QQuestionChemistry
QuestionChemistry
If 0.035pC of charge is transferred via the movement of Al^3 + ions, how many of these ions must have been transferred in total?
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Answer
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Step 1:
When it loses three electrons to become Al$^{3+}$, it has 10 electrons.
Aluminum has an atomic number of 13, so it has 13 protons and 13 electrons when neutral.
Step 2:: Calculate the charge of one Al$^{3 +}$ ion:
q_{\text{Al}^{3 +}} = -e \times |q_{\text{electron}}| \times \text{number of electrons} Substituting the values, we get: q_{\text{Al}^{3 +}} = - 1.602 \times 10^{- 19} \, \text{C} \times 1.602 \times 10^{- 19} \, \text{C} \times 3 = 4.806 \times 10^{- 18} \, \text{C}
Step 3:
Now, we can find the number of Al$^{3+}$ ions that have been transferred by dividing the total charge transferred by the charge of one Al$^{3+}$ ion.
Step 4:: Calculate the number of Al$^{3 +}$ ions:
n_{\text{Al}^{3 +}} = \frac{q_{\text{transferred}}}{q_{\text{Al}^{3 +}}} Substituting the values, we get:
Step 5:
Since we cannot have a fraction of an ion, we round up to the nearest whole number to ensure that the total charge transferred is accounted for.
Step 6:: Round up to the nearest whole number:
n_{\text{Al}^{3 +}} = 7,281,000,000,000,00
Final Answer
n_{\text{Al}^{3 +}} = 7,281,000,000,000,00
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