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Yanka [14]
3 years ago
12

The formula for binary ionic compound formed between cesium and fluorine

Chemistry
1 answer:
OleMash [197]3 years ago
7 0

Answer:

Cesium fluoride(CsF)

Explanation:

A binary compound is a compound that is composed of 2 distinct element. An ionic compound is composed of ions, usually one is a metal why the other is a non metal. One element gives out electron to form cation and the other receives electron to form anion in a binary compound.

Cesium is a group 1 element and it has one valence electron and it can easily donate this 1 electron to form a bond with other element. Group 1 element are generally very reactive. Cesium is a metal

Fluorine is in  group 7 of the periodic table and is a non metal .Fluorine have 7 valency electron and requires  1 electron to form a stable octet.

When cesium and fluorine bond to form a binary compound cesium donate 1 electron and fluorine receives the 1 electron for both element to form a stable octet. The formula for the binary ionic compound of cesium and fluorine can therefore be expressed as Cesium fluoride(CsF)

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Which way of dissolving changes the chemical identity of the solute?
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What is the significance of molecular formula​
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The equilibrium constant for the gas-phase isomerization of borneol (c10h17oh) to isoborneol at 503 k is 0.106. a mixture consis
Dimas [21]
The solution is as follows:

K = [Partial pressure of isoborneol]/[Partial pressure of borneol] = 0.106

The molar mass of isoborneol/borneol is 154.25 g/mol

Mol isoborneol = 15 g/154.25 = 0.0972 mol
Mol borneol = 7.5 g/154.25 = 0.0486 mol

Use the ICE approach

        borneol  →  isoborneol
I         0.0972           0.0486
C         -x                     +x
E     0.0972 - x        0.0486 + x

Total moles = 0.1458

Using Raoult's Law,
Partial Pressure = Mole fraction*Total Pressure
[Partial pressure of isoborneol] = [(0.0972-x)/0.1458]*P
[Partial pressure of borneol] = [(0.0486+x/0.1458)]*P

0.106 = [(0.0972-x)/0.1458]*P/ [(0.0486+x/0.1458)]*P
Solving for x,
x = 0.0832 

Thus,
<em>Mol fraction of borneol = (0.0486+0.0832)/0.1458 = 0.904</em>
<em>Mol fraction of isoborneol = (0.0972-0.0832)/0.1458 = 0.096</em>
6 0
3 years ago
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