<span>CO is the limiting reactant
( 25.0 x 3 = 75 moles of CO are required)
Moles Fe = 30.0 x 2 / 3 = 20.0
mass Fe = 20.0 x 55.847 g/mol=1117 g </span><span>
I'm just saying</span>
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: What is the mass, in grams, of 135 mL of ethanol? d=0.789 g/mL - the ethanol density. V=135 mL - the volume of ethanol. m=0.789g/mL*135mL=106.515g ~ 106.5g- the mass of ethanol.
Explanation:
Hope this helps :)
Answer:
Li2O+H2O---->(2)LiOH
Explanation:
you have to balance the equation and not all the blanks have to be filled all the time but if it makes it easier for you in the first 2 question marks you can put a 1 which isnt necessary but if ur a visual person it will help.
hope this helps im litterally learning the same thing as u lol
Answer: what would swimming in mathane lake on titan feel like.
Explanation: