1: Explain how you can use Boyle’s Law to determine the new volume of
gas when its pressure is increased from 270 kPa to 540 kPa? The original
volume of gas was 1 L. Assume the temperature and number of particles are constant. What is the new volume?
Boyle's Law of gases: At constant temperature, the volume of a gas is inversely proportional to the pressure of the gas.
This is pV = constant. Then 1 L * 270 kPa = x * 540 kPa => x = 1L*270 kPa /540 kPa = 0.5 L
2: What are three common clues that a chemical change has occurred?
- Change of color
- Production of bubles
- Change of temperature
- Production of odors / smell
- Formation of precipitates (solids)
3: You have a sealed glass jar full of air. If you put it in the freezer, what happens to the gas pressure in the jar?
The pressure decreases, according to Gay-Lussac, at constant volume, the pressure and the temperaure of a gas are proportional.
Then at lower temperature (inside the freezer) the pressure in the jar will decrase.
4: Name and describe the phase change that occurs when solid carbon
dioxide (dry ice) is placed in an open container at room temperature.
This change of phase is called Sublimation. The solid passes directly to gas state, without passing by the liquid state.
Answer:
Explanation:
There are two hypotheses she could test:
A cat's heart rate changes while it is napping.
A cat's heart rate does not change while it is napping.
Answer:
w=255
Explanation:
The change in internal energy is given by the first law:
ΔE = Q - w
where ΔE is the change in internal energy of the system
q is the heat added to the system
w is the work done *by* the system on the surroundings
So, for the first phase of this process:
ΔE = Q - w
Q=160J
w=309J
ΔE = 160J - 309J = -149J
To bring the system back to its initial state after this, the internal energy must change by +149J (the system myst gain back the 149 J of energy it lost). We are told that the system loses 106 J of heat in returning to its initial state, so the work involved is given by:
ΔE = Q - w
+149J = -106J - w
255J = -w
w = -255J
Because molecules in the air scatter blue light from the sun more than they scatter red light
Answer:
V = 4.6 × 10^20 m/s
Explanation:
Given that the magnitude of the drift speed V in a wire that carries a current I of 1A. The wire has radius 2mm and the number of electrons per unit volume of free electrons is n= 10 28 m -3 Cross-sectional area of the wire is A = πr 2
Convert the radius from mm to m by dividing it by 1000. Substitute it in the below formula
The cross sectional area A = π × (0.002)^2
A = 1.257 × 10^-5 m^2
The formula to use is:
I = VneA
Substitutes all the parameters into the formula
1 = V × 10^23 × 1.67×10^-19 × 1.257×10^-5
1 = 2.16 × 10^-21V
V = 1 / 2.16 × 10^-21
V = 4.6 × 10^20 m/s