false is the answer to this question.
Because the reaction occurs between a metal and non metal
CL=2,8,7
C
The titrant for this exercise. suppose Ca(OH)₂ were used as the titrant, instead of NaOH. This will make the titrant twice as concentrated in hydroxide ion. the analyte will still be HC₂H₃O₂. the stoichiometry ratio of HC₂H₃O₂ to Ca(OH)₂ is 1 : 2.
The balanced reaction of the given condition as follow :
Ca(OH)₂ + 2HC₂H₃O₂ ------> Ca(C₂H₃O₂)₂ + 2H₂O
from the equation it is clear that stoichiometry of Ca(OH)₂ is 1 and the stoichiometry of HC₂H₃O₂ is 2. therefore the stoichiometry ratio of HC₂H₃O₂ to Ca(OH)₂ is 1 : 2.
Thus, The titrant for this exercise. suppose Ca(OH)₂ were used as the titrant, instead of NaOH. This will make the titrant twice as concentrated in hydroxide ion. the analyte will still be HC₂H₃O₂. the stoichiometry ratio of HC₂H₃O₂ to Ca(OH)₂ is 1 : 2.
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<span>Zn⁰(s) + CuSO4(aq) --> Zn²⁺SO4²⁻(aq) + Cu(s)
</span>Zn⁰ --- in free elements oxidation number is 0.
Zn²⁺SO4²⁻ ---- ionic compound , it has ion Zn²⁺, so oxidation number here is +2.
<span>The oxidation number of Zn changes from 0 to +2.</span>