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2 years ago

CH3CN major species present when dissolved in water

2 answers:

5 0

The major species present when CH3CN is dissolved in water are CH3+ and CN-. It doesn't completely dissociate in water since it is not a strong acid or strong base. Only a strong acid or base will dissociate completely in water. For example HCl dissociates into H+ and Cl-, and NaOH dissociates into Na+ and OH-.

Margaret [11]2 years ago

3 0

Answer: CH₃CN and H₂O.

Explanation:

1) The spieces present in a solution may be either the molecules, in case of covalent compounds, or ions, in case of ionic compounds that dissociate (ionize).

2) Both, CH₃CN and H₂O are covalent (polar covalent) substances, so they do not ionize and the spieces in the solution are the molecules per se.

3) In solution, the molecules of H₂O will solvate the molecules of CH₃CN, meaning that H₂O molecules are able to separate the molecules of CH₃N from each other, and so every molecule of CH₃CN will end surrounded by many molecules of H₂O.

This happens because the interaction between the polar molecules of the two different compounds is strong enough to overcome the intermolecular forces between the molecules of the same compound.

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Are the strengths of the interactions between the particles in the solute and between the particles in the solvent before the so

Colt1911 [192]

Answer:

Less than

Explanation:

The process of dissolution occurs as a kind of "tug of war". On one side are the solute-solute and solvent-solvent interaction forces, while on the other side are the solute-solvent forces.

Only when the solute-solvent forces are strong enough to overcome the pre-mixing forces do they overcome the "tug of war", and thus dissolution occurs.

Thus, it is concluded that the interaction forces between solute particles and solvent particles before they are combined are less than the interaction forces after dissolution.

5 0

2 years ago

Draw a structure for the product of nucleophilic substitution obtained on solvolysis of tert−butyl bromide in methanol, and arra

Elena-2011 [213]

Answer:

2-methoxy-2-methylpropane

Explanation:

The first step for this reaction is the carbocation formation. In this step, a tertiary carbocation is formed. Also, we will have a good leaving group so bromide will be formed. Then the methanol acts as a nucleophile and attacks the carbocation. Next, a positive charge is generated upon the oxygen, this charge can be removed when the hydrogen leaves the molecule as $H^+$ . (See figure)

6 0

2 years ago

A group of students measured the average monthly temperature of 1000 cities around the world and plotted the cumulative frequenc

mr Goodwill [35]

Answer:

The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

Explanation:

For comparing the mean absolute deviations of both data sets we have to calculate the mean absolute deviation for both data sets first,

So for city 1:

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored as the deviation is the distance of value from the mean and it cannot be negative. For this purpose absolute is used)

The deviations will be added then.

So the mean absolute deviation for city 1 is 24 ..

For city 2:

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored)

The deviations will be added then.

So the MAD for city 2 is 11.33 ..

So,

The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

8 0

2 years ago

The mole fraction of hcl in a solution prepared by dissolving 5.5 g of hcl in 200 g of c2h6o is ________. the density of the sol

Thepotemich [5.8K]

The answer:
all that we search for is the number of mole of HCl and the number of mole of C2H6O

M(HCl) = 5.5g/ mole of HCl , so mole of HCl = 5.5/M(HCl), where M(HCl) is the molar mass.
M(HCl) = 1+ 36.5= 37.5

moles of HCl = 5.5/37.5=0.14

M(C2H6O) = 200g / moles of C2H6O, so moles of C2H6O=200g / M(C2H6O)

M(C2H6O)= 2x12+ 6 + 16=46,

moles of C2H6O=200g / 46 =4.35 moles

the sum of the moles is 0.14 + 4.35 = 4.501 moles

finally, The mole fraction of hcl in a solution prepared by dissolving 5.5 g of hcl in 200 g of c2h6o is 0.031

because it can be found by 0.14 / 4.501= 0.031

5 0

2 years ago

Hydrogen gains stability as the H2 molecule by forming the noble gas configuration of______.

grandymaker [24]

I believe that it is a. I tried...

7 0

2 years ago