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:
(1) Rate = k[NO2]^2[CO]
(2) The rate of the reaction is 1.195 Ms^-1
Explanation:
(1) Equation of reaction
NO2 + CO = NO + CO2
The reaction is second-order in NO2 and zero-order in CO
Therefore, rate = k[NO2]^2[CO]^0
(2) Rate = k[NO2]^2[CO]^0
k = 0.790 M^-1s^-1
[NO2] = 1.23 M
[CO] = 0.221 M
Rate = 0.79×1.23^2×0.221^0 = 1.195 Ms^-1
"The reaction is exothermic and ΔH is negative" can be understood about the reaction and the enthalpy change (ΔH) during the reaction.
<u>Option: D</u>
<u>Explanation:</u>
When the reaction is positive, the process becomes endothermic, i.e. heat appears to be consumed by the system because the reaction products are more enthalpic than the reactants. When the reaction is negative, on the other hand, the process is exothermic, which is the total decrease in enthalpy is caused by heat production. Here the initial temperature is 21.0 C but increase in final temperature to 38.8 C, because if some processes require heat, others must give off heat when they take place.
Answer:
The volume increases,
Final volume is 2.1L
Explanation:
Based on Boyle's law, the absolute temperature of a gas is directly proportional to the volume of the gas.
As the balloon change increases its temperature, the volume increases
The Boyle's equation is:
V1T2 = V2T1
<em>Where V is volume and T is absolute temperature of 1, initial state and 2, final state of the gas.</em>
<em />
Replacing:
V1 = 2.0L
T2 = 35°C + 273.15 = 308.15K
V2 = ?
T1 = 25°C + 273.15 = 298.15K
2.0L*308.15K = V2*298.15K
2.1L = V2
<h3>Final volume is 2.1L</h3>
<em />
Saturated hydrocarbons consists of C-C single bond whereas Unsaturated hydrocarbons consists C-C double/triple bond.
