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.
Answer:
Can you please tell us what the following are?
The mass of the atom is equal to the sum of the number of protons and the number of neutrons. In a neutral atom, the number of protons is equal to the number of electrons. The atomic number meanwhile of an atom is equal to the number of protons of the atom.
Two subatomic particles that are located in the nucleus of an atom are Protons and Neutrons.
5.512 litres is the volume of 15.2 grams of sulphur dioxide gas at STP.
Explanation:
Data given:
mass of sulphur dioxide = 15.2 grams
conditions is at STP whech means volume = 22.4 litres
atomic mass of sulphur dioxide = 64.06 grams/mole
Number of moles is calculated as:
number of moles = 
Putting the values in the equation:
number of moles = 
= 0.23 moles
Assuming that sulphur dioxide behaves as an ideal gas, we can calculate the volume as:
When 1 mole of sulphur dioxide occupies 22.4 litres at STP
Then 0.23 moles of sulphur dioxide occupies 22.4 x 0.23
= 5.152 litres is the volume.