S + O2 →	SO2
Moles of S:
11/32.06 = 0.3431 mol
Moles of O2:
44/15.999 = 2.75 mol
Sulfur will be the limiting reagent.
Oxygen will be the excess reagent
2.75 - 0.3431 = 2.4071 mols of oxygen leftover
2.4071 * 15.999 = 38.511
38.511 grams of oxygen will be leftover
 
        
                    
             
        
        
        
There was no equipment to create temperature difference in the water
hope this helps and can i get a brainly?
btw i make deals, if thi is wrong i own you ten questions you can ask me on my pf for free and i'll do it! 
        
                    
             
        
        
        
Answer:
I think the answer is Prokaryote and Eukaryote 
Explanation: