Explanation:
The given data is as follows.
Pressure (P) = 760 torr = 1 atm
Volume (V) = 
= 0.720 L
Temperature (T) = 
= (25 + 273) K = 298 K
Using ideal gas equation, we will calculate the number of moles as follows.
PV = nRT
Total atoms present (n) = 
= 
= 0.0294 mol
Let us assume that there are x mol of Ar and y mol of Xe.
Hence, total number of moles will be as follows.
x + y = 0.0294
Also, 40x + 131y = 2.966
x = 0.0097 mol
y = (0.0294 - 0.0097)
= 0.0197 mol
Therefore, mole fraction will be calculated as follows.
Mol fraction of Xe = 
= 
= 0.67
Therefore, the mole fraction of Xe is 0.67.
Answer:
The coefficients are 6, 1, 3
Explanation:
HNCO →C3N3(NH2)3 + CO2
From the above equation, there are a total of 6 atoms of nitrogen on the right side and 1atom on the left. It can be balance by putting 6 in front of HNCO as shown below:
6HNCO → C3N3(NH2)3 + CO2
Now there are 6 atoms of carbon on the left side and 4 atoms on the right side. It can be balance by putting 3 in front of CO2 as shown below:
6HNCO → C3N3(NH2)3 + 3CO2
Now the equation is balanced as the numbers of atoms of the different elements on both sides of the equation are the same.
The coefficients are 6, 1, 3
Answer:
b. precipitate is your answer
Answer:
= 72.73%
Explanation:
The percentage by mass of an element is given by;
% element = total mass of element in compounds/molar mass of compound × 100
The mass of oxygen in carbon dioxide = 32 g
Molar mass of CO2 = 44 g
Therefore;
% of O2 = 32/44 × 100%
<u>= 72.73%</u>
