Is this a answer multiple choice or u just answer it
The correct options would be
OPTIONS 1 & 2
The state which a person lives in has nothing to do with the experiment, although it would most likely make it easier to observe. Wheter they develop heart disease or not is the results of the experiment.
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%
Answer: option D. the ability of a base to react with a soluble metal salt.
Justification:
NaOH is a strong base, which means that in water it will dissociate according to this reaction:
- NaOH(aq) → Na⁺ (aq) + OH⁻ (aq)
On the other hand, CuSO₄ is a soluble ionic salt which in water will dissociate into its ions according to this other reaction:
Hence, in solution, the sodium ion (Na⁺) will react with the metal salt in a double replacement reaction, where the highly reactive sodium ion (Na⁺) will substitute the Cu²⁺ in the CuSO₄ to form the sodium sulfate salt, Na₂SO₄ (water soluble), and the copper(II) hydroxide, Cu(OH)₂ (insoluble).
That is what the given reaction represents:
CuSO₄ (aq) + 2NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)
↑ ↑ ↑ ↑
soluble metal salt strong base insoluble base solube salt
B) they have no charges and are inside an atom.*
