Answer:
Gases have no definite shape or volume. They are fluid, allowing particles/molecules to move freely.
The behavior of a gas is that the volume changes directly with temperature. With a constant volume, the pressure will be directly proportional to the amount of gas.
Explanation:
These are some of the properties I can think of
B. Heating up the reaction will increase the entropy of a reaction.
<h3>
What is entropy?</h3>
Entropy is the measure of the degree of disorderliness of a system.
Entropy is also the measure of a system's thermal energy per unit temperature that is unavailable for doing useful work.
S = ΔH/T
where;
- S is entropy
- ΔH is energy input
- T is temperature
Entropy increases in reactions in which the total number of product molecules is greater than the total number of reactant molecules.
However, entropy increases as temperature increases. Thus, heating up the reaction will increase the entropy of a reaction.
Learn more about entropy here: brainly.com/question/6364271
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Answer:
The answer is D if it's wrong let me know pls
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>
<span>The metalloids; boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), polonium (Po) and astatine (At) are the elements found along the step like line between metals and non-metals of the periodic table.</span>
