1) The element that will most likely lose electrons to form positive ions when bonding with other elements is rubidium (Rb).
2) The correct statement about sodium atoms is; "The sodium atom transfers electrons to the chlorine atoms to form ionic bonds."
3) Based on their location in the periodic table, nitrogen (N) and oxygen (O) are most likely to form covalent bonds with each other
4) Electronegativity is best described by the phrase; "the relative strength with which an element attracts electrons in a chemical bond"
Metals of group 1 and 2 are highly electropositive and are more likely to loose electrons in a bonding situation. Therefore, the element that will most likely lose electrons to form positive ions when bonding with other elements is rubidium (Rb).
Sodium chloride is an ionic compound. It is formed by transfer of electrons from sodium to chlorine. Sodium is highly electropositive while chlorine is highly electronegative. Therefore, sodium chloride is formed when sodium atom transfers electrons to the chlorine atoms to form ionic bonds.
Covalent bonds are formed between two nonmetals. Nitrogen and oxygen are non metals hence they form covalent bonds.
According to Linus Pauling, electronegativity refers to the ability of an element in a compound to draw electrons towards itself.
Learn more: brainly.com/question/14077687
I think you forgot to give the options along with the question. I am answering the question based on my knowledge and research.All scientific theories <span>must be based on careful and rational examinations of the facts. I hope that this is the answer that you were looking for and it has actually come to your help.</span>
I think the family of Group 2 metals are most likely to form a 2+ion.
In a combustion of a hydrocarbon compound, 2 reactions are happening per element:
C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O
Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.
1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C
Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H
The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%
2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:
Amount of C = 0.01138 mol
Amount of H = 0.014244 mol
Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25
The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5
Thus, the empirical formula of the hydrocarbon is C₄H₅.
3. The molar mass of the empirical formula is
Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.
Molecular Formula = C₈H₁₀
