(For a bit of context I will use the reaction between HCl and Mg as an example)
The larger the surface area of the magnesium metal, the more particles are exposed to collide with the aqueous HCl particles to cause the reaction to occur. This increases the frequency per second of collisions, speeding up the rate of reaction.
The effect of a catalyst is to reduce the minimum collision energy which allows the reaction to happen. This does not increase the number of collisions per second, but increases the percentage of successful collisions, which consequently causes the rate of reaction to increase .
I have drawn diagrams showing the effect of surface area, but there isn't really a meaningful diagram that I know of to show the impact of a catalyst (at least not at GCSE level).
The titrant for this exercise. suppose Ca(OH)₂ were used as the titrant, instead of NaOH. This will make the titrant twice as concentrated in hydroxide ion. the analyte will still be HC₂H₃O₂. the stoichiometry ratio of HC₂H₃O₂ to Ca(OH)₂ is 1 : 2.
The balanced reaction of the given condition as follow :
Ca(OH)₂ + 2HC₂H₃O₂ ------> Ca(C₂H₃O₂)₂ + 2H₂O
from the equation it is clear that stoichiometry of Ca(OH)₂ is 1 and the stoichiometry of HC₂H₃O₂ is 2. therefore the stoichiometry ratio of HC₂H₃O₂ to Ca(OH)₂ is 1 : 2.
Thus, The titrant for this exercise. suppose Ca(OH)₂ were used as the titrant, instead of NaOH. This will make the titrant twice as concentrated in hydroxide ion. the analyte will still be HC₂H₃O₂. the stoichiometry ratio of HC₂H₃O₂ to Ca(OH)₂ is 1 : 2.
To learn more about stoichiometry here
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Answer:
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The question is incomplete. Complete question is attached below:
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Answer:
Given: conc. of HBr = 1.4 M
Volume of HBr = 15.4 mL
Volume of KOH = 22.10 mL
We know that, M1V1 = M2V2
(HBr) (KOH)
Therefore, M2 = M1V1/V2
= 1.4 X 15.4/22.10
= 0.9756 M
Concentration of KOH is 0.9756 M.