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.
In human beings fertilization occurs in fallopian tube
Hey there!
No of hybrid orbitals , H = ( V +S - C + A ) / 2
Where H = no . of hybrid orbitals
V = Valence of the central atom = 5
S = No . of single valency atoms = 4
C = No . of cations = 1
A = No . of anions = 0
For PCl4 +
Plug the values we get H = ( 5+4-1+0) / 2
H = 4 ---> sp3 hybridization
sp3 hybrid orbitals are used by phosphorous in the PCl4+ cations
Answer C
Hope that helps!
Answer : The molar mass of unknown substance is, 39.7 g/mol
Explanation : Given,
Mass of unknown substance = 9.56 g
Volume of solution = 100.0 mL
Molarity = 2.41 M
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :
Now put all the given values in this formula, we get:
Therefore, the molar mass of unknown substance is, 39.7 g/mol
