Here I found some info at Yahoo answers: https://answers.yahoo.com/question/index?qid=20090119191941AAB7oAb
The more electronegative an atom is the more unwilling it is to lose its electrons in a compound. If you do try to take a very EN atom away from a compound you'll need to apply a lot of energy for that to happen. I can give an example of a single atom though
<span>Cl has 7 valence electron filled and every atom wants to be like nobles (noble gases), so it's not going to give an electron away b/c it's really close to being like a noble gas. Noble gases are the most stable atoms, which is why I say stability counts.</span>
When the same species undergoes both oxidation and reduction in a single redox reaction, this is referred to as a disproportionation. Therefore, divide it into two equal reactions.
NO2→NO^−3
NO2→NO
and do the usual changes
First, balance the two half reactions:
3. NO2 +H2O →NO^−3 + 2 H^+ + e−
4. NO2 +2 H^+ + 2e− → NO + H2O
Now multiply one or both half-reactions to ensure that each has the same number of electrons. Here, Eqn (3) x 2 results in each half-reaction having two electrons:
5. 2 NO2 + 2 H2O → 2 NO^−3 + 4H^+ + 2e−
Now add Eqn 4 and 5 (the electrons now cancel each other):
3NO2 + 2H^+ + 2H2O → NO + 2 NO−3 + H2O + 4H+
and cancel terms that’s common to both sides:
3NO2 + H2O → NO + 2NO^−3 + 2H+
This is the net ionic equation describing the oxidation of NO2 to NO3 in basic solution.
Learn more about balancing equation here:
brainly.com/question/26227625
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Moles of Oxygen= 2.8075 moles
<h3>Further explanation</h3>
Given
29.2 grams of acetylene
Required
moles of Oxygen
Solution
Reaction(Combustion of Acetylene) :
2 C₂H₂ (g) + 5 O₂ (g) ⇒ 4CO₂ (g) + 2H₂O (g)
Mol of Acetylene :
= mass : MW Acetylene
= 29.2 g : 26 g/mol
= 1.123
From equation, mol ratio of Acetylene(C₂H₂) : O₂ = 2 : 5, so mol O₂ :
= 5/2 x mol C₂H₂
= 5/2 x 1.123
= 2.8075 moles
<span>Based on your information 1000 times greater than pH 13 is the best I can come up </span>with.
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