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Soloha48 [4]
3 years ago
13

Acids hydrolyze or dissolve in solutions to form what? acids hydrolyze or dissolve in solutions to form what? hydroxide ions sol

utes protons precipitates
Chemistry
1 answer:
nikklg [1K]3 years ago
4 0
<h3>Answer:</h3>

                      Acids hydrolyze or dissolve in solutions to form <u>Protons</u>.

<h3>Explanation:</h3>

                      Acid are defined according to three different theories and concepts.

                       i)  Arrhenius Concept

                       ii) Lowry Bronsted Theory

                      iii) Lewis Theory

According to Arrhenius Concept of Acid and Base, "Acid is any specie which when dissolved in water ionizes to produce H⁺ ions".

Example:

                                  HCl    →    H⁺₍aq₎  +  Cl⁻₍aq₎

                                  H₂SO₄    →    2 H⁺₍aq₎  +  SO₄²⁻₍aq₎

Proton (H⁺):

                   Hydrogen atom with atomic mass of 1 amu contains one proton and one electron in its first shell. When this single electron is removed from hydrogen atom it is left with only one proton hence, H⁺ is called proton.

Hydrolysis:  

                 In simple, hydrolysis is that reaction in which water molecules are utilized to breakdown a chemical compound.

<h3>Conclusion:</h3>

                     Hence, we can conclude that Arrhenius Acids when added to water are being hydrolyzed by water molecule and yield H⁺ (proton) along with corresponding negative specie.

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A 32.2 g iron rod, initially at 21.9 C, is submerged into an unknown mass of water at 63.5 C. in an insulated container. The fin
lorasvet [3.4K]

Answer : The mass of the water in two significant figures is, 3.0\times 10^1g

Explanation :

In this case the heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron metal = 0.45J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of iron metal = 32.3 g

m_2 = mass of water = ?

T_f = final temperature of mixture = 59.2^oC

T_1 = initial temperature of iron metal = 21.9^oC

T_2 = initial temperature of water = 63.5^oC

Now put all the given values in the above formula, we get

32.3g\times 0.45J/g^oC\times (59.2-21.9)^oC=-m_2\times 4.18J/g^oC\times (59.2-63.5)^oC

m_2=30.16g\approx 3.0\times 10^1g

Therefore, the mass of the water in two significant figures is, 3.0\times 10^1g

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