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3 years ago

What type of intermolecular force contributes to the high boiling point of water??

2 answers:

8 0

Hydrogen bonding, a particularly strong dipole-dipole force, causes a powerful attraction between water molecules, which results in a high boiling point.

____ [38]3 years ago

4 0

Hydrogen bonding, a particularly strong dipole-dipole force, causes a powerful attraction between water molecules, which results in a high boiling point.

You might be interested in

A gold atom in its neutral state has how many electrons

riadik2000 [5.3K]

To find the number of electrons in a neutral gold atom look at it's atomic number. The atomic number is the number of protons in the element. For an atom to be neutral it needs the same number of electrons as protons, therefore the positive and negative charges will be balanced out. Gold's atomic number is 79 therefore it has 79 electrons.

4 0

4 years ago

13. The equation used to find acceleration is a =

lubasha [3.4K]

Acceleration represented by the equation = Δv/Δt.

6 0

3 years ago

The second messenger cyclic AMP (cAMP) is synthesized from ATP by the activity of the enzyme adenylyl cyclase. Cyclic AMP, in tu

scoundrel [369]

Answer: 1. Cyclic AMP(cAMP) concentration increases;

2. cAMP then binds to the two regulatory subunits;

3. A change in the conformation of enzyme results. The autoinhibitory sequences move away from the active site of the catalytic subunits;

4. The catalytic sites are now free to catalyse the various reactions of PKA.

(Note: The question did not list the steps to be placed in the correct order)

Explanation:

The inactive form of Protein Kinase A (PKA) contains two identical sub-units (C) and two identical regulatory sub-units (R). Two auto-inhibitory sequences of the regulatory subunits occupy the catalytic sites of the catalytic subunits. After the reaction of adenylyl cyclase, this inactive form of PKA is activated in the following steps:

1. Cyclic AMP(cAMP) concentration increases;

2. cAMP then binds to the two regulatory subunits;

3. A change in the conformation of enzyme results. The autoinhibitory sequences move away from the active site of the catalytic subunits;

4. The catalytic sites are now free to catalyse the reactions various of PKA which includes the regulation of glycogen, sugar, and lipid metabolism.

8 0

3 years ago

If you have 45.6mL of a liquid and its

tankabanditka [31]

The density of the liquid is 1.027g/cm³

D=M/V
D=46.83/45.6
D=1.027g/cm³

4 0

3 years ago

1. If you have a sample of gas at a pressure of 16 atm, what will the pressure be if the volume is halved?

vagabundo [1.1K]

Answer:

1. The pressure will be 32 atm, twice the initial pressure.

2. The pressure will be 1.83 atm, one third of the initial pressure.

Explanation:

Boyle's law is one of the gas laws that relates the volume and pressure of a certain quantity of gas kept at a constant temperature.

This law says that "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

or P * V = k

Ahora es posible suponer que tienes un cierto volumen de gas V1 que se encuentra a una presión P1 al comienzo del experimento. Si varias el volumen de gas hasta un nuevo valor V2, entonces la presión cambiará a P2, y se cumplirá:

P1*V1=P2*V2

1. In this case:

P1= 16 atm
V1
P2= ?
V2= V1÷2= $\frac{V1}{2}$ because the volume is halved.

So:

16 atm*V1= P2* $\frac{V1}{2}$

Solving:

$\frac{16 atm*V1*2}{V1}$ =P2

16 atm*2= P2

32 atm= P2

The pressure will be 32 atm, twice the initial pressure.

2. Now

P1= 5.5 atm
V1
P2= ?
V2= V1*3 because the volume is tripled.

So:

5.5 atm*V1= P2* V1*3

Solving:

$\frac{5.5 atm*V1}{3*V1}$ =P2

$\frac{5.5 atm}{3}$ = P2

1.83 atm= P2

The pressure will be 1.83 atm, one third of the initial pressure.

7 0

3 years ago