Assuming stomach acid is 0.1 m hcl, how many grams of stomach acid would have been neutralized by your tablet (use the average value)?

HCl = 0.1 M. 

find grams of Acid to neutralize the tablet. 

n = 0.00739 mol of CO3-2. 

D = 1.02 g/ml. 

Find the moles of CO3-2.

CO3-2 + 2HCl <---> H2CO3 + 2Cl-.

therefore,

we need 2 mol of CO3-2 to neutralize 1 mol of HCl.

since we have 0.00739 mol of CO3-2, that will neutralize 2X mol of HCl.

2*0.00739 = 0.01478 mol of HCl will be neutralized.

Now, find the mass of Acid :

M = mol/V,

V = mol/M.

Substitute data of acid :

V = 0.01478 mol / 0.1 M = 0.1478 Liters of stomach acid

or 147.8 ml of stomach acid

now find mass:

D = mass/V.

mass = D*V = 1.02*147.8 = 150.76 grams of Stomach Acid will be neutralized

8 m/s because for him to get the same amount of energy as ray is putting out, bobby, who weighs half as much as ray, needs to run twice as fast.

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.