Answer:
10.87 g of Ethyl Butyrate
Solution:
The Balance Chemical Equation is as follow,
H₃C-CH₂-CH₂-COOH + H₃C-CH₂-OH → H₃C-CH₂-CH₂-COO-CH₂-CH₃ + H₂O
According to equation,
88.11 g (1 mol) Butanoic Acid produces = 116.16 g (1 mol) Ethyl Butyrate
So,
8.25 g Butanoic Acid will produce = X g of Ethyl Butyrate
Solving for X,
X = (8.25 g × 116.16 g) ÷ 88.11 g
X = 10.87 g of Ethyl Butyrate
Answer:
Average density for method A = 2.4 g/cm³
Average density for method B = 2.605 g/cm³
Explanation:
In order to calculate the average density for each method, we need to add the data for each method, and then divide the result by the number of measurements (in this case is 4 for both methods):
Σ = 2.2 + 2.3 + 2.7 + 2.4 = 9.6
Average = 9.6/4 = 2.4 g/cm³
Σ = 2.603 + 2.601 + 2.605 + 2.611 = 10.420
Average = 10.420/4 = 2.605 g/cm³
Answer:
HCl(aq) + KOH(aq) ===> H2O(l) + KCl(aq)
Note the stoichiometry of the balanced equations shows us that HCl and KOH react in a 1:1 mole ratio. So, let us find moles of HCl and moles of KOH that are present:
moles HCl = 250.0 ml x 1 L/1000 ml x 0.25 mol/L = 0.06250 moles HCl
moles KOH = 200.0 ml x 1 L/1000 ml x 0.40 mol/L = 0.0800 moles KOH
You can see that there are more moles of KOH than there are of HCl, meaning that KOH is in excess and after neutralizing all of the HCl, the solution will be left with excess KOH making the pH > 7 = BASIC
Answer:
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Explanation:
Answer:
Ea=5.5 Kcal/mole
Explanation:
Let rate constant are
and
at temperature
and 
By using Arrhenius equation at two different two different temperature,
By putting value of R=2 cal/mole.K

By rounding off upto 2 significant figure;
