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shusha [124]
2 years ago
7

Write the ionic charges (such as Ca2+) and chemical formulas and fill-in the table below.

Chemistry
1 answer:
Vikentia [17]2 years ago
4 0

1) Lithium and fluorine:

Ionic charges: lithium cation Li⁺ and fluorine anion F⁻.

Chemical formula LiF.

In ionic salt lithium fluoride (LiF), fluorine has electronegativity approximately χ = 4 and lithium χ = 1 (Δχ = 4 - 1; Δχ = 3).

Fluorine attracts electron and it has negative charge and lithium has positive charge.

2) Beryllium and oxygen:

Ionic charges cation Be²⁺ and anion O²⁻.

Chemical formula is BeO.

Beryllium is metal from group 2 and oxygen is nonmetal from group 16.

Electron configuration of beryllium: ₄Be: 1s² 2s², it has two valence electrons in 2s orbital.

Beryllium lose two electrons and to gain electron configuration as noble gas helium (He).

Electron configuration of oxygen atom: ₈O 1s² 2s² 2p⁴.

Oxygen gain two valence electron to form anion with stable electron configuration as noble gas neon (atomic number 10).

3) Magnesium and fluorine:  

Ionic charges cation Mg²⁺ and anion F⁻.

Chemical formula is MgF₂.

Magnesium fluoride (MgF₂) is salt, ionic compound.  

Magnesium (Mg) is metal from 2. group of Periodic table of elements and has low ionisation energy and electronegativity, which means it easily lose valence electons (two valence electrons).  

Magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.  

Fluorine (F) is nonmetal with greatest electronegativity, which means it easily gain electrons.  

Fluorine has atomic number 9, which means it has 9 protons and 9 electrons. It gain one electron to form fluorine anion (F⁻) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.  

4) Aluminum and chlorine:  

Ionic charges cation Al³⁺ and anion Cl⁻.

Chemical formula is AlCl₃.

The right name for AlCl₃ is aluminium chloride.

Aluminium chloride is a salt with ionic bonds.

Aluminium (metal from group 13) has oxidation number +3 and chlorine (nonmetal from group 17) has oxidation number -1, chemical compound has neutral charge (+3 + 3 · (-1) = 0).

5) Beryllium and nitrogen:  

Ionic charges cation Be²⁺ and anion N³⁻.

Chemical formula is Be₃N₂.

Atomic number of nitrogen is 7, it has 7 protons and 7 electrons.

Electron configuration of nitrogen atom: ₇N 1s² 2s² 2p³.

Nitrogen gain three electrons to form anion with stable electron configuration as noble gas neon (atomic number 10).

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A 50.0 mL solution of 0.129 M KOH is titrated with 0.258 M HCl. Calculate the pH of the solution after the addition of each of t
kobusy [5.1K]

Answer:

A- pH = 13.12

B- pH = 12.91

C- pH = 12.71

D- pH = 12.43

E- pH = 11.55

F- pH = 7

G- pH = 2.46

H- pH = 1.88

Explanation:

This is a titration of a strong base with a strong acid. The neutralization reaction is: KOH (aq) + HCl (aq) →  H₂O(l) + KCl(aq)

Our pH at the equivalence point is 7, because we have made a neutral salt.

To determine the volume at that point we state the formula for titration:

mmoles of base = mmoles of acid

Volume of base  . M of base = Volume of acid . M of acid

50mL . 0.129M = 0.258 M . Volume of acid

Volume of acid = (50mL . 0.129M) / 0.258 M →  25 mL (Point <u>F</u>)

When we add 25 mL of HCl, our pH will be 7.

A- At 0 mL of acid, we only have base.

KOH → K⁺ + OH⁻

[OH⁻] = 0.129 M

To make more easy the operations we will use, mmol.

mol . 1000 = mmoles → mmoles / mL = M

- log 0.129 = 0.889

14 - 0.889 = 13.12

B-  In this case we are adding, (7 mL . 0.258M) = 1.81 mmoles of H⁺

Initially we have  0.129 M . 50 mL = 6.45 mmoles of OH⁻

1.81 mmoles of H⁺ will neutralize, the 6.45 mmoles of OH⁻ so:

6.45 mmol - 1.81 = 4.64 mmoles of OH⁻

This mmoles of OH⁻ are not at 50 mL anymore, because our volume has changed. (Now, we have 50 mL of base + 7 mL of acid) = 57 mL of total volume.

[OH⁻] = 4.64 mmoles / 57 mL = 0.0815 M

- log 0.0815 M = 1.09 → pOH

pH = 14 - pOH → 14 - 1.09 = 12.91

C- In this case we add (12.5 mL . 0.258M) = 3.22 mmoles of H⁺

<em>Our initial mmoles of OH⁻ would not change through all the titration. </em>

Then 6.45 mmoles of OH⁻ are neutralized by 3.22 mmoles of H⁺.

6.45 mmoles of OH⁻ - 3.22 mmoles of H⁺ = 3.23 mmoles of OH⁻

Total volume is: 50 mL of base + 12.5 mL = 62.5 mL

[OH⁻] = 3.23 mmol / 62.5 mL = 0.0517 M

- log  0.0517 = 1.29 → pOH

14 - 1.11 = 12.71

D- We add (18 mL . 0.258M) = 4.64 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 4.64 mmoles of H⁺.

6.45 mmoles of OH⁻ - 4.64 mmoles of H⁺ = 1.81 mmoles of OH⁻

Total volume is: 50 mL of base + 18 mL = 68 mL

[OH⁻] = 1.81 mmol / 68 mL = 0.0265 M

- log  0.0265 = 1.57 → pOH

14 - 1.57 = 12.43

E- We add (24 mL . 0.258M) = 6.19 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 6.19 mmoles of H⁺.

6.45 mmoles of OH⁻ - 6.19 mmoles of H⁺ = 0.26 mmoles of OH⁻

Total volume is: 50 mL of base + 24 mL = 74 mL

[OH⁻] = 0.26 mmol / 74 mL = 3.51×10⁻³ M

- log  3.51×10⁻³  = 2.45 → pOH

14 - 2.45 = 11.55

F- This the equivalence point.

mmoles of OH⁻ = mmoles of H⁺

We add (25 mL . 0.258M) = 6.45 mmoles of H⁺

All the OH⁻ are neutralized.

OH⁻  +  H⁺  ⇄   H₂O              Kw

[OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷  →  pOH = 7

pH → 14 - 7 = 7

G- In this case we have an excess of H⁻

We add (26 mL . 0.258M ) = 6.71 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

6.71 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 0.26 mmoles of H⁺

[H⁺] = 0.26 mmol / Total volume

Total volume is: 50 mL + 26 mL → 76 mL

[H⁺] = 0.26 mmol / 76 mL → 3.42×10⁻³ M

- log 3.42×10⁻³ = 2.46 → pH

H- Now we add (29 mL . 0.258M) = 7.48 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

7.48 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 1.03 mmoles of protons

Total volume is 50 mL + 29 mL = 79 mL

[H⁺] = 1.03 mmol / 79 mL → 0.0130 M

- log 0.0130 = 1.88 → pH

After equivalence point, pH will be totally acid, because we always have an excess of protons. Before the equivalence point, pH is basic, because we still have OH⁻ and these hydroxides, will be neutralized through the titration, as we add acid.

5 0
3 years ago
describe how the current modern atomic theory and model differs from the model jj Thompson proposed ?
myrzilka [38]
J.J Thompson’s model shows a sphere with electrons that are moving around freely. However, Thompson’s model does not show protons or neutrons. The model that we have today gives a clearer structure showing protons, neutrons, and electrons inside an atom.
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