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Sagot :
Sure, let's determine the molar mass of calcium carbonate [tex]\((CaCO_3)\)[/tex].
To solve for the molar mass, we need to sum up the atomic masses of all the atoms in the compound:
1. Identify the atomic masses:
- Calcium (Ca): 40.08 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol
2. Count the number of each type of atom in the formula [tex]\(CaCO_3\)[/tex]:
- There is 1 atom of Calcium (Ca).
- There is 1 atom of Carbon (C).
- There are 3 atoms of Oxygen (O).
3. Multiply the atomic masses by the number of each type of atom in the compound and sum them up:
- Contribution from Calcium: [tex]\(1 \times 40.08 = 40.08\)[/tex] g/mol
- Contribution from Carbon: [tex]\(1 \times 12.01 = 12.01\)[/tex] g/mol
- Contribution from Oxygen: [tex]\(3 \times 16.00 = 48.00\)[/tex] g/mol
4. Add these contributions together to get the molar mass of [tex]\(CaCO_3\)[/tex]:
[tex]\[ 40.08 \text{ g/mol (from Ca)} + 12.01 \text{ g/mol (from C)} + 48.00 \text{ g/mol (from O)} = 100.09 \text{ g/mol} \][/tex]
Thus, the molar mass of calcium carbonate [tex]\((CaCO_3)\)[/tex] is 100.09 g/mol.
To solve for the molar mass, we need to sum up the atomic masses of all the atoms in the compound:
1. Identify the atomic masses:
- Calcium (Ca): 40.08 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol
2. Count the number of each type of atom in the formula [tex]\(CaCO_3\)[/tex]:
- There is 1 atom of Calcium (Ca).
- There is 1 atom of Carbon (C).
- There are 3 atoms of Oxygen (O).
3. Multiply the atomic masses by the number of each type of atom in the compound and sum them up:
- Contribution from Calcium: [tex]\(1 \times 40.08 = 40.08\)[/tex] g/mol
- Contribution from Carbon: [tex]\(1 \times 12.01 = 12.01\)[/tex] g/mol
- Contribution from Oxygen: [tex]\(3 \times 16.00 = 48.00\)[/tex] g/mol
4. Add these contributions together to get the molar mass of [tex]\(CaCO_3\)[/tex]:
[tex]\[ 40.08 \text{ g/mol (from Ca)} + 12.01 \text{ g/mol (from C)} + 48.00 \text{ g/mol (from O)} = 100.09 \text{ g/mol} \][/tex]
Thus, the molar mass of calcium carbonate [tex]\((CaCO_3)\)[/tex] is 100.09 g/mol.
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