Chapter 21, Problem 21.12P

Interpretation Introduction

Interpretation:

The mean annual flux of each metal has to be calculated.

Explanation of Solution

The metals that are present in the snow was identified by atomic fluorescence.  The mean annual flux for the metals can be calculated as shown below,

For $\text{Pb}$:

$\begin{array}{l}\left(104\pm 17\frac{\text{pg}\text{Pb}}{\text{g}\text{snow}}\right)\left(11.5\frac{\text{g}\text{snow}}{{\text{cm}}^{\text{2}}}\right)=1196\pm 196\frac{\text{pg}\text{Pb}}{{\text{cm}}^2}\\ \left(1196\pm 196\frac{\text{pg}\text{Pb}}{{\text{cm}}^2}\right)\left(\frac{1\text{ng}}{1000\text{pg}}\right)=1.2\pm 0.2\frac{\text{ng}\text{Pb}}{{\text{cm}}^2}\end{array}$

The mean annual flux of $\text{Pb}$ is $1.2\pm 0.2\text{ng}{\text{Pb/cm}}^{\text{2}}$.

For $\text{Tl}$:

$\begin{array}{l}\left(0.43\pm 0.087\frac{\text{pg}\text{Tl}}{\text{g}\text{snow}}\right)\left(11.5\frac{\text{g}\text{snow}}{{\text{cm}}^{\text{2}}}\right)=4.945\pm 1.0005\frac{\text{pg}\text{Tl}}{{\text{cm}}^2}\\ \left(4.945\pm 1.0005\frac{\text{pg}\text{Tl}}{{\text{cm}}^2}\right)\left(\frac{1\text{ng}}{1000\text{pg}}\right)=0.005\pm 0.001\frac{\text{ng}\text{Tl}}{{\text{cm}}^2}\end{array}$

The mean annual flux of $\text{Tl}$ is $0.005\pm 0.001\text{ng}{\text{Tl/cm}}^{\text{2}}$.

For $\text{Cd}$:

$\begin{array}{l}\left(3.5\pm 0.087\frac{\text{pg}\text{Cd}}{\text{g}\text{snow}}\right)\left(11.5\frac{\text{g}\text{snow}}{{\text{cm}}^{\text{2}}}\right)=40.25\pm 1.0005\frac{\text{pg}\text{Cd}}{{\text{cm}}^2}\\ \left(40.25\pm 1.0005\frac{\text{pg}\text{Cd}}{{\text{cm}}^2}\right)\left(\frac{1\text{ng}}{1000\text{pg}}\right)=0.04\pm 0.01\frac{\text{ng}\text{Cd}}{{\text{cm}}^2}\end{array}$

The mean annual flux of $\text{Cd}$ is $0.04\pm 0.01\text{ng}{\text{Cd/cm}}^{\text{2}}$.

For $\text{Zn}$:

$\begin{array}{l}\left(174\pm 26\frac{\text{pg}\text{Zn}}{\text{g}\text{snow}}\right)\left(11.5\frac{\text{g}\text{snow}}{{\text{cm}}^{\text{2}}}\right)=2001\pm 299\frac{\text{pg}\text{Zn}}{{\text{cm}}^2}\\ \left(2001\pm 299\frac{\text{pg}\text{Zn}}{{\text{cm}}^2}\right)\left(\frac{1\text{ng}}{1000\text{pg}}\right)=2.0\pm 0.3\frac{\text{ng}\text{Zn}}{{\text{cm}}^2}\end{array}$

The mean annual flux of $\text{Zn}$ is $2.0\pm 0.3\text{ng}{\text{Zn/cm}}^{\text{2}}$.

For $\text{Al}$:

$\begin{array}{l}\left(6100\pm 1700\frac{\text{pg}\text{Al}}{\text{g}\text{snow}}\right)\left(11.5\frac{\text{g}\text{snow}}{{\text{cm}}^{\text{2}}}\right)=70150\pm 19550\frac{\text{pg}\text{Al}}{{\text{cm}}^2}\\ \left(70150\pm 19550\frac{\text{pg}\text{Al}}{{\text{cm}}^2}\right)\left(\frac{1\text{ng}}{1000\text{pg}}\right)=7\left(\pm 2\right)\times {10}^1\frac{\text{ng}\text{Al}}{{\text{cm}}^2}\end{array}$

The mean annual flux of $\text{Al}$ is $7\left(\pm 2\right)\times {10}^1{\text{ng/cm}}^{\text{2}}$.