Answer:
3.1atm
Explanation:
Given parameters:
Volume of gas = 2L
Number of moles = 0.25mol
Temperature = 25°C = 25 + 273 = 298K
Unknown:
Pressure of the gas = ?
Solution:
To solve this problem, we use the ideal gas equation.
This is given as;
PV = nRT
P is the pressure
V is the volume
n is the number of moles
R is the gas constant = 0.082atmdm³mol⁻¹K⁻¹
T is the temperature
P =
Now insert the parameters and solve;
P = = 3.1atm
Answer:
The concentration of this solution in units of pounds per gallon is
Explanation:
Units of measurement are established models for measuring different quantities. The conversion of units is the transformation of a quantity, expressed in a certain unit of measure, into an equivalent one, which may or may not be of the same system of units.
In this case, the conversion of units is carried out knowing that 1 μg are equal to 2.205*10⁻⁹ Lb and 1 mL equals 0.00022 Gallons. So
If 1 μg equals 2.205*10⁻⁹ lb, 2.77 μg how many lb equals?
lb=6.10785*10⁻⁹
So, 2.77 μg= 6.10785*10⁻⁹ lb
Then:
You get:
<u><em>The concentration of this solution in units of pounds per gallon is </em></u><u><em></em></u>
The coefficients in the order CH₄, O₂, CO₂, and H₂O, are 1,2,1,2
The equation becomes:
CH₄ (g) + 2O₂ (g) -> CO₂ (g) + 2H₂0 (g)
<h3><em>Further explanation</em></h3>
Complete combustion of hydrocarbon with oxygen will be obtained by CO₂ and H₂O compounds.
If O is insufficient there will be incomplete combustion produced by CO and H and O
Hydrocarbon combustion reactions (specifically alkanes)
Equalization of chemical reaction equations can be done using variables.
Steps in equalizing the reaction equation:
For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:
Balancing C atoms, H and last atoms O atoms
Methana combustion reaction
CH₄ (g) + O₂ (g) -> CO₂ (g) + H₂0 (g)
We give the most complex compounds, namely Methane, we give the number 1
So the reaction becomes
CH₄ (g) + aO₂ (g) -> bCO₂ (g) + cH₂0 (g)
C atom on the left 1, right b, so b = 1
H atom on the left 4, right 2c, so 2c = 4 ---> c = 2
Atom O on the left 2a, right 2b + c, so 2a = 2b + c
2a = 2.1 + 2
2a = 4
a = 2
The equation becomes:
CH₄ (g) + 2O₂ (g) -> CO₂ (g) + 2H₂0 (g)
<h3><em>Learn more</em></h3>
the combustion of octane in gasoline
hydrogen and excess oxygen
Keywords: methane, combustion, hydrocarbons, equalization of reaction equations
