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Step2247 [10]
3 years ago
11

To draw a complete lewis structure of one molecule of water, how many electrons must be shown?

Chemistry
2 answers:
Ipatiy [6.2K]3 years ago
6 0

Answer:

Total 8 electrons.

Explanation:

To draw a complete lewis structure in water molecule we will calculate the total number of valence electrons

Valence electrons:

Two hydrogen atoms: 2 valence electrons (one each)

One oxygen atom : 6 valence electrons

So total there will be 8 electrons that must be shown in the structure.

Out of these 8, four will be involved in two covalent bonds

Rest four will be in the form of two lone pair of electrons on oxygen atom.

love history [14]3 years ago
5 0
In the Lewis structure, each single bond drawn between elements contain two electrons. The electron dots represent the lone pairs. These are the electrons that take part in the reaction. To obey the octet rule, the oxygen must have eight electrons around it. The hydrogen is exempted from this rule. Therefore, you have to show two electrons.

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Explanation:

Given parameters:

 Wavelength of photon  = 827nm  = 827  x 10⁻⁹m

Unknown:

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The energy of a photon can be derived using the expression below:

      E = \frac{h c}{wavelength}

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Insert the parameters and solve;

      E  = \frac{6.63 x 10^{-34} x 3 x 10^{8}  }{827x 10^{-9} }  

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4 years ago
A sheet of polyethylene 1.5-mm thick is being used as an oxygen barrier at a temperature of 600K. If the flux is 2.48 x 10-5 kg/
Eddi Din [679]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The concentration of  high-pressure side is  C_1 =   0.722 \ kg/m^3

Explanation:

From the question we are told that

    The thickness of the polyethylene is  d  =  1.5 \ mm = 0.0015 \ m

     The  temperature is  T  =  600 \   K

      The flux is  JA  =  2.48 *10^{-5} \  kg/m^2\cdot s

      The concentration on the low-pressure side is  C_2 =  0.5 \ kg/m^3

       The initial diffusivity  is  D_o  =  6.2 *10^{-4} \ m^2 /s

       The activation energy for  diffusion is   Q_d  =  41 \ kJ /mol  =  41*10^3 J /mol

Generally the diffusivity  of the oxygen at 600 K can be  mathematically evaluated  as

        D   = D_o * e^{- \frac{Q_d}{R * T  } }  

substituting values  

         D   = 6.2*10^{-4} * e^{- \frac{41 *10^3}{8.314 * 600  } }  

          D   = 1.671 *10^{-7} \ m^2 /s  

Generally the flux is mathematically represented as

          JA  =  D  *  \frac{C_1 -C_2}{d}

Where C_1 is the concentration of oxygen at the higher side

       So  

             C_1 =    d  * \frac{JA}{D}  + C_2

substituting values  

             C_1 =    0.0015   * \frac{2.48*10^{-5}}{1.671*10^{-7}}  + 0.5

              C_1 =   0.722 \ kg/m^3

 

6 0
3 years ago
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