Answer:
A. Can A will make a louder and stronger fizz than can B.
Explanation:
The solubility of a gas in a liquid decreases as the temperature increases, so the warmer can will have more undissolved carbon dioxide.
The warmer can will be under greater pressure, so it will make a louder and stronger fizz.
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>
Answer:
D
Explanation: Decreasing the temperature of the system would lower the reaction rate.
<em>Same group element have same</em><em><u> Valence electron</u></em><em> and behave similarly in </em><em><u>Chemistry.</u></em>
<u>Explanation:</u>
For example. First group elements Alkali metals:- H, Li, K, Rb, Cs, Fr
Valance electron will take part in forming a bond with other elements and compound will form. All the above-given elements (H-Fr) have valence electron 1 in outer most 'S' shell. All have electronic configuration S1
Behavior: Since valence electrons are the same so the behavior of all the elements in this group is the same. All are metal (from Li-Fr, except Hydrogen), all are very reactive, does not found in native state in the environment, and all react with water.
Its described as a Straight Branch, hope this helps :)
