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dalvyx [7]
3 years ago
7

With what compound will nh3 experience only dispersion intermolecular forces? with what compound will nh3 experience only disper

sion intermolecular forces? hcn ch3oh bf3 lif ch3f
Chemistry
2 answers:
zhannawk [14.2K]3 years ago
8 0

Answer : NH3 will experience only dispersion forces with BF3.

Explanation :

There are different types of intermolecular forces (IMF) which are experienced by the molecules.

1) Hydrogen Bonds : These are present when a molecule has hydrogen atom attached to N , F or O atoms.

Among the given molecules, CH3OH can have hydrogen bonds.

2) Dipole-dipole interactions : These are experienced by polar molecules. Among the given molecules, CH3F and HCN being polar can have dipole dipole interactions.

3) Electrostatic forces of attraction : These are present among the oppositely charged ions of ionic compounds. Here, LiF can have these force.

4) Dispersion forces : London dispersion forces are dominant in case of non polar molecules.

In case BF3, the trigonal planar geometry cancels out the dipoles of individual B-F bond making the overall molecule non polar.

Therefore NH3 will experience only dispersion forces with BF3.

Burka [1]3 years ago
6 0
Answer is: ammonia experience only dispersion intermolecular forces with BF₃ (boron trifluoride) because BF₃ is only nonpolar molecule (vectors of dipole moments cansel each other, dipole moment is zero).
The London dispersion force (intermolecular force) <span>is a temporary attractive </span>force between molecules.
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A plot of binding energy per nucleon Eb A versus the mass number (A) shows that nuclei with a small mass number have a small bin
Illusion [34]

Answer:

6He =   4.90 MeV/Nucleon

8Li =     5.18  MeV / Nucleon

62Ni =   8.82 MeV / Nucleon

115 In =  8.54 MeV / Nucleon

Explanation:

Our strategy here is to remember that when the mass of a given nuclei is calculated from the sum of the mass of its  protons and neutrons, this mass is greater than  the actual  value . This is the mass defect.

Now this mass defect we can convert to energy  by utilizing Einstein´s equation, E = mc².This is  the binding energy.

For 6He with actual mass 6.0189 u ( He has Z = 2, that is 2 protons )

mass protons =   2 x 1.0078 u  =  2.0516 u

mass neutrons = 4 x 1.0087 u  =  4.0348 u

predicted mass = (2.0516 + 4.0348) u = 6.0504 u

mass defect = (6.0504 - 6.0189) u = 0.0315 u

Now we need to convert this mass expressed in atomic mass units to kilograms ( 1 u = 1.66054 x 10⁻²⁷ Kg )

0.0315 u x 1.66054 x 10⁻²⁷ Kg =5.231 x 10⁻²⁹ Kg

E =5.231x 10⁻²⁸ Kg x (3 x 10⁸ m/s )² = 4.707 x 10⁻¹²J

Finally we will convert this energy in Joules to eV

E = 4.707 x 10⁻¹²  J x 6.242 x 10¹⁸ eV/J = 2.94 x 10⁷ eV = 29.4 MeV

E per nucleon for 6He = 29.4 MeV / 6 =4.90 MeV / Nucleon

Now the calculations for the rest of the nuclei are performed in similar manner with the following results:

8Li = 5.18 MeV / Nucleon

62Ni = 8.82 MeV / Nucleon

115 In =  8.54 MeV / Nucleon

8 0
4 years ago
PLZ HELP
nika2105 [10]
Mjdksksksnsjdkjdjsksksjdndnd
8 0
4 years ago
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Many classic experiments have given us indirect evidence of the nature of the atom. Which of the experiments below did not give
taurus [48]

The Rutherford experiment proved the Thomson “plum-pudding” model of the atom to be essentially correct did not give the results described and is denoted as option A.

<h3>What is Thomson “plum-pudding” model?</h3>

This model was proposed by J.J Thomson in which referred an atom as a sphere of positive charge, and negatively charged electrons are embedded in it to balance the total positive charge.

This model was incorrect and the Rutherford atomic model was adopted in which he described the electrons orbits about a tiny positive nucleus.

The nucleus contains protons and neutrons instead thereby making it the correct choice.

Read more about Atom here brainly.com/question/6258301

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The options include the following:

a.The Rutherford experiment proved the Thomson “plum-pudding” model of the atom to be essentially correct.

b.The Rutherford experiment was useful in determining the nuclear charge on the atom.

c.Milikan’s oil-drop experiment showed that the charge on any particle was a simple multiple of the charge on the electron.

d.The electric discharge tube proved that electrons have a negative charge

4 0
2 years ago
Which of the following lists elements in order of increasing malleability?
hodyreva [135]

Malleability is the property due to which substances tend to hammered into thin sheets when pressure is applied. Malleability is observed in metal as the metal atoms are bound by metallic bonds. The layers of metal atoms can roll over each other without breaking the metal bond when pressure is applied. In the periodic table malleability decreases as the metallic nature decreases. Among the given metals Al is the most metallic element, followed by Zinc an Carbon is a non metal. Therefore, the order of increasing malleability will be C, Zn, Al


8 0
3 years ago
Read 2 more answers
NaHCO3 (s) + HC2H3O2 (aq) = NaC2H3O2 (aq) + H2O (I) + CO2 (g)
Misha Larkins [42]

Moles=volume*concentration   
         =0.1*.83
         =.083 Moles of HC2H3O2
Mole ratio between HC2H3O2 and CO2 is 1:1
This means .083 Moles of CO2

Mass =Moles*Rfm of CO2
         =.083*(12+16+16)
         =3.7grams
8 0
4 years ago
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