Data:
V1 = 6.7 liter
T1 = 23° = 23 + 273.15 K = 300.15 K
P1 = 0.98 atm
V2 = 2.7 liter
T2 = 125° = 125 + 273.15 K = 398.15 K
P2 = ?
Formula:
Combined law of ideal gases: P1 V1 / T1 = P2 V2 / T2
=> P2 = P1 V1 T2 / (T1 V2)
P2 = 0.98 atm * 6.7 liter * 398.15 K / (300.15K * 2.7 liter)
P2 = 3.22 atm
Answer: The balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is
.
Explanation:
When a substance tends to gain oxygen atom in a chemical reaction and loses hydrogen atom then it is called oxidation reaction.
For example, chemical equation for oxidation of methane is as follows.

Number of atoms present on reactant side are as follows.
Number of atoms present on product side are as follows.
To balance this equation, multiply
by 2 on reactant side. Also, multiply
by 2 on product side. Hence, the equation can be rewritten as follows.

Now, the number of atoms present on reactant side are as follows.
Number of atoms present on product side are as follows.
Since, the atoms present on both reactant and product side are equal. Therefore, this equation is now balanced.
Thus, we can conclude that balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is
.
Answer:

Step-by-step explanation:

Data:
n = 5 mol
V = 2.5 L
Calculation:

The molar concentration of the solution is
.
It’s a chemical reaction I guess? Or maybe it hits the water and unlike a solid and it being a liquid it goes into the water causing it to spread or something like that.