Oxygen and carbon dioxide
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:
2Na⁺ (aq) and 2OH⁻(aq)
Explanation:
Spectator ions:
Spectator ions are those ions which are same on both side of chemical reaction. These ions are same in the reactant side and product side. Their presence can not effect the chemical equilibrium that's why when we write the net ionic equation these ions are neglect or omitted.
Given ionic equation:
Ba⁺²(aq) + 2OH⁻(aq) + 2Na⁺ (aq) + CO²⁻₃(aq) → BaCO₃(s) + + 2Na⁺ (aq) + 2OH⁻(aq)
In given ionic equation by omitting the spectator ions i.e, 2Na⁺ (aq) and 2OH⁻(aq) net ionic equation can be written as,
Net ionic equation:
Ba⁺²(aq) + CO²⁻₃(aq) → BaCO₃(s)