Answer:
The time taken for the cross to become invisible decreases.
Explanation:
We know that one of the factors affecting the rate of reaction is the concentration of reactants. From the collision theory, we know that the higher the concentration of reactants, the greater the possibility of effective collision between reactants leading ultimately to an increase in the rate of reaction. Increase in the rate of reaction implies that the reaction takes a shorter time to reach completion.
In the case of the reaction shown in the question, the point when the reaction is completed is observed by the time take for the cross mark to become invisible. If we look at the given data closely, we will notice that the volume of acid was held constant, the volume of thiosulphate was increased gradually while the volume of water was decreased accordingly. This implies that the concentration of the reactants was increased. Decreasing the volume of water increases reactant concentration.
As explained above, increase in reactant concentration increases the rate of reaction. Hence, the rate of reaction of the acid and thiosulphate increases as reactant concentration increases and the cross mark becomes invisible faster. This implies that in the last column for time taken for the cross to become invisible, the values of time decreases steadily as concentration of reactants increases.
What do you mean by that Oh wait do you mean like 2×2=4 or 2+2=4 something like that maby?? I am still confused?? XD
Answer:
See below.
Step-by-step explanation:
Ethers react with HI at high temperature to produce an alky halide and an alcohol.
R-OR' + HI ⟶ R-I + H-OR'
<em>Benzylic ethers</em> react by an Sₙ1 mechanism by forming the stable benzyl cation.
- PhCH₂-OR + HI ⟶ PhCH₂-O⁺(H)R + I⁻ Protonation of the ether
- PhCH₂-O⁺(H)R ⟶ PhCH₂⁺ + HOR Sₙ1 ionization of oxonium ion
- PhCH₂⁺ + I⁻ ⟶ PhCH₂-I Nucleophilic attack by I⁻
If there is excess HI, the alcohol formed in Step 2 is also converted to an alkyl iodide:
ROH +HI ⟶ R-I + H-OH
Thus, benzyl ethyl ether reacts to form benzyl iodide (a) and ethanol (b).
The ethanol reacts with excess HI in an Sₙ2 reaction to form ethyl iodide (c).
Answer:
72.8 % (But verify explanation).
Explanation:
Hello,
In this case, with the following obtained results, the percent error is computed as follows:
Volume of vinegar= 7.0 mL
Volume of NaOH= (7+6.6+6.4)/3= 6.7 mL
Used concentration of NaOH= 1.5M
Concentration of acetic acid= (concentration of NaOH*volume of NaOH)/volume of vinegar= (6.7mL*1.5M)/7.0M= 1.44M
Assuming we have 100 mL (0.100L) of vinegar, moles of acetic acid in vinegar = 1.44M x 0.100 L= 0.144 mol
Mass of acetic acid in 100g of vinegar = 0.144 mol x 60.0g/mol= 8.64 g
% of acetic acid in vinegar=8.64 %
% error in percentage of acetic acid = [(8.64% - 5.0%)/5.0] x 100=72.8 %
Clearly, this result depend on your own measurements, anyway, you can change any value wherever you need it.
Regards.
