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ra1l [238]
3 years ago
14

Which of the following is an example of ionization of an acid?

Chemistry
1 answer:
Amiraneli [1.4K]3 years ago
4 0

Answer:

NH3(g) + H2O(1) → NH4+(aq) + OH (aq)

HF(aq) + H2O(1) → H3O+(aq) + F (aq)

Explanation:

Acid-base reactions are chemical reactions involving acids and bases. Acids tend to ionize/dissociate in water, a property which determines their strength. Ionization of an acid refers to the acid losing its hydrogen ion (H+) in water solution. An acid ionizes or dissociates to form a conjugate base.

A strong acid is so because it ionizes completely in water i.e. loses all its hydrogen ion (H+) while a weak acid partially ionizes in water.

In the chemical reactions;

1) NH3(g) + H2O(1) → NH4+(aq) + OH (aq)

H20 loses its hydrogen ion (H+) in this reaction to form an anion (OH-). Hence, water (H20) is an acid in this case which ionizes to form a conjugate base (OH-). This is an example of ionization of acid.

2) HF(aq) + H2O(1) → H3O+(aq) + F (aq)

Hydrogen fluoride (HF) loses its hydrogen ion (H+) in the presence of water to form anion (F-). The HF is the acid while F- is it's conjugate base. Thus, an example of ionization of acid

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1. When the following oxidation-reduction reaction in acidic solution is balanced, what is the
Bogdan [553]

Answer:

2Rb(s) + Sr^+(aq) → 2Rb^+ (aq) + Sr(s)

Explanation:

Rubidium has a more negative reduction potential (-2.98 V) compared to strontium (-2.89 V).

Hence, in a redox reaction involving rubidium and strontium, rubidium will be oxidized while strontium is reduced.

The balanced redox reaction equation is obtained from;

Oxidation half equation;

2Rb(s) ---->2Rb^+(aq) + 2e

Reduction half equation;

Sr^2+(aq) + 2e ----> Sr(s)

Overall reaction equation;

2Rb(s) + Sr^+(aq) → 2Rb^+ (aq) + Sr(s)

5 0
3 years ago
The data were collected over five spring seasons. What is the difference between the number of frogs in the pond when the rainfa
Lunna [17]

Answer: they like deep water

Explanation: they do

3 0
3 years ago
Given pH = 3.50 Find: [H3O+] and [OH-] Is this acidic, basic or neutral?
lys-0071 [83]

Answer:

Explanation:

Given parameters:

           pH = 3.50

Unknown:

    concentration of [H₃0⁺] = ?

    concentration of [OH⁻] = ?

Solution:

In order to find the unknown, we use some simple expressions which best explains the pH scale and the equilibrium systems of aqueous solutions.

         pH = -log₁₀[H₃O⁺]

         [H₃O⁺] = inverse log₁₀ (-pH) = 10^{-pH} = 10^{-3.5}

          [H₃O⁺] = 3.2 x 10⁻⁴moldm⁻³

       

For the  [OH⁻]:

       we use : pOH = -log₁₀ [OH⁻]

     Recall: pOH + pH = 14

                  pOH = 14 - pH = 14 - 3.5 = 10.5

  Now we plug the value of pOH into pOH = -log₁₀ [OH⁻]

                                   [OH⁻] = 10^{-pOH}

                        [OH⁻] = 10^{-10.5} = 3.2 x 10⁻¹¹moldm⁻³

The solution is acidic as the concentration of H₃0⁺ is more than that of the OH⁻ ions.

                   

8 0
4 years ago
What is the volume of an object with a density of 2.6 g/cm³ and a mass of 30.5g?
inessss [21]

Answer:

The answer is

<h2>11.73 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density}

From the question

mass of object = 30.5 g

Density = 2.6 g/cm³

The volume is

volume =  \frac{30.5}{2.6}  \\  = 11.7307692...

We have the final answer as

<h3>11.73 mL</h3>

Hope this helps you

5 0
3 years ago
How many moles of oxygen are formed when 58.6 g of KNO3 decomposes according to the following reaction? 4 KNO3(s) → 2 K2O(s) + 2
Leya [2.2K]

Answer:

0.725 mol

Explanation:

Moles are calculated as the given mass divided by the molecular mass.

i.e. ,

moles = ( mass / molecular mass )

since,

mass of KNO₃ = 58.6 g  ( given )

Molecular mass of KNO₃ = 101 g / mol

Therefore,

moles of KNO₃ = 58.6 g / 101 g / mol

moles of KNO₃ = 0.58 mol

From the balanced reaction ,

4 KNO₃ (s) ---> 2K₂O (s) + 2N₂ (g) + 5O₂ (g)

By the decomposition of 4 mol of KNO₃ , 5 mol of O₂ are formed ,

hence, unitary method is used as,

1  mol of KNO₃  gives 5 / 4 mol O₂

Therefore,

0.58 mol of KNO₃ , gives , 5 / 4  * 0.58 mol of O₂

Solving,

0.58 mol of KNO₃ , gives , 0.725 mol of O₂

Therefore,

58.6g of KNO₃ gives 0.725 mol of O₂.

3 0
3 years ago
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