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svetoff [14.1K]
3 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]3 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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Which statement correctly compares the masses of subatomic particles in an atom such as carbon?
iogann1982 [59]

Answer:

C- A proton has about the same mass as a neutron .

Explanation:

In an atom such as a carbon atom, the masses of the proton and neutrons are the same.

The mass of the electrons is very negligible.

  • Protons are the positively charged particles in an atom
  • Neutrons do not carry any charges
  • Both protons and neutrons have similar masses.
  • They contribute the bulk of the mass of the atom.
  • The electrons carry negative charges and they have negligible masses.

The mass of protons and neutrons are similar.

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A 0.1510 gram sample of a hydrocarbon produces 0.5008 gram CO2 and 0.1282 gram H2O in combustion analysis. Its
Over [174]
In a combustion of a hydrocarbon compound, 2 reactions are happening per element:

C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O

Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.

1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C

Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H

The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%

2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:

Amount of C = 0.01138 mol
Amount of H = 0.014244 mol

Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25

The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5

Thus, the empirical formula of the hydrocarbon is C₄H₅.

3. The molar mass of the empirical formula is

Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.

Molecular Formula = C₈H₁₀

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n(Al(OH)₃) = 33 g ÷ 78 g/mol.
n(Al(OH)₃) = 0,423 mol.
From chemical reaction: n(H₂SO₄) : n(Al(OH)₃) = 3 : 2.

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