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3 years ago

At a certain temperature, 0.900 mol of SO3 is placed in a 2.00-L container.

Chemistry

1 answer:

Goryan [66]3 years ago

3 0

Answer:

Kc = 2.34 mol*L

Explanation:

The calculation of the Kc of a reaction is performed using the values of the concentrations of the participants in the equilibrium.

A + B ⇄ C + D

Kc = [C] * [D] / [A] * [B]

According to the reaction

Kc = [SO2]^2 * [O2]^2 / [SO3]^2

Knowing the 0.900 mol of SO3 is placed in a 2.00-L it means we have a 0.450 mol/L of SO3

0.450 --> 0 + 0 (Beginning of the reaction)

0.260 --> 0.260 + 0.130 (During the reaction)

0.190 --> 0.260 + 0.130 (Equilibrium of the reaction)

Kc = [0.260]^2 + [0.130]^2 / [0.190]^2

Kc = 2.34 mol*L

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The average density of a carbon-fiber-epoxy composite is 1.615 g/cm3. the density of the epoxy resin is 1.21 g/cm3 and that of t

ra1l [238]

The composite material is composed of carbon fiber and epoxy resins. Now, density is an intensive unit. So, to approach this problem, let's assume there is 1 gram of composite material. Thus, mass carbon + mass epoxy = 1 g.

Volume of composite material = 1 g / 1.615 g/cm³ = 0.619 cm³
Volume of carbon fibers = x g / 1.74 g/cm³
Volume of epoxy resin = (1 - x) g / 1.21 g/cm³

a.) V of composite = V of carbon fibers + V of epoxy resin
0.619 = x/1.74 + (1-x)/1.21
Solve for x,
x = 0.824 g carbon fibers
1-x = 0.176 g epoxy resins

Vol % of carbon fibers = [(0.824/1.74) ÷ 0.619]*100 = 76.5%

b.) Weight % of epoxy = 0.176 g epoxy/1 g composite * 100 = 17.6%
Weight % of carbon fibers = 0.824 g carbon/1 g composite * 100 = 82.4%

4 0

3 years ago

Why is the reaction of a strong acid and a strong base classified as a neutralization reaction?

Lera25 [3.4K]

Answer:

Answer choice C

Explanation:

Basicly for the Arrhenius and Bornsted-Lowery theories of acids & bases, acid-base reactions can be divided into 4 forms...

a. Strong Acid + Strong Base (HCl/NaOH) => pH = 7 at Eqv. Pt.

b. Weak Acid + Strong Base (HOAc/NaOH) => pH > 7 at Eqv. Pt.

c. Strong Acid + Weak Base (HCl & NH₄OH) => pH < 7 at Eqv. Pt.

d. Weak Acid + Weak Base (HOAc & NH₄OH) => pH ∝ Stronger Electrolyte

*HOAc = Acetic Acid

*NH₄OH = Ammonium Hydroxide

For each type reaction the pH at equivalence point depends upon the salt generated by the acid-base reaction. Ions of the salt, if they react with water, (hydrolysis) will shift the pH up or down depending upon which ion reacts. If there is no reaction by the salt ions then the pH will depend only upon autoionization of water which gives pH = 7. Typically Strong Acids + Strong Bases will give a pH = 7 at equivalence point because the ions of the salt will not undergo hydrolysis in water.

Example:

Strong Acid + Strong Base

HCl(aq) + NaOH(aq) => NaCl(aq) + H₂O(l)

NaCl(aq) => Na⁺(aq) + Cl⁻(aq)

Na⁺(aq) + H₂O(l) => No Rxn (theoretically NaOH, but NaOH is a strong base which prefers to remain 100% ionized in water).

Cl⁻(aq) + H₂O(l) => No Rxn (theoretically HCl, but HCl is a strong acid which prefers to remain 100% ionized in water).

Therefore, the net rxn is H⁺ + OH⁻ => H₂O & pH = 7.0

Weak Acid + Strong Base

HOAc(aq) + NaOH(aq) => NaOAc(aq) + H₂O(l)

NaOAc(aq) => Na⁺(aq) + OAc⁻(aq)

Na⁺(aq) + H₂O(l) => No Rxn

OAc⁻(aq) + H₂O(l) => HOAc(aq) + OH⁻(aq) => (Excess OH⁻ functions to increase pH>7 at eqv. pt.)

Strong Acid + Weak Base

HCl(aq) + NH₄OH(aq) => NH₄Cl(aq) + H₂O(l)

NH₄Cl(aq) => NH₄⁺(aq) + Cl⁻(aq)

Cl⁻(aq) + H₂O(l) => No Rxn

NH₄⁺(aq) + H₂O(l) => NH₄OH(aq) + H⁺(aq) => (Excess H⁺ functions to decrease pH < 7 at eqv. pt. )

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Answer:

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k0ka [10]

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