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brilliants [131]
3 years ago
14

Identify the conjugate base in each pairs

Chemistry
1 answer:
DiKsa [7]3 years ago
8 0

Answer: 1) RCOO^-

2) H_2PO_4^-

3) RNH_2

4) HCO_3^-

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

1) RCOOH\rightarrow RCOO^-+H^+

Here, RCOOH is loosing a proton, thus it is considered as an acid and after losing a proton, it forms RCOO^- which is a conjugate base.

2) H_3PO_4\rightarrow H_2PO_4^-+H^+

Here, H_3PO_4 is loosing a proton, thus it is considered as an acid and after losing a proton, it forms H_2PO_4^- which is a conjugate base.

3) RNH_3^+\rightarrow RNH_2+H^+

Here, RNH_3^+ is loosing a proton, thus it is considered as an acid and after losing a proton, it forms RNH_2 which is a conjugate base.

4) H_2CO_3\rightarrow HCO_3^-+H^+

Here, H_2CO_3 is loosing a proton, thus it is considered as an acid and after losing a proton, it forms HCO_3^- which is a conjugate base.

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Given the ion C2O4-2, what species would you expect to form with each of the following ions?
Ksivusya [100]

Answer:

A. K₂C₂O₄          Potassium oxalate

B. CuC₂O₄          Copper oxalate

C. Bi₂(C₂O₄)₃         Bismuth (III) oxalate

D. Pb(C₂O₄)₂         Lead (IV) oxalate

E. (NH₄)₂C₂O₄       Ammonium oxalate

F. HC₂O₄⁻             Acid oxalate

Explanation:

C₂O₄⁻²  → oxalate anion

This is the conjugate base from the H₂C₂O₄ which is the oxalic acid. A weak dyprotic acid that can release 2 protons.

A. 2K⁺  +  C₂O₄⁻²  → K₂C₂O₄          Potassium oxalate

It can be formed by the neutralization of the acid with the base

H₂C₂O₄  + 2KOH  → K₂C₂O₄  +  2H₂O

B. Cu²⁺ +  C₂O₄⁻²   ⇄  CuC₂O₄  ↓

This is a precipitate.

C.  2Bi³⁺  +  3C₂O₄⁻²   ⇄  Bi₂(C₂O₄)₃  ↓

This is a precipitate.

D.  Pb⁴⁺ +  2C₂O₄⁻²   ⇄  Pb(C₂O₄)₂  ↓

This is a precipitate.

E. 2NH₄⁺  +  C₂O₄⁻²   ⇄  (NH₄)₂C₂O₄  ↓

This is a precipitate.

F. This is the conjugate strong base, for the weak acid

H⁺  +  C₂O₄⁻²   ⇄  HC₂O₄⁻

HC₂O₄⁻  + H₂O  ⇄  C₂O₄⁻²  +  H₃O⁺    Ka

HC₂O₄⁻  + H₂O  ⇄  H₂C₂O₄  +  OH⁻    Kb

HC₂O₄⁻  is an amphoteric compound

6 0
3 years ago
Mason notices that his boat sinks lower in the water in a freshwater lake than in the ocean. what could explain this
tino4ka555 [31]

Answer:

Mason notices that his boat sinks lower into the water in a freshwater lake than in the ocean. What could explain this?

Explanation:

The freshwater has less density then the ocean!

6 0
3 years ago
The vapor pressure of water at 65oC is 187.54 mmHg. What is the vapor pressure of a ethylene glycol (CH2(OH)CH2(OH)) solution ma
Pavlova-9 [17]

Answer:

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

Explanation:

Vapor pressure of water at 65 °C=p_o= 187.54 mmHg

Vapor pressure of the solution at 65 °C= p_s

The relative lowering of vapor pressure of solution in which non volatile solute is dissolved is equal to mole fraction of solute in the solution.

Mass of ethylene glycol = 22.37 g

Mass of water in a solution = 82.21 g

Moles of water=n_1=\frac{82.21 g}{18 g/mol}=4.5672 mol

Moles of ethylene glycol=n_2=\frac{22.37 g}{62.07 g/mol}=0.3603 mol

\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}

\frac{187.54 mmHg-p_s}{187.54 mmHg}=\frac{0.3603 mol}{0.3603 mol+4.5672 mol}

p_s=173.83 mmHg

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

6 0
3 years ago
compare the boiling point of water at sea leavel to the boiling point in denver (altitude = 1 mile) Explain
steposvetlana [31]

Answer:

boiling point decreases in denver

Explanation:

in higher places

theres less atmospheric pressure

it takes less energy to bring water to the boiling point.

Less energy means less heat

which means water will boil at a lower temperature

wonderopolisorg

6 0
2 years ago
What is the molar mass, in grams, of a mole of an element equal to?
insens350 [35]

Answer:

B. the atomic weight of the element

Explanation:

since it talks about mass i think weight

5 0
3 years ago
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