11) Methane and oxygen react to form carbon dioxide and water. What mass of water is formed if 0.80 g of methane reacts with 3.2
1 answer:
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Answer:The correct answer is ;
The oxidation state of nitrogen in NO changes from +2 to 0, and the oxidation state of carbon in CO changes from +2 to +4 as the reaction proceeds.
Explanation:
In an oxidation recation addition of oxygen atom takes place or loss of electrons takes place.
In an reduction reaction removal of oxygen atom takes place or gain of electrons takes place.
In the given reaction , the nitrogen atom is present in +2 oxidation state in NO molecule and present in 0 oxidation state in molecule. Hence, nitrogen is getting reduced that is reduction reaction. NO is oxidizing agent
In the given reaction , the carbon atom is present in +2 oxidation state in CO molecule and present in +4 oxidation state in molecule. Hence ,carbon is getting oxidized that is oxidation reaction. CO is a reducing agent.
Answer:
Volume of NaOH required = 3.61 L
Explanation:
H2SO3 is a diprotic acid i.e. it will have two dissociation constants given as follows:
--------(1)
where, Ka1 = 1.5 x 10–2 or pKa1 = 1.824
--------(2)
where, Ka2 = 1.0 x 10–7 or pKa2 = 7.000
The required pH = 6.247 which is beyond the first equivalence point but within the second equivalence point.
Step 1:
Based on equation(1), at the first eq point:
moles of H2SO3 = moles of NaOH
Step 2:
For the second equivalence point setup an ICE table:
Initial 1.98 ? 0
Change -x -x x
Equil 1.98-x ?-x x
Here, ?-x =0 i.e. amount of OH- = x
Based on the Henderson buffer equation:
Volume of NaOH required is:
Step 3:
Total volume of NaOH required = 3.22+0.389 =3.61 L