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ExtremeBDS [4]
3 years ago
8

If the empirical formula of a compound is C2OH4, and the molar mass is 88 g/mol, what is the molecular formula

Chemistry
1 answer:
kondor19780726 [428]3 years ago
8 0

Answer:

               Molecular Formula = C₄O₂H₈

Explanation:

Molecular formula is calculated from following formula.

Molecular Formula = n(Empirical formula)

The formula for n is given as,

n = Molecular Formula Mass / Empirical Formula Mass

Molecular Formula mass = 88 g/mol

Empirical formula mass = 44

Putting values,

n = 88 / 44 = 2

Now,

Molecular Formula = 2(C₂OH₄)

Molecular Formula = C₄O₂H₈

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D = 1 x 10⁴ / 2

D = 5000 kg/m³ in g/cm³:

1 kg/m³ ---------------- 0.001 g/cm³
5000 kg/m³ ----------- ??

5000 x 0.001 / 1 =>  5.0 g/cm³

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0.90 g of sodium hydroxide ( NaOH ) pellets are dissolved in water to make 3.0 L of solution. What is the pH of this solution
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2 years ago
Which characteristic of a strong acid
olga nikolaevna [1]

Answer:

Strong acid breaks up into ions

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7 0
3 years ago
Determine if the bond between each pair of atoms is pure covalent, polar covalent, or ionic. drag the appropriate items to their
dsp73

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

 

                Less than 0.4 then it is Non Polar Pure Covalent

                

                Between 0.4 and 1.7 then it is Polar Covalent 

            

                Greater than 1.7 then it is Ionic

 

For Br and Br,

                    E.N of Bromine      =   2.96

                    E.N of Bromine      =   2.96

                                                   ________

                    E.N Difference             0.00         (Non Polar/Pure Covalent)

 

For N and O,

                    E.N of Oxygen      =   3.44

                    E.N of Nitrogen     =   3.04

                                                   ________

                    E.N Difference             0.40           (Non Polar/Pure Covalent)

 

For P and H,

                    E.N of Hydrogen       =   2.20

                    E.N of Phosphorous  =   2.19

                                                              ________

                    E.N Difference                  0.01          (Non Polar/Pure Covalent)

 

For K and O,

                    E.N of Oxygen          =   3.44

                    E.N of Potassium      =   0.82

                                                   ________

                    E.N Difference                2.62              (Ionic)

6 0
4 years ago
The force of attraction between a divalent cation and a divalent anion is 1.64 x 10-8 N. If the ionic radius of the cation is 0.
Elden [556K]

The radius of the anion is 7.413 nm

<h3>How to calculate the force of attraction between charges</h3>

The force of attraction (F) is given by the formula:

  • F = (1/4π∈r²)(Zc*e)(Za*e)

where:

∈ = permittivity of free space = 8.85*10⁻¹⁵ F/m

Zc = charge on the cation = +2

Zc = charge on the anion = -2

e = charge on an electron = 1.602 * 10⁻¹⁹ C

r = interionic distance

r = rc + ra

where rc and ra are the radius of the cation and anion respectively

F = 1.64 * 10⁻⁸ N

Therefore based on the equation of force of attraction:

1.64 *10⁻⁸ = [1/4π(8.85*10⁻¹⁵)r²](2 * 1.602*10⁻¹⁹)²

r² = 5.63 * 10⁻¹⁷

r = 7.50 nm

Since r = rc + ra

where rc = 0.087 nm

thus, ra = r - rc = 7.50 - 0.087

ra = 7.413 nm

Therefore, the radius of the anion is 7.413 nm

Learn more about ionic radius at: brainly.com/question/2279609

6 0
2 years ago
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