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:
T₁ = 135.41 K
Explanation:
Given data:
Initial pressure = 1.12 atm
Finial temperature = 36.5 °C (36.5 +273 = 309.5 K)
Initial temperature = ?
Final pressure = 2.56 atm
Formula:
P₁/T₁ = P₂/T₂
P₁ = Initial pressure
T₁ = Initial temperature
P₂ = Final pressure
T₂ = Final temperature
Solution:
P₁/T₁ = P₂/T₂
T₁ = P₁T₂ /P₂
T₁ = 1.12 atm × 309.5 K / 2.56 atm
T₁ = 346.64 atm . K / 2.56 atm
T₁ = 135.41 K
Answer:
3
Explanation:
The first election shell can only hold 2 electrons, but the next one can hold up to 8
Viscosity is the rate at which liquids flow.
