon $(Ra)$ atoms in $15 mg$ is $3.99×1019 atoms_$ .

Given

Mass of radon atoms is $15 mg$ .

Half life of radon is $1.60×103 years$ .

Atomic mass of radon is $226 g$ .

The conversion of milligram $(mg)$ into gram $(g)$ is done as,

$1 mg=10−3 g$

Hence,

The conversion of $15 mg$ into gram is,

$15 mg=(15×10−3) g=15×10−3 g$

Formula

The number of radon atoms is calculated as,

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra$

Substitute the values of mass and atomic mass of radon in the above equation.

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra=15×10−3 g×1 mol Ra226.0 g×6.022×1023 atoms 1 mol Ra=3.99×1019 atoms_$

Number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ .

Half life of radon is $1.60×103 years$ .

Number of radon $(Ra)$ atoms in $15 mg$ is $3.99×1019 atoms$ .

Total time from $1925$ to $2025$ is,

$2025−1925=100 years$

Formula

Amount of radon left is calculated using the formula,

$AE=A00.5t/t1/2$

Where,

$AE$ is the amount of radon left.
$A0$ is the original amount of radon.
$t1/2$ is the half-life.
$t$ is the total time.

Substitute the values of $t1/2,t$ and $A0$ in the above equation.

$AE=A00.5t/t1/2=3.99×1019 Ra atoms×(0.5)100 years1600 years=3.82×1019 Ra atoms_$

Conclusion

Conclusion

The calculated value of number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ . Number of radon $(Ra)$ atoms in $25 g$ is $6.66×1022 atoms_$

Answer 6

Chapter 20, Problem 116IP

Interpretation Introduction

Interpretation: It is given that, radon is present as $1.0 g$ per $7.0 metric ton$ of a pitch ball. The number of radon atoms that can be isolated from given mass of pitch-blende is to be calculated. The number of radon atoms remain in $2025$ is to be calculated if $15 mg$ radon is manufactured in $1925$ .

Concept introduction: Amount of radon left is calculated using the formula,

$AE=A00.5t/t1/2$

To determine: The number of radon atoms that can be isolated from $1.75×108 g$ pitch-blende; the number of radon atoms remain in $2025$ , if $15 mg$ radon is manufactured in $1925$ .

Expert Solution & Answer

Answer to Problem 116IP

Answer

Number of radon $(Ra)$ atoms that can be isolated from $1.75×108 g$ pitch-blende is $6.66×1022 atoms_$ . Number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ .

Explanation of Solution

Mass of radon $(Ra)$ is $25 g_$ .

Given

Mass of pitch-blende is $1.75×108 g$ .

The conversion of gram $(g)$ into kilogram $(kg)$ is done as,

$1 g=10−3 kg$

Hence,

The conversion of $1.75×108 g$ into kilogram is,

$1.75×108 g=(1.75×108×10−3) kg=1.75×105 kg$

Since, radon is present as $1.0 g$ per $7.0 metric ton$ of a pitch ball, mass of radon is calculated as,

$Mass of Ra=Mass of pitch-blende×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)$

Substitute the value of mass of pitch-blende in the above equation.

$Mass of Ra=Mass of pitch-blende×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)=1.75×105 kg×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)=25 g_$

Number of radon $(Ra)$ atoms that can be isolated from $1.75×108 g$ pitch-blende is $6.66×1022 atoms_$ .

Mass of radon is $25 g$ .

Atomic mass of radon is $226 g$ .

Formula

The number of radon atoms is calculated as,

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra$

Substitute the values of mass and atomic mass of radon in the above equation.

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra=25 g×1 mol Ra226.0 g×6.022×1023 atoms 1 mol Ra=6.66×1022 atoms_$

Number of radon $(Ra)$ atoms in $15 mg$ is $3.99×1019 atoms_$ .

Given

Mass of radon atoms is $15 mg$ .

Half life of radon is $1.60×103 years$ .

Atomic mass of radon is $226 g$ .

The conversion of milligram $(mg)$ into gram $(g)$ is done as,

$1 mg=10−3 g$

Hence,

The conversion of $15 mg$ into gram is,

$15 mg=(15×10−3) g=15×10−3 g$

Formula

The number of radon atoms is calculated as,

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra$

Substitute the values of mass and atomic mass of radon in the above equation.

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra=15×10−3 g×1 mol Ra226.0 g×6.022×1023 atoms 1 mol Ra=3.99×1019 atoms_$

Number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ .

Half life of radon is $1.60×103 years$ .

Number of radon $(Ra)$ atoms in $15 mg$ is $3.99×1019 atoms$ .

Total time from $1925$ to $2025$ is,

$2025−1925=100 years$

Formula

Amount of radon left is calculated using the formula,

$AE=A00.5t/t1/2$

Where,

$AE$ is the amount of radon left.
$A0$ is the original amount of radon.
$t1/2$ is the half-life.
$t$ is the total time.

Substitute the values of $t1/2,t$ and $A0$ in the above equation.

$AE=A00.5t/t1/2=3.99×1019 Ra atoms×(0.5)100 years1600 years=3.82×1019 Ra atoms_$

Conclusion

Conclusion

The calculated value of number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ . Number of radon $(Ra)$ atoms in $25 g$ is $6.66×1022 atoms_$

Answer 7

Chapter 20, Problem 116IP

Interpretation Introduction

Interpretation: It is given that, radon is present as $1.0 g$ per $7.0 metric ton$ of a pitch ball. The number of radon atoms that can be isolated from given mass of pitch-blende is to be calculated. The number of radon atoms remain in $2025$ is to be calculated if $15 mg$ radon is manufactured in $1925$ .

Concept introduction: Amount of radon left is calculated using the formula,

$AE=A00.5t/t1/2$

To determine: The number of radon atoms that can be isolated from $1.75×108 g$ pitch-blende; the number of radon atoms remain in $2025$ , if $15 mg$ radon is manufactured in $1925$ .

Answer 8

Expert Solution & Answer

Answer to Problem 116IP

Answer

Number of radon $(Ra)$ atoms that can be isolated from $1.75×108 g$ pitch-blende is $6.66×1022 atoms_$ . Number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Mass of radon $(Ra)$ is $25 g_$ .

Given

Mass of pitch-blende is $1.75×108 g$ .

The conversion of gram $(g)$ into kilogram $(kg)$ is done as,

$1 g=10−3 kg$

Hence,

The conversion of $1.75×108 g$ into kilogram is,

$1.75×108 g=(1.75×108×10−3) kg=1.75×105 kg$

Since, radon is present as $1.0 g$ per $7.0 metric ton$ of a pitch ball, mass of radon is calculated as,

$Mass of Ra=Mass of pitch-blende×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)$

Substitute the value of mass of pitch-blende in the above equation.

$Mass of Ra=Mass of pitch-blende×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)=1.75×105 kg×1.0 g Ra7.0 metric ton×(1 metric ton1000 kg)=25 g_$

Number of radon $(Ra)$ atoms that can be isolated from $1.75×108 g$ pitch-blende is $6.66×1022 atoms_$ .

Mass of radon is $25 g$ .

Atomic mass of radon is $226 g$ .

Formula

The number of radon atoms is calculated as,

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra$

Substitute the values of mass and atomic mass of radon in the above equation.

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra=25 g×1 mol Ra226.0 g×6.022×1023 atoms 1 mol Ra=6.66×1022 atoms_$

Number of radon $(Ra)$ atoms in $15 mg$ is $3.99×1019 atoms_$ .

Given

Mass of radon atoms is $15 mg$ .

Half life of radon is $1.60×103 years$ .

Atomic mass of radon is $226 g$ .

The conversion of milligram $(mg)$ into gram $(g)$ is done as,

$1 mg=10−3 g$

Hence,

The conversion of $15 mg$ into gram is,

$15 mg=(15×10−3) g=15×10−3 g$

Formula

The number of radon atoms is calculated as,

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra$

Substitute the values of mass and atomic mass of radon in the above equation.

$Atoms of Ra= Mass of Ra×1 mol RaAtomic mass of Ra×6.022×1023 atoms 1 mol Ra=15×10−3 g×1 mol Ra226.0 g×6.022×1023 atoms 1 mol Ra=3.99×1019 atoms_$

Number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ .

Half life of radon is $1.60×103 years$ .

Number of radon $(Ra)$ atoms in $15 mg$ is $3.99×1019 atoms$ .

Total time from $1925$ to $2025$ is,

$2025−1925=100 years$

Formula

Amount of radon left is calculated using the formula,

$AE=A00.5t/t1/2$

Where,

$AE$ is the amount of radon left.
$A0$ is the original amount of radon.
$t1/2$ is the half-life.
$t$ is the total time.

Substitute the values of $t1/2,t$ and $A0$ in the above equation.

$AE=A00.5t/t1/2=3.99×1019 Ra atoms×(0.5)100 years1600 years=3.82×1019 Ra atoms_$

Answer 12

Conclusion

Conclusion

The calculated value of number of radon $(Ra)$ atoms left in $2025$ is $3.82×1019 Ra atoms_$ . Number of radon $(Ra)$ atoms in $25 g$ is $6.66×1022 atoms_$

Answer 13

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Answer to Problem 116IP

Explanation of Solution

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