Following are the possible isomers of secondary alcohol and ketones for six carbon molecules. In order to distinguish between sec. alcohol and ketone we can simply treat the unknown compound with acidified Potassium Dichromate (VI) in the presence of acid. If with treatment with unknown compound the colour of K2Cr2O7 (potassium dichromate VI) changes from orange to green then it is confirmed that the unknown compound is sec. alcohol, or if no change in colour is detected then ketone is confirmed. This is because ketone can not be further oxidized while, sec. alcohol can be oxidized to ketones as shown below,
N₂H₄ + 2H₂O₂ → N₂ + 4H₂O
mol = mass ÷ molar mass
If mass of hydrazine (N₂H₄) = 5.29 g
then mol of hydrazine = 5.29 g ÷ ((14 ×2) + (1 × 4))
= 0.165 mol
mole ratio of hydrazine to Nitogen is 1 : 1
∴ if moles of hydrazine = 0.165 mol
then moles of nitrogen = 0.165 mol
Mass = mol × molar mass
Since mol of nitrogen (N₂) = 0.165
then mass of hydrazine = 0.165 × (14 × 2)
= 4.62 g
M=70.0 g
p=0.70 g/mL
v=m/p
v=70.0/0.70=100.00 mL
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Hey there! Hello!
Not sure if you still need the answer to this question, but I'd love to help out if you do.
So, the way to balance this equation is pretty simple. First, you need to keep in mind that molecules of hydrogen and oxygen do not come in single molecules, but in bonded pairs, represented by H2 and O2.
But, that's incorrect. The combination of 2 hydrogen molecules with 1 oxygen molecule yields water, but that leaves one oxygen molecule leftover. When broken down, this is how many of each molecule is on each side of the previously stated equation:
Left:
H: 2
O: 2
Right:
H: 2
O: 1
So we have to multiply H2O on the right side by 2 in order to get this:
Left:
H: 2
O: 2
Right:
H: 4
O: 2
The last step is to multiply H2 on the left by two to make it match up with the right side, balancing the equation:
Left:
H: 4
O: 2
Right:
H: 4
O: 2
That makes our equation balanced! I hope this helped you out, feel free to ask any additional questions if you need further clarification. :-)
