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Brilliant_brown [7]
3 years ago
8

. The nucleophile in the reaction is _______ b. The Lewis acid catalyst in the reaction is ______ c. This reaction proceeds_____

______(faster or slower) than benzene. d. Draw the structure of product D
Chemistry
1 answer:
strojnjashka [21]3 years ago
6 0

Answer:

a. eletrophile

b. able to impose regioselectivity and stereo selectivity.

c. faster

Explanation:

Necleophile reaction is chemical reaction in which electron rich chemical specie replaces functional group with another electron deficient molecule. Lewis acid catalyst is organic chemical reaction which lewis acid act as electron pair acceptor. Nucleophile reaction proceeds about 25 times more faster than benzene.

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Calculate the volume (liters) of solution that can be prepared from 180. grams of K2Cr2O7 for a 0.200M solution.
Sladkaya [172]

Answer:

The volume of the solution is 0.305 liters.

Explanation:

Molar mass is the amount of mass that a substance contains in one mole. The molar mass of K₂Cr₂O₇ is 294 g / mole. Then you can apply the following rule of three: if by definition of molar mass 294 grams of the compound are contained in 1 mole, 180 grams are contained in how many moles?

moles=\frac{180 grams*1mole}{294 grams}

moles= 0.61

Molarity is a measure of the concentration of a substance that is defined as the number of moles contained in a certain volume. So, the molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

molarity=\frac{number of moles of solute}{volume}

Molarity is expressed in units \frac{moles}{liter}.

In this case:

  • molarity= 2 M
  • number of moles of solute= 0.61 moles
  • volume= ?

Replacing in the definition of molarity:

2 M=\frac{0.61 moles}{volume}

Solving:

volume=\frac{0.61 moles}{2 M}

volume= 0.305 liters

<u><em>The volume of the solution is 0.305 liters.</em></u>

8 0
3 years ago
I don’t know if I’m right??
allochka39001 [22]

Answer:

what is this for? A-level..?!

Explanation:

3 0
3 years ago
Read 2 more answers
What is homogeneous mixture?
Nataly [62]

A mixture is a portion of matter made up of two or more substances called constituents. Mixtures are the product of the mechanical joining of substances without change in chemical nature, and therefore, each constituent retains its properties.

the option 4

4 0
3 years ago
Given the Henry’s law constant for O2(4.34*109Pa) at 25° C, calculate the molar concentration of oxygen in air-saturated and O2s
Flauer [41]

This is an incomplete question, here is a complete question.

The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.

Answer : The molar concentration of oxygen is, 2.67\times 10^2mol/L

Explanation :

As we know that,

C_{O_2}=k_H\times p_{O_2}

where,

C_{O_2} = molar solubility of O_2 = ?

p_{O_2} = partial pressure of O_2 = 0.2 atm  = 1.97×10⁻⁶ Pa

k_H = Henry's law constant  = 4.34 × 10⁹ g/L.Pa

Now put all the given values in the above formula, we get:

C_{O_2}=(4.34\times 10^9g/L.Pa)\times (1.97\times 10^{-6}Pa)

C_{O_2}=8.55\times 10^3g/L

Now we have to molar concentration of oxygen.

Molar concentration of oxygen = \frac{8.55\times 10^3g/L}{32g/mol}=2.67\times 10^2mol/L

Therefore, the molar concentration of oxygen is, 2.67\times 10^2mol/L

8 0
3 years ago
A closed, frictionless piston-cylinder contains a gas mixture with the following composition on a mass basis: 40% carbon dioxide
Hitman42 [59]

Answer:

W=-37.6kJ, therefore, work is done on the system.

Explanation:

Hello,

In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe

Next, the total moles:

n_T=0.00727kmol+0.00625kmol+0.0139kmol=0.02742kmol

After that, since the process is isobaric, we can compute the work as:

W=P(V_2-V_1)

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3

Thereby, the magnitude and direction of work turn out:

W=450kPa(0.19m^3-0.27m^3)\\\\W=-37.6kJ

Thus, we conclude that since it is negative, work is done on the system (first law of thermodynamics).

Regards.

7 0
3 years ago
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