Thats allllllll otttt but ill help
step by step
.2 moles of Carbon Dioxide because you are given .38 moles of water and you convert moles of water to grams and then into moles of CO2
Answer:
P₂ = 5.550213 Pa
Explanation:
Given data:
Heat of vaporization = 376.6 KJ/mol ( 376600 j/mol)
R = 8.3143 j mol⁻¹ K⁻¹
Temperature = T1 = 100 °C = 100 + 273 = 373 K
Temperature = T2 = 508.0 °C = 508.0+273 = 781 K
Pressure at 100°C= P1 = 1.6 × 10⁻³² Pa
Pressure at 508.0 °C = ?
Solution:
ln P₁/P₂ = ΔH /R ( 1/T₂ - 1/T₁)
ln (1.6 × 10⁻³² Pa) - ln (P₂) = 376600 j. mol⁻¹ / 8.3143 j mol⁻¹ K⁻¹ ( 1/T₂ - 1/T₁)
-73.212719 - ln (P₂) = 45295.45 (0.001280 - 0.002680)
-73.212719 - ln (P₂) = 45295.45 (-0.00414)
- ln (P₂) = -63.41363 + 73.212719
- ln (P₂) = 9.799089
P₂ = e⁻⁹°⁷⁹⁹⁰⁸⁹
P₂ = 5.550213 Pa