1) W(Ca) в Ca_3P_2 2)W(C) в C_6H_12O_6 3)W(Ag)=? W(CI)=?

To find the weight fractions of specific elements in compounds, you can use the following formula:

\[ W(X) = \frac{n(X) \times M(X)}{M(\text{compound})} \]

Where:

• \( W(X) \) is the weight fraction of element \( X \),
• \( n(X) \) is the number of atoms of element \( X \) in the compound,
• \( M(X) \) is the molar mass of element \( X \),
• \( M(\text{compound}) \) is the molar mass of the compound.

Let's calculate each of the weight fractions step-by-step.

1) Weight Fraction of Calcium (Ca) in \( Ca_3P_2 \)

• Molar mass of Ca: approximately 40.08 g/mol
• Molar mass of P: approximately 30.97 g/mol

Calculate molar mass of \( Ca_3P_2 \): \[ M(Ca_3P_2) = 3 \times M(Ca) + 2 \times M(P) = 3 \times 40.08 + 2 \times 30.97 = 120.24 + 61.94 = 182.18 \, \text{g/mol} \]

Calculate weight fraction of Ca: \[ W(Ca) = \frac{3 \times 40.08}{182.18} \approx \frac{120.24}{182.18} \approx 0.659 \]

2) Weight Fraction of Carbon (C) in \( C_6H_{12}O_6 \)

• Molar mass of C: approximately 12.01 g/mol
• Molar mass of H: approximately 1.008 g/mol
• Molar mass of O: approximately 16.00 g/mol

Calculate molar mass of \( C_6H_{12}O_6 \): \[ M(C_6H_{12}O_6) = 6 \times M(C) + 12 \times M(H) + 6 \times M(O) = 6 \times 12.01 + 12 \times 1.008 + 6 \times 16.00 \] \[ = 72.06 + 12.096 + 96.00 = 180.156 \, \text{g/mol} \]

Calculate weight fraction of C: \[ W(C) = \frac{6 \times 12.01}{180.156} \approx \frac{72.06}{180.156} \approx 0.400 \]

3) Weight Fraction of Silver (Ag) and Chlorine (Cl)

To calculate \( W(Ag) \) and \( W(Cl) \), we need to know the chemical formula of the compound containing them. A common example is \( AgCl \) (silver chloride).

Assuming \( AgCl \):

• Molar mass of Ag: approximately 107.87 g/mol
• Molar mass of Cl: approximately 35.45 g/mol

Calculate molar mass of \( AgCl \): \[ M(AgCl) = M(Ag) + M(Cl) = 107.87 + 35.45 = 143.32 \, \text{g/mol} \]

Calculate weight fraction of Ag: \[ W(Ag) = \frac{107.87}{143.32} \approx 0.751 \]

Calculate weight fraction of Cl: \[ W(Cl) = \frac{35.45}{143.32} \approx 0.248 \]

Summary of Results:

1. \( W(Ca) \) in \( Ca_3P_2 \approx 0.659 \)
2. \( W(C) \) in \( C_6H_{12}O_6 \approx 0.400 \)
3. Assuming \( AgCl \):
   • \( W(Ag) \approx 0.751 \)
   • \( W(Cl) \approx 0.248 \)

If you meant a different compound for silver and chlorine, please specify!