"a 0.2720 g sample of the acid ha, was titrated with standard sodium hydroxide, naoh, solution. determine the molecular weight of the acid if the sample required 45.00 ml of 0.1000 m naoh to neutralize."

60.44 g

Explanation:

45.00 mL x 0.1000 M NaOH x     1 mol HA    

                    1000 mL NaOH      1 mol NaOH

= 0.0045 mol HA

 0.2720 g HA  

0.0045 mol HA

= 60.44 g/mol

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.

it would be a gas formation