The Gay-Lussac's law or Amonton's law states that the pressure of a given amount of a gas is directly propotional to its temperature if its volume is kept constant .
P∝T
and
The Charles Law states that volume of given amount of gas at constant pressure is directly propotional to temperature.
V∝T
So, by Gay-Lussac's law if we increase the temperature the Pressure will increase and by Charles Law, if we increase the temperature the volume will increase.
Therefore, if the temperature of gas increases either the pressure of the gas, the volume of the gas, or both, will increase.
Hence,
Answer is option C
The correct answer is option C. <span>This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger.
</span><span>
Keisha follows the instructions for a demonstration on gas laws.
1. Place a small marshmallow in a large plastic syringe.
2. Cap the syringe tightly.
3. Pull the plunger back to double the volume of gas in the syringe.
Now, this activity is being done at the same temperature, because there is no mention of the temperature change. Thus, when the plunger is pulled back, the volume doubles, so pressure will decrease. Therefore, </span>This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger.
Answer:
t = 13.7 s or t = 14 s with proper significant figures
Explanation:
The initial speed is 0 m/s since the car starts from rest, acceleration is 5.5 m/s2 and distance is 523 m.
Since we have initial speed, acceleration and distance we can use the following formula to find the time. We can now use algebra to work out our answer.
d = vt + 
at²
523 = (0)t + ()(5.5)t²
523 = 2.8t²
186.8 = t²
13.7 s = t
(t = 14 s with proper significant figures)
Answer:
Potential difference = 6.0 V
I for 1.0Ω = 6 A
I for 2.0Ω = 3 A
I for 3.0Ω = 2 A
Explanation:
Potential difference (ΔV) = Current (I) x Resistance (R)
The potential difference is constant and equals 6.0 V, hence;
I = ΔV/R
When R = 1.0, I =6/1 = 6 amperes
When R = 2.0, I = 6/2 = 3 amperes
When R = 3.0, I = 6/3 = 2 amperes
<em>The potential difference is 6.0 V and the current is 6, 3, and 2 amperes for a resistance of 1.0, 2.0 and 3.0Ω respectively.</em>
A. krypton (atomic mass 83.8 amu)
B. argon (atomic mass 39.95 amu)
C. xenon (atomic mass 131.3 amu)
D. radon (atomic mass 222 amu)
hope this helps!
