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Firlakuza [10]
3 years ago
12

A catalyst decreases the activation energy of a particular exothermic reaction by 56 kJ/mol, to 35 kJ/mol. Assuming that the mec

hanism has only one step, and that the products are 78 kJ lower in energy than the reactants, sketch approximate energy-level diagrams for the catalyzed and uncatalyzed reactions. What is the activation energy for the uncatalyzed reverse reaction
Chemistry
1 answer:
-Dominant- [34]3 years ago
5 0

Answer:

Activation energy for the uncatalyzed reverse reaction = 103 kJ/mol

Explanation:

Activation energy decreases from = 56 kj/mol to 35 kj/mol

products = 78 KJ lower in energy than reactants

Activation energy for the uncatalyzed reverse reaction = 103 kJ/mol

attached below are the sketches of approximate energy-level for both catalyzed and uncatalyzed reactions

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Give all possible ml values for orbitals that have each of the following: (a) l = 2; (b) n = 1; (c) n = 4, l = 3.
Olin [163]

Answer:

(a) ml = 0, ±1, ±2

(b) ml = 0

(c) ml = 0, ±1, ±2, ±3, ±4

Explanation:

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n

2. Subshell number, 0 ≤ l ≤ n − 1

3. Orbital energy shift, -l ≤ ml ≤ l

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So in our exercise,

(a) l = 2; equivalent with with sublevel <em>d</em>

-l ≤ ml ≤ l, ml = 0, ±1, ±2, equivalent with dxy, dxz, dyz, dx2-y2, dz2

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7 0
3 years ago
What is the percent by mass of oxygen in mg(oh)2
kakasveta [241]

54.868% because well la la la (sorry had to be 20 characters)

3 0
3 years ago
Rango carefully mixed 15.6 g of sodium with chlorine. The reaction produced 39.7 g of sodium chloride. How many grams of chlorin
Evgen [1.6K]

Answer:

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Explanation:

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5 0
3 years ago
A researcher studying the nutritional value of a new candy places a 4.90 g sample of the candy inside a bomb calorimeter and com
snow_lady [41]

Answer:

449730.879 cal/g

Explanation:

Given data:

Mass of sample = 4.9 g

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Q = C (calorimeter) × ΔT

C(candy) = C (calorimeter) × ΔT / m

C(candy) =  33.50 KJ . K⁻¹ × 275.23 K / 4.90 g

C(candy) = 9220.205 KJ / 4.90 g

C(candy) =  1881.674 KJ / g

It is known that,

1 KJ /g = 239.006 cal/g

1881.674 × 239.006 = 449730.879 cal/g

8 0
3 years ago
How much heat energy is required to raise the temperature of 0.358 kg of copper from 23.0 ∘C to 60.0 ∘C? The specific heat of co
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In order to calculate how much heat is needed to raise the temperature you need to use the formula q =mass x specific heat x (final temperature- initial temperature) where q represents heat being absorbed or released. Before you begin you would convert kg to g because the specific heat is measure in g. So you would set up the equation as q = 358 g x .092 x (60-23 degrees Celsius) which would give you 1218.6
7 0
3 years ago
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