A. 0.0105 kg
b. 1570 m
c. 3.5x10^-6
d. 3500000
e. 1000 mL
f. 0.000358 m3
g. 548.6 cm3
Answer:
Element X has valency 2 while Y Valency 2, element Y has high tendency of gaining electrons for its stability thus the compound formed will be B
elements X is positively charged while Y will be negatively charged
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%
58.7 %
Please correct me if I’m wrong. :)
This is a neutron induced fission, therefore a neutron will be added to the U²³⁵ to cause the reaction, and thus it will be added to the left side. There will be unknown number of neutrons produced and thus we put this on the right hand side.
n₁ + U²³⁵ = Te¹³⁷ + Zr ⁹⁷ + xn1 ( n1 to mean a neutron of mass 1)
To balance the masses on both sides of the equation;
1 + 235 = 137 +97+ x
x = 2
the end reaction will be
n₁ +U²³⁵ = Te¹³⁷ + Zr⁹⁷ + 2 n₁