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3 years ago

An ionic bond is a bond between ____.

Chemistry

2 answers:

dlinn [17]3 years ago

7 0

1.) 2

2.) k+

3.) It gains electrons

4.) The attraction of metal ions to mobile electrons

5.) A cation ans an anion

6.) body-centered cubic

Just took the quiz. You're welcome c:

exis [7]3 years ago

4 0

An ionic bond is a bond which occurs when there is a difference in the electronegativity of the atoms. One atom which has a stronger electronegativity tends to steal the electron of the other less electronegative atom hence making it more negative (anion), while the other becomes a cation. So answer is:

b. a cation and an anion

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You have a 1.153 g sample of an unknown solid acid, HA, dissolved in enough water to make 20.00 mL of solution. HA reacts with K

seropon [69]

Answer:

HA + KOH → KA + H₂O

Explanation:

The unknown solid acid in water can release its proton as this:

HA + H₂O → H₃O⁺ + A⁻

As we have the anion A⁻, when it bonded to the cation K⁺, salt can be generated, so the reaction of HA and KOH must be a neutralization one, where you form water and a salt

HA + KOH → KA + H₂O

It is a neutralization reaction because H⁺ from the acid and OH⁻ from the base can be neutralized as water

7 0

3 years ago

If a 100. -g sample of a hydrated compound contains 37.07-g sodium, 48.39-g carbonate and 14.54-g water, find the empirical form

Mumz [18]

he required empirical formula based on the data provided is Na2CO3.H2O.

<h3>What is empirical formula?</h3>

The term empirical formula refers to the formula of a compound which shows the ratio of each specie present.

We have the following;

Mass of sodium = 37.07-g

Mass of carbonate = 48.39 g

Mass of water = 14.54-g

Number of moles of sodium = 37.07-g/23 g/mol = 2 moles

Number of moles of carbonate = 48.39 g/61 g/mol = 1 mole

Number of moles of water = 14.54/18 g/mol = 1 mole

The mole ratio is 2 : 1: 1

Hence, the required empirical formula is Na2CO3.H2O

Learn more about empirical formula : brainly.com/question/11588623

3 0

2 years ago

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

3 years ago

Sodium carbonate reacts with silver nitrate according to the following balanced equation: Na2CO3 (s) + 2 AgNO3 (aq) → Ag2CO3 (s)

klemol [59]

Answer:

a) 2.01 g

Explanation:

Na₂CO₃ (s) + 2AgNO₃ (aq) → Ag₂CO₃ (s) + 2NaNO₃

First we convert 0.0302 mol AgNO₃ to Na₂CO₃ moles, in order to calculate how many Na₂CO₃ moles reacted:

0.0302 mol AgNO₃ * $\frac{1molNa_2CO_3}{2molAgNO_3}$ = 0.0151 mol Na₂CO₃

So the remaining Na₂CO₃ moles are:

0.0340 - 0.0151 = 0.0189 moles Na₂CO₃

Finally we convert Na₂CO₃ moles into grams, using its molar mass:

0.0189 moles Na₂CO₃ * 106 g/mol = 2.003 g Na₂CO₃

The closest answer is option a).

8 0

3 years ago

List the 3 mole equalities.

zloy xaker [14]

12.000 g of 12c is one

5 0

3 years ago

Read 2 more answers