All categories

3 years ago

Of the following substances, only _____ has London dispersion forces as its only intermolecular force.

Chemistry

1 answer:

Aleksandr [31]3 years ago

5 0

Answer:

Explanation:

a) C6H14, C8H18 – London-dispersion force

Since both substances have the same intermolecular force, the substances with the

larger molecular mass with have the higher boiling point. In this case, C8H18 will

have the higher boiling point.

b) C3H8 – London-dispersion force

CH3OCH3 –Dipole-dipole

The higher boiling point belongs to CH3OCH3 because is has the stronger

intermolecular force, dipole-dipole interaction.

c) CH3OH – Hydrogen bonding

CH3SH – Dipole-dipole interaction

Hydrogen bonding is the strongest intermolecular force, so CH3OH will have the

higher boiling point.

d) NH2NH2 – Hydrogen bonding

CH3CH3 – London-dispersion force

Hydrogen bonding is the strongest intermolecular force, so NH2NH2 will have the

higher boiling point.

You might be interested in

Summarize the process of alcoholic fermentation in yeast

AysviL [449]

Alcoholic fermentation is mainly used by various yeast species to make energy.

If there is no oxygen available, the yeasts have in the alcoholic fermentation another possibility of energy supply. But they can - as compared with cellular respiration - recover substantially less energy from glucose, in the form of adenosine triphosphate (ATP): by complete oxidation, a molecule of glucose provides 36 molecules of ATP, but by alcoholic fermentation only 2 molecules of ATP. These two molecules are obtained in glycolysis, the first step in the chain of reactions for both cellular respiration and fermentation.

The two additional steps of the fermentation, and thus the production of ethanol serve not to make energy, but the regeneration of the NAD + cofactor used by the enzymes of glycolysis. As NAD + is available in limited quantities, it is converted by the NADH reduced state fermentation enzymes to the NAD + oxidized state by reduction of acetaldehyde to ethanol.

6 0

4 years ago

How to draw Hess' Cycle for this question ?

NISA [10]

Answer : The standard enthalpy of formation of ethylene is, 51.8 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of $C_2H_4$ will be,

$2C(s)+2H_2(g)\rightarrow C_2H_4(g)$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$ $\Delta H_1=-1411kJ/mole$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.7kJ/mole$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_3=-285.9kJ/mole$

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equations, we get :

(1) $2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)$ $\Delta H_1=+1411kJ/mole$

(2) $2C(s)+2O_2(g)\rightarrow 2CO_2(g)$ $\Delta H_2=2\times (-393.7kJ/mole)=-787.4kJ/mole$

(3) $2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=2\times (-285.9kJ/mole)=-571.8kJ/mole$

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+1411kJ/mole)+(-787.4kJ/mole)+(-571.8kJ/mole)$

$\Delta H=51.8kJ/mole$

Therefore, the standard enthalpy of formation of ethylene is, 51.8 kJ/mole

7 0

3 years ago

Kp for the following reaction is 0.16 at 25 degree C. 2 NOBr(g) 2 NO(g) Br_2(g) The enthalpy change for the reaction at standard

finlep [7]

Answer:

Explanation:

Given that:

$2 NOBr_{(g)} \iff 2 NO_{(g)} + Br_{2(g)}$

From above:

$K_p = 0.16 = \dfrac{(P_{NO})^2 (P_{Br})}{(P_{NOBr})^2}$

To predict the effect of the addition of Br₂(g);

The addition of Br₂(g) will favor the equilibrium to shift to the left i.e. formation of NOBr

The removal of some NOBr will cause the equilibrium position to shift to the left side. This is because concentration on the left side is decreased and the concentration on the right side will be increased. Thus, the equilibrium will shift towards where the concentration is reduced which is the left side.

5 0

3 years ago

Help, please? will mark brainliest

alexira [117]

Answer:

Sulfur would gain electrons

Explanation:

Atoms want to have a complete out valence shell and because sulfur only needs 2 more electrons to complete the outer shell it would take 2 more.

8 0

3 years ago

What particles form the nucleus of an atoms?

melomori [17]

Answer:

Protons and neutrons

Explanation:

The particles to forms the nucleus of an atom are the protons and neutrons of the atoms.

These materials are located in the tiny nucleus and contributes the most mass of the atom.

Protons are the positively charged particles in an atom
Neutrons do not carry any charges.
Sum of the protons and neutrons gives the mass number of the atom.

7 0

3 years ago