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3 years ago

If two atoms have an electronegativity difference of 1.3, what type of bond would form between them?

Chemistry

2 answers:

Debora [2.8K]3 years ago

5 0

Answer:

The correct answer is b polar covalent

Explanation:

When two atoms joined by covalent bond has difference in their electronegativities at that time polarity arise.

When the electronegativity difference is low such as 1.3 then the polar bond formed by the two atoms are called polar covalent bond.For example H2O

on the other hand polar bond formed by two atoms having high difference in their electronegativities is called ionic bond.For example NaCl.

zloy xaker [14]3 years ago

3 0

Answer:

a. polar covalent

Explanation:

Electronegativity is the tendency of an atom to attract a shared pair of electrons towards itself.

The higher the electronegativity, the more an atom attracts electrons towards itself.

The most commonly used method of calculation of electronegativity is done by Pauling's scale.

Cesium is the least electronegative (0.79), and fluorine is most electronegative (3.98)

Atoms that form bonds with high electronegativity difference among them form ionic bonds (metal and non-metal). If their electronegativity difference (Δ)=0, then they form non polar covalent bond.

Δ=0 : Non- polar covalent bond.
0 < Δ < 1.7 : Polar covalent bond.
Δ = 1.7 : 50% ionic and 50% covalent bond

Δ > 1.7 : Ionic bond

The answer is a. polar covalent

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For the following example, list the given and unknown information (including gratis or moles)

Setler79 [48]

Answer:

9.6 moles O2

Explanation:

I'll assume it is 345 grams, not gratis, of water. Hydrogen's molar mass is 1.01, not 101.

The molar mass of water is 18.0 grams/mole.

Therefore: (345g)/(18.0 g/mole) = 19.17 or 19.2 moles water (3 sig figs).

The balanced equation states that: 2H20 ⇒ 2H2 +02

It promises that we'll get 1 mole of oxygen for every 2 moles of H2O, a molar ratio of 1/2.

get (1 mole O2/2 moles H2O)*(19.2 moles H2O) or 9.6 moles O2

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2 years ago

In Part III, the phenolphthalein indicator is used to monitor the equilibrium shifts of the ammonia/ammonium ion system. The phe

AnnZ [28]

The pink color in the solution fades. Some of the colored indicator ion converts to the colorless indicator molecule.

<h3>Explanation</h3>

What's the initial color of the solution?

$\text{NH}_4\text{Cl}$ is a salt soluble in water. $\text{NH}_4\text{Cl}$ dissociates into ions completely when dissolved.

$\text{NH}_4\text{Cl} \; (aq)\to {\text{NH}_4}^{+} \; (aq) +{\text{Cl}}^{-} \; (aq)$ .

The first test tube used to contain $\text{NH}_4\text{OH}$ . $\text{NH}_4\text{OH}$ is a weak base that dissociates partially in water.

$\text{NH}_4\text{OH} \; (aq) \rightleftharpoons {\text{NH}_4}^{+} \;(aq)+ {\text{OH}}^{-} \; (aq)$ .

There's also an equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ ions.

${\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)$ .

$\text{OH}^{-}$ ions from $\text{NH}_4\text{OH}$ will shift the equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ to the right and reduce the amount of ${\text{H}_3\text{O}}^{+}$ in the solution.

The indicator equilibrium will shift to the right to produce more ${\text{H}_3\text{O}}^{+}$ ions along with the colored indicator ions. The solution will show a pink color.

What's the color of the solution after adding NH₄Cl?

Adding $\text{NH}_4\text{Cl}$ will add to the concentration of ${\text{NH}_4}^{+}$ ions in the solution. Some of the ${\text{NH}_4}^{+}$ ions will combine with $\text{OH}^{-}$ ions to produce $\text{NH}_4\text{OH}$ .

The equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ ions will shift to the left to produce more of both ions.

${\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)$

The indicator equilibrium will shift to the left as the concentration of ${\text{H}_3\text{O}}^{+}$ increases. There will be less colored ions and more colorless molecules in the test tube. The pink color will fade.

7 0

3 years ago

In a 0.737 m solution, a weak acid is 12.5% dissociated. calculate ka of the acid.

r-ruslan [8.4K]

Hello! Let me try to answer this :)

Thanks and please correct if there are any mistakes ^ ^