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3 years ago

Chemistry Help (Naming Oxyacids Chart)

Chemistry

1 answer:

guapka [62]3 years ago

7 0

Answer:

HClO₃ /chloric acid /suffix -ic/ ClO₃⁻ (chlorate)

HClO₂/ chlorous acid/ suffix -ous/ ClO₂⁻ (chlorite)

HNO₃ /nitric acid /suffix -ic/ NO₃⁻ (nitrate)

HNO₂/ nitrous acid/ suffix -ous/ NO₂⁻ (nitrite)

Explanation:

Chlorine has 4 positive oxidation numbers to form oxyacids: +1, +3, +5 and +7.

When it uses the oxidation number +5, it forms HClO₃, which is named chloric acid, with the suffix -ic. When it loses an H⁺, it forms the oxyanion ClO₃⁻ (chlorate).

When it uses the oxidation number +3, it forms HClO₂, which is named chlorous acid, with the suffix -ous. When it loses an H⁺, it forms the oxyanion ClO₂⁻ (chlorite).

Nitrogen has 2 positive oxidation numbers to form oxyacids: +3 and +5.

When it uses the oxidation number +5, it forms HNO₃, which is named nitric acid, with the suffix -ic. When it loses an H⁺, it forms the oxyanion NO₃⁻ (nitrate).

When it uses the oxidation number +3, it forms HNO₂, which is named nitrous acid, with the suffix -ous. When it loses an H⁺, it forms the oxyanion NO₂⁻ (nitrite).

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Given the following thermodynamic data, calculate the lattice energy of LiCl:

tiny-mole [99]

Answer:

$\boxed{\text{-862 kJ/mol}}$

Explanation:

One way to calculate the lattice energy is to use Hess's Law.

The lattice energy U is the energy released when the gaseous ions combine to form a solid ionic crystal:

Li⁺(g) + Cl⁻(g) ⟶ LiCl(s); U = ?

We must generate this reaction rom the equations given.

(1) Li(s) + ½Cl₂ (g) ⟶ LiCl(s); ΔHf° = -409 kJ·mol⁻¹

(2) Li(s) ⟶ Li(g); ΔHsub = 161 kJ·mol⁻¹

(3) Cl₂(g) ⟶ 2Cl(g) BE = 243 kJ·mol⁻¹

(4) Li(g) ⟶Li⁺(g) +e⁻ IE₁ = 520 kJ·mol⁻¹

(5) Cl(g) + e⁻ ⟶ Cl⁻(g) EA₁ = -349 kJ·mol⁻¹

Now, we put these equations together to get the lattice energy.

E/kJ

(5) Li⁺(g) +e⁻ ⟶ Li(g) 520

(6) Li(g) ⟶ Li(s) -161

(7) Li(s) + ½Cl₂(g) ⟶ LiCl(s) -409

(8) Cl(g) ⟶ ½Cl₂(g) -121.5

(9) Cl⁻(g) ⟶ Cl(g) + e⁻ +349

Li⁺(g) + Cl⁻(g) ⟶ LiCl(s) -862

The lattice energy of LiCl is $\boxed{\textbf{-862 kJ/mol}}$ .

3 0

3 years ago

Write the isotopic symbol for the following (show your work) a) An isotope of iodine whose atoms have 78 neutrons b) An isotope

morpeh [17]

Answer:

For a: The isotopic representation of iodine is $_{53}^{131}\textrm{I}$

For b: The isotopic representation of cesium is $_{55}^{137}\textrm{Cs}$

For c: The isotopic representation of strontium is $_{38}^{52}\textrm{Sr}$

Explanation:

The isotopic representation of an atom is: $_Z^A\textrm{X}$

where,

Z = Atomic number of the atom

A = Mass number of the atom

X = Symbol of the atom

For a:

We are given:

Number of neutrons = 78

Atomic number of iodine = 53 = Number of protons

Mass number = 53 + 78 = 131

Thus, the isotopic representation of iodine is $_{53}^{131}\textrm{I}$

For b:

We are given:

Number of neutrons = 82

Atomic number of cesium = 55 = Number of protons

Mass number = 55 + 82 = 137

Thus, the isotopic representation of cesium is $_{55}^{137}\textrm{Cs}$

For c:

We are given:

Number of neutrons = 52

Atomic number of strontium = 38 = Number of protons

Mass number = 38 + 52 = 90

Thus, the isotopic representation of strontium is $_{38}^{52}\textrm{Sr}$

3 0

3 years ago

Calculate the maximum volume in ml of 0.15M HCl that each of the following antacid formulations would be expected to neutralize.

vlada-n [284]

a. 34 mL; b. 110 mL

a. A tablet containing 150 Mg(OH)₂

Mg(OH)₂ + 2HCl ⟶ MgCl₂ + 2H₂O

Moles of Mg(OH)₂ = 150 mg Mg(OH)₂ × [1 mmol Mg(OH)₂/58.32 mg Mg(OH)₂

= 2.572 mmol Mg(OH)₂

Moles of HCl = 2.572 mmol Mg(OH)₂ × [2 mmol HCl/1 mmol Mg(OH)₂]

= 5.144 mmol HCl

Volume of HCl = 5.144 mmol HCl × (1 mmol HCl/0.15 mmol HCl) = 34 mL HCl

b. A tablet containing 850 mg CaCO₃

CaCO₃ + 2HCl ⟶ CaCl₂ + CO₂ + H₂O

Moles of CaCO₃ = 850 mg CaCO₃ × [1 mmol CaCO₃/100.09 mg CaCO₃

= 8.492 mmol CaCO₃

Moles of HCl = 8.492 mmol CaCO₃ × [2 mmol HCl/1 mmol CaCO₃]

= 16.98 mmol HCl

Volume of HCl = 16.98 mmol HCl × (1 mL HCl/0.15 mmol HCl) = 110 mL HCl

5 0

3 years ago

Define an Arrhenius base and describe properties of bases. Use an example to explain how an Arrhenius base will behave in water.

frozen [14]

An Arrhenius base is a molecule that when dissolved in water will break down to yield an $OH^-$ or hydroxide ion in solution.

<h3>What is Arrhenius base?</h3>

An Arrhenius base is a compound that increases the $OH^-$ ion concentration in aqueous solution.

An Arrhenius base is a substance that, when dissolved in an aqueous solution, increases the concentration of hydroxide, or , $OH^-$ ions in the solution.

Bases Properties

Arrhenius bases that are soluble in water can conduct electricity.

Bases often have a bitter taste and are found in foods less frequently than acids. Many bases, like soaps, are slippery to the touch.

Bases also change the colour of indicators. Red litmus turns blue in the presence of a base (see figure below), while phenolphthalein turns pink.

Some bases react with metals to produce hydrogen gas.

Acids (pH < 7.0) react with bases (pH > 7.0) to produce a salt and water. When equal moles of an acid and a base are combined, the acid is neutralized by the base. The resulting mixture will have a more neutral pH.

An Arrhenius acid is a substance that dissociates in water to form hydrogen ions or protons. In other words, it increases the number of $H^+$ ions in the water. In contrast, an Arrhenius base dissociates in water to form hydroxide ions $OH^-$ .

Example, sodium hydroxide, is added to an aqueous solution. NaOH dissociates into sodium, $Na^+$ , and hydroxide, $OH^-$ ions.

Learn more about the Arrhenius bases here:

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8 0

2 years ago

You are pushing on a heavy door, trying to slide it open. If your friend stands behind you and helps you push, how have the forc

NeTakaya

Answer:

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Explanation:

a egur oguta nestáincompel ta

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3 years ago