What are the effects on the bodies of water? use terms algal bloom, dead zone and hypotoxic
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i. The given compound X is 2-methyl pent-2-ene. When it is reacted with ozone it forms an ozonide in the first step. In the second step the reduce to forms acetone and propanal.
ii. The structural formula of Y is and Z is
.
iii. Alkenes, upon catalytic hydrogenation, form alkanes. This will occur in the presence of Nickel as the catalyst.
iv. The process of ozonolysis is useful in the field of pharmaceutics.
v. The test of unsaturation can be performed by passing a compound through Bromine water.
<h3>What is ozonolysis?</h3>Ozonolysis is a reaction used in organic chemistry to determine the position of a carbon-carbon double bond in unsaturated compounds.
i. The given alkene X, that is subject to ozonolysis would be 2-methyl-2-pentene. Upon exposure to ozone, an ozonide is initially formed, after which it is broken down into 2 products - acetone and propanal, both with the molecular formula C₃H₆O.
The given compound X is 2-methyl pent-2-ene. When it is reacted with ozone it forms an ozonide in the first step. In the second step the reduce to forms acetone and propanal.
ii. The formulas of Y is and Z is
. They are functional isomers as they have the same molecular formula but different functional groups - ketone and aldehyde.
iii. When alkenes undergo catalytic hydrogenation, they form alkanes, X will form 2 methyl petane on reaction with hydrogen gas in presence of Ni.
iv. The ozonolysis is used for the industrial-scale synthesis of pharmaceuticals.
v. The unsaturation of compound X can be proved by the bromine water test. As on reaction with it, the brown colour of bromine water becomes colourless due to the formation of dibromo alkane.
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The freezing point depression of a solution containing 30.7 g of glycerin is calculated as -1.65°C
Equating :
It is given that,
Given mass of glycerin is = 30.7 grams (Solute)
Volume of water = 376 mL
or molar -freezing-depression point is = 1.86°C/m
Molar mass of glycerin = 92.09 g/mole
Now, to work out the value, the mass of water should be known. Thus, to calculate, the formula used will be:
Mass = Density X Volume
Mass = 1.0 g/mL X 376 mL
Mass = 376 g or 0.376 Kg
Using the formula of melting point depression, the equation becomes:
Δ = i ×
×m
T⁰- =
in which,
Δ = change in freezing point
Δ = freezing point of solution that has to be find
ΔT° = freezing point of water ()
Since, glycerin is a non-electrolyte, the Van't Hoff factor will be 1.
Substituting the values in the above equation:
0⁰C₋T = 1 ×1.86°C/m ×
= -1.65°C
Thus, the freezing point depression of a solution is -1.65°C
<h2 />Freezing point depression
Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all less than that of the pure solvent and is directly proportional to the molality of the solute
Is melting point elevation or depression?
Boiling point elevation is that the raising of a solvent's boiling point due to the addition of a solute. Similarly, melting point depression is the lowering of a solvent's freezing point due to the addition of a solute. In fact, because the boiling point of a solvent increases, its melting point decreases
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Answer:
b . 0.351 L.
Explanation:
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In this case, since diluted solutions are prepared by adding an extra amount of diluent to a stock-concentrated solution, we infer that the number of moles of solute remains the same, therefore we can write:
Thus, solving for the volume of the stock solution, V1, we obtain:
Now, by plugging in the given data we obtain:
Therefore, the answer is b . 0.351 L.
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