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
The dilution formula can be used to find the volume needed
c1v1 = c2v2
Where c1 is concentration and v1 is volume of the concentrated solution
And c2 is concentration and v2 is volume of the diluted solution to be prepared
c1 - 0.33 M
c2 - 0.025 M
v2 - 25 mL
Substituting these values in the equation
0.33 M x v1 = 0.025 M x 25 mL
v1 = 1.89 mL
Therefore 1.89 mL of the 0.33 M solution needs to be diluted up to 25 mL to make a 0.025 M solution
Answer:
Einstein tells us that the relationship between energy and matter is defined by the equation E
=
M
C
2
, where E is energy, and M is mass.
Explanation:
By looking at Einstein's equation we can see that matter (or something that has mass) and energy are actually the same thing
What kind of question is that it’s Hot
Answer:
B) 271 g.
Explanation:
- It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
- Half-life time is the time needed for the reactants to be in its half concentration.
- If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
- Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
- The half-life of As-76 = 26.0 hours.
- For, first order reactions:
<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>
Where, k is the rate constant of the reaction.
t1/2 is the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(26.0 hours) = 0.02665 hour⁻¹.
- Also, we have the integral law of first order reaction:
<em>kt = ln([A₀]/[A]),</em>
where, k is the rate constant of the reaction (k = 0.02665 hour⁻¹).
t is the time of the reaction (t = 538 min = 8.97 hour).
[A₀] is the initial concentration of (As-76) ([A₀] = 344 g).
[A] is the remaining concentration of (As-76) ([A] = ??? g).
∴ (0.02665 hour⁻¹)(8.97 hour) = ln((344 g)/[A])
∴ 0.239 = ln((344 g)/[A]).
- Taking exponential for both sides:
∴ 1.27 = ((344 g)/[A]).
∴ [A] = (344 g)/(1.27) = 270.88 g ≅ 271 g.
- So, the right choice is: B) 271 g.