Answer:
+1
Explanation:
Na₂O₂
NOTE: the oxidation number of oxygen is always –2 except in peroxides where it is –1.
Thus, we can obtain the oxidation number of sodium (Na) in Na₂O₂ as illustrated below:
Na₂O₂ = 0 (oxidation number of ground state compound is zero)
2Na + 2O = 0
O = –1
2Na + 2(–1) = 0
2Na – 2 = 0
Collect like terms
2Na = 0 + 2
2Na = 2
Divide both side by 2
Na = 2/2
Na = +1
Thus, the oxidation number of sodium (Na) in Na₂O₂ is +1
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The molarity of the acid is
0.086 00 mol/L.
HC₂H₃O₂ + NaOH → NaC₂H₃O₂ + H₂O
Since 1 mol of acid reacts with 1 mol of base, we can use the formula
M₁V₁ = M₂V₂
M₁ = ?; V₁ = 50.00 mL
M₂ = 0.1000 mol/L; V₂ = 43.00 mL
M₁ = M₂ ×

= 0.1000 mol/L ×

= 0.086 00 mol/L
A was the rate of reaction in trial 1.