Answer:
15.95
Explanation:
This question is a modification of the calculation of the empirical formula of a compound given its percent composition and atomic weights of the elements in the compound.
Here we are given the formula and the percent composition, so we know that there are 4 atoms of E per 2 atoms of N so lets solve using the information given.
In 100 grams of the binary compound we have
30.46 g N
69.54 g E
The number of moles is the mass divided by atomic weight:
mol N = 30.46 g / A.W N = 30.46 g / 14.00 g/mol = 2.18 mol N
mol E = 65.54 g / A.W E
Thus,
4 mol E/ 2 mol N = ( 69.54 g/ A.W E ) / 2.18
2 A.E = 65.54 g / 2.18 ⇒ A.W E = 69.54 g / ( 2 x 2.18 ) = 15.94 g
So the A.W is 15.94 g/mol which is close the atomic weight of O.
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).
Sedimentary rocks are formed when different layers of debris ( usually sand ) are forced together under pressure.
Metamorphic rock forms from other types of rock that have been changed by a combination of high heat and pressure.
Igneous rock forms when Lava/Magma cools down
