Answer:
C.
will precipitate out first
the percentage of
remaining = 12.86%
Explanation:
Given that:
A solution contains:
![[Ca^{2+}] = 0.0440 \ M](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%20%3D%200.0440%20%5C%20M)
![[Ag^+] = 0.0940 \ M](https://tex.z-dn.net/?f=%5BAg%5E%2B%5D%20%3D%200.0940%20%5C%20M)
From the list of options , Let find the dissociation of 

where;
Solubility product constant Ksp of
is 
Thus;
![Ksp = [Ag^+]^3[PO_4^{3-}]](https://tex.z-dn.net/?f=Ksp%20%3D%20%5BAg%5E%2B%5D%5E3%5BPO_4%5E%7B3-%7D%5D)
replacing the known values in order to determine the unknown ; we have :
![8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}]](https://tex.z-dn.net/?f=8.89%20%5Ctimes%2010%20%5E%7B-17%7D%20%20%3D%20%280.0940%29%5E3%5BPO_4%5E%7B3-%7D%5D)
![\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}]](https://tex.z-dn.net/?f=%5Cdfrac%7B8.89%20%5Ctimes%2010%20%5E%7B-17%7D%7D%7B%280.0940%29%5E3%7D%20%20%3D%20%5BPO_4%5E%7B3-%7D%5D)
![[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}](https://tex.z-dn.net/?f=%5BPO_4%5E%7B3-%7D%5D%20%3D%5Cdfrac%7B8.89%20%5Ctimes%2010%20%5E%7B-17%7D%7D%7B%280.0940%29%5E3%7D)
![[PO_4^{3-}] =1.07 \times 10^{-13}](https://tex.z-dn.net/?f=%5BPO_4%5E%7B3-%7D%5D%20%3D1.07%20%5Ctimes%2010%5E%7B-13%7D)
The dissociation of 
The solubility product constant of
is 
The dissociation of
is :

Thus;
![Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2](https://tex.z-dn.net/?f=Ksp%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5E3%20%5BPO_4%5E%7B3-%7D%5D%5E2)
![2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2](https://tex.z-dn.net/?f=2.07%20%5Ctimes%2010%5E%7B-33%7D%20%3D%20%280.0440%29%5E3%20%20%5BPO_4%5E%7B3-%7D%5D%5E2)
![\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2](https://tex.z-dn.net/?f=%5Cdfrac%7B2.07%20%5Ctimes%2010%5E%7B-33%7D%20%7D%7B%280.0440%29%5E3%7D%3D%20%20%20%5BPO_4%5E%7B3-%7D%5D%5E2)
![[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}](https://tex.z-dn.net/?f=%5BPO_4%5E%7B3-%7D%5D%5E2%20%3D%20%5Cdfrac%7B2.07%20%5Ctimes%2010%5E%7B-33%7D%20%7D%7B%280.0440%29%5E3%7D)
![[PO_4^{3-}]^2 = 2.43 \times 10^{-29}](https://tex.z-dn.net/?f=%5BPO_4%5E%7B3-%7D%5D%5E2%20%3D%202.43%20%5Ctimes%2010%5E%7B-29%7D)
![[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}](https://tex.z-dn.net/?f=%5BPO_4%5E%7B3-%7D%5D%20%3D%20%5Csqrt%7B2.43%20%5Ctimes%2010%5E%7B-29%7D)
![[PO_4^{3-}] =4.93 \times 10^{-15}](https://tex.z-dn.net/?f=%5BPO_4%5E%7B3-%7D%5D%20%3D4.93%20%5Ctimes%2010%5E%7B-15%7D)
Thus; the phosphate anion needed for precipitation is smaller i.e
in
than in

Therefore:
will precipitate out first
To determine the concentration of
when the second cation starts to precipitate ; we have :
![Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2](https://tex.z-dn.net/?f=Ksp%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5E3%20%5BPO_4%5E%7B3-%7D%5D%5E2)
![2.07 \times 10^{-33} = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2](https://tex.z-dn.net/?f=2.07%20%5Ctimes%2010%5E%7B-33%7D%20%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5E3%20%281.07%20%5Ctimes%2010%5E%7B-13%7D%29%5E2)
![[Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%5E3%20%3D%20%20%5Cdfrac%7B2.07%20%5Ctimes%2010%5E%7B-33%7D%20%7D%7B%281.07%20%5Ctimes%2010%5E%7B-13%7D%29%5E2%7D)
![[Ca^{2+}]^3 =1.808 \times 10^{-7}](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%5E3%20%3D1.808%20%5Ctimes%2010%5E%7B-7%7D)
![[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%20%3D%5Csqrt%5B3%5D%7B1.808%20%5Ctimes%2010%5E%7B-7%7D%7D)
![[Ca^{2+}] =0.00566](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%20%3D0.00566)
This implies that when the second cation starts to precipitate ; the concentration of
in the solution is 0.00566
Therefore;
the percentage of
remaining = concentration remaining/initial concentration × 100%
the percentage of
remaining = 0.00566/0.0440 × 100%
the percentage of
remaining = 0.1286 × 100%
the percentage of
remaining = 12.86%
<h2>The required option d) "specific heat" is correct.</h2>
Explanation:
- To raise the temperature of any substance or material of certain mass to respective temperature it requires some amount of heat.
- Specific heat is the amount of heat necessary to raise the temperature of the substance of 1 gram to 1 Kelvin.
- It is the amount of heat which is required to raise the temperature per unit mass to per unit temperature.
- Thus, the required "option d) specific heat" is correct.
The _____melting point________ is the temperature at which a substance changes from solid to liquid; _______boiling point_________ is the temperature at which a substance changes from a liquid to as gas; _______vapourisation_________ is the process by which atoms of molecules leave a liquid and become a gas.
It's a mixture of H2O + a soluble substance like, salt.
Answer:
4060
Explanation:
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