For each trial, compute the mol of titrant; (molarity x l) and keep the number of significant figures to 4. trial 1: 12.49 ml = naoh trial 2: 12.32 ml = mol naoh trial 3: 11.87 ml = naoh

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.