Answer:
Explanation:
<u>1) Data:</u>
a) n = 2 moles
b) T = 373 K
c) V = 2.5 liter
d) P = ?
<u>2) Chemical principles and formula</u>
You need to calculate the pressure of the propane gas in the mixture before reacting. So, you can apply the partial pressure principle which states that each gas exerts a pressure as if it occupies the entire volume.
Thus, you just have to use the ideal gas equation: PV = nRT
<u>3) Solution:</u>
P = 2 mol × 0.08206 atm-liter /K-mol × 373K / 2.5 liter = 24.5 atm
Since the number of moles are reported with one significant figure, you must round your answer to one significant figure, and that is 20 atm (20 is closer to 24.5 than to 30).
Displacement = √(3² + 4²)
Displacement = 5 meters north east
Velocity = displacement / time
Velocity = 5 / 35
Velocity = 0.14 m/s northeast
<u>Answer:</u>
P2 = 778.05 mm Hg = 1.02 atm
<u>Explanation:</u>
We are to find the final pressure (expressed in atm) of a 3.05 liter system initially at 724 mm hg and 298 K which is compressed to a final volume of 2.60 liter at 273 K.
For this, we would use the equation:

where P1 = 724 mm hg
V1 = 3.05 L
T1 = 298 K
P2 = ?
V2 = 2.6 L
T2 = 173 K
Substituting the given values in the equation to get:

P2 = 778.05 mm Hg = 1.02 atm
Answer:
Explanation:
In 150 ml of .06 g / ml solution , gram of iodine = 150 x .06 g = 9 g
Let volume of given concentration of .12 g / ml required be V
In volume V , gram of iodine = V x .12 g
According to question
V x .12 = 9 g
V = 9 / .12 = 75 ml
So, 75 ml of .12 g/ml will be taken and it is diluted to the volume of 150 ml to get the solution of required concentration .
At STP, also known as standard temperature and pressure, 1 mole of a gas occupies 22.4 L. Since we are given with the volume of 6.3L, we calculate the amount of gas in mol.
n = (6.3L)/ (22.4L/mol) = 0.28125 mol
We are given with the mass of 6.7 g. Therefore, the molar mass or molecular weight of the gas is equal to,
6.7g/0.28125 mol = 23.82 g/mol