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LekaFEV [45]
3 years ago
14

The four major attractive forces between particles are ionic bonds, dipole-dipole attractions, hydrogen bonds, and dispersion fo

rces. Consider the compounds below, and classify each by its predominant attractive or intermolecular force among atoms or molecules of the same type.
Chemistry
2 answers:
svetlana [45]3 years ago
7 0

Answer:

NaCl \rightarrow ionic bonds.

N_2 \rightarrow dispersion forces.

HCl \rightarrow dipole-dipole attractions.

HF \rightarrow hydrogen bonds

Explanation:

NaCl is formed between ions Na^+ and Cl^- which makes it an ionic compound having ionic bonding between it's atoms.

N_2 is a covalent non-polar molecule with only week dispersion forces caused by temporary dipoles in the binary molecule.

HCl is a covalent polar molecule with permanent dipoles Hδ+ and Clδ-.

HF consists of both Hydrogen and Fluoride

Anna007 [38]3 years ago
3 0

Answer:

NaCl: ionic, HF: hydrogen bond,  HCl: dipole dipole , F2: dispersion force

Explanation:

complete question is:

The four major attractive forces between particles are ionic bonds, dipole-dipole attractions, hydrogen bonds, and dispersion forces. Consider the compounds below, and classify each by its predominant attractive or intermolecular force among atoms or molecules of the same type.Identify each of the following ( NaCl, HF, HCl, F2) as Ionic, H Bonding, Dipole or Dispersion.

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How many atoms are in .45 moles of P4010
Bogdan [553]

Answer:

5×6.02×1023

Explanation:

there are 5×6.02×1023 molecules of p4010 in 5mole. there are four P atoms in a single molecule of p4010

5 0
3 years ago
Iron (III) oxide - what does<br> the III mean?
Aleonysh [2.5K]

Answer:

The iron is in the +3 oxidation state, which is what the III means. 

7 0
3 years ago
Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1= 1.0x10–5 and Ka2= 5x10–12) found
Igoryamba

Answer:

The concentrations are :

[HAsc^-]=0.000702 M

[Asc^{2-}]=5.92\times 10^{-8} M

The pH of the solution is 3.15.

Explanation:

H_2Asc\rightleftharpoons HAs^-+H^+         K_{a1}=1.0\times 10^{-5}

Initial

c                0              0

Equilibrium

c-x                x          x

K_{a1}=\frac{[HAs^-][H^+]}{[H_2Asc]}

1.0\times 10^{-5}=\frac{x\times x}{(c-x)}

1.0\times 10^{-5}=\frac{x^2}{(0.050-x)}

Solving for x:

x = 0.000702 M

[HAsc^-]=0.000702 M

HAsc^-\rightleftharpoons As^{2-}+H^+        K_{a2}=5\times 10^{-12}

Initially

x                0          0

At equilibrium ;

(x - y)            y         y

K_{a2}=\frac{[As^{2-}][H^+]}{[HAsc^-]}

5\times 10^{-12}=\frac{y\times y}{(x-y)}

5\times 10^{-12}=\frac{y^2}{(x-y)}

Putting value of x = 0.000702 M

5\times 10^{-12}=\frac{y^2}{(0.000702 -y)}

y=5.92\times 10^{-8} M

[Asc^{2-}]=5.92\times 10^{-8} M

Total concentration of [H^+]=x+y=0.000702 M+5.92\times 10^{-8} M=7.0206\times 10^{-4} M

The pH of the solution :

pH=-\log[H^+]

pH=-\log[7.0206\times 10^{-4} M}=3.15

7 0
3 years ago
Please help if you can !
iogann1982 [59]

Answer:

7630

Explanation:

8 0
3 years ago
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So if there is 16 protons and 16 neutrons what would be the charge right now?​
sveta [45]

Answer:

16

Explanation:

Protons have a positive net charge,

Neutrons being neutral don't have a net charge.

Each proton is one extra net charge if you have 16 of them and the neutrons don't affect your net charge you will have 16.

6 0
3 years ago
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