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svetoff [14.1K]
4 years ago
6

Dicarbon monoxide, C2O, is found in dust clouds in space. Analysis of it shows that the sequence of atoms in this molecule is C–

C–O. All bonds are double bonds and there are no unpaired electrons. How many lone pairs of electrons are present in a molecule of C2O?
Chemistry
1 answer:
Sunny_sXe [5.5K]4 years ago
3 0
Carbon dioxide has 4 lone pairs of electrons. During bonding, the bonds accounts for 8 electrons of the 16 valence electrons the molecule has. The remaining 8 valence electrons will be placed as lone pairs, two on each oxygen atom. As a result carbon dioxide molecule has a total 4  lone pairs of electrons. Dicarbon monoxide is a molecule that contains two carbon atoms and one oxygen atom.  A lone pair refers to a pair of valence electrons that are not shared with another atom and is some times called a non-bonding pair. Lone pairs are found in the outermost electron shell of atoms. 
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Consider the following reaction: CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at25 ∘C under these conditions: PCO2PCC
Salsk061 [2.6K]

<u>Answer:</u> The \Delta G for the reaction is 54.6 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

We are given:

\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol

  • To calculate \Delta G^o_{rxn} for the reaction, we use the equation:

\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]

For the given equation:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J

Conversion factor used = 1 kJ = 1000 J

  • The expression of K_p for the given reaction:

K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}

We are given:

p_{COCl_2}=0.760atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.180atm

Putting values in above equation, we get:

K_p=\frac{(0.760)^2}{0.140\times 0.180}\\\\K_p=22.92

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

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

where,

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

\Delta G^o = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 22.92

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(22.92))\\\\\Delta G=54659.78J/mol=54.6kJ/mol

Hence, the \Delta G for the reaction is 54.6 kJ/mol

7 0
3 years ago
What are moles in chemistry?
Firlakuza [10]
A mole is a unit to describe an amount of something/specific substance. Typically standard for measuring a large quantity of small entities like atoms, molecules, or other particles.
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Triss [41]

Answer:

1s2 2s2

Explanation:

7 0
3 years ago
The kw for water at 0 °c is 0.12× 10–14 m2. Calculate the ph of a neutral aqueous solution at 0 °c.
jekas [21]

Answer:

pH = 7.46.

Explanation:

  • The ionization of water is given by the equation :

<em>H₂O(l) ⇄ H⁺(aq) + OH⁻(aq),</em>

<em></em>

  • The equilibrium constant (Kw) expression is:

<em>Kw = [H⁺][OH⁻] = 0.12 x 10⁻¹⁴.  </em>

<em></em>

in pure water and neutral aqueous solution, [H⁺] = [OH⁻]  

So, Kw = [H⁺]²

∴ 0.12 x 10⁻¹⁴ = [H⁺]²

∴ [H⁺] = 3.4 x 10⁻⁸ M.

∵ pH = - log [H⁺]  

pH = - log (3.4 x 10⁻⁸) = 7.46.

8 0
3 years ago
When you shine a white light through a prism the light is split into the different colors of the rainbow what are the colors tha
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Answer:

Red, orange, yellow, green, blue, indigo, and violet.

Explanation:

These are the colors that light takes on when split using a prism. These are the same colors that are in a rainbow. You can remeber these colors using this acronym:

R.O.Y G. B.I.V

It's kinda like a name!

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4 years ago
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