Answer:
nope its a myth don't worry :)
Since we are only asked for the number of moles, we don't need the information of density. The concentration is expressed in terms of 0.135 M AgCl or 0.135 moles of AgCl per liter solution. The solution is as follows:
Moles AgCl = Molarity * Volume
Moles AgCl = 0.135 mol/L * 244 mL * 1 L/1000 mL
<em>Moles AgCl = 0.03294 mol </em>
The answer is 19.9 grams cadmium.
Assuming there was no heat leaked from the system, the heat q lost by cadmium would be equal to the heat gained by the water:
heat lost by cadmium = heat gained by the water
-qcadmium = qwater
Since q is equal to mcΔT, we can now calculate for the mass m of the cadmium sample:
-qcadmium = qwater
-(mcadmium)(0.850J/g°C)(38.6°C-98.0°C)) = 150.0g(4.18J/g°C)(38.6°C-37.0°C)
mcadmium = 19.9 grams
