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%
If a substance is not chemically pure, it is either a heterogeneous mixture or a homogeneous mixture. If its composition is uniform throughout, it is a homogeneous mixture.
Answer:
Maintain constant velocity
Answer:
Here's what I get.
Explanation:
At the end of the reaction you will have a solution of the alcohol in THF.
The microdistillation procedure will vary, depending on the specific apparatus you are using, but here is a typical procedure.
- Transfer the solution to a conical vial.
- Add a boiling stone.
- Attach a Hickman head (shown below) and condenser.
- Place the assembly in in the appropriate hole of an aluminium block on top of a hotplate stirrer.
- Begin stirring and heating at a low level so the THF (bp 63 °C) can distill slowly.
- Use a Pasteur pipet to withdraw the THF as needed.
- When all the THF has been removed, raise the temperature of the Al block and distill the alcohol (bp 143 °C).
False; an ionic bond generally occurs between a nonmetal and a metal; for example, NaCl.