What is 9872002 in scientific notation?
2 answers:
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Answer:
pH = 13.1
Explanation:
Hello there!
In this case, according to the given information, we can set up the following equation:
Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:
Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:
And the resulting concentration of KOH and OH ions as this is a strong base:
And the resulting pH is:
Regards!
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:
- 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.
- 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product
- 3. Select the coefficient of the substance with the most complex chemical formula equal to 1
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
