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>
2NH₂ + O₂ → N₂ + 2H₂O
<u>Explanation:</u>
Balancing the equation means, the number of atoms on both sides of the equation must be the same.
In the case of the given equation, we have to find out whether it is balanced or not.
2NH₂ + O₂ → N₂ + 2H₂O
Atoms Number of atoms before balancing after balancing
LHS RHS LHS RHS
N 1 2 2 2
H 2 2 4 4
O 2 1 2 2
To balance the N atoms, we have to put 2 in front of NH₂, and then to balance the H, O atoms, we have to put 2 in front of H₂O, so that each atom in left hand as well as right hand side of the equation was balanced.
Characteristic properties can be used to describe and identify the substances, while non-characteristic properties, although can be used to describe the substances, cannot be used to identify them.
Temperature, mass, color, shape and volume are examples of non-characteristic properties.
Density, boiling point, melting point, chemical reactivity are examples of characteristic properties.
List of the properties observed by the scientist:
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Property Type of property
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Volume: 5 ml non-characteristic
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Color: blue non-characteristic
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State: liquid characteristic
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density: 1.2 g/cm characteristic
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Reaction: reacts with CO2 characteristic
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Answer:
[Co(NH3)5CO3]I3
Explanation:
The naming of coordination compounds follows certain rules specified by IUPAC. Usually, the name of the complex makes it quite easy to deduce its structure.
"Pentaamine" means that there are five NH3 ligands as shown in the structure. The ligand carbonato is CO3^2-. It has no prefix attached to it in the IUPAC name of the complex hence there is only one carbonato ligand present(recall that the complex has a coordination number of six). I did not enclose it within parenthesis as required in the question.
Lastly the III that appeared after the metal name "cobalt" shows its oxidation state. The iodide counter ions must then be 3 in number in order to satisfy this primary valency of the metal hence the inclusion of I3 in the structure of the complex.
Hard question thx for the points