1) At tne same temperature and with the same volume, initially the chamber 1 has the dobule of moles of gas than the chamber 2, so the pressure in the chamber 1 ( call it p1) is the double of the pressure of chamber 2 (p2)
=> p1 = 2 p2
Which is easy to demonstrate using ideal gas equation:
p1 = nRT/V = 2.0 mol * RT / 1 liter
p2 = nRT/V = 1.0 mol * RT / 1 liter
=> p1 / p2 = 2.0 / 1.0 = 2 => p1 = 2 * p2
2) Assuming that when the valve is opened there is not change in temperature, there will be 1.00 + 2.00 moles of gas in a volumen of 2 liters.
So, the pressure in both chambers (which form one same vessel) is:
p = nRT/V = 3.0 mol * RT / 2liter
which compared to the initial pressure in chamber 1, p1, is:
p / p1 = (3/2) / 2 = 3/4 => p = (3/4)p1
So, the answer is that the pressure in the chamber 1 decreases to 3/4 its original pressure.
You can also see how the pressure in chamber 2 changes:
p / p2 = (3/2) / 1 = 3/2, which means that the pressure in the chamber 2 decreases to 3/2 of its original pressure.
Answer:
4.4×10² cm³
Explanation:
From the question given above, the following data were obtained:
Diameter (d) = 68.3 mm
Height (h) = 0.120 m
Volume (V) =?
Next, we shall convert the diameter (i.e 68.3 mm) to cm.
This can be obtained as follow:
10 mm = 1 cm
Therefore
68.3 mm = 68.3 mm / 10 mm × 1 cm
68.3 mm = 6.83 cm
Therefore, the diameter 68.3 mm is equivalent 6.83 cm.
Next, we shall convert the height (i.e 0.120 m) to cm. This can be obtained as follow:
1 m = 100 cm
Therefore,
0.120 m = 0.120 m/ 1 m × 100 cm
0.120 m = 12 cm
Therefore, the height 0.120 m is equivalent 12 cm.
Next, we shall determine the radius of the cylinder. This can be obtained as follow:
Radius (r) is simply half of a diameter i.e
Radius (r) = Diameter (d) /2
r = d/2
Diameter (d) = 6.83 cm
Radius (r) =?
r = d/2
r = 6.83/2
r = 3.415 cm
Finally, we shall determine the volume of the cylinder as follow:
Radius (r) = 3.415 cm
Height (h) = 12 cm
Volume (V) =?
Pi (π) = 3.14
V = πr²h
V = 3.14 × (3.415) ² × 12
V = 440 cm³
V = 4.4×10² cm³
Therefore, the volume of the cylinder is 4.4×10² cm³
The correct answer to your question is: <span>C) tin (IV) bromide, SnBr₄</span>
Answer:
2.8 g/mL is the density of the mineral.
Explanation:
The mass, 
Volume of the object = Volume of water level rose - Initial volume of water = 30.5 mL - 20.3 mL = 10.2 mL
The expression for the calculation of density is shown below as:-
<u>2.8 g/mL is the density of the mineral.</u>
You didn’t show the cylinder containing water, so I created one that you can use as a model (see image).
The water level was originally at 37 mL.
Then you added the ball, and it displaced its volume of water.
The new volume reading is 52 mL, so
Volume of ball = volume of displaced water = 52 mL – 37 mL = 15 mL.
