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Harlamova29_29 [7]
3 years ago
7

This is a specific type of dipole-dipole force that exists between molecules with hydrogen atoms and molecules with nitrogen, ox

ygen or fluorine atoms
Chemistry
2 answers:
RoseWind [281]3 years ago
5 0
Answer:  This description refers to:  "hydrogen bonding" .
______________________________________________________

         
<u>  Hydrogen bonding  </u> is a specific type of "dipole-dipole force" that exists <u /><em><u>between</u> </em> [molecules with hydrogen atoms]  <em><u>and</u> </em> [molecules with nitrogen, oxygen, or fluorine atoms.] .
______________________________________________________
cricket20 [7]3 years ago
5 0

Answer:

This is a specific type of dipole-dipole force is known as H<u><em>ydrogen bonding.</em></u>

Explanation:

Hydrogen bonding (H-bonding) is an intermolecular force having partial ionic-covalent character. H-bonding takes place between a hydrogen atom attached with an electronegative atom( like : O, N and F) and lone pairs of electrons on an neighboring atoms.

Generally this type of bonding is observed where atoms like nitrogen , oxygen, fluorine attached to another molecule are present in neighbors of a hydrogen atom attached to other molecule. This bonding arises due to the interaction between the  developed partial positive and negative charges on the hydrogen and electronegative atoms.

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PLS 10 points!!!
nekit [7.7K]
50m^2



Explanation:

10*5=50
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7 0
3 years ago
Read 2 more answers
The equilibrium constant, K, for the following reaction is 5.10X10 at 548 K. NH_CH(s) 2 NH3(E) + HC1(2) Calculate the equilibriu
Serga [27]

Answer:

The equilibrium concentration of HCl is 0.01707 M.

Explanation:

Equilibrium constant of the reaction = K_c=5.10\times 10^{-6}

Moles of ammonium chloride = 0.573 mol

Concentration of ammonium chloride = \frac{0.573 mol}{1.00 L}=0.573 M

     NH_4HCl(s)\rightleftharpoons 2 NH_3(g) + HCl(g)

Initial:            0.573     0           0

At eq'm:      (0.573-x)   x           x

We are given:

[NH_4Cl]_{eq}=(0.573-x)

[HCl]_{eq}=x

[NH_3]_{eq}=x

Calculating for 'x'. we get:

The expression of K_{c} for above reaction follows:

K_c=\frac{[HCl][NH_3]}{[NH_4Cl]}

Putting values in above equation, we get:

5.10\times 10^{-6}=\frac{x\times x}{(0.573-x)}

2.9223\times 10^{-6}-5.10x\times 10^{-6}=x^2

x^2-2.9223\times 10^{-6}+(5.10\times 10^{-6})x=0

On solving this quadratic equation we get:

x = 0.01707 M

The equilibrium concentration of HCl is 0.01707 M.

3 0
3 years ago
Identify the unit cell that has a = B = y = 90°.
elixir [45]

Monoclinic I think. I'm not sure

8 0
3 years ago
light frequency 1.40x10^15 Hz strike a surface causing photoelectrons to leave with a KE of 1.05 eV what is the work function in
guapka [62]

Answer:

The Work Function is 4.74eV.

Explanation:

The photoelectric effect is a phenomenon in which a beam of light is projected towards a metal plate and this leads to the ejection of electrons from the metal. The minimum amount of energy required for this ejection of electrons is known as the work function. When the energy of the light beam is less than the work function, then the electrons are not emitted.

The photoelectric effect is represented by the equation:

E = φ + KE  where

E = energy of the incident photons

KE = kinetic energy of photoelectrons

φ = Work function of the metal

This can also be written in terms of frequency as:

h = h₀ + \frac{1}{2} mv²

= frequency of the incident light beam

₀ = frequency of the photoelectrons

m = mass of the electron

v = velocity of the electron

h = Plank's constant 6.623 x 10⁻³⁴Js

Conversion factor: 1eV = 1.602 x 10⁻¹⁹ J

So, 1.05eV = 1.682 10⁻¹⁹ J

Using the formulas,

(6.623 x 10⁻³⁴Js)(1.40x10¹⁵ s⁻¹) = φ + 1.682 x 10⁻¹⁹ J

(9.272 x 10⁻¹⁹) - (1.682 x 10⁻¹⁹) = φ

φ = 7.590 x 10⁻¹⁹ J = 4.74 eV

Learn more about work function here:

brainly.com/question/9757301

#SPJ2

3 0
2 years ago
Giving brainiest i need help with number 3
olganol [36]

Answer:

answer 2 I believe bbcc x

3 0
3 years ago
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