Answer:
1.43 * 10^-2
Explanation:
Firstly, we need to get the ionization equation for lead I chloride. We can then get the Initial, change and equilibrium table of the lead and chloride ions that can help us calculate the molar solubility. This is shown as follows:
PbCl2(s) --> Pb2+(aq) + 2 Cl-(aq)
This shows that one mole of lead ii chloride will yield 1 mole of the lead ion and 2 moles of the chloride ion.
The ICE table is shown below:
Pb2+ 2Cl-
Initial 0.00 0.00
Change +s +2s
Equilibrium s 2s
Ksp = [Pb2+][Cl-]^2
Ksp = [s][2s]^2
Ksp = 4s^3
s^3 = Ksp/4
s = cube.root [Ksp/4]
s = cube.root[1.17 * 10^-5/4]
s = 1.43 * 10^-2
Their molecules move at different speeds
You have to use the equation PV=nRT.
P=pressure (in this case 1.89x10^3 kPa which equals 18.35677 atm)
1V=volume (in this case 685L)
n=moles (in this case the unknown)
R=gas constant (0.08206 (L atm)/(mol K))
T=temperature (in this case 621 K)
with the given information you can rewrite the ideal gas law equation as n=PV/RT.
n=(18.35677atm x 685L)/(0.08206atmL/molK x 621K)
n=246.8 moles
Covalent bonding is the sharing of electrons while ionic bonding is the transferring of electrons.
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