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).
The correct answer is B) Basic. Hope this helps.
Answer:
The answer is (H30+) =3,55e-8M and (OH-)=2,82e-7M
Explanation:
We use the formulas:
pH= - log(H30+) and Kwater=(H30+)x(OH-)
pH= - log(H30+) ----< (H30+)= antilog- pH=antilog- 7,45=3,55E-8M
Kwater=(H30+)x(OH-)
(OH-)=Kwater/(H30+)= 1,00e-14/3,55e-8 = 2,82e-7
Answer:
hope this help !
Explanation:
Use the given functions to set up and simplify 173 ° C .
1.5 =
CH4 = CH4
4.4 = CH4
173 ° C = CH4
