Answer:
E - Be and O
A - Mg and N
E - Li and Br
F - Ba and Cl
B - Rb and O
Explanation:
Be and O
Be is a metal that loses 2 e⁻ to form Be²⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form BeO (E-MX).
Mg and N
Mg is a metal that loses 2 e⁻ to form Mg²⁺ and N is a nonmetal that gains 3 e⁻ to form O³⁻. For the ionic compound to be neutral, it must have the form Mg₃N₂ (A-M₃X₂).
Li and Br
Li is a metal that loses 1 e⁻ to form Li⁺ and Br is a nonmetal that gains 1 e⁻ to form Br⁻. For the ionic compound to be neutral, it must have the form LiBr (E-MX).
Ba and Cl
Ba is a metal that loses 2 e⁻ to form Ba²⁺ and Cl is a nonmetal that gains 1 e⁻ to form Cl⁻. For the ionic compound to be neutral, it must have the form BaCl₂ (F-MX₂).
Rb and O
Rb is a metal that loses 1 e⁻ to form Rb⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form Rb₂O (B-M₂X).
All the following are equal to Avogadro's number EXCEPT a. the number of atoms of bromine in 1 mol Br₂.
1 mol Br₂ contains Avogadro’s number of molecules of Br₂.
However, each molecule contains two atoms of Br, so there are
<em>2 × Avogadro’s number of Br atoms </em>in 1 mol Br₂.
Answer:
See explanation
Explanation:
The question is incomplete but i will try to give you all the necessary guide that you need in order to answer the question.
When compounds are formed, atoms exchange valency. The valency of nitrogen is three while that of the metal is two. The exchange yields M3N2.
If the compound has been specifically mentioned to be a metal, then it must be a group two element. It could be any of Mg, Ca, Sr, Ba or Ra. I did not mention Be here because most of its compounds are covalent.
This will help you to answer the complete question.
Answer:
Orbitals of equal energy are each occupied by one electron before any orbital is occupied by a second electron, and all electrons in singly occupied orbitals must have the same spin state. Noble-gas configuration refers to an outer main energy level occupied, in most cases, by eight electrons.
Explanation:
I hope this helps you.
The mole fraction of pentane in the vapor is 0.291
Vapour pressure rises with temperature and is a measurement of a substance's propensity to transform into a gaseous or vapour state. The boiling point of a liquid is the temperature at which the pressure exerted by its surroundings equals the vapour pressure present at the liquid's surface.
The number of moles of a particular component in the solution divided by the overall number of moles in the sample solution is known as the mole fraction.
Using the formula for vapour pressure,
vapour pressure = 
°
× 
+ °
× 
vapour pressure = 151 × 
+ 425 ×
240 = 151 - 151 + 425
240 - 151 = - 151 + 425
89 = 274
=
0.291 =
Therefore, the mole fraction of pentane in the vapor is 0.291
Learn more about vapour pressure here;
brainly.com/question/25699778
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