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Arte-miy333 [17]
3 years ago
11

In substitution reactions, (CH3)3C-I reacts at the same rate with Br- and Cl- even though Br- is a more reactive nucleophile tha

n Cl-. Why?A. (CH3)3C-Br reacts by SN1 mechanism whose rate is independent of nucleophile reactivity. B. (CH3)3C-Br reacts by SN2 mechanism and therefore all nucleophiles react at the same rate regardless of their reactivity. C. The t-butyl carbocation is so reactive, the measurable rate of it's reaction with different nucleophiles is imperceptible. D. The t-butyl group sterically hinders nucleophiles, making different nucleophiles appear to react at the same rate.
Chemistry
1 answer:
Kamila [148]3 years ago
5 0

Answer:

A. (CH3)3C-I reacts by SN1 mechanism whose rate is independent of nucleophile reactivity.

Explanation:

We must recall that (CH3)3C-I is a tertiary alkyl halide. Tertiary alkyl halides preferentially undergo substitution reaction via SN1 mechanism.

In SN1 mechanism, the rate of reaction depends solely on the concentration of the alkyl halide (unimolecular mechanism) and is independent of the concentration of the nucleophile. As a result of this, both Br^- and Cl^- react at the same rate.

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CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is
BaLLatris [955]

Answer : The moles of O_2 left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of CH_4.

Using ideal gas equation:

PV=nRT

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = 27^oC=273+27=300K

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)

n=0.406mole

Now we have to calculate the moles of O_2.

The balanced chemical reaction will be:

CH_4+2O_2\rightarrow CO_2+2H_2O

From the balanced reaction we conclude that,

As, 1 mole of CH_4 react with 2 moles of O_2

So, 0.406 mole of CH_4 react with 2\times 0.406=0.812 moles of O_2

Now we have to calculate the excess moles of O_2.

O_2 is 20 % excess. That means,

Excess moles of O_2 = \frac{(100 + 20)}{100} × Required moles of O_2

Excess moles of O_2 = 1.2 × Required moles of O_2

Excess moles of O_2 = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of O_2 left in the products.

Moles of O_2 left in the products = Excess moles of O_2 - Required moles of O_2

Moles of O_2 left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of O_2 left in the products are 0.16 moles.

7 0
3 years ago
19. Which of the following best describes the role of the spark from the spark plug in an automobile engine?
sergey [27]

The role of a spark plug is to supply some of the energy of activation for the combustion reaction.

<u>Explanation:</u>

  • A Spark plug is a tiny bolt of lightning in which a spark of electricity is emitted across a gap creating the ignition of the combustion chamber thereby starting the engine. By putting an engine piston in motion we can power up which produces a smooth burn of the compressed air-fuel mixture.
  • An electrical device that fits into the cylinder head and ignites compressed aerosol gasoline by an electric spark. They have an insulated electrode connected to a coil that ignites thereby producing sparks.
  • The spark plug works as a heat exchanger. They tend to pull unwanted thermal energy from the combustion chamber and heat is transferred to the engine's cooling system. Thus they supply some of the energy for the activation of engines.
4 0
3 years ago
Identify the variables in this hypothesis.
Zanzabum

Answer:

AJJJJJJJJJJJJJJ

Explanation:

5 0
3 years ago
Read 2 more answers
You need to prepare 1 L of the citric acid/citrate buffer. You have chosen to use Method 1 (see lab presentation). Calculate the
prisoha [69]

Answer:

3.11 is the pH of the buffer

Explanation:

The pH of a buffer is obtained using H-H equation:

pH = pKa + log [Conjugate base] / [Weak acid]

<em>Where pH is the pH of the buffer, pKa = -log Ka = 3.14 for the citric buffer and [] could be taken as the moles of each species.</em>

The citric acid,HX (Weak acid), reacts with NaOH to produce sodium citrate, NaX (weak base) and water:

HX + NaOH → H2O + NaX

That means the moles of NaOH added = Moles of sodium citrate produced

And the resulitng moles of HX = Initial moles - Moles NaOH added

<em>Moles HX and NaX:</em>

Moles NaOH = 0.100L * (0.65mol / L) = 0.065 moles NaOH = Moles NaX

Moles HX = 0.300L * (0.45mol / L) = 0.135 moles HX - 0.065 moles NaOH = 0.070 moles HX

Replacing in H-H equation:

pH = 3.14 + log [0.065mol] / [0.070mol]

pH = 3.11 is the pH of the buffer

8 0
2 years ago
How much time would it take for 336 mg of copper to be plated at a current of 5.6 A ? Express your answer using two significant
schepotkina [342]

Answer:

1.8 × 10² s

Explanation:

Let's consider the reduction that occurs upon the electroplating of copper.

Cu²⁺(aq) + 2 e⁻ ⇒ Cu(s)

We will establish the following relationships:

  • 1 g = 1,000 mg
  • The molar mass of Cu is 63.55 g/mol
  • When 1 mole of Cu is deposited, 2 moles of electrons circulate.
  • The charge of 1 mole of electrons is 96,486 C (Faraday's constant).
  • 1 A = 1 C/s

The time  that it would take for 336 mg of copper to be plated at a current of 5.6 A is:

336mgCu \times \frac{1gCu}{1,000mgCu} \times \frac{1molCu}{63.55gCu} \times \frac{2mole^{-} }{1molCu} \times \frac{94,486C}{1mole^{-}} \times \frac{1s}{5.6C} = 1.8 \times 10^{2} s

3 0
3 years ago
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