Answer:
Increasing the temperature
Increasing particle size
Explanation:
Increasing the temperature allows for particles and molecules to move faster (because there is an increase in energy) and allows for more collisions.
∴ Increasing the temperature would be an answer.
Increasing the particle size allows for more surface area and a greater chance for particles to collide with each other.
∴ Increasing particle size would be an answer.
Topic: AP Chemistry
Unit: Kinetics
Answer: 67 mmHg
Explanation:
According to Dalton's Gas Law, the total pressure of a mixture of gases is the sum of the pressure of each individual gas.
i.e Ptotal = P1 + P2 + P3 + .......
In this case,
Ptotal = 512 mmHg
P(oxygen) = 332 mmHg
P(carbon mono-oxide) = 113 mmHg
Remaining pressure (P3) = ?
To get P3, apply Dalton's Gas Law formula
Ptotal = P(oxygen) + P(carbon mono-oxide) + P3
512 mmHg = 332 mmHg + 113 mmHg + P3
512 mmHg = 445 mmHg + P3
P3 = 512 mmHg - 445 mmHg
P3 = 67 mmHg
Thus, the remaining pressure is 67 mmHg
Well glass and water or more heavy then air
Answer: 0.55L
Explanation:
Given that,
Original volume of balloon (V1) = 0.3L
Original temperature of balloon (T1) = 25.0°C
[Convert 25.0°C to Kelvin by adding 273
25.0°C + 273 = 298K]
New volume of balloon (V2) = ?
New temperature of balloon (T2# = 275.0°C
[Convert 25.0°C to Kelvin by adding 273
275.0°C + 273 = 548K]
Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law
V1/T1 = V2/T2
0.3L/298K = V2/548K
To get the value of V2, cross multiply
0.3L x 548K = 298K x V2
164.4L•K = 298K•V2
Divide both sides by 298K
164.4L•K /298K = 298K•V2/298K
0.55L = V2
Thus, the new volume of the balloon would be 0.55 litres