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Whitepunk [10]
3 years ago
6

According to current theory, a major reason for bonding is

Chemistry
2 answers:
tatiyna3 years ago
6 0
Covalent bond   a chemical bond formed by two electrons that are shared between two atoms
ionic bond<span>    a chemical bond formed by the electrostatic attraction between ions</span>
metallic bond<span>    a bond characteristic of metals in which mobile valence electrons are shared among positive nuclei in the metallic crystal</span>
Kisachek [45]3 years ago
6 0
According to current theory, a major reason for bonding is to gain the stability of the octet formation. In order for an octet (8 parts) formation to be stable, there has to be a bond achieved among the elements which comprise it. 
 Read more on Brainly - brainly.com/sf/question/2781693
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What is a dark mysterious and non-nicotine mineral containing sample of a very common silicate? It's geology actually.
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Orthoclase & Plagioclase...
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3 years ago
The isomerization of methylisonitrile to acetonitrileCH3NC(g)→CH3CN(g)is first order in CH3NC . The rate constant for the reacti
lisabon 2012 [21]

Answer:

Option E, Half life = 2.96\times 10^3\ s

Explanation:

For a first order reaction, rate constant and half-life is related as:

            t_{1/2}=\frac{0.693}{k}

Where,

t_{1/2} = Half life

k = Rate constant

Rate constant given = 2.34\times 10^{-4}\ s^{-1}

t_{1/2}=\frac{0.693}{k}

=\frac{0.693}{2.34 \times 10^{-4}}=2.96\times 10^3\ s

So, the correct option is option E.

4 0
3 years ago
Phenolphthalein has a pKa of 9.7 and is colorless in its acid form and pink in its basic form. calculate [In-}/{HIn} for the fol
zvonat [6]

Answer:

1.58x10⁻⁵

2.51x10⁻⁸

0.0126

63.10

Explanation:

Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:

pH = pKa + log[In-]/[HIn]

pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,

i) pH = 4.9

4.9 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = - 4.8

[In-]/[HIn] = 10^{-4.8}

[In-]/[HIn] = 1.58x10⁻⁵

ii) pH = 2.1

2.1 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -7.6

[In-]/[HIn] = 10^{-7.6}

[In-]/[HIn] = 2.51x10⁻⁸

iii) pH = 7.8

7.8 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -1.9

[In-]/[HIn] = 10^{-1.9}

[In-]/[HIn] = 0.0126

iv) pH = 11.5

11.5 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = 1.8

[In-]/[HIn] = 10^{1.8}

[In-]/[HIn] = 63.10

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3 years ago
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