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:
a substance that dissolves another to form a solution: Water is a solvent for sugar.
Answer:
gamma ray
Explanation:
Gamma emission results in the release of energy without any apparent change in mass or nuclear charge.
Properties of Gamma radiations:
Gamma radiations are high energy radiations having no mass.
These radiations are travel at the speed of light.
Gamma radiations can penetrate into the many materials.
These radiations are also used to treat the cancer.
Lead is used for the protection against gamma radiations because of its high molecular density.
The lead apron are used by the person when treated with gamma radiations.
Lead shields are also used in the wall, windows and doors of the room where gamma radiations are treated, in-order to protect the surroundings.
MM Zn(NO₃)₂ = 189.36 g/mol
mass 1 mol Zn(NO₃)₂ = 189.36 g
mass hydrate = 100 / 63.67 x 189.36 = 297.409 g
mass 1 mol hydrate = 297.409 g
MM hydrate = 297.409 g/mol
MM hydrate = MM Zn(NO₃)₂ + MM xH₂O
297.409 = 189.36 + x(18)
x = 6
