you start with the 5.00g PbCl2 then use a conversion factor for the molar mass of pbcl2 (1mol pbcl2/278.1gpbcl2) now from mol pbcl2 use another conversion factor to get g cl2 (70.90gCl2/1 mol PbCl2) do the math and you wind up with 1.27 g Cl2
PbCl2 --> Pb2+ + Cl-
80g approx, if the density = 1
64 moles of chlorine atoms would be 64 x 35.5 grams, i.e. 2272 grams. If you meant the substance chlorine, which consists of Cl2 molecules, it would be 4544 grams.
A gravimetric factor converts grams of a compound into grams of a single element. For example, we'll find the gravimetric factor of Cl in AgCl. Use the atomic mass of Ag(107.868) and the atomic mass of Cl(35.453) and add them together to get 143.3. Then divide 35.453 by 143.3 to get .2474. .2474 is the gravimetric factor of Cl in AgCl.
35.488 cl.
PbCl2 --> Pb2+ + Cl-
50 cL is 500 mL.
Since Lead (II) Chloride has the formula PbCl2, the equilibrium equation for its dissolution is: PbCl2 <=> Pb+2+2Cl- so the equilibrium-constant expression is Ksp= [Pb+2][Cl-]
Pb = lead Cl = chlorineThis compound has 1 lead atom and 2 chlorine atoms.
80g approx, if the density = 1
how do you fix brakes on a 2002 Cl 500 Benz
500 cl 1 centiliter = 10 milliliters 1 milliliter = 0.1 centiliter
1.4 g*convert g to mol*Pb=207.2 amu (atomic mass)Cl=35.453 amu (atomic mass)since it's PbCl2, 207.2 + 35.453 + 35.453=278.1061.4/278.106=.005 mol*use Avogadro's number to find formula units*.005 x 6.02E10^23 --> 3.03E10^22 formula units PbCl2
15cL equals 0.15L* There are 100cL per liter
Silver chloride - AgClAg (107.89 grams) + Cl (35.45 grams) = 143.34 grams
Balanced equation. 2K + Cl2 >> 2KCl 39 grams K (1mol K/39.10g )(1mol Cl/2mol K )(35.45g/1 mol Cl ) = 17.7 grams
Balanced equation. 2Na + Cl2 >> 2NaCl 46 grams sodium = 2 mol 23 grams Chlorine = 0.65 mol ( I think Chlorine is limiting ) 0.65 mol Cl (2mol Na/1mol Cl ) = 1.3 mol ( you do not have that; Cl limits ) 0.65 mol Cl (2mol NaCl/1mol Cl2 )(58.44g/1mol NaCl ) = 75.9 grams