Answer :
To determine the number of moles in 144 grams of calcium bromide (CaBr₂), we can follow these steps:
1. Calculate the molar mass of CaBr₂:
- The atomic mass of calcium (Ca) is approximately 40.08 g/mol.
- The atomic mass of bromine (Br) is approximately 79.904 g/mol.
- Since CaBr₂ contains one calcium atom and two bromine atoms, the molar mass of CaBr₂ can be calculated as:
[tex]\[ \text{Molar mass of CaBr}_2 = (\text{Atomic mass of Ca}) + 2 \times (\text{Atomic mass of Br}) \][/tex]
Substituting the values:
[tex]\[ \text{Molar mass of CaBr}_2 = 40.08 \, \text{g/mol} + 2 \times 79.904 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of CaBr}_2 = 40.08 \, \text{g/mol} + 159.808 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of CaBr}_2 = 199.888 \, \text{g/mol} \][/tex]
2. Calculate the number of moles:
- The number of moles ([tex]\(n\)[/tex]) can be calculated using the formula:
[tex]\[ n = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \][/tex]
Given the mass of CaBr₂ is 144 grams:
[tex]\[ n = \frac{144 \, \text{g}}{199.888 \, \text{g/mol}} \][/tex]
3. Perform the division:
[tex]\[ n \approx \frac{144}{199.888} \approx 0.7204 \, \text{moles} \][/tex]
Therefore, there are approximately 0.7204 moles in 144 grams of calcium bromide, CaBr₂.
1. Calculate the molar mass of CaBr₂:
- The atomic mass of calcium (Ca) is approximately 40.08 g/mol.
- The atomic mass of bromine (Br) is approximately 79.904 g/mol.
- Since CaBr₂ contains one calcium atom and two bromine atoms, the molar mass of CaBr₂ can be calculated as:
[tex]\[ \text{Molar mass of CaBr}_2 = (\text{Atomic mass of Ca}) + 2 \times (\text{Atomic mass of Br}) \][/tex]
Substituting the values:
[tex]\[ \text{Molar mass of CaBr}_2 = 40.08 \, \text{g/mol} + 2 \times 79.904 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of CaBr}_2 = 40.08 \, \text{g/mol} + 159.808 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of CaBr}_2 = 199.888 \, \text{g/mol} \][/tex]
2. Calculate the number of moles:
- The number of moles ([tex]\(n\)[/tex]) can be calculated using the formula:
[tex]\[ n = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \][/tex]
Given the mass of CaBr₂ is 144 grams:
[tex]\[ n = \frac{144 \, \text{g}}{199.888 \, \text{g/mol}} \][/tex]
3. Perform the division:
[tex]\[ n \approx \frac{144}{199.888} \approx 0.7204 \, \text{moles} \][/tex]
Therefore, there are approximately 0.7204 moles in 144 grams of calcium bromide, CaBr₂.
