Answer:
![[Pb^{2+}]=3.9 \times 10^{-2}M](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%3D3.9%20%5Ctimes%2010%5E%7B-2%7DM)
this is the concentration required to initiate precipitation
Explanation:
⇄ 
Precipitation starts when ionic product is greater than solubility product.
Ip>Ksp
Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.
This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.
![Ip=[Pb^{2}][2Cl^-]^2=Ksp](https://tex.z-dn.net/?f=Ip%3D%5BPb%5E%7B2%7D%5D%5B2Cl%5E-%5D%5E2%3DKsp)
lets solubility=S
![[Pb^{2+}] = S](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%20%3D%20S)
![[Cl^-]=2S](https://tex.z-dn.net/?f=%5BCl%5E-%5D%3D2S)
![Ksp=[Pb^{2+}]\times [Cl^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BPb%5E%7B2%2B%7D%5D%5Ctimes%20%5BCl%5E-%5D%5E2)
![S=\sqrt[3]{\frac{Ksp}{4} }](https://tex.z-dn.net/?f=S%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BKsp%7D%7B4%7D%20%7D)
this is the concentration required to initiate precipitation
Answer:
I think it will option B it will retain enough heat
I believe the answer is 50.5 molecules
Answer:
10437calories
Explanation:
The following data were obtained from the question given:
M = 347.9g
C = 4.2J/g°C
T1 = 25°C
T2 = 55°C
ΔT = 55 — 25 = 30°C
Q =?
Q = MCΔT
Q = 347.9 x 4.2 x 30
Q = 43835.4J
Converting this to calories, we obtained the following:
4.2J = 1 calorie
43835.4J = 43835.4/ 4.2 = 10437calories
Answer:
True
Explanation:
The majority of the reactions happened with a flow of heat. When there's no heat, the reaction is adiabatic.
For no adiabatic reactions, the heat can be released (evolution) by the system, so the reaction will be exothermic, or absorbed by the system (absorption), then the reaction is called endothermic.
