Identify the two most common anions present in antacids

CO32- and OH- are the two most common anions in antacids.

negative ions.:

Hydroxyde ion (OH-)

Carbonate ion. (CO3 -- )

Explanation:

Acids generally, when are mixed in water, produces a lot of positive hydrogen protons, also knew as protons (H +) are responsible for raising the pH.

pH is (power of hydrogen), it means the more positive hydrogens ions more acid will be that substance. So the antacid's role, to neutralize these positive charges is to add more negative ions to the solutions.

In the stomach, when the stomach acid is liberated to degrade the food, it produces a lot of these protons (H +). The antacids who helps to neutralize this, most commonly used are, (OH-) and (CO3--).Acids generally, wher ar mixed in water, produce a lot of positive hydrogen protons, also knowed as protons (H+) are responsable for rise the pH.

Actually, pH are (power of hydrogen), it means the more positive hydrogens ions more acid will be that substance. So the antacids role, to neutralize this positive charges is add more negative ions to the solutios.

In the stomach, when the stomach acid are liberate to degradate the food, produces a lot of these protons (H+). The antacids who hepls to neutralize this, most commos used are, (OH-) and (CO3-- ).

This anions can be provide for substances like aluminum hydroxyde, magnesium hydroxyde, aluminium carbonate, and others..

These anions can be provided for substances like aluminum hydroxide, magnesium hydroxide, aluminum carbonate, and others.

the correct answer is boyle

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml