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Scrat [10]
4 years ago
14

Consider the reaction below. Which species are conjugate acid/base pairs? H2SO3 (aq) + CN (aq) HSO3 (aq)HCN (aq) A) HSO3, CN B)

HSO3, H2SO3 C) H2SO 3, CN D) HCN, H2SO
Chemistry
1 answer:
SpyIntel [72]4 years ago
7 0

Answer : The correct option is, (B) H_2SO_3,HSO_3^-

Explanation :

According to the Bronsted Lowry concept, Bronsted Lowry-acid is a substance that donates one or more hydrogen ion in a reaction and Bronsted Lowry-base is a substance that accepts one or more hydrogen ion in a reaction.

Or we can say that, conjugate acid is proton donor and conjugate base is proton acceptor.

The given equilibrium reaction is,

H_2SO_3(aq)+CN^-\rightleftharpoons HSO_3^-(aq)+HCN(aq)

Here, H_2SO_3 is loosing a proton, thus it is considered as an acid and after losing a proton, it forms HSO_3^{-} which is a conjugate base. That means, H_2SO_3/HSO_3^- are act as a conjugate acid-base pairs.

Hence, correct option is, (B) H_2SO_3,HSO_3^-

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State the type of enthalpy in the followinv equation Koh+Hcl-kcl+h2o ∆h+=-57kj\mol​
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Answer:

The Enthalpy of neutralization

Explanation:

The reaction of a base (KOH) with an acid (HCl) produce water and its salt (KCl) is called <em>Neutralization Reaction.</em>

This neutralization releases 57kJ/mol.

As the type of enthalpy is due the type of reaction. This enthalpy is:

<h3>The Enthalpy of neutralization</h3>
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Write the autoionization reaction for methanol, ch3oh.
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4 0
4 years ago
. In an experiment, 1.90 g of NH3 reacts with 4.96 g of O2. 4NH3(g) + 5O2(g) ⟶ 4NO(g) + 6H2O(g) (i) Which is a limiting reactant
Simora [160]

1. The limiting reactant in the reaction is NH₃

2. The mass of the excess reactant remaining is 0.49 g

3. The mass of NO produced from the reaction is 3.35 g

<h3>Balanced equation </h3>

4NH₃ + 5O₂ —> 4NO + 6H₂O

Molar mass of NH₃ = 14 + (3×1) = 17 g/mol

Mass of NH₃ from the balanced equation = 4 × 17 = 68 g

Molar mass of O₂ = 16 × 2 = 32 g/mol

Mass of O₂ from the balanced = 5 × 32 = 160 g

Molar mass of NO = 14 + 16 = 30 g/mol

Mass of NO from the balanced equation = 4 × 30 = 120 g

SUMMARY

From the balanced equation above,

68 g of NH₃ reacted with 160 g of O₂ to produce 120 g of NO

<h3>1. How to determine the limiting reactant </h3>

From the balanced equation above,

68 g of NH₃ reacted with 160 g of O₂

Therefore,

1.90 g of NH₃ will react with = (1.90 × 160) / 68 = 4.47 g of O₂

From the calculation made above, we can see that only 4.47 g out of 4.96 g of O₂ given is needed to react completely with 1.90 g of NH₃.

Therefore, NH₃ is the limiting reactant.

<h3>2. How to determine the mass of the excess reactant remaining </h3>
  • Mass of excess reactant (O₂) given = 4.96 g
  • Mass of excess reactant (O₂) that reacted = 4.47 g
  • Mass of excess reactant (O₂) remaining =?

Mass of excess reactant (O₂) remaining = 4.96 – 4.47

Mass of excess reactant (O₂) remaining = 0.49 g

<h3>3. How to determine the mass of NO produced </h3>

In this case, the limiting reactant (NH₃) will be used.

From the balanced equation above,

68 g of NH₃ reacted to produce 120 g of NO

Therefore,

1.90 g of NH₃ will react to produce = (1.90 × 120) / 68 = 3.35 g of NO.

Thus, 3.35 g of NO were obtained from the reaction.

Learn more about stoichiometry:

brainly.com/question/14735801

5 0
2 years ago
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