Since the question manages to include moles, pressure, volume, and temperature, then it is evident that in order to find the answer we will have to use the Ideal Gas Equation: PV = nRT (where P = pressure; V = volume; n = number of moles; R = the Universal Constant [0.082 L·atm/mol·K]; and temperature.
First, in order to work out the questions, there is a need to convert the volume to Litres and the temperature to Kelvin based on the equation:
250 mL = 0.250 L
58 °C = 331 K
Also, based on the equation P = nRT ÷ V
⇒ P = (2.48 mol)(0.082 L · atm/mol · K)(331 K) ÷ 0.250 L
⇒ P = (67.31 L · atm) ÷ 0.250 L
⇒ P = 269.25 atm
Thus the pressure exerted by the gas in the container is 269.25 atm.
Answer:
geez I have no clue sorry
One way of knowing that oxygen was the gas removed from the volume of air and not another is to know what the volume of air is made of first. When the composition of the volume of air is already identified, then next would be the process of separating these elements from each other and as to which is to be separated first. This would usually lead to knowing their masses, their boiling and freezing points, the temperatures at which they condense, and so on. This is to identify their differences to each other and use those differences to successfully separate those elements to each other.
The number of students is your independent variable
Answer:
At a front, the two air masses have different densities, based on temperature, and do not easily mix. One air mass is lifted above the other, creating a low pressure zone.
Explanation:
Hope this helps!
