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2 years ago

I need help. Will mark brainliest

Chemistry

2 answers:

Ugo [173]2 years ago

8 0

Answer:

17. HC₂H₃O₂ + H₂O ⇄ C₂H₃O₂⁻ + H₃O⁺ Ka

A B CB CA

18. H₂SO₄ + 2H₂O → 2H₃O⁺ + SO₄⁻²

A B CA CB

19. CO₃⁻² + H₂O ⇄ HCO₃⁻ + OH⁻ Kb

B A AC BC

20. HCO₃⁻ + NH₃ ⇄ CO₃⁻² + NH₄⁺ Ka

A B BC AC

21. NH₄⁺ + OH⁻ ⇄ NH₃ + H₂O Ka

A B BC AC

Explanation:

The acetic acid release a proton to water, to produce acetate and hydronium.

This is a weak acid.

The sulfuric acid release 2 protons to water to produce sulfate and hydronium.

This acid is considered as strong, but it is only strong in the first dissociation. The second dissociation is weak, with a Ka.

Carbonate takes a proton from water, to become bicarbonate (In this case, the carbonate behaves as a base).

Bicarbonate release a proton to ammonia, to make ammonium and carbonate anion.

Amonium cation release a proton to the hydroxide to make ammonia and water.

exis [7]2 years ago

5 0

Answer:

See explanation

Explanation:

Step 1: Data given

Bronsted-Lowry says the following:

Acid dissociation: HX + H2O → H3O+ + X-

Base dissociation: B + H2O → OH- + HB+

HC2H3O2 + H2O →

HC2HO2 is acetic acid, it's a weak acid. (CH3COOH)

This will be following the reaction HX + H2O → H3O+ + X-

CH3COOH + H2O → CH3COO- + H3O+

H2SO4 + H2O →

H2SO4 is a strong acid

This will be following the reaction HX + H2O → H3O+ + X-

H2SO4 + H2O → HSO4- + H3O+

CO3^2- + H2O →

CO3^2_ is a conjugate base

This will be following the reaction B + H2O → OH_ + HB+

CO3^2- + H2O → HCO3- + OH-

HCO3- + NH3 →

HCO3- (known as bicarbonate) is the conjugate base of H2CO3, a weak acid, and the conjugate acid of the carbonate ion.

NH3 is a weak base

HCO3- + NH3 → CO3^2- + NH4+

NH4+ + OH- →

The hydrogen on the ammonium ion (NH4+) can go back to the hydroxide ion (OH-) to form NH3 and H2O (ammonia and water) again. In this case, because the ammonium ion is donating a proton, it is called a conjugate acid.

NH4+ + OH- → NH3 + H2O

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Answer:

Option D. 4.02 kJ

Explanation:

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Q = m . C . ΔT

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2 years ago

At 85°C, the vapor pressure of A is 566 torr and that of B is 250 torr. Calculate the composition of a mixture of A and B that b

Phantasy [73]

Answer:

Composition of the mixture:

$x_{A} =0.652=65.2$ %

$x_{B} =0.348=34.8$ %

Composition of the vapor mixture:

$y_{A} =0.809=80.9$ %

$y_{B} =0.191=19.1$ %

Explanation:

If the ideal solution model is assumed, and the vapor phase is modeled as an ideal gas, the vapor pressure of a binary mixture with $x_{A}$ and $x_{B}$ molar fractions can be calculated as:

$P_{vap}=x_{A}P_{A}+x_{B}P_{B}$

Where $P_{A}$ and $P_{B}$ are the vapor pressures of the pure compounds. A substance boils when its vapor pressure is equal to the pressure under it is; so it boils when $P_{vap}=P$ . When the pressure is 0.60 atm, the vapor pressure has to be the same if the mixture is boiling, so:

$0.60*760=P_{vap}=x_{A}P_{A}+x_{B}P_{B}\\456=x_{A}P_{A}+(1-x_{A})P_{B}\\456=x_{A}*(P_{A}-P_{B})+P_{B}\\\frac{456-P_{B}}{P_{A}-P_{B}}=x_{A}\\\\\frac{456-250}{566-250}=x_{A}=0.652$

With the same assumptions, the vapor mixture may obey to the equation:

$x_{A}P_{A}=y_{A}P$ , where P is the total pressure and y is the fraction in the vapor phase, so:

$y_{A} =\frac{x_{A}P_{A}}{P}=\frac{0.652*566}{456} =0.809=80.9$ %

The fractions of B can be calculated according to the fact that the sum of the molar fractions is equal to 1.

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