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Elza [17]
3 years ago
15

Form a bond: Each electron has a charge of 1–, and each proton has a charge of 1+. You can calculate the charge of an atom by su

btracting the number of electrons from the number of protons. Move an electron from the sodium to the chlorine atom
Chemistry
2 answers:
dimaraw [331]3 years ago
7 0

Answer:

A.What are the charges of each atom now?

Sodium: +1___ Chlorine: __-1___

Turn on Show charge to check. These charged atoms are called ions.

B.Is each ion stable? Explain.

-Electromagnetic force pulls them together

Shtirlitz [24]3 years ago
5 0

Explanation:

Bond forming:

   Given species:

             Na - Sodium              Cl - Chlorine

 Na has a valence electron  

 Cl has 7 valence electron

To form a bond of this nature, there must be a transfer of electrons.

 Here, the bond is between a metal and non-metal

The metal is sodium and the non-metal is chlorine;

  Na         →     Na⁺  +   e⁻                                           loss of 1 electron

2,8,1               2,8

                  isoelectronic with Neon

Cl  + e  →   Cl⁻                                                              gain of one electron

               isoelectronic with Argon

  Na⁺  +  Cl⁻   →   Na⁺.Cl⁻

  This bond type is an ionic bond

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18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c
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Answer:

a) Percentage by mass of carbon: 18.3%

   Percentage by mass of hydrogen: 0.77%

b)  Percentage by mass of chlorine: 80.37%

c) Molecular formula: C_{2} H Cl_{3}

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}

The same process is used to calculate the amount of hydrogen:

0.089g H_{2}O*\frac{2g H}{18g H_{2}O }  = 0.010g H

The percentage by mass of carbon and hydrogen are calculated as follows:

%C\frac{0.238g}{1.3g} *100%= 18.3%

%H\frac{0.010g}{1.3g} *100%=0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}= 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl=\frac{0.43g}{0.535g} * 100%= 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

18.3g C*\frac{1mol C}{12g C} = 1.52mol C

0.77g H*\frac{1mol H}{1g H} = 0.77mol H

80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl

Dividing each of the quantities above by the smallest (0.77mol), the  subscripts in a tentative formula would be

C=\frac{1.52}{0.77} = 1.97 ≈ 2

H = \frac{0.77}{0.77} = 1

Cl =\frac{2.27}{0.77}=2.94≈3

The empirical formula for the compound is:

C_{2} H Cl_{3}

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

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