First you figure out how many grams of sulfur has been used.
2.080g - 1.659g = 0.421g of S used
Now change all of your grams to mols.
1.659g of Cu / 63.546g/mol = 0.02611mol of Cu
0.421g of S / 32.065g/mol = 0.01313 mol of S
From this you can see that S has less so we divide both numbers by the number of moles of S so we can get a ratio of S to Cu.
0.01313/0.01313 = 1 for S
0.02611/0.01313 = 2 for Cu
So the empirical formula would be Cu2S or copper sulfide.
<span>Supporting combustion is a chemical property because it involves a chemical reaction that can only be reversed with another chemical reaction.</span>
First Set of Answers:
Heat Released Reaction 1: 3700
Heat Released Reaction 2: 3200
Second Set of Answers:
Moles Reactant 1: .00823
Moles Reactant 2: .0397
Third Set of Answers:
Enthalpy 1: -450
Enthalpy 2: -81
FOR EDG!!!!!!!!!!!!
( Record in your data table (; )
Yes because the metal on the iron is heated but by that happening the shirt starts to heat up and then i will start to get warm <span />
