Answer: The final concentration of aluminum cation is 0.335 M.
Explanation:
Given: 
= 47.8 mL (1 mL = 0.001 L) = 0.0478 L
= 0.321 M, 
= 21.8 mL = 0.0218 L, 
= 0.366 M
As concentration of a substance is the moles of solute divided by volume of solution.
Hence, concentration of aluminum cation is calculated as follows.
![[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}](https://tex.z-dn.net/?f=%5BAl%5E%7B3%2B%7D%5D%20%3D%20%5Cfrac%7BM_%7B1%7DV_%7B1%7D%20%2B%20M_%7B2%7DV_%7B2%7D%7D%7BV_%7B1%7D%20%2B%20V_%7B2%7D%7D)
Substitute the values into above formula as follows.
![[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}\\= \frac{0.321 M \times 0.0478 L + 0.366 M \times 0.0218 L}{0.0478 L + 0.0218 L}\\= \frac{0.0153438 + 0.0079788}{0.0696}\\= 0.335 M](https://tex.z-dn.net/?f=%5BAl%5E%7B3%2B%7D%5D%20%3D%20%5Cfrac%7BM_%7B1%7DV_%7B1%7D%20%2B%20M_%7B2%7DV_%7B2%7D%7D%7BV_%7B1%7D%20%2B%20V_%7B2%7D%7D%5C%5C%3D%20%5Cfrac%7B0.321%20M%20%5Ctimes%200.0478%20L%20%2B%200.366%20M%20%5Ctimes%200.0218%20L%7D%7B0.0478%20L%20%2B%200.0218%20L%7D%5C%5C%3D%20%5Cfrac%7B0.0153438%20%2B%200.0079788%7D%7B0.0696%7D%5C%5C%3D%200.335%20M)
Thus, we can conclude that the final concentration of aluminum cation is 0.335 M.
Elements cannot be broken down in this fashion.
Compounds, homogeneous mixtures, and heterogeneous mixtures can be broken down or separated with chemical processes, or in some cases, just by manual separation.
I know the answer to your question but if you want me to answer this question first, I need your help with my question "How do I solve number 12? Quick please I need help ASAP!!!"
To find the empirical formula you would first need to find the moles of each element:
58.8g/ 12.0g = 4.9 mol C
9.9g/ 1.0g = 9.9 mol H
31.4g/ 16.0g = 1.96 O
Then you divide by the smallest number of moles of each:
4.9/1.96 = 2.5
9.9/1.96 = 6
1.96/1.96 = 1
Since there is 2.5, you find the least number that makes each moles a whole number which is 2.
So the empirical formula is C5H12O2.
