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melisa1 [442]
2 years ago
6

A reaction rate increases by a factor of 655 in the presence of a catalyst at 37°C. The activation energy of the original pathwa

y is 106 kJ/mol. What is the activation energy of the new pathway, all other factors being equal? Group of answer choices 89.3 kJ/mol 16,600 kJ/mol 89.3 J/mol 16,600 J/mol
Chemistry
1 answer:
irinina [24]2 years ago
5 0

For a reaction rate that increases by a factor of 655 in the presence of a catalyst at 37°C, the activation energy of the new pathway  is mathematically given as

Ea2=89.28JKoule/mol

<h3>What is the activation energy of the new pathway, all other factors being equal?</h3>

Generally, the equation for the rate constant  is mathematically given as

K=Ae^{Ea1/Rt}

Therefore

\frac{1}{655}=\frac{1}{8.31*3.10}(Ea2-105000)

Ea2=89286.083Joule/mol

In conclusion, the activation energy

Ea2=89.28JKoule/mol

Read more about Energy

brainly.com/question/13439286

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3. A compound consists of 91.63 grams of carbon, 7.69 grams of hydrogen and 40.81 grams of
Vikentia [17]

Answer:

Molecular formula = C₁₂H₁₂O₄

Empirical formula is C₃H₃O.

Explanation:

Given data:

Mass of C = 91.63 g

Mass of H = 7.69 g

Mass pf O = 40.81 g

Molar mass of compound = 220 g/mol

Empirical formula = ?

Molecular formula = ?

Solution:

Number of gram atoms of H = 7.69 / 1.01 = 7.61

Number of gram atoms of O = 40.81 / 16 = 2.55

Number of gram atoms of C = 91.63 / 12 = 7.64

Atomic ratio:

            C                      :      H                :         O

           7.64/2.55          :    7.61 /2.55    :       2.55/2.55

               3                     :          3               :        1

C : H : O = 3 : 3 : 1

Empirical formula is C₃H₃O.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass  = 3×12+ 3×1.01 +16 = 55.03

n = 220 / 55.03

n = 4

Molecular formula = 4 (empirical formula)

Molecular formula = 4 (C₃H₃O)

Molecular formula = C₁₂H₁₂O₄

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3 years ago
Ammonia can be produced in the laboratory by heating ammonium chloride
AnnyKZ [126]

Answer:

Mass = 2.89 g

Explanation:

Given data:

Mass of NH₄Cl = 8.939 g

Mass of Ca(OH)₂ = 7.48 g

Mass of ammonia produced = ?

Solution:

2NH₄Cl   +  Ca(OH)₂     →    CaCl₂ + 2NH₃ + 2H₂O

Number of moles of NH₄Cl:

Number of moles = mass/molar mass

Number of moles = 8.939 g / 53.5 g/mol

Number of moles = 0.17 mol

Number of moles of Ca(OH)₂ :

Number of moles = mass/molar mass

Number of moles = 7.48 g / 74.1 g/mol

Number of moles = 0.10 mol

Now we will compare the moles of ammonia with both reactant.

                      NH₄Cl          :          NH₃

                          2              :           2

                         0.17          :          0.17

                   Ca(OH)₂         :          NH₃

                        1                :           2

                    0.10              :          2/1×0.10 = 0.2 mol

Less number of moles of ammonia are produced by ammonium chloride it will act as limiting reactant.

Mass of ammonia:

Mass = number of moles × molar mass

Mass = 0.17 mol × 17 g/mol

Mass = 2.89 g

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pH is an important parameter for many reactions to take place in solution and in biological systems. It is related to the concentration of H⁺ ions through the following expression:

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