Answer:- Volume of the gas in the flask after the reaction is 156.0 L.
Solution:- The balanced equation for the combustion of ethane is:
From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.
Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:
= 126 L 
126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.
let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:
= 60.0 L 
Similarly, let's calculate the volume of water vapors formed:
= 90.0 L 
Since ethane is present in excess, the remaining volume of it would also be present in the flask.
Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:
= 30.0 L 
Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L
Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L
Hence. the answer is 156.0 L.
Smaller atoms ; free neutrons and energy
Answer:
4. The combined volume of the Ar atoms is too large to be negligible compared with the total volume of the container.
Explanation:
Deviations from ideality are due to intermolecular forces and to the nonzero volume of the molecules themselves. At infinite volume, the volume of the molecules themselves is negligible compared with the infinite volume the gas occupies.
However, the volume occupied by the gas molecules must be taken into account. Each <u>molecule does occupy a finite, although small, intrinsic volume.</u>
The non-zero volume of the molecules implies that instead of moving in a given volume V they are limited to doing so in a smaller volume. Thus, the molecules will be closer to each other and repulsive forces will dominate, resulting in greater pressure than the one calculated with the ideal gas law, that means, without considering the volume occupied by the molecules.
Answer:
2.4 ×10^24 molecules of the herbicide.
Explanation:
We must first obtain the molar mass of the compound as follows;
C3H8NO5P= [3(12) + 8(1) + 14 +5(16) +31] = [36 + 8 + 14 + 80 + 31]= 169 gmol-1
We know that one mole of a compound contains the Avogadro's number of molecules.
Hence;
169 g of the herbicide contains 6.02×10^23 molecules
Therefore 669.1 g of the herbicide contains 669.1 × 6.02×10^23/ 169 = 2.4 ×10^24 molecules of the herbicide.
Explanation:
The given data is as follows.
Moles of propylene = 100 moles, 
= 300 K
= 800 K, 
, 
of propylene = 100 J/mol
Now, we assume the following assumptions:
Since, it is a compression process therefore, work will be done on the system. And, work done will be equal to the heat energy liberating without any friction.
W = 
= 
= 5 MJ
Thus, we can conclude that a minimum of 5 MJ work is required without any friction.
