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Zina [86]
3 years ago
8

What is the velosity of a 72.3 kg jogger with a kinetic energy of 1080.0 J?

Chemistry
1 answer:
Svet_ta [14]3 years ago
7 0

Answer: 5.47m/s

Explanation:

Mass = 72.3kg

K.E = 1080.0J

V =?

K.E = 1 /2MV^2

V^2 = 2K.E /M = (2x1080)/72.3

V = sqrt [(2x1080)/72.3]

V = 5.47m/s

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The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the r
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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;

K=Ae^{\frac{-E}{RT}}

If K_1=Ae^{\frac{-E_1}{RT}} -------equation 1     &

K_2=Ae^{\frac{-E_2}{RT}} -------equation 2

\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}

E_1= E_2-RT*In(\frac{K_1}{K_2})

E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})

E_1 = 28957.39292  J/mol

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

8 0
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What goes in the rest of the atoms diagram ?
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Answer:

c. Number of electrons

f. Isotope

Explanation:

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Which of these elements has four total electrons in its outer p orbitals?
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g The following equation is given for the dissociation of the complex ion Ag(NH3) 2. The dissociation constant, Kd, is the equil
Sveta_85 [38]

Answer:

Kd = [Ag⁺] × [NH₃]² / [Ag(NH₃)₂⁺]

Explanation:

Let's consider the dissociation reaction of the complex ion Ag(NH₃)₂⁺.

Ag(NH₃)₂⁺(aq) ⇄ Ag⁺(aq) + 2 NH₃(aq)

The dissociation constant, Kd, is the equilibrium constant for the dissociation of the complex ion, that is, it is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients.

The dissociation constant for this reaction is:

Kd = [Ag⁺] × [NH₃]² / [Ag(NH₃)₂⁺]

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