Answer:
E₁ ≅ 28.96 kJ/mol
Explanation:
Given that:
The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,
Let the activation energy for a catalyzed biochemical reaction = E₁
E₁ = ??? (unknown)
Let the activation energy for an uncatalyzed biochemical reaction = E₂
E₂ = 50.0 kJ/mol
= 50,000 J/mol
Temperature (T) = 37°C
= (37+273.15)K
= 310.15K
Rate constant (R) = 8.314 J/mol/k
Also, let the constant rate for the catalyzed biochemical reaction = K₁
let the constant rate for the uncatalyzed biochemical reaction = K₂
If the rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:
K₁ = 3.50 × 10³
K₂ = 1
Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;
we can use the formula for Arrhenius equation;

If
&





E₁ ≅ 28.96 kJ/mol
∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol
The copper atom is oxidized while the silver is reduced.
<h3>What is a redox reaction?</h3>
A redox reaction is one that occurs between an oxidizing agent and a reducing agent. One specie is oxidized while the other specie is reduced in the reaction.
In this case, the oxidizing agent is the silver and the reducing agent is the copper atom. The copper atom is oxidized while the silver is reduced.
The complete equation is;
Cu(s) + 2AgNO3(aq) -----> Cu(NO3)2(aq) + Ag(s)
Ionically;
Cu(s) + 2Ag^+ -----> Cu^2+(aq) + Ag(s)
Learn more about redox reaction:brainly.com/question/13293425
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Explanation:
Sodium stearate , is the soap , it a ionic compound , hence , can easily break to cation and anion , and therefore is soluble in water .It works on the principle of micelle formation .
Tristearin and stearic acid are fatty acids , and have are carboxyl group and have a very huge hydrocarbon chain , making it non polar in nature ,hence is is insoluble in water , since , water is a polar molecule .
Answer:
49.5J/°C
Explanation:
The hot water lost some energy that is gained for cold water and the calorimeter.
The equation is:
Q(Hot water) = Q(Cold water) + Q(Calorimeter)
<em>Where:</em>
Q(Hot water) = S*m*ΔT = 4.184J/g°C*54.56g*(80.4°C-59.4°C) = 4794J
Q(Cold water) = S*m*ΔT = 4.184J/g°C*47.24g*(59.4°C-40°C) = 3834J
That means the heat gained by the calorimeter is
Q(Calorimeter) = 4794J - 3834J = 960J
The calorimeter constant is the heat gained per °C. The change in temperature of the calorimeter is:
59.4°C-40°C = 19.4°C
And calorimeter constant is:
960J/19.4°C =
<h3>49.5J/°C</h3>
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