<u>Answer:</u> The solubility product of magnesium phosphate tribasic is 
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:

Given mass of magnesium phosphate = 1.24 g
Molar mass of magnesium phosphate = 262.85 g/mol
Volume of solution = 1 L
Putting values in above equation, we get:

The equation for the ionization of the magnesium phosphate is given as:

Expression for the solubility product of
will be:
![K_{sp}=[Mg^{2+}]^3[PO_4^{3-}]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BMg%5E%7B2%2B%7D%5D%5E3%5BPO_4%5E%7B3-%7D%5D%5E2)
We are given:
![[Mg^{2+}]=(3\times 4.72\times 10^{-3})=1.416\times 10^{-2}M](https://tex.z-dn.net/?f=%5BMg%5E%7B2%2B%7D%5D%3D%283%5Ctimes%204.72%5Ctimes%2010%5E%7B-3%7D%29%3D1.416%5Ctimes%2010%5E%7B-2%7DM)
![[PO_4^{3-}]=(2\times 4.72\times 10^{-3})=9.44\times 10^{-3}M](https://tex.z-dn.net/?f=%5BPO_4%5E%7B3-%7D%5D%3D%282%5Ctimes%204.72%5Ctimes%2010%5E%7B-3%7D%29%3D9.44%5Ctimes%2010%5E%7B-3%7DM)
Putting values in above expression, we get:

Hence, the solubility product of magnesium phosphate tribasic is 
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Answer:
GRAPHENE is a single layer of graphite
Explanation:
each carbon atom bonded to 3 other carbon atoms forming a hexagonal lattice
Answer:
The volume of the gas is determined, which will allow you to calculate the temperature.
Explanation:
According to Charles law; the volume of a given mass of an ideal gas is directly proportional to its temperature at constant pressure.
This implies that, when the volume of an ideal gas is measured at constant pressure, the temperature of the ideal gas can be calculated from it according to Charles law.
Hence in the Ideal Gas Law lab, the temperature of an ideal gas is measured by determining the volume of the ideal gas.
Electronic Configuration of Cu⁴⁺ is written using these two steps;
Step I:
Write the Electronic Configuration of Copper (Cu).
As Atomic Number of Cu = 29
So,
1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰, 4s¹
Rearranging it in order of increasing orbital energy,
1s², 2s², 2p⁶, 3s², 3p⁶, 4s¹, 3d¹⁰
Or,
Cu = [Ar] 4s¹, 3d¹⁰
Step II:
Now, the Cu⁴⁺ ion has lost 4 electrons, So one electron from 4s and 3 electrons from 3d orbital are removed, The resulting electronic configuration becomes,
Cu⁴⁺ = [Ar] 3d⁷