Caffeine is more soluble in dichloromethane and the both are separated by evaporating the solvent.
Caffeine is an organic plant material which is more soluble in non-polar solvents than in polar solvents. As such, caffeine is more soluble in dichloromethane than in pure water.
In order to carry out a liquid-liquid exaction of dichloromethane from a commercial teabag, the dichloromethane is mixed with water. The caffeine is found to be more soluble in the organic dichloromethane layer than in water.
The two solvents can now be separated using a separating funnel and the solution is evaporated to obtain the caffeine.
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Answer:
3.925 mol.
Explanation:
- From the balanced equation:
<em>2 Na₂O₂(s) + 2 H₂O(l) → 4 NaOH(s) + O₂(g)
,</em>
It is clear that 2 moles of Na₂O₂ react with 2 moles of H₂O to produce 4 moles of NaOH and 1 mole of O₂
.
<em>Using cross multiplication:</em>
4 moles of NaOH produced with → 1 mole of O₂
.
15.7 moles of NaOH produced with → ??? mole of O₂
.
<em>∴ The no. of moles of O₂ made =</em> (1 mole)(15.7 mole)/(4 mole) = <em>3.925 mol.</em>
Answer:
1. When observing a positive test for the jones reagent and negative for the Lucas test, it indicates that it is in the presence of a primary alcohol.
Jones reagent behaves like a strong oxidant, where it transforms the primary alcohols into carboxylic acids and the secondary alcohols into ketones. Tertiary alcohols do not react.
With the Lucas test, tertiary alcohols react immediately producing turbidity, while secondary alcohols do so in five minutes. Primary alcohols do not react significantly with Lucas reagent at room temperature.
2. No reaction (See the attached drawing)
3. (see the attached drawing)
D! the molecules of the warmer water would have a higher average kinetic energy :)
Answer:
the change in energy of the gas mixture during the reaction is 227Kj
Explanation:
THIS IS THE COMPLETE QUESTION BELOW
Measurements show that the enthalpy of a mixture of gaseous reactants increases by 319kJ during a certain chemical reaction, which is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that -92kJ of work is done on the mixture during the reaction. Calculate the change of energy of the gas mixture during the reaction in kJ.
From thermodynamics
ΔE= q + w
Where w= workdone on the system or by the system
q= heat added or remove
ΔE= change in the internal energy
q=+ 319kJ ( absorbed heat is + ve
w= -92kJ
If we substitute the given values,
ΔE= 319 + (-92)= 227 Kj
With the increase in enthalpy and there is absorbed heat, hence the reaction is an endothermic reaction.
