QuestionFind the abundance of ${ }^{10} \mathrm{~B}$ and ${ }^{11} \mathrm{~B}$ given their masses and average atomic mass of boron, $10.81 \mathrm{~u}$ .

Studdy Solution

STEP 1

Assumptions1. There are two naturally occurring isotopes of boron ${ }^{10} \mathrm{~B}$ with a mass of $10.0129 \mathrm{~u}$ and ${ }^{11} \mathrm{~B}$ with a mass of $11.0093 \mathrm{~u}$ . . The average atomic mass of boron is $10.81 \mathrm{~u}$ .
3. The sum of the abundances of the two isotopes is100%.

STEP 2

Let's denote the abundance of ${ }^{10} \mathrm{~B}$ as $x$ and the abundance of ${ }^{11} \mathrm{~B}$ as $y$ . Since the sum of the abundances is100%, we can write the following equation $x + y =1$

STEP 3

The average atomic mass is calculated as the sum of the mass of each isotope times its relative abundance. We can write this as $Average\, atomic\, mass = x \cdot Mass_{10} + y \cdot Mass_{11}$

STEP 4

Substitute the given values into the equation from3 $10.81 = x \cdot10.0129 + y \cdot11.0093$

STEP 5

We now have a system of two equations\begin{align} x + y &=1 \\ 10.81 &= x \cdot10.0129 + y \cdot11.0093\end{align}

STEP 6

We can solve this system of equations by substitution. First, solve the first equation for $y$ $y =1 - x$

STEP 7

Substitute $y$ into the second equation $10.81 = x \cdot10.0129 + (1 - x) \cdot11.0093$

STEP 8

implify the equation $10.81 =10.012x +11.0093 -11.0093x$

STEP 9

Rearrange the equation to solve for $x$ $x = \frac{.81 -11.0093}{.0129 -11.0093}$

STEP 10

Calculate the value of $x$ $x = \frac{10.81 -.0093}{10.0129 -.0093} \approx0.199$

STEP 11

Substitute $x$ into the first equation to find $y$ $y = - x = -0.199 =0.801$ The abundance of ${ }^{10} \mathrm{~B}$ is approximately19.9% and the abundance of ${ }^{11} \mathrm{~B}$ is approximately80.%.

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QuestionFind the abundance of ${ }^{10} \mathrm{~B}$ and ${ }^{11} \mathrm{~B}$ given their masses and average atomic mass of boron, $10.81 \mathrm{~u}$ .

Studdy Solution

STEP 1

STEP 2

Let's denote the abundance of ${ }^{10} \mathrm{~B}$ as $x$ and the abundance of ${ }^{11} \mathrm{~B}$ as $y$ . Since the sum of the abundances is100%, we can write the following equation $x + y =1$

STEP 3

The average atomic mass is calculated as the sum of the mass of each isotope times its relative abundance. We can write this as $Average\, atomic\, mass = x \cdot Mass_{10} + y \cdot Mass_{11}$

STEP 4

Substitute the given values into the equation from3 $10.81 = x \cdot10.0129 + y \cdot11.0093$

STEP 5

We now have a system of two equations\begin{align} x + y &=1 \\ 10.81 &= x \cdot10.0129 + y \cdot11.0093\end{align}

STEP 6

We can solve this system of equations by substitution. First, solve the first equation for $y$ $y =1 - x$

STEP 7

Substitute $y$ into the second equation $10.81 = x \cdot10.0129 + (1 - x) \cdot11.0093$

STEP 8

implify the equation $10.81 =10.012x +11.0093 -11.0093x$

STEP 9

Rearrange the equation to solve for $x$ $x = \frac{.81 -11.0093}{.0129 -11.0093}$

STEP 10

Calculate the value of $x$ $x = \frac{10.81 -.0093}{10.0129 -.0093} \approx0.199$

STEP 11

Substitute $x$ into the first equation to find $y$ $y = - x = -0.199 =0.801$ The abundance of ${ }^{10} \mathrm{~B}$ is approximately19.9% and the abundance of ${ }^{11} \mathrm{~B}$ is approximately80.%.

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QuestionFind the abundance of 10 B{ }^{10} \mathrm{~B}10 B and 11 B{ }^{11} \mathrm{~B}11 B given their masses and average atomic mass of boron, 10.81 u10.81 \mathrm{~u}10.81 u.

Studdy Solution

STEP 1

STEP 2

Let's denote the abundance of 10 B{ }^{10} \mathrm{~B}10 B as xxx and the abundance of 11 B{ }^{11} \mathrm{~B}11 B as yyy. Since the sum of the abundances is100%, we can write the following equationx+y=1x + y =1x+y=1

STEP 3

The average atomic mass is calculated as the sum of the mass of each isotope times its relative abundance. We can write this asAverage atomic mass=x⋅Mass10+y⋅Mass11Average\, atomic\, mass = x \cdot Mass_{10} + y \cdot Mass_{11}Averageatomicmass=x⋅Mass10​+y⋅Mass11​

STEP 4

Substitute the given values into the equation from310.81=x⋅10.0129+y⋅11.009310.81 = x \cdot10.0129 + y \cdot11.009310.81=x⋅10.0129+y⋅11.0093

STEP 5

We now have a system of two equations\begin{align*} x + y &=1 \\ 10.81 &= x \cdot10.0129 + y \cdot11.0093\end{align*}

STEP 6

We can solve this system of equations by substitution. First, solve the first equation for yyyy=1−xy =1 - xy=1−x

STEP 7

Substitute yyy into the second equation10.81=x⋅10.0129+(1−x)⋅11.009310.81 = x \cdot10.0129 + (1 - x) \cdot11.009310.81=x⋅10.0129+(1−x)⋅11.0093

STEP 8

implify the equation10.81=10.012x+11.0093−11.0093x10.81 =10.012x +11.0093 -11.0093x10.81=10.012x+11.0093−11.0093x

STEP 9

Rearrange the equation to solve for xxxx=.81−11.0093.0129−11.0093x = \frac{.81 -11.0093}{.0129 -11.0093}x=.0129−11.0093.81−11.0093​

STEP 10

Calculate the value of xxxx=10.81−.009310.0129−.0093≈0.199x = \frac{10.81 -.0093}{10.0129 -.0093} \approx0.199x=10.0129−.009310.81−.0093​≈0.199

STEP 11

Substitute xxx into the first equation to find yyyy=−x=−0.199=0.801y = - x = -0.199 =0.801y=−x=−0.199=0.801The abundance of 10 B{ }^{10} \mathrm{~B}10 B is approximately19.9% and the abundance of 11 B{ }^{11} \mathrm{~B}11 B is approximately80.%.

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QuestionFind the abundance of ${ }^{10} \mathrm{~B}$ and ${ }^{11} \mathrm{~B}$ given their masses and average atomic mass of boron, $10.81 \mathrm{~u}$ .

Let's denote the abundance of ${ }^{10} \mathrm{~B}$ as $x$ and the abundance of ${ }^{11} \mathrm{~B}$ as $y$ . Since the sum of the abundances is100%, we can write the following equation $x + y =1$

The average atomic mass is calculated as the sum of the mass of each isotope times its relative abundance. We can write this as $Average\, atomic\, mass = x \cdot Mass_{10} + y \cdot Mass_{11}$

Substitute the given values into the equation from3 $10.81 = x \cdot10.0129 + y \cdot11.0093$

We now have a system of two equations\begin{align} x + y &=1 \\ 10.81 &= x \cdot10.0129 + y \cdot11.0093\end{align}

We can solve this system of equations by substitution. First, solve the first equation for $y$ $y =1 - x$

Substitute $y$ into the second equation $10.81 = x \cdot10.0129 + (1 - x) \cdot11.0093$

implify the equation $10.81 =10.012x +11.0093 -11.0093x$

Rearrange the equation to solve for $x$ $x = \frac{.81 -11.0093}{.0129 -11.0093}$

Calculate the value of $x$ $x = \frac{10.81 -.0093}{10.0129 -.0093} \approx0.199$

Substitute $x$ into the first equation to find $y$ $y = - x = -0.199 =0.801$ The abundance of ${ }^{10} \mathrm{~B}$ is approximately19.9% and the abundance of ${ }^{11} \mathrm{~B}$ is approximately80.%.