The Octet rule is a general rule of thumb that applies to most atoms. Basically, it states that every atom wants to have eight valence electrons in its outermost electron shell.
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: See below
Explanation:
1. a) 0.15 moles calcium carbonate (15g/100g/mole)
b) 0.15 moles CaO (molar ratio of CaO to CaCO3 is 1:1)
c) 8.4 grams CaO (0.15 moles)*(56 grams/mole)
2. a) 0.274 moles Na2O (17g/62 grams/mole)
b) 46.6 grams NaNO3 (2 moles NaNO3/1 mole Na2O)*(0.274 moles Na2O)*(85 g/mole NaNO3)
It has a double C=C bond so that means it's unsaturated, but it can also be a cyclic compound with only simple C-C bonds
Answer : 0.0392 grams of Zn metal would be required to completely reduced the vanadium.
Explanation :
Let us rewrite the given equations again.
On adding above equations, we get the following combined equation.
We have 12.1 mL of 0.033 M solution of VO₂⁺.
Let us find the moles of VO₂⁺ from this information.
From the combined equation, we can see that the mole ratio of VO₂⁺ to Zn is 2:3.
Let us use this as a conversion factor to find the moles of Zn.
Let us convert the moles of Zn to grams of Zn using molar mass of Zn.
Molar mass of Zn is 65.38 g/mol.
We need 0.0392 grams of Zn metal to completely reduce vanadium.
