Answer:
![\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}](https://tex.z-dn.net/?f=%5Cmathbf%7Bs%20%3D%5Csqrt%20%5B3%5D%7B%5Cdfrac%7BK_%7Bsp%7D%7D%7B4%7D%7D%7D)
Less than the concentration of Pb2+(aq) in the solution in part ( a )
Explanation:
From the question:
A)
We assume that s to be the solubility of PbI₂.
The equation of the reaction is given as :
PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹
[Pb²⁺] = s
Then [I⁻] = 2s
![K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} = 4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BPb%24%5E%7B2%2B%7D%24%5D%5BI%24%5E%7B-%7D%24%5D%7D%5E%7B2%7D%20%3D%20s%5Ctimes%20%282s%29%5E%7B2%7D%20%3D%20%204s%5E%7B3%7D%5C%5Cs%5E%7B3%7D%20%3D%20%5Cdfrac%7BK_%7Bsp%7D%7D%7B4%7D%5C%5C%5C%5Cs%20%3D%5Cmathbf%7B%20%5Csqrt%20%5B3%5D%7B%5Cdfrac%7BK_%7Bsp%7D%7D%7B4%7D%7D%7D%5C%5C%5C%5C%5Ctext%7BThe%20mathematical%20expressionthat%20can%20be%20used%20to%20determine%20the%20value%20of%20%20%7D%5Cmathbf%7Bs%20%3D%5Csqrt%20%5B3%5D%7B%5Cdfrac%7BK_%7Bsp%7D%7D%7B4%7D%7D%7D)
B)
The Concentration of Pb²⁺ in water is calculated as :
![\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}](https://tex.z-dn.net/?f=%5Cmathbf%7Bs%20%3D%5Csqrt%20%5B3%5D%7B%5Cdfrac%7B7%2A10%5E%7B-9%7D%7D%7B4%7D%7D%7D)
![\mathbf{s} =\sqrt[3]{1.75*10^{-9}}](https://tex.z-dn.net/?f=%5Cmathbf%7Bs%7D%20%3D%5Csqrt%5B3%5D%7B1.75%2A10%5E%7B-9%7D%7D)
The Concentration of Pb²⁺ in 1.0 mol·L⁻¹ NaI
The equilibrium constant:
![K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \ m/L](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5BPb%5E%7B2%2B%7D%7D%5D%5BI%5E-%5D%5E2%20%5C%5C%20%5C%5C%20K_%7Bsp%7D%20%3D%20s%2A%281.0%2A2s%29%5E2%20%3D7%2A1.0%5E%7B-9%7D%20%5C%5C%20%5C%5C%20s%20%3D%207%2A10%5E%7B-9%7D%20%5C%20%5C%20%20m%2FL)
It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.
Answer:
In the calorimeter, water is the <u>exothermic</u>. The salt LiCI, which will dissolve, is the <u>endothermic</u>. The final temperature of the water after the dissolution of LiCI was <u>lower</u> than the initial temperature, meaning the process is <u>exothermic</u>. In the microscopic view of the disspolution of LiCI, water molecules were seen to move <u>slowly</u> as they <u>gained </u>energy.
Explanation:
Exothermic is a process in which heat is released during the process. Endothermic reactions absorbs heat from surrounding during a chemical process. The dissolution of salt into water is an exothermic reaction. During this process heat is release and water molecules are broken down which are surrounded by salt ions.
Bonds of two atoms of equal electronegativity are nonpolar covalent bonds.
Your second sentence is identical to the first sentence; I'll bet the second sentence is "Bonds between two atoms that are unequally electronegative are polar covalent bonds."
The kinetic energy of the products is equal to the energy liberated which is 92.2 keV. But let's convert the unit keV to Joules. keV is kiloelectro volt. The conversion that we need is: 1.602×10⁻¹⁹ <span>joule = 1 eV
Kinetic energy = 92.2 keV*(1,000 eV/1 keV)*(</span>1.602×10⁻¹⁹ joule/1 eV) = 5.76×10²³ Joules
From kinetic energy, we can calculate the velocity of each He atom:
KE = 1/2*mv²
5.76×10²³ Joules = 1/2*(4)(v²)
v = 5.367×10¹¹ m/s
Answer:inner Diameter =9.19cm
Explanation:
Density is calculated as Mass/ Volume
therefore
Volume= Mass/ Density = 1360g/ 0.953g/ml=1,427 ml
1ml = 1cm³
1,427ml = 1,427cm³
Also We know that Volume of a cylinder = πr²h or πr²l
1,427cm³ = 3.142 x r² x 21.5 cm
r² = 1,427cm³/ (3.142 x 21.5cm)
r² =21.124cm²
= r²
r= 4.596cm
Diameter= 2 x radius
=2 x 4.596
=9.19cm
