Mole ratio:
<span>2 H2 + O2 = 2 H2O
</span>
2 mol H2 ---------- 1 mol O2
1.67 mol H2 ------ ??
1.67 x 1 / 2 => 0.835 moles of O2
Molar mass of solute is 30 g/mol
molality is defined as the number of moles of solute in 1 kg of solvent
number of moles of solute - 5.0 g / 30 g/mol = 0.17 mol
molality of solution is 0.70 mol/kg
the mass of solvent when there are 0.7 mol of solute - 1 kg
then the mass of solvent when there's 0.17 mol of solute - 0.17 mol / 0.7 mol/kg
= 0.242 kg
therefore mass of solvent is 0.242 kg
Mass of the of sample using the balance is 433.2 g
You can estimate the decimal part because it not exactly 3
400 + 30 + 3.2
To find the density of the sample you divide mass / volume = density
Mass = 433.2 and Volume = 65.6ml 433.2 g / 65.6
6. 6 g/ml^3
Density = 6.6 g / ml^3 and Mass = 433.2 g
Answer:
Explanation:
The balanced equation is
CaCO₃(s) ⇌ CaO(s) + CO₂(g); Kp = 1.16
Calculate Kc
T= 800 °C = 1073 K; Δn= 1
Equilibrium concentration of CO₂
![K_{c} = [\text{CO}_{2}] = 1.233 \times 10^{-4}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5B%5Ctext%7BCO%7D_%7B2%7D%5D%20%3D%201.233%20%5Ctimes%2010%5E%7B-4%7D)
Moles of CO₂ formed
Moles of CaCO₃ used up
Moles of CaCO₃ used up = moles of CO₂ formed = 6.155 × 10⁻⁴ mol
Mass of CaCO₃ used up
Moles of CaO formed
Moles of CaO formed = moles of CO₂ formed = 6.155 × 10⁻⁴ mol
Mass of CaO formed
Mass of solid at equilibrium
m = 10.0 g – 0.0616 g + 0.0345 g = 10.0 g
