To prove that they were tested to see if they were correct or not.
I think:/
Answer:
a. 63.2%
b. 11.7%
c. 73.3%
d. 0.995%
e. 55.5%
Explanation:
An ionic compound is a compound that is formed by ions, so one of the elements must donate electrons (which is the cation, the positive ion), and the other will receive these electrons (which is the anion, the negative ion).
The power of an element has to attract the electrons is called electronegativity, and so, as higher is the difference of electronegative of the elements, it is more probable that one of them will "still" the electrons and will form an ionic compound. The percent of this ionic character can be found by the Pauling's equation:
*100%
Where 
is the electronegativity difference of the elements. Thus, consulting an electronegativity table:
a. 
= 1.5
= 3.5
*100%
%IC = 63.2%
b. 
= 1.6
= 2.1
*100%
%IC = 11.7%
c. 
= 0.7
= 3.0
*100%
%IC = 73.3%
d. 
= 1.7
= 1.9
*100%
%IC = 0.995 %
e. 
= 1.2
= 3.0
*100%
%IC = 55.5%
To convert minutes to hours we divide the minutes by 60. So if we divide 3 by 60 we get 0.05 hours.
<h3>How to convert minutes into hour?</h3>
We know that in hour, there are 60 minutes so if we go from minutes to hours then we have to divide the number by 60 and when we go from hours to minutes we multiply with the same 60 number.
So we can conclude that to convert minutes to hours we divide the minutes by 60. So if we divide 3 by 60 we get 0.05 hours.
Learn more about hour here: brainly.com/question/291457
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Answer:
Explanation:
Cu²⁺ + 2e⁻ → Cu ( copper gets reduced )
Cu → Cu²⁺ + 2e⁻ ( copper gets oxidized )
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
Na₂CO₃ + H₃PO₄ → Na₂HPO₄ + CO₂ + H₂O
The oxidation state of carbon on reactant side is +4. while on product side is also +4 so it neither oxidized nor reduced.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.
