Answer:
B) Transportation and photosynthesis
C) Cellular respiration and photosynthesis
Answer:
B. Na and Li
Both are group I elements.
Answer:
0.8078 Kg
Explanation:
Pressure of water = 0.15 MPa = 1.5 bar
At critical point of water ,temperature = 647 K=374°C
From the ideal gas equation
P×V= m×R×T
Let us assume volume = 1 m^3
1.5 x 105 x 1 = m x 287 x 647
m= 0.8078 kg
the fraction of mass of liquid at 25°C.
Answer:
63.53% yield
Explanation:
The balanced equation for this reaction is 2NaCl + H2O -> 2NaOH +Cl2
First we must find the limiting reactant
From NaCl we can only produce 6.06 grams of Cl2 in <u>theory</u>
From H20 we can only produce 38.995 grams in theory
so we know NaCl is the limiting
% yield is (Actual/Theoretical) x100 so
(3.85/6.06)x100= 63.53% yield
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>
