According to the kinetic molecular theory, the pressure of a gas in a container will increase if the "number of collisions with the container wall increases".
<u>Option: A</u>
<u>Explanation:</u>
Pressure is characterized as force acting by area per unit. It is triggered by gas molecules bombarding on the container walls. Kinetic molecular theory suggests that gases are made up of large numbers of small particles with continuous movement. Here there are elastic collisions between gas particles and between particles and container walls. There are no forces of attraction among them because of the large spaces between the gas particles. The gas temperature relies on the particle's average kinetic energy.
Answer:
Partial pressure O₂ = 1.78 atm
Explanation:
We can apply the mole fraction to solve the question:
Moles of gas / Total moles = Partial pressure of gas / Total pressure
Total moles = 1 H₂ + 2.5 He + 2O₂ = 4.5 moles
2 mol O₂ / 4.5 mol = Partial pressure O₂ / 4 atm
(2 mol O₂ / 4.5 mol ) . 4 atm = Partial pressure O₂ → 1.78 atm
Answer:
The answer to your question is ΔHrxn = 0 kJ
Explanation:
Process
1.- Multiply Equation 1 by 2
2C(coal) + 2H₂O ⇒ 2CO (g) + 2H₂ ΔH rxn = 259.4 kJ
2.- Sum equation 2
CO(g) + H₂O ⇒ CO₂ (g) + H₂ (g) ΔHrxn = -41 kJ
Result
2C + 3H₂O + CO ⇒ 2CO + 3H₂ + CO₂ ΔHrxn = 218.4 kJ
Simplification
2C + 3H₂O ⇒ CO + 3H₂ + CO₂
3.- Sum equation 3
CO(g) + 3H₂ (g) ⇒ CH₄ (g) + H₂O (g) ΔHrxn = -218.4 kJ
Result
2C + 3H₂O + 3H₂ + CO ⇒ CO + 3H₂ + CO₂ + CH₄ + H₂O
Simplification
2C + 2H₂O ⇒ CO₂ + CH₄ ΔHrxn = 0 kJ
Van der Waals equation is as follows:
(P + a(n/V)²) * (V -nb) = n RT
Moles of H₂ is calculated by dividing 25 g by 2 (molecular weight of H₂) = 12.5 moles
Values of a and b are:
a = 0.02444 atm L² / mol
b = 0.0266 L / mol
(P + 0.02444 (12.5/1)²) * (1 -(12.5 *0.0266)) = 12.5 * 0.0821 * 293 K
If we solve this equation we get pressure of 412.29 atm
With ideal gas equation we get:
P V = n R T
P = n * R * T / V = (12.5 * 0.0821 * 293 / 1) = 300.69 atm
The doctor required a 15% 750mL solution. To lower the concentration of the solution, you have to add more diluting liquid.
750mL * 0.15 = 112.5mL. This 112.5mL signifies the pure chemical or compound in the solution. This 112.5mL can be taken from the stock solution.
112.5 * 2 = 250mL of stock solution would be needed and 500 mL of the diluting agent should be added. Adding the two together would result to a 750mL solution of 15% concentration.
