Answer:
if 12 moles are produced then it came 24 moles of Al(OH)3
Explanation:
So the molarity equation is moles of solute/liters of solution. so i’m pretty sure the answer should be 0.63/0.70= .9
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Answer:
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
Explanation:
One colligative property is the freezing point depression due the addition of a solute. The equation is:
ΔT=Kf*m*i
<em>Where ΔT is change in temperature = 0.400°C</em>
<em>Kf is freezing point constant of the solvent = 1.86°C/m</em>
<em>m is molality of the solution (Moles of solute / kg of solvent)</em>
<em>And i is Van't Hoff constant (1 for a nonelectrolyte)</em>
Replacing:
0.400°C =1.86°C/m*m*1
0.400°C / 1.86°C/m*1 = 0.215m
As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:
0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.
The mass of ethylene glycol must be added is:
0.0602 moles * (62.10g / mol) =
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
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Answer:Free radical mono-halogenation of an alkane is typically conducted using bromine versus chlorine because the bromine radical is less reactive and therefore more selective.
Explanation: Halogenation occurs when a halogen replaces one or more hydrogen atoms in an organic compound ie chlorine or bromine with the reactivity of the halogens decreasing in the order of F2 > Cl2 > Br2 > I2
Since fluorine reacts explosively making it is difficult to control, and iodine is unreactive. Free radical mono-halogenation of an alkane is typically conducted using bromine versus chlorine with Chlorination ie chlorine radical being more reactive and not selective and the Bromination of alkanes ie bromine radical occurring similarly but slower and less reactive but more selective which is due to the fact that a bromine atom is less reactive in the hydrogen abstraction than a chlorine atom evidence in the higher bond energy of H-Cl than H-Br.
