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Scrat [10]
3 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]3 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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what volume of a 0.138 m potassium hydroxide solution is required to neutralize 26.0 ml of a 0.205 m nitric acid solution?
Law Incorporation [45]

A neutralization titration is a chemical response this is used to decide the composition of an answer and what kind of acid or base is in it. This is a way of volumetric analysis and the formula is (M1V1 = M2V2).

Utilize the titration method of M1V1 = M2V2 in view that we're given the concentrations of every compound and the quantity of KOH. Let: M1 = 0.138M, V1 = x, M2 = 0.205M, V2 = 26.0 ML.

  • M1 = initial mass
  • V1= initial volume
  • M2 = final mass
  • V2= final volume
  • (M1V1 = M2V2)
  • (0.138)(V1) = (0.205)x(26.0)
  • V2=(0.205)x(26.0)\ 0.138
  • V2 = 47.10 M/L
  • The final value of Volume needed for neutralization of nitric acid solution is  V2 = 47.10 M/L

Read more about the neutralization:

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4 0
1 year ago
Scientists have been able to find a full record of Earth's fossil history, thus making the fossil record complete.
Arada [10]

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5 0
3 years ago
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Answer:

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Explanation:

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3 years ago
If earth is 1.50 x 10^8km from the sun what is the distance in Mm?
Readme [11.4K]

The distance of the earth to the sun in Mm = 1.5 x 10⁵

<h3>Further explanation</h3>

Given

The distance of the earth to the sun : 1.50 x 10⁸ km

Required

The distance in Mm

Solution

In converting units we must pay attention to the conversion factor.

the conversion factor :

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