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PilotLPTM [1.2K]
3 years ago
15

Drag each tile to the correct location.

Chemistry
1 answer:
Yuri [45]3 years ago
6 0

Answer:

ujhbvcx

Explanation:

fc

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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
rodikova [14]

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
2 years ago
Which of the following statements best describes the procedure that should be followed by the person first responding to a hazar
brilliants [131]
Make sure that there an exit and no over head power lines and wate for a backup
3 0
3 years ago
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Amphetamine (C9H13N)(C9H13N) is a weak base with a pKbpKb of 4.2. You may want to reference (Pages 710 - 713) Section 16.8 while
Soloha48 [4]

Answer:

pH = 10.38

Explanation:

  • C9H13N ↔ C9H20O3N+  +  OH-

∴ molar mass C9H13N = 135.21 g/mol

∴ pKb = - log Kb = 4.2

⇒ Kb = 6.309 E-5 = [OH-][C9H20O3N+] / [C9H13N]

∴ <em>C</em> sln = (205 mg/L )*(g/1000 mg)*(mol/135.21 g) = 1.516 E-3 M

mass balance:

⇒ <em>C</em> sln = 1.516 E-3 = [C9H20O3N+] + [C9H13N]......(1)

charge balance:

⇒ [C9H20O3N+] + [H3O+] = [OH-]; [H3O+] is neglected, come from water

⇒ [C9H20O3N+] = [OH-].......(2)

(2) in (1):

⇒ [C9H13N] = 1.516 E-3 - [OH-]

replacing in Kb:

⇒ Kb = 6.3096 E-5 = [OH-]² / (1.516 E-3 - [OH-])

⇒ [OH-]² + 6.3096 E-5[OH] - 7.26613 E-8 = 0

⇒ [OH-] = 2.3985 E-4 M

∴ pOH = - Log [OH-]

⇒ pOH = 3.62

⇒ pH = 14 - pOH = 14 - 3.62 = 10.38

5 0
2 years ago
How many molecules of ammonia are produced from the reaction of
vivado [14]

Answer:

9.5X10^23 molecules N2 X (2 molecules NH3 / 1 molecule N2) = 1.9X10^24 molecules NH3

Explanation:

.

8 0
3 years ago
#11 plz help must need would appreciate it
mario62 [17]

Explanation:

325 g HgO

your answer

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7 0
2 years ago
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