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:
the other stars are much farther away from Earth than our sun
Covalent compounds are generally not very hard because they are formed by two or more nonmetallic atoms.
<h3>COVALENT COMPOUNDS:</h3>
Covalent compounds are compounds whose constituent elements are joined together by covalent bonds.
Covalent bonding occurs when two or more nonmetallic atoms of an element share valence electrons. This means that covalent compounds will not be physically hard since they constitute non-metals.
Examples of covalent compounds are:
- H2 - hydrogen
- H2O - water
- HCl - hydrogen chloride
- CH4 - methane
Learn more about covalent compounds at: brainly.com/question/21505413
The number of H atoms in 3(NH₄)₂CrO₄ = 24
<h3>Further explanation </h3>
The empirical formula is the smallest comparison of atoms of compound forming elements.
A molecular formula is a formula that shows the number of atomic elements that make up a compound.
(empirical formula) n = molecular formula
Subscripts in the chemical formula indicate the number of atoms
The compound of 3(NH₄)₂CrO₄ ( 3 molecules of (NH₄)₂CrO₄ ) :
Number of H :

The fraction of Earth's radius (6371 km) relative to the thickness of the oceanic (7.5 km) and continental crust (35 km) is 0.12 and 0.55, respectively.
What we know:
- The average radius of Earth (E) = 6371 km
- The average thickness of oceanic crust (O) = 7.5 km
- The average thickness of continental crust (C) = 35 km
We need to convert all the above units from kilometers to miles:

Now, we can calculate the fraction of Earth's radius relative to each type of crust, with the given equation:

- <u>For the oceanic crust (O)</u>:

- <u>For the continental crust (C)</u>:

Therefore, the fraction of Earth's radius relative to the oceanic and continental crust is 0.12 and 0.55, respectively.
You can see another example of calculation of fractions of Earth's radius here: brainly.com/question/4675868?referrer=searchResults
I hope it helps you!