<span>Oxidation is the loss of electrons and corresponds to an increase in oxidation state. The reduction is the gain of electrons and corresponds to a decrease in oxidation state. Balancing redox reactions can be more complicated than balancing other types of reactions because both the mass and charge must be balanced. Redox reactions occurring in aqueous solutions can be balanced by using a special procedure called the half-reaction method of balancing. In this procedure, the overall equation is broken down into two half-reactions: one for oxidation and the other for reduction. The half-reactions are balanced individually and then added together so that the number of electrons generated in the oxidation half-reaction is the same as the number of electrons consumed in the reduction half-reaction.</span>
Moles of NaN3 at STP = volume of gas / 22.4 = 11.5/22.4 = 0.5mole. Massof NaN3 = moles of NaN3 x molecular weight = 0.5 x 65 = 32.5 grams.
Answer:
4804.5 g of SO₂ are needed to the reaction
Explanation:
The reaction to produce sulfuric acid is:
2SO₂ + O₂ + 2H₂O → 2H₂SO₄
Ratio is 1:2. 1 mol of oxygen needs 2 moles of sulfur dioxide in order to react. We can propose this rule of three.
If 1 mol of O₂ react to 2 moles of SO₂
Then, 37.50 moles of O₂ will react with (37.5 . 2) /1 = 75 moles of SO₂
We convert the moles to mass, to know the answer:
75 mol . 64.06 g / 1 mol = 4804.5 g of SO₂
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.
