8.97 g/cm^
D=m/v
D=43.5/4.85cm^3
D=8.97 g/cm^3
1 answer · Chemistry
Best Answer
Water steam condenses if its pressure is equal to vapor saturation vapor pressure.
Use the Clausius-Clapeyron relation.
I states the temperature gradient of the saturation pressure is equal to the quotient of molar enthalpy of phase change divided by molar volume change due to phase transition time temperature:
dp/dT = ΔH / (T·ΔV)
Because liquid volume is small compared to vapor volume
ΔV in vaporization is approximately equal to to the vapor volume. Further assume ideal gas phase:
ΔV ≈ V_v = R·T/p
Hence
dp/dT = ΔHv / (R·T²/p)
<=>
dlnp/dT = ΔHv / (R·T²)
If you solve this DE an apply boundary condition p(T₀)= p₀.
you get the common form:
ln(p/p₀) = (ΔHv/R)·(1/T₀ - 1/T)
<=>
p = p₀·exp{(ΔHv/R)·(1/T₀ - 1/T)}
For this problem use normal boiling point of water as reference point:
T₀ =100°C = 373.15K and p₀ = 1atm
Therefore the saturation vapor pressure at
T = 350°C = 623.15K
is
p = 1atm ·exp{(40700J / 8.314472kJ/mol)·(1/373.15K - 1/623.15K)} = 193 atm
hope this helps
Mass and volume are two factors that affect that density of matter!
Sun and hydrogen are the answers
Hey there!
In order to solve for the percentage of water in the compound, you will first need to find its total molar mass. You can do this by adding up the molar masses of each individual element in the compound. Then, you will divide the mass that you find of the water molecules by the total mass to get the percentage.
→ Na₂CO₃ ×<span> 10 H</span>₂<span>O
</span>→ Na₂ = 22.9898 × 2 = 45.9796
→ C = 12.0107
→ O₃ = 15.999 × 3 = 47.997
→ 10 H₂O = 18.015 × 10 = 180.15
Now, just add all of those numbers up for the total molar mass.
→ 45.9796 + 12.0107 + 47.997 + 180.15 = <span>286.1373
</span>
The last step is to divide the molar mass of the 10 water molecules by the total mass.
→ 180.15 ÷ 286.1373 = <span>0.62959 </span>≈ 0.63
Your answer will be about 63%.
Hope this helped you out! :-)
