184_notes:examples:week2_moleoelectrons

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184_notes:examples:week2_moleoelectrons [2017/08/24 17:11] – [Example: How much total charge is in one mole of electrons?] tallpaul184_notes:examples:week2_moleoelectrons [2018/05/17 15:16] curdemma
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- +[184_notes:charge|Return to Electric Charge Page] 
-===== Example: How much total charge is in one mole of electrons===== +===== Example: Find the total charge for a mole of electrons ===== 
-How much total charge (in coulombs) is in one mole ($n=6.022*10^{23} \text{ particles/mole}$) of electrons?+How much total charge (in coulombs) is in one mole of electrons?
  
 ===Facts=== ===Facts===
-  * $1 \text{ mol} = 6.022 \cdot 10^{23} \text{ particles}$ +  * The Avogadro constant is $N_A = 6.022 \cdot 10^{23} \text{ mol}^{-1}$. This is easy to look up, which is what we did. 
-  * All electrons have the same charge, which is $e$-1.602\cdot10^{-19} \text{ C}$.+    * Note: When we write the unit as $\text{ mol}^{-1}$, we mean particles per mole. We could also write this unit as $mol^{-1}=\frac{1}{mol}$. 
 +  * All electrons have the same charge, which is $e = -1.602\cdot10^{-19} \text{ C}$.
  
-===Lacking=== +===Goal=== 
-  * Total Charge+  * Find the amount of charge in 1 mole of electrons.
  
-===Approximations & Assumptions=== 
-  * None here, we have all the information we need. 
  
-===Representations=== 
-  * The total charge $Q$ can be written as the number of particles $N$ times the charge of each particle ($e$, for electrons): $Q=N\cdot e$. 
 ====Solution==== ====Solution====
-The total charge $Q$ is given by+The total charge $Q$ can be written as the number of particles $N$ times the charge of each particle ($e$, for electrons): $Q=N\cdot e$. We know $e$, and since we know we are interested in exactly 1 mole, we can find $N$: 
 +\begin{align*} 
 +N &= 1 \text{ mol} \cdot 6.022 \cdot 10^{23} \text{ mol}^{-1} \\ 
 +  &= 6.022 \cdot 10^{23} 
 +\end{align*} 
 +We now have $N$ and $e$. The total charge $Q$ is then given by
 \begin{align*} \begin{align*}
 Q &= N \cdot e \\ Q &= N \cdot e \\
-  &\textmol} \cdot 1.602*10^{19} \text{ C} \\ +  &6.022 \cdot 10^{23} \cdot -1.602 \cdot 10^{-19} \text{ C} \\ 
-  &+  &-9.647 \cdot 10^4 \text{ C}
 \end{align*} \end{align*}
-The number of electrons in one mole is obtained by multiplying the number of moles by Avogodro's number. 
-$$N=(1 mole)*6.022*10^{23}$$ 
-$$N=6.022*10^{23} electrons$$ 
-Therefore, the total charge $Q$ is given by... 
-$$Q=N*e$$ 
-$$Q=(6.022*10^{23})*(1.602*10^{19} C)$$ 
-$$Q=96472.44 C$$ 
  • 184_notes/examples/week2_moleoelectrons.txt
  • Last modified: 2018/05/17 15:16
  • by curdemma