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juin [17]
3 years ago
13

What change would you expect on the rate of theSN2 reaction of 1-iodo-2-methylbutane withcyanide ion if the nucleophile concentr

ation isunchanged and the alkyl halide concentration isdoubled?
What change would you expect on the rate of thereaction of ethanol with 2-iodo-2-methylbutane ifthe nucleophile concentration is halved by adding diethylether as an inert solvent and the alkyl halideconcentration is unchanged?
Chemistry
1 answer:
Ivenika [448]3 years ago
6 0

Answer:

The reaction rate of the both questions remain unchanged.

Explanation:

      For question 1:  The reaction 1-iodo -2- methylbutane with cyanide ion is an SN2 reaction because the Alkyl halide is a primary alkyl halide. The rate of reaction is dependent on concentration of the nucleophile and the alkyl halide at the same. For the rate of reaction to be affected (increased or decreased), the concentration of nucleophile and the alkyl halide have to be altered.

      For question 2: The reaction of 2-iodo -2- methylbutane with ethanol is an SN1 reaction because the Alkyl halide is a tertiary alkyl halide. There are two-step reaction mechanism in this reaction. The first step is the rate determining step which determines the extent of the reaction and hence the rate of reaction. For the rate of reaction to be affected (increased or decreased), the concentration of the Alkyl halide alone will be altered. The rate of reaction is independent of the concentration of the nucleophile.

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Which is generally stronger, intermolecular interactions or intramolecular interactions? Which is generally stronger, intermolec
Inessa [10]

Answer:

A. Intramolecular interactions are generally stronger.

B. a. Only intermolecular interactions are broken when a liquid is converted to a gas.

Explanation:

<em>A. Which is generally stronger, intermolecular interactions or intramolecular interactions?</em>

Intramolecular interactions, in which electrons are gained, lost or shared, constitute true bonds and are one or two orders of magnitude stronger than intermolecular interactions.

<em>B. Which of these kinds of interactions are broken when a liquid is converted to a gas?</em>

When a liquid vaporizes, the intermolecular attractions are broken, that is, molecules get more separated. However, true bonds are not broken which is why the molecules keep their chemical identity.

7 0
4 years ago
1. What happens when like charges are brought closer to each other?<br><br><br>​
deff fn [24]
Like charges repel each other; unlike charges attract. Thus, two negative charges repel one another, while a positive charge attracts a negative charge. The attraction or repulsion acts along the line between the two charges. The size of the force varies inversely as the square of the distance between the two charges.
5 0
3 years ago
Read 2 more answers
You perform a distillation to separate a mixture of propylbenzene and cyclohexane, and you obtain 2.9949 grams of cyclohexane (d
Romashka-Z-Leto [24]

Answer:

66.67%

Explanation:

From the given information:

mass of cyclohexane = 2.9949 grams

density of cyclohexane = 0.779 g/mL

Recall that:

Density = mass/volume

∴

Volume = mass/density

So, the volume of cyclohexane = 2.9949 g/ 0.779 g/mL

= 3.8445 mL

Also,

mass of propylbenzene = 1.6575 grams

density of propylbenzene = 0.862 g/mL

Volume of propylbenzene =  1.6575 g/ 0.862 g/mL

= 1.9229 mL

The volume % composition of cyclohexane from the mixture is:

= (\dfrac{v_{cyclohexane}}{v_{cyclohexane}+v_{propylbenzene}})\times 100

= (\dfrac{3.8445}{3.8445+1.9229})\times 100

= (\dfrac{3.8445}{5.7674})\times 100

= 66.67%

6 0
3 years ago
What is the total number of moles represented by 20 grams of CACO3
Nutka1998 [239]

Answer:

B. 0.2.

Explanation:

  • We can use the relation:

<em>n = mass/molar mass</em>

mass of CaCO₃ = 20 g, molar mass of CaCO₃ = 100.0869 g/mol.

<em>∴ n = mass/molar mass = </em>(20 g)/(100.0869 g/mol) <em>= 0.1998 ≅ 0.2 mol.</em>

<em></em>

<em>So, the right choice is: B. 0.2.</em>

8 0
3 years ago
The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g
ra1l [238]

Answer:

1) When 6.97 grams of sodium(s) react with excess water(l), 56.0 kJ of energy are evolved.

2) When 10.4 grams of carbon monoxide(g) react with excess water(l), 1.04 kJ of energy are absorbed.

Explanation:

1) The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g).

2 Na(s) + 2H₂O(l) ⇒ 2NaOH(aq) + H₂(g) ΔH = -369 kJ

The enthalpy of the reaction is negative, which means that 369 kJ of energy are evolved per 2 moles of sodium. The energy evolved for 6.97 g of Na (molar mass 22.98 g/mol) is:

6.97g.\frac{1mol}{22.98g} .\frac{-369kJ}{2mol} =-56.0kJ

2) The following thermochemical equation is for the reaction of carbon monoxide(g) with water(l) to form carbon dioxide(g) and hydrogen(g).

CO(g) + H₂O(l) ⇒ CO₂(g) + H₂(g)  ΔH = 2.80 kJ

The enthalpy of the reaction is positive, which means that 2.80 kJ of energy are absorbed per mole of carbon monoxide. The energy evolved for 10.4 g of CO (molar mass 28.01 g/mol) is:

10.4g.\frac{1mol}{28.01g} .\frac{2.80kJ}{mol} =1.04kJ

3 0
4 years ago
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