Talk:CH 101/Keys 2

From WolfWikis

Missing Solution for Exercise 4, Problem 2

The key seems to be missing the solution steps for this problem. The way I did it:

   6 mg of Sc2Te3
       6 / 472.712 * 6.202 * 1023 * 10-3[mg][mol/g][1/mole][mg/g] = 7.87 X 1018.

Missing Solution on Answer of Exercise 1, Problem 4

The key only seems to have the first line of the solution,

   2 mg of Ti2Te3
   
       M for this titanium telluride is 2*48+3*127.6= 478.8[g/mol].

I got 4.179 μmoles via:

   M for this titanium telluride is 2*48+3*127.6= 478.8[g/mol] 
   2mg / 478.8  [g] .[mol/g] = .00000417 mol = 4.17 μmoles.

Possible Error on Answer of Exercise 1, Problem 1

The key states:

   1.2 gram of B2H6
       M for this borane is 2*10.8+6*1=23.6 [g/mol] 
       So putting it upside down we get: 1.2/23.6 [g] .[mol/g]= = 0.0508 [mol] = 50.8 [mmol]

Should the mass of the B₂H₆ be 27.6 [g/mol]?

Yes, absolutely, thank you!! Jcfolmer 16:10, 18 August 2008 (EDT)