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tigry1 [53]
3 years ago
13

Draw a sodium formate molecule. The structure has been supplied here for you to copy. To add formal charges, click the button be

low (which will turn yellow when activated) before clicking on the molecule. Draw sodium formate by placing atoms on the grid and connecting them with bonds. Include all lone-pair electrons. View available hint(s)

Chemistry
1 answer:
Karo-lina-s [1.5K]3 years ago
5 0

The Molecule of Sodium Formate along with Formal Charges (in blue) and lone pair electrons (in red) is attached below.

Sodium Formate is an ionic compound made up of a positive part (Sodium Ion) and a polyatomic anion (Formate).

Nomenclature:

                       In ionic compounds the positive part is named first. As sodium ion is the positive part hence, it is named first followed by the negative part i.e. formate.

Name of Formate:

                             Formate ion has been derived from formic acid ( the simplest carboxylic acid). When carboxylic acids looses the acidic proton of -COOH, they are converted into Carboxylate ions.

E.g.

                    HCOOH (formic acid)    →     HCOO⁻ (formate)  +  H⁺

                H₃CCOOH (acetic acid)     →      H₃CCOO⁻ (acetate)  +  H⁺

Formal Charges:

                           Formal charges are calculated using following formula,

          F.C  =  [# of Valence e⁻] - [e⁻ in lone pairs + 1/2 # of bonding electrons]

For Oxygen:

                    F.C  =  [6] - [6 + 2/2]

                    F.C  =  [6] - [6 + 1]

                    F.C  =  6 - 7

                    F.C  =  -1

For Sodium:

                    F.C  =  [1] - [0 + 0/2]

                    F.C  =  [1] - [0]

                    F.C  =  1 - 0

                    F.C  =  +1

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The metallic radius of an aluminum atom is 143 pm. What is the volume of an aluminum atom in cubic meters?
Marina86 [1]

Answer:

6.61 × 10∧-29 m³

Explanation:

Given data:

Atomic radius= 143 pm = 143 × 10∧-12 m

volume = ?

Formula:

r = a/2√2

143 × 10∧-12 m = a/ 2√2

a= 143 × 10∧-12 m × 2√2

a= 404.4 × 10∧-12 m

where a is edge length, so we can calculate the volume by using following formula:

volume= a³

V= (404.4 × 10∧-12 m)³

v= 6.61 × 10∧-29 m³

8 0
3 years ago
What is the empirical formula of a compound with 35.94% aluminum and 64.06% sulfur
Sunny_sXe [5.5K]

                        Al                                               S

    1)            <u>  35.94  </u>    =1.33111                     <u>  64.06 </u>     = 2.001875

                       27                                               32

    2)          =    <u> 1.33111 </u>                                 = <u>  2.001875  </u>

                     1.33111                                         1.33111

                 

    3)         =   ( 1 ) × 2                                     =    ( 1.5 ) ×2

    4)         =  2                                                =   3

                     Empirical Formula = Al2S3

1) Divide the percentage given in question by the Relative Atomic Mass (RAM)

of the given elements.

2) When you find the answers of the first part of question, divide these once again but this time, by the lowest number you found in part 1.

3) and 4) Write down the values. If you get a decimal which is in between 0.3-0.7 (including the 0.3 and 0.7), you cannot make it a whole number by rounding of. Therefore, multiply the decimal with a whole number until you get a whole number as your answer. In this question, when you multiply 1.5 by 2, the answer is 3 which is a whole number. Multiply the other whole number by the same number as that you multiplied for 1.5. And use these numbers in part 4 to make the empirical formula which is Aluminium Sulfide (Al2S3)

Hope this helps you :)))

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3 years ago
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7 0
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mezya [45]

4. Rubidium

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4 0
2 years ago
If an ideal gas has a pressure of 4.03 atm,
kkurt [141]

Answer:

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Explanation:

7 0
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