Answer:
Sal, cobre, aluminio, anillo de oro puro Oxígeno ------ suspensión.
Cerveza, carbonato de sodio, oxígeno, carbono y helio ------ solución.
Explicación:
Sal, Cobre, Aluminio, Anillo de oro puro Oxígeno son los ejemplos de suspensión porque permanecen suspendidos en la solución mientras que por otro lado, Cerveza, Carbonato de sodio, Oxígeno, Carbono y helio son los ejemplos de solución porque formaron una mezcla homogénea en la que ambas soluciones no se pueden diferenciar entre sí pero en suspensión sí se pueden diferenciar.
Answer:
the answer on acellus is 2, 3 , 2
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
Answer:
A i. Internal energy ΔU = -4.3 J ii. Internal energy ΔU = -6.0 J B. The second system is lower in energy.
Explanation:
A. We know that the internal energy,ΔU = q + w where q = quantity of heat and w = work done on system.
1. In the above q = -7.9 J (the negative indicating heat loss by the system). w = 3.6 J (It is positive because work is done on the system). So, the internal energy for this system is ΔU₁ = q + w = -7.9J + 3.6J = -4.3 J
ii. From the question q = +1.5 J (the positive indicating heat into the system). w = -7.5 J (It is negative because work is done by the system). So, the internal energy for this system is ΔU₂ = q + w = +1.5J + (-7.5J) = +1.5J - 7.5J = - 6.0J
B. We know that ΔU = U₂ - U₁ where U₁ and U₂ are the initial and final internal energies of the system. Since for the systems above, the initial internal energies U₁ are the same, then we say U₁ = U. Let U₁ and U₂ now represent the final energies of both systems in A i and A ii above. So, we write ΔU₁ = U₁ - U and ΔU₂ = U₂ - U where ΔU₁ and ΔU₂ are the internal energy changes in A i and A ii respectively. Now from ΔU₁ = U₁ - U, U₁ = ΔU₁ + U and U₂ = ΔU₂ + U. Subtracting both equations U₁ - U₂ = ΔU₁ - ΔU₂
= -4.3J -(-6.0 J)= 1.7 J. Since U₁ - U₂ > 0 , U₂ < U₁ , so the second system's internal energy increase less and is lower in energy and is more stable.
The poly atomic ion formula for ammonium would be NH4+
