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Alexeev081 [22]
3 years ago
15

Part a identify the compound that does not have dipole-dipole forces as its strongest force. Identify the compound that does not

have dipole-dipole forces as its strongest force. Co2 ch2 br2 ch3 br hccl3 ch3och3

Chemistry
2 answers:
posledela3 years ago
6 0

CO2 from the given compounds does not have dipole-dipole forces as its strongest force. While the other given compounds including CH2Br2, CH3Br, CHCl3 and CH3OCH3 have dipole-dipole forces as their strongest forces.  

Further explanation:

CO2:  

CO2 is a non polar molecule although it has polar bonds between its central carbon atom and two oxygen atoms. The dipoles are pointing in opposite direction so they cancel out each other and hence it only has London dispersion forces. The carbon and oxygen both are non metals so carbon dioxide has covalent bond in it.  

CH2Br2, CH3Br and CHCl3:  

In the compound CH2Br2, CH3Br and CHCl3, there exists an electronegative difference between the carbon atom and the halogen atom (Br or Cl). So these compounds have dipole-dipole forces in them these are polar compounds as well. In compound CH2Br2 both the dipole-dipole and London dispersion forces are present.  

CH3OCH3:

The given compound is dimethyl ether. The carbon and oxygen bonds in methoxy methane (or dimethyl ether) are polar. It has angular geometry. There are dipole-dipole and London dispersion forces present in this molecule.  

Dipole-dipole forces:

These are the attractive forces which are present between positive end of one polar molecule and the negative end of the other polar molecule. These are one percent as effective as a covalent bond.  

Answer details

Subject: Chemistry

Level: College

Keywords

  • CO2
  • CH2Br2, CH3Br and CHCl3  
  • CH3OCH3
  • Dipole-dipole forces

Learn more to evaluate

brainly.com/question/10885719  

brainly.com/question/5031462

Solnce55 [7]3 years ago
5 0

CO₂ does not have dipole-dipole forces as its strongest force.

Draw the 3D structures of each molecule.  All have C-O, C-Br, or C-Cl <em>bond dipoles</em>.

In CO₂ the C-O bond dipoles point in opposite directions, so they cancel.

CO₂ has <em>no molecular dipole</em>. Its strongest intermolecular forces are <em>London dispersion forces</em>.

In all the other molecules, the C-Br, C-Cl, and C-O bond dipoles do not cancel. These molecules all have intermolecular dipole-dipole forces.


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How many moles of carbon dioxide gas should be produced when 10.0 g of C2H6 are combusted at STP?
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Answer:

                    0.665 moles of CO₂

Explanation:

                     The balance chemical equation for the combustion of Ethane is as follow:

                            2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

Step 1: <u>Calculate moles of C₂H₆ as;</u>

                              Moles  =  Mass  /  M.Mass

Putting values,

                              Moles  =  10.0 g / 30.07 g/mol

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Step 2: <u>Calculate Moles of CO₂ as;</u>

According to balance chemical equation,

                    2 moles of C₂H₆ produced  =  4 moles of CO₂

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Solving for X,

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Formula used :

A=\epsilon \times c\times l

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c = concentration of solution = ?

l = length of the cell = 1.20 cm

\epsilon = molar absorptivity of this solution =5.56\times 10^4 M^{-1} cm^{-1}

0.945=5.56\times 10^4 M^{-1} cm^{-1}\times 1.20 \times c

c=1.4163\times 10^{-5} M=14.16 \mu M

(1\mu M=10^{-6} M)

14.16 μM is the molarity of the red dye solution at the optimal wavelength 519nm and absorbance value 0.945.

b) c=1.4163\times 10^{-5} mol/L

1 L of solution contains 1.4163\times 10^{-5} moles of red dye.

Mass of 1.4163\times 10^{-5} moles of red dye:

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c) In order to dilute red dye solution by 5 times, we will need to add 1 L of water to solution of given concentration.

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Concentration of red solution after dilution = c'

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Explanation:

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This can be seen from the following illustration:

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Collect like terms

2Cu = 0 + 2

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Collect like terms

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From the above illustrations,

We can see that the roman numeral in both copper(I) oxide, Cu₂O and copper(II) oxide, CuO are different because the oxidation number of Cu in both cases are different.

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