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cricket20 [7]
4 years ago
5

Carbon atoms form ____ (2,3,4) bonds with other atoms by _____ (gaining, losing, sharing) valence electrons in order to attain a

stable configuration
Chemistry
2 answers:
Dimas [21]4 years ago
5 0

Answer : Carbon atoms form 4 bonds with other atoms by sharing valence electrons in order to attain a stable configuration.

Explanation :

The given element is, Carbon

The element carbon with symbol (C) having the atomic number 6 and the atomic mass 12.

The number of electrons in carbon atom is, 6

The electronic configuration of carbon will be, 1s^22s^22p^2

The number of valence electrons in carbon atom = 2 + 2 = 4

That means carbon atom have the ability to form four bonds with the other atoms by the sharing of four valence electrons in order to achieve a stable electronic configuration.

For example : Carbon atom share their four valence electrons with four hydrogen atoms that has one valence electron to form a methane molecule, CH_4 and carbon and hydrogen are bonded by the covalent bond.

Levart [38]4 years ago
3 0

carbon atoms form 2 bonds with sharing valence electrons

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Methane, a component in natural gas, can be used as a fuel in combustion reactions. What is the value for ΔGnon (in kJ) for the
Oksanka [162]

<u>Answer:</u> The \Delta G for the reaction is -806.86 kJ

<u>Explanation:</u>

We are given:

\Delta H^o_{rxn}=-803kJ=-803000J      (Conversion factor:  1 kJ = 1000)

\Delta S^o_{rxn}=-4.05J/K

Temperature of the reaction = 293 K

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

\Delta G^o_{rxn}=\Delat H^o_{rxn}-T\Delta S^o_{rxn}

Putting values in above equation, we get:

\Delta G^o_{rxn}=-803000J-[(293K)\times (-4.05J/K)]=-801813.35J

For the given chemical equation:

CH_4(g)+2O_2(g)\rightleftharpoons CO_2(g)+2H_2O(g)

The expression for K_c is given as:

K_{c}=\frac{[H_2O]^2[CO_2]}{[CH_4][O_2]^2}

We are given:

[H_2O]=6.41M

[CO_2]=3.83M

[CH_4]=14.51M

[O_2]=9.27M

Putting values in above equation, we get:

K_c=\frac{(6.41)^2\times 3.83}{14.51\times (9.27)^2}

K_c=0.126

To calculate the Gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_c

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = -801813.35 J

R = Gas constant = 8.314J/K mol

T = Temperature = 293 K

K_c = equilibrium constant in terms of concentration = 0.126

Putting values in above equation, we get:

\Delta G=-801813.35J+(8.314J/K.mol\times 293K\times \ln(0.126))

\Delta G=-806859.46J=-806.86kJ

Hence, the \Delta G for the reaction is -806.86 kJ

8 0
3 years ago
How many moles of naoh are present in 29.0 ml of 0.290 m naoh?
Dovator [93]
29.0 mL in liters:

29.0 / 1000 => 0.029 L

n = M x V

n = 0.290 x 0.029

n = 0.00841 moles of NaOH

hope this helps!
7 0
3 years ago
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