Answer:

Explanation:
Hello!
In this case, according to the ideal gas equation ratio for two states:

Whereas both n and R are cancelled out as they don't change, we obtain:

Thus, by solving for the final pressure, we obtain:

Now, since initial conditions are 1.00 atm, 273.15 K and 17 L and final temperature and volume are 94 + 273 = 367 K and 12 L respectively, the resulting pressure turns out to be:

Best regards!
Answer:
.✓is related to the solute content
✓gives information about potential changes in cell volume when cells are placed in that solution
√is related to membrane permeability to solutes.
Explanation:
Tonicity of a solution can be explained as how an extracellular solution can give room for the liquid to move in and out of the cell through osmosis.
It should be noted that Tonicity of a solution is
.✓is related to the solute content
✓gives information about potential
changes in cell volume when cells are placed in that solution
√is related to membrane permeability to solutes.
Answer:
1) The value of Kc:
C. remains the same.
2) The value of Qc:
A. is greater than Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium.
4) The concentration of N2 will:
B. decrease.
Explanation:
Hello,
In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:
1) The value of Kc:
C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:

2) The value of Qc:
A. is greater than Kc, since the reaction quotient is:
![Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}](https://tex.z-dn.net/?f=Qc%3D%5Cfrac%7B%5BN_2%5D%5BH_2%5D%5E3%7D%7B%5BNH_3%5D%5E2%7D)
Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.
4) The concentration of N2 will:
B. decrease, since less reactant is forming the products.
Best regards.
Answer:
612 K
Explanation:
From the question given above, the following data were obtained:
Initial temperature (T₁) = 306 K
Initial pressure (P₁) = 150 kPa
Final pressure (P₂) = 300 kPa
Volume = 4 L = constant
Final temperature (T₂) =?
Since the volume is constant, the final (i.e the new) temperature of the gas can be obtained as follow:
P₁ / T₁ = P₂ / T₂
150 / 306 = 300 / T₂
Cross multiply
150 × T₂ = 306 × 300
150 × T₂ = 91800
Divide both side by 150
T₂ = 91800 / 150
T₂ = 612 K
Thus, the new temperature of the gas is 612 K