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:
2.8 L
Explanation:
From the question given above, the following data were obtained:
Number of mole (n) = 0.109 mole
Pressure (P) = 0.98 atm
Temperature (T) = 307 K
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) =?
The volume of the helium gas can be obtained by using the ideal gas equation as follow:
PV = nRT
0.98 × V = 0.109 × 0.0821 × 307
0.98 × V = 2.7473123
Divide both side by 0.98
V = 2.7473123 / 0.98
V = 2.8 L
Thus, the volume of the helium gas is 2.8 L.
Answer:
The first question is 4
The second one is 1
Explanation:
Please mark brainliest! Hope it helped!
1 mol of any gas or mix of gases at STP conditions will have a volume of 22.4 L. Since the problem doesn’t said what are the conditions I will asume that are STP condition and the volume of one mole of the mix will have a volume of 22.4 L.
You may know that density is
D=m/v
In one mole of air I will have 80% of Nitrogen (N2) and 20% oxygen (O2).
So the mass of one mole of air will be
14 x2x0.80+16x2x0.20 = 22.4 g + 6.4 g = 28.8 g
D= 28.8/22.4 = 1.28 g/L
Of course if the temperature is higher the density will be smaller because the volume of one mole will be bigger and viceversa if the temperature decrease. Also if the pressure is different than one atm the volume of a mol will change.
