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Sagot :
Given that the concentration of stomach acid is 0.10M and the
1. Neutralizing capacity of the acid is 0.013 mol/g,
Therefore number of moles required to neutralize 1 gram of antiacid is , n = 0.013 mol
2. Concentartion of an acid is 0.1M
3. We know that C = n/V
∴ Volume = n/C
= 0.013mol/g/ 0.1M
=0.13 L
or Volume = 130ml
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