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2 years ago

Which statement best describes the effect of low ionization energies and low electronegativities on metallic bonding?

2 answers:

kramer2 years ago

6 0

The statement that best describes the effect of low ionization energies and low electronegativities on metallic bonding is the first one - the valence electrons are easily delocalized.
Due to these low energies and negativities, valence electrons can be moved around quite easily and their positions may be altered quite drastically.

Rudik [331]2 years ago

3 0

Answer:

the valence electrons are easily delocalized.

Explanation:

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Caffeine (C8H10N4O2) is a weak base with a pKb of 10.4. Calculate the pH of a solution containing a caffeine concentration of 41

RSB [31]

Answer:

pH → 7.47

Explanation:

Caffeine is a sort of amine, which is a weak base. Then, this pH should be higher than 7.

Caffeine + H₂O ⇄ Caffeine⁺ + OH⁻ Kb

1 mol of caffeine in water can give hydroxides and protonated caffeine.

We convert the concentration from mg/L to M

415 mg = 0.415 g

0.415 g / 194.19 g/mol = 2.14×10⁻³ mol

[Caffeine] = 2.14×10⁻³ M

Let's calculate pH. As we don't have Kb, we can obtain it from pKb.

- log Kb = pKb → 10^-pKb = Kb

10⁻¹⁰'⁴ = 3.98×10⁻¹¹

We go to equilibrium:

Caffeine + H₂O ⇄ Caffeine⁺ + OH⁻ Kb

Initially we have 2.14×10⁻³ moles of caffeine, so, after the equilibrium we may have (2.14×10⁻³ - x)

X will be the amount of protonated caffeine and OH⁻

Caffeine + H₂O ⇄ Caffeine⁺ + OH⁻ Kb

(2.14×10⁻³ - x) x x

We make the expression for Kb:

3.98×10⁻¹¹ = x² / (2.14×10⁻³ - x)

We can missed the -x in denominator, because Kb it's a very small value.

So: 3.98×10⁻¹¹ = x² / 2.14×10⁻³

√(3.98×10⁻¹¹ . 2.14×10⁻³) = x → 2.92×10⁻⁷

That's the [OH⁻]. - log [OH⁻] = pOH

- log 2.92×10⁻⁷ = 6.53 → pOH

14 - pOH = pH → 14 - 6.53 = 7.47

4 0

2 years ago

Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang

Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

$\Delta (H_{f})_{MnO_{2}}$ = -520.0 KJ/mole

$\Delta (H_{f})_{Al_{2}O_{3}}$ = -1699.8 KJ/mole

The balanced chemical reaction is,

$3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)$

Formula used :

$\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}$

$\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))$

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

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

$\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]$

= (-3399.6) + (1560)

= -1839.6 KJ

5 0

2 years ago

The region in which an electron is likely to be found is known as a(n) ____________.

Zinaida [17]

Answer:

orbital

Explanation:

electrons are found in an orbital

3 0

2 years ago

Describe the factors which make an organism living

Anvisha [2.4K]

All living organisms share several key characteristics or functions: order, sensitivity or response to the environment, reproduction, adaptation, growth and development, homeostasis, energy processing, and evolution. When viewed together, these characteristics serve to define life.

Hey Mate! ! ! I hope this was helpful if yes please mark me brainliest.

6 0

2 years ago

50 pts With 11 electrons how many electron shells does sodium have? 3 4 2 1

barxatty [35]

I think the answer is three

3 0

2 years ago

Read 2 more answers