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Naya [18.7K]
3 years ago
5

Choose the substance with the lowest viscosity. choose the substance with the lowest viscosity. f2chchf2 f2chch2f fch2ch2f f3cch

f2 f3ccf3
Chemistry
2 answers:
Maksim231197 [3]3 years ago
7 0

Answer : FCH_{2}CH_{2}F

Explanation : In the molecule of FCH_{2}CH_{2}F  the hydrogen bonding forces tend to resist any phase transition form liquid state to gaseous phase, and these hydrogen bonding forces are greater  usually when the hydrogen is bound to an electronegative atom like chlorine, fluorine,oxygen,etc. This then lowers the viscosity of the compound.

Viktor [21]3 years ago
7 0

The substance with lowest viscosity is \boxed{{\text{FC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{F}}}.

Further explanation:

Viscosity is defined as the resistance of any liquid to flow. Stronger the intermolecular forces of attraction within the liquid, larger will be its viscosity.

Intermolecular forces:

The forces that exist between the molecules are known as intermolecular forces (IMF). They are electrostatic in nature and determine the bulk properties of the substances like melting and boiling points. Molecules are held in any substance due to these forces.

The various types of intermolecular forces are as follows:

1. Hydrogen bonding.

2. Ion-dipole forces.

3. Ion-induced dipole forces.

4. Dispersion forces

Hydrogen bonding:  

It is an attractive force that exists between hydrogen and more electronegative elements like N, O, F. It can either be intermolecular or intramolecular. Intermolecular hydrogen bonding is the one that occurs between different molecules. For example, the bond between HF and {{\text{H}}_{\text{2}}}{\text{O}} is an intermolecular hydrogen bond. Intramolecular hydrogen bonding occurs between various parts of the same molecule. Ortho-nitro phenol and salicylaldehyde show this type of bonding.

With an increase in the number of fluorine atoms in the compound, the polarity of the compound increases. Due to increase in polarity of compound, the tendency to form hydrogen bonds with water molecules also increases and that is the reason for increase in viscosity or the compound.

{{\text{F}}_{\text{3}}}{\text{CHCH}}{{\text{F}}_{\text{3}}} has the highest number of fluorine atoms among all the compounds, so it has highest tendency to form hydrogen bonding that attributes to its highest viscosity.

{\text{FC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{F}} has lowest number of fluorine atoms among all the compounds, so it has lowest tendency to form hydrogen bonding that attributes to its lowest viscosity.

Learn more:

1. Determination of kinetic energy of the emitted electrons when cesium is exposed to UV rays:  

shttps://brainly.com/question/5031462

2. At what gas temperature would the average translational kinetic energy of a helium atom be equal to that of an oxygen molecule: brainly.com/question/9924094

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Kinetic Energy

Keywords: Kinetic energy, viscosity, intermolecular forces, hydrogen bonding, electronegative, fluorine, polarity, and hydrogen.

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Answer:

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Explanation:

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Q = m.c. ΔT

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A. Diethyl ether will react with the alkenes that were formed in the experiment.

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According to stoichiometry :

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8 0
3 years ago
When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing poin
LenaWriter [7]

The question is incomplete, the complete question is:

When 177. g of alanine (C_3H_7NO_2) are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is 5.9^oC lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is 7.2^oC lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X.

<u>Answer:</u> The van't Hoff factor for potassium bromide in X is 1.63

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

\Delta T_f=i\times K_f\times m

OR

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}} ......(1)

  • <u>When alanine is dissolved in mystery liquid X:</u>

\Delta T_f=5.9^oC

i = Vant Hoff factor = 1 (for non-electrolytes)

K_f = freezing point depression constant

m_{solute} = Given mass of solute (alanine) = 177. g

M_{solute} = Molar mass of solute (alanine) = 89 g/mol

w_{solvent} = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

5.9=1\times K_f\times \frac{177\times 1000}{89\times 800}\\\\K_f=\frac{5.9\times 89\times 800}{1\times 177\times 1000}\\\\K_f=2.37^oC/m

  • <u>When KBr is dissolved in mystery liquid X:</u>

\Delta T_f=7.2^oC

i = Vant Hoff factor = ?

K_f = freezing point depression constant = 2.37^oC/m

m_{solute} = Given mass of solute (KBr) = 177. g

M_{solute} = Molar mass of solute (KBr) = 119 g/mol

w_{solvent} = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

7.2=i\times 2.37\times \frac{177\times 1000}{119\times 800}\\\\i=\frac{7.2\times 119\times 800}{2.37\times 177\times 1000}\\\\i=1.63

Hence, the van't Hoff factor for potassium bromide in X is 1.63

7 0
3 years ago
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