1. **Problem:** Calculate the formula mass of Al₂(PO₄)₃. 2. **Formula:** Formula mass is the sum of the atomic masses of all atoms in the formula. 3. **Step 1:** Identify the number of each atom in Al₂(PO₄)₃: - Al: 2 atoms - P: 3 atoms (since PO₄ is multiplied by 3) - O: 12 atoms (4 oxygens per PO₄ times 3) 4. **Step 2:** Use atomic masses (approximate): - Al = 26.98 g/mol - P = 30.97 g/mol - O = 16.00 g/mol 5. **Step 3:** Calculate total mass: $$\text{Mass} = 2 \times 26.98 + 3 \times 30.97 + 12 \times 16.00$$ 6. **Step 4:** Calculate each term: $$2 \times 26.98 = 53.96$$ $$3 \times 30.97 = 92.91$$ $$12 \times 16.00 = 192.00$$ 7. **Step 5:** Sum all: $$53.96 + 92.91 + 192.00 = 338.87$$ 8. **Answer:** The formula mass of Al₂(PO₄)₃ is approximately **338.87 g/mol**. 1. **Problem:** Calculate the number of moles in 500 g of C₄H₁₀. 2. **Formula:** Number of moles $n = \frac{\text{mass}}{\text{molar mass}}$ 3. **Step 1:** Calculate molar mass of C₄H₁₀: - C: 12.01 g/mol, H: 1.008 g/mol - Molar mass = $4 \times 12.01 + 10 \times 1.008 = 48.04 + 10.08 = 58.12$ g/mol 4. **Step 2:** Calculate moles: $$n = \frac{500}{58.12}$$ 5. **Step 3:** Simplify fraction: $$n = \frac{\cancel{500}}{\cancel{58.12}} \approx 8.60$$ 6. **Answer:** There are approximately **8.60 moles** in 500 g of C₄H₁₀. 1. **Problem:** Use the ideal gas law to find volume $V$ of 40.6 moles of CO₂ at 205°C and 1.5 atm. 2. **Formula:** $PV = nRT$ 3. **Step 1:** Convert temperature to Kelvin: $$T = 205 + 273.15 = 478.15 K$$ 4. **Step 2:** Rearrange formula to solve for $V$: $$V = \frac{nRT}{P}$$ 5. **Step 3:** Use $R = 0.0821$ atm·L/(mol·K), plug in values: $$V = \frac{40.6 \times 0.0821 \times 478.15}{1.5}$$ 6. **Step 4:** Calculate numerator: $$40.6 \times 0.0821 = 3.333$$ $$3.333 \times 478.15 = 1593.5$$ 7. **Step 5:** Calculate volume: $$V = \frac{1593.5}{1.5} = 1062.3$$ 8. **Answer:** The volume is approximately **1062 liters**. 1. **Problem:** Balance the reaction: Na₂CO₃ + Cd(NO₃)₂ → CdCO₃ + NaNO₃ 2. **Step 1:** Write unbalanced: $$\_ Na_2CO_3 + \_ Cd(NO_3)_2 \rightarrow \_ CdCO_3 + \_ NaNO_3$$ 3. **Step 2:** Balance Na: - Left has 2 Na, right has 1 Na in NaNO₃, so put 2 in front of NaNO₃. 4. **Step 3:** Balance Cd: - 1 Cd on each side, balanced. 5. **Step 4:** Balance CO₃: - 1 CO₃ on each side, balanced. 6. **Step 5:** Balance NO₃: - Left has 2 NO₃ groups in Cd(NO₃)₂, right has 2 NO₃ in 2 NaNO₃, balanced. 7. **Balanced equation:** $$1 Na_2CO_3 + 1 Cd(NO_3)_2 \rightarrow 1 CdCO_3 + 2 NaNO_3$$ 8. **Reaction type:** Double displacement (metathesis) reaction. **Final answers:** - Formula mass Al₂(PO₄)₃ = 338.87 g/mol - Moles in 500 g C₄H₁₀ = 8.60 moles - Volume of CO₂ gas = 1062 L - Balanced reaction: 1,1,1,2 coefficients - Reaction type: Double displacement