Which of these is NOT a property of water?
2 answers:
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The organic product formed when 1−hexyne is treated with H₂O, H₂SO₄, and HgSO₄ will be 2-hexanone (structure attached).
This reaction is an example of an oxymercuration reaction of the organic product 1−hexyne.
Oxymercuration is shown in three steps to the right. The nucleophilic double bond attacks the mercury ion, releasing an acetoxy group. The mercury ion's electron pair attacks carbon on the double bond, generating a positive-charged mercuronium ion. Mercury's dxz and 6s orbitals give electrons to the double bond's lowest unoccupied molecular orbitals.
In the second stage, the nucleophilic H₂O attacks the highly modified carbon, freeing its mercury-bonding electrons. Electrons neutralize mercury ions by collapsing. Water molecules have positive-charged oxygen.
In the third stage, the negatively charged acetoxy ion released in the first step attacks the hydrogen of the water group, generating the waste product HOAc. The two electrons in the oxygen-hydrogen link collapse into oxygen, neutralizing its charge and forming alcohol.
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Answer : The mole fraction and partial pressure of and
gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.
Explanation : Given,
Moles of = 1.79 mole
Moles of = 1.20 mole
Moles of = 3.71 mole
Now we have to calculate the mole fraction of and
gases.
and,
and,
Thus, the mole fraction of and
gases are, 0.267, 0.179 and 0.554 respectively.
Now we have to calculate the partial pressure of and
gases.
According to the Raoult's law,
where,
= partial pressure of gas
= total pressure of gas = 5.78 atm
= mole fraction of gas
and,
and,
Thus, the partial pressure of and
gases are, 1.54, 1.03 and 3.20 atm respectively.
We know that to relate solutions of with the factors of molarity and volume, we can use the equation: 
**NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.
So now we can assign values to these variables. Let us say that the 18 M
is the left side of the equation. Then we have:
We can then solve for
:
and 
or 
We now know that the total amount of volume of the 4.35 M solution will be 210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.
**NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.
So now we can assign values to these variables. Let us say that the 18 M
We can then solve for
We now know that the total amount of volume of the 4.35 M solution will be 210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.