Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄
Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.
For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:
Kp = ![\frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }](https://tex.z-dn.net/?f=%5Cfrac%7BP%28N_%7B2%7DO_%7B4%7D%20%29%7D%7BP%28NO_%7B2%7D%20%5E%7B2%7D%29%20%7D)
where:
P(N₂O₄) and P(NO₂) are the partial pressure of each gas.
Calculating constant:
Kp = ![\frac{38.8}{61.2^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B38.8%7D%7B61.2%5E%7B2%7D%20%7D)
Kp = 0.0104
After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.
P(N₂O₄) + P(NO₂) = 200
P(N₂O₄) = 200 - P(NO₂)
Kp = ![\frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }](https://tex.z-dn.net/?f=%5Cfrac%7BP%28N_%7B2%7DO_%7B4%7D%20%29%7D%7BP%28NO_%7B2%7D%20%5E%7B2%7D%29%20%7D)
0.0104 = ![\frac{200 - P(NO_{2}) }{[P(NO_{2} )]^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B200%20-%20P%28NO_%7B2%7D%29%20%20%7D%7B%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D%7D)
0.0104
+
- 200 = 0
Resolving the second degree equation:
=
= 98.7
Find partial pressure of N₂O₄:
P(N₂O₄) = 200 - P(NO₂)
P(N₂O₄) = 200 - 98.7
P(N₂O₄) = 101.3
The partial pressures are
= 98.7 MPa and P(N₂O₄) = 101.3 MPa
Answer:
the correct answer is b antifreeze protects the radiator during summer in winter.
Explanation:
Antifreeze is (typically mixed with water) to help manage the engine during extreme temperatures. As the temperature fluctuates between hot and cold coolant is distributed throughout the engine block to sustain even operating temperature.
100.133 degree celsius is the boiling point of the solution formed when 15.2 grams of CaCl2 dissolves in 57.0 g of water.
Explanation:
Balanced eaquation for the reaction
CaCl2 + 2H20 ⇒ Ca(OH)2 + HCl
given:
mass of CaCl2 = 15.2 grams
mass of the solution = 57 grams
Kb (molal elevation constant) = 0.512 c/m
i = vont hoff factor is 1 as 1 mole of the substance is given as product.
Molality is calculated as:
molality = ![\frac{grams of solute}{grams of solution}](https://tex.z-dn.net/?f=%5Cfrac%7Bgrams%20of%20solute%7D%7Bgrams%20of%20solution%7D)
= ![\frac{15.2}{57}](https://tex.z-dn.net/?f=%5Cfrac%7B15.2%7D%7B57%7D)
= 0.26 M
Boiling point is calculated as:
ΔT = i x Kb x M
= 1 x 0.512 x 0.26
= 0.133 degrees
The boiling point of the solution will be:
100 degrees + 0.133 degrees (100 degrees is the boiling point of water)
= 100.133 degree celcius is the boiling point of mixture formed.
Answer:
![Kp = 0.0410](https://tex.z-dn.net/?f=Kp%20%3D%200.0410)
Explanation:
The reaction of gaseous sulfur dioxide and oxygen to form SO3 (g) is:
2SO₂(g) + O₂(g) ⇄ 2SO₃(g)
Kp is defined as the ratio of pressure of products and pressure of reactants:
![Kp = \frac{P_{SO_3}^2}{P_{SO_2}^2P_{O_2}}](https://tex.z-dn.net/?f=Kp%20%3D%20%5Cfrac%7BP_%7BSO_3%7D%5E2%7D%7BP_%7BSO_2%7D%5E2P_%7BO_2%7D%7D)
I hope it helps!