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Marizza181 [45]
2 years ago
8

Nonmetals gain electrons under certain conditions to attain a noble-gas electron configuration. how many electrons must be gaine

d by the element c?
Chemistry
1 answer:
r-ruslan [8.4K]2 years ago
8 0
System is said to have achieved noble-gas configuration, when it's valance shell is completely filled.

Atomic number of carbon is 6. Thus, it has 6 electrons.

The electronic configuration of carbon is 1s2 2s2 2p2

Now, the inert gas closest to C is Ne, whose atomic number is 10.

Thus, there are excess of 4 electrons in Ne as compared to C.

Hence, carbon must gain 4 electrons to achieve noble-gas configuration.

Alternatively,  C can also lose 4 electron to achieve noble gas configuration of He.
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What happened to the size of the sun one carbon and oxygen
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Explanation:

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4 0
3 years ago
If the K a Ka of a monoprotic weak acid is 7.3 × 10 − 6 , 7.3×10−6, what is the pH pH of a 0.40 M 0.40 M solution of this acid?
olga_2 [115]

Answer:

pH =3.8

Explanation:

Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:

HA + H₂O   ⇄ H₃O⁺ + A⁻    with  Ka = [ H₃O⁺] x [A⁻]/ [HA]

The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.

In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:

                          HA                                   H₃O⁺                          A⁻          

Initial, M             0.40                                   0                              0

Change , M          -x                                     +x                            +x

Equilibrium, M    0.40 - x                              x                               x

Lets express these concentrations in terms of the equilibrium constant:

Ka = x² / (0.40 - x )

Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,

7.3 x 10⁻⁶ = x² / 0.40  ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³

[H₃O⁺] = 1.71 x 10⁻³

Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.

pH = - log ( 1.71 x 10⁻³ ) = 3.8

Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.

4 0
3 years ago
The isomerization of methylisonitrile to acetonitrileCH3NC(g)→CH3CN(g)is first order in CH3NC . The rate constant for the reacti
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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
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