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Answer :
100 mg of aluminum carbonate contains 4 mEq of aluminum and 10 mEq of carbonate ions.
To determine the milliequivalents (mEq) of aluminum and carbonate ions in 100 mg of aluminum carbonate, we need to consider the molecular weight and the valence of each ion.
The molecular weight of aluminum carbonate (Al₂(CO₃)₃) is calculated by summing the atomic weights of each element. Aluminum (Al) has an atomic weight of 27, and each carbonate ion (CO₃) has a molecular weight of 60. So, the molecular weight of aluminum carbonate is:
Al: 27
C: 12 (carbon)
O: 16 (oxygen) × 3 (three oxygen atoms in each carbonate ion) = 48
Total molecular weight: 27 + 48 = 75
Now, let's calculate the milliequivalents (mEq) of aluminum and carbonate ions.
The valence of aluminum is 3+, meaning it can donate or accept three electrons. So, the number of milliequivalents (mEq) of aluminum in 100 mg of aluminum carbonate can be calculated as follows:
mEq of aluminum = (mass of aluminum in mg / molecular weight of aluminum) × valence of aluminum
mEq of aluminum = (100 mg / 75) × 3 = 4 mEq
The valence of each carbonate ion is 2-, meaning it can accept two electrons. Since there are three carbonate ions in aluminum carbonate, the number of milliequivalents (mEq) of carbonate ions can be calculated as follows:
mEq of carbonate ions = (mass of aluminum carbonate in mg / molecular weight of carbonate ions) × valence of carbonate ions × number of carbonate ions
mEq of carbonate ions = (100 mg / 60) × 2 × 3 = 10 mEq
Therefore, in 100 mg of aluminum carbonate, there are 4 mEq of aluminum and 10 mEq of carbonate ions.
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