<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 
The ZnCl₂ solution have a molarity of 1.33 M.
Explanation:
We have the following chemical reaction:
Zn (s) + CuCl₂ (aq) → ZnCl₂ (aq) + Cu (s)
number of moles = mass / molar weight
number of moles of Zn = 25 / 65.4 = 0.38 moles
From the chemical equation we see that 1 moles of Zn produces 1 mole of ZnCl₂, so 0.38 moles of Zn will produce 0.38 moles of ZnCl₂.
molarity = number of moles / volume (L)
molarity of the ZnCl₂ solution = 0.38 / 0.285 = 1.33 M
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molarity
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Na2CO3 Na= 23*2=46 C=12 O=16*3=48. Na2CO3=106. 46/106*0.175=0=403
Answer:
a) find attached image 1
b) find attached image 2
Explanation :
The more stable radical is formed by a reaction with smaller bond dissociation energy.
since the bond dissociation for cleavage of the bond to form primary free radical is higher, more energy must be added to form it. This makes primary free radical higher in energy and therefore less stable than secondary free radical.
Derivatization means adding fluorescent labels or combining the analyte with chiral reagents or other chemicals to increase detectability.
Some analytes must be derivatized to increase their column retention or detectability.
Retention time can be referred to as the amount of time a solute spends in the stationary and mobile phases of a column.
Detectability is the ability of an analyte to get detected in the mobile phase of chromatography.
The refractometer, fluorescence detector, and UV detector are the three most popular liquid chromatography detectors. These detectors increase the detectability.
For derivatization, the fluorescence detector are used.
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