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3 years ago

10

After hydrogen bonding the strongest intermolecular force would be ___

1 answer:

mihalych1998 [28]3 years ago

6 0

The answer is van der Waal force, or to be precise, London dispersion forces.

This type of force is present between all molecules and they are caused by electrons that are orbiting the atom cores. The moving of the electrons can result in making the atom temporary dipole due to the asymmetry of electron distribution around the core. This temporary dipole now can interact with other atoms leading to an electrostatic interaction. The more electrons an atom has, the stronger are these intermolecular forces.

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Which of the following is true of solids?

Over [174]

<span>Answer: B. Ionic solids have higher melting points than molecular solids.
</span>
This is because the rest are false, as solids are able to melt, and do have melting points. Also, not all solids have the same melting points.

5 0

3 years ago

Hydrogen and iodine react to form hydrogen iodide, like this: H_2 (g) + I_2 (g) rightarrow 2 HI(g) Also, a chemist finds that at

telo118 [61]

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

Hydrogen and iodine react to form hydrogen iodide, like this:

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen, iodine, and hydrogen iodide has the following composition:

Compound Pressure at equilibrium

$H_2$ 61.8 atm

$I_2$ 46.5 atm

$HI$ 52.3 atm

Calculate the value of the equilibrium constant $K_p$ for this reaction. Round your answer to 2 significant digits.

Answer : The value of equilibrium constant $K_p$ for this reaction is, 0.952

Explanation :

The given chemical reaction :

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

The expression of $K_p$ for above reaction follows:

$K_p=\frac{(P_{HI})^2}{P_{H_2}\times P_{I_2}}$

We are given:

$P_{H_2}=61.8atm$

$P_{I_2}=46.5atm$

$P_{HI}=52.3atm$

Putting values in above equation, we get:

$K_p=\frac{(52.3)^2}{61.8\times 46.5}\\\\K_p=0.952$

Therefore, the value of equilibrium constant $K_p$ for this reaction is, 0.952

7 0

3 years ago

He diagram below represents part of the oxygen cycle.

salantis [7]

Answer:

The answer is C

Explanation:

7 0

3 years ago

Please can any one help me by answering this question!!+ignore my answer!,

den301095 [7]

I have posted the answetr in this picture.

8 0

3 years ago

What is the volume of 33.25g of butane gas at 293 C and 10.934 kPa?

PSYCHO15rus [73]

Answer:

V = 240.79 L

Explanation:

Given data:

Volume of butane = ?

Temperature = 293°C

Pressure = 10.934 Kpa

Mass of butane = 33.25 g

Solution:

Number of moles of butane:

Number of moles = mass/ molar mass

Number of moles = 33.25 g/ 58.12 g/mol

Number of mole s= 0.57 mol

Now we will convert the temperature and pressure units.

293 +273 = 566 K

Pressure = 10.934/101 = 0.11 atm

Volume of butane:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

V = nRT/P

V = 0.57 mol × 0.0821 atm.L/ mol.K ×566 K / 0.11 atm

V = 26.49 L/0.11

V = 240.79 L

6 0

3 years ago