The type of molecule that is shown is D. Alcohol
The correct answer is A.
Hunger, chemical imbalance, and blood levels are all internal issues. Cat dander on the other hand is not something the body creates or and is something in the environment.
Thus, your answer is A.
A food chain contains a producer and 3 consumers. Primary Consumers (East producers) the arrows in a food chain show the flow of energy, from the sun or hydrothermal vent your a top predator.
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%