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UNO [17]
2 years ago
10

1-Bromobutane reacts with water to form 1-butanol according to the following equation.

Chemistry
1 answer:
photoshop1234 [79]2 years ago
6 0

1-Bromobutane reacts with water to form 1-butanol according to the following equation overall rate law for this proposed mechanism is rate = k [C₄H₉Br]

The reaction intermediate in the chemical species that is formed in one elementary step and consumed in subsequent step and the slowest step in the reaction mechanism is rate determining step and the rate determining step limit the overall rate and therefore determines the rate law for the overall reaction

Here given reaction

C₄H₉Br(aq) + 2H₂O(l) → C₄H₉OH(aq) + H₃O⁺(aq) + Br⁻(aq)

And the following is a proposed mechanism for the reaction

C₄H₉Br(aq) → C₄H₉⁺(aq) + Br⁻(aq)     slow

C₄H₉⁺(aq) + H₂O(l) → C₄H₉OH₂⁺(aq)   fast

C₄H₉OH₂⁺(aq) + H₂O(l) → C₄H₉OH(aq) + H₃O⁺(aq)   fast

So in that 1st step is the rate determining step means slow step

So rate expressions take the general form: rate = k . [A]a . [B]b

So given equation rate expressions is k [C₄H₉Br]

Know more about rate expression

brainly.com/question/14207841

#SPJ1

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A hurricane is MOST LIKELY to occur in an area
Flura [38]

Answer:

Western Pacific, such as the Philippines, Guam, southeast Asia (including China and Taiwan) and Japan.

Explanation:

6 0
3 years ago
Find the amount of heat energy needed to convert 400 grams of ice at -38°C to steam at 160°C.
Marianna [84]

The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)

<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
  • Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
  • Initial temperature (T₁) = –25 °C
  • Final temperature (T₂) = 0 °
  • Change in temperature (ΔT) = 0 – (–38) = 38 °C
  • Specific heat capacity (C) = 2050 J/(kg·°C)
  • Heat (Q₁) =?

Q = MCΔT

Q₁ = 0.4 × 2050 × 38

Q₁ = 31160 J

<h3>How to determine the heat required to melt the ice at 0 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
  • Heat (Q₂) =?

Q = mL

Q₂ = 0.4 × 334000

Q₂ = 133600 J

<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 0 °C
  • Final temperature (T₂) = 100 °C
  • Change in temperature (ΔT) = 100 – 0 = 100 °C
  • Specific heat capacity (C) = 4180 J/(kg·°C)
  • Heat (Q₃) =?

Q = MCΔT

Q₃ = 0.4 × 4180 × 100

Q₃ = 167200 J

<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
  • Heat (Q₄) =?

Q = mHv

Q₄ = 0.4 × 2260000

Q₄ = 904000 J

<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 100 °C
  • Final temperature (T₂) = 160 °C
  • Change in temperature (ΔT) = 160 – 100 = 60 °C
  • Specific heat capacity (C) = 1996 J/(kg·°C)
  • Heat (Q₅) =?

Q = MCΔT

Q₅ = 0.4 × 1996 × 60

Q₅ = 47904 J

<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>
  • Heat for –38 °C to 0°C (Q₁) = 31160 J
  • Heat for melting (Q₂) = 133600 J
  • Heat for 0 °C to 100 °C (Q₃) = 167200 J
  • Heat for vaporization (Q₄) = 904000 J
  • Heat for 100 °C to 160 °C (Q₅) = 47904 J
  • Heat for –38 °C to 160 °C (Qₜ) =?

Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Qₜ = 31160 + 133600 + 167200 + 904000 + 47904

Qₜ = 1.28×10⁶ J

Learn more about heat transfer:

brainly.com/question/10286596

#SPJ1

7 0
2 years ago
2. What is the average height in cm for plants in group 2? * 38 cm 56 cm 97 cm 114 cm
Nina [5.8K]

Answer:

76.25cm

Explanation:

38cm + 56cm +97cm +114cm = 305cm

305cm÷4 = 76.25cm

6 0
3 years ago
The equilibrium constant is given for two of the reactions below. Determine the value of the missing equilibrium constant. 2A(g)
monitta

Answer:

The value of the missing equilibrium constant ( of the first equation) is 1.72

Explanation:

First equation: 2A + B ↔ A2B   Kc = TO BE DETERMINED

 ⇒ The equilibrium expression for this equation is written as: [A2B]/[A]²[B]

Second equation: A2B + B ↔ A2B2   Kc= 16.4

⇒ The equilibrium expression is written as: [A2B2]/[A2B][B]

Third equation:  2A + 2B ↔ A2B2     Kc = 28.2

⇒ The equilibrium expression is written as: [A2B2]/ [A]²[B]²

If we add the first to the second equation

2A + B + B ↔ A2B2   the equilibrium constant Kc will be X(16.4)

But the sum of these 2 equations, is the same as the third equation ( 2A + 2B ↔ A2B2)   with Kc = 28.2

So this means: 28.2 = X(16.4)

or X = 28.2/16.4

X = 1.72

with X = Kc of the first equation

The value of the missing equilibrium constant ( of the first equation) is 1.72

7 0
3 years ago
Changing the number of protons in an atom ONLY changes the weight of an atom.<br><br> Need help ASAP
likoan [24]

Answer:

no

Explanation:

protons and electrons are equal to atomic number

3 0
3 years ago
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