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stira [4]
2 years ago
9

What is Markanikov rule?​

Chemistry
2 answers:
lana [24]2 years ago
5 0
Markovnikov rule, in organic chemistry, a generalization, formulated by Vladimir Vasilyevich Markovnikov in 1869, stating that in addition reactions to unsymmetrical alkenes, the electron-rich component of the reagent adds to the carbon atom with fewer hydrogen atoms bonded to it, while the electron-deficient component ...
iren [92.7K]2 years ago
3 0

Answer:

Regla de Markovnikov En fisicoquímica orgánica, la regla de Markovnikov es una observación respecto a la reacción de adición electrófila.Regla de Markovnikov En fisicoquímica orgánica, la regla de Markovnikov es una observación respecto a la reacción de adición electrófila.

Explanation:

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If 90.0 grams of ethane reacted with excess chlorine,how many grams of dicarbon hexachloride would form
tigry1 [53]

Answer:

709 g  

Step-by-step explanation:

a) Balanced equation

Normally, we would need a balanced chemical equation.

However, we can get by with a partial equation, as log as carbon atoms are balanced.

We know we will need an equation with masses and molar masses, so let’s <em>gather all the information</em> in one place.  

M_r:    30.07          236.74

           C₂H₆ + … ⟶ C₂Cl₆ + …  

m/g:    90.0

(i) Calculate the moles of C₂H₆

n = 90.0 g C₂H₆  × (1 mol C₂H₆ /30.07 g C₂H₆)

  = 2.993 mol C₂H₆

(ii) Calculate the moles of C₂Cl₆

The molar ratio is (1 mol C₂Cl₆/1 mol C₂H₆)

n = 2.993 mol C₂H₆ × (1 mol C₂Cl₆/1 mol C₂H₆)

  = 2.993 mol C₂Cl₆

(iii) Calculate the mass of C₂Cl₆

m = 2.993 mol C₂Cl₆ × (236.74 g C₂Cl₆/1 mol C₂Cl₆)

m = 709 g C₂Cl₆

The reaction produces 709 g C₂Cl₆.

6 0
3 years ago
33. Why does Fluorine exert a stronger pull than Chlorine on the valence electrons of another
mariarad [96]

Answer:

Fluorine is much more reactive than chlorine (despite the lower electron affinity) because the energy released in other steps in its reactions more than makes up for the lower amount of energy released as electron affinity.

Explanation:

7 0
3 years ago
In the laboratory you dissolve 22.0 g of nickel(II) nitrate in a volumetric flask and add water to a total volume of 125
Ahat [919]

Answer:

<u>1.364 M</u>.

Explanation:

Molarity formula: M= n/v, where n is moles of solute, and v is liters of solution.

Now, we need to convert the grams of nickel to moles and the volume of water to liters.

125mL/1000= 0.125 L.

To convert nickel grams to moles, we need to take a look at it's chemical formula, which is:

Ni(NO_{3} )_{2}

Now we count how many molecules of each element we have:

Ni= 1

N= 2

O= 6

Calculate the weight (g) of each element (the values of g/mol can be found on the periodic table and they may vary slightly between one table and the other):

Ni: (1) (58.6934)= 58.6934

N= (2) (14.007)= 28.014

O= (6) (15.999)= 95.994

Sum all the values to obtain the total weight of 1 mole of this compound:

58.6934+28.014+95.994= 129(g/mole)

Now that we know the that 129 grams equal 1 mole of nickel(II) nitrate, we can convert the 22.0 g to moles:

129g ------- 1 mole

22.0g ----- x

x= (22*1)/129= 0.1705 moles.

Now, we have all the values needed to calculate the molarity of this solution. All we have to do is substitute the values in the formula:

M= (0.1705 moles) / (0.125 L)= <u>1.364 M</u>.

3 0
2 years ago
A chemistry student needs 55.0g of carbon tetrachloride for an experiment. by consulting the crc handbook of chemistry and physi
Monica [59]

The volume of a substance is simply the ratio of mass and density. Therefore:

volume = mass / density

 

Calculating for volume of Carbon Tetrachloride that the student has to pour out:

volume = 55.0 g / (1.59 g / cm^3)

<span>volume = 34.60 cm^3</span>

6 0
3 years ago
The formation of SO3 from SO2 and O2 is an intermediate step in the manufacture of sulfuric acid, and it is also responsible for
jeka57 [31]

Answer:

118.22 atm

Explanation:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)      

KP = 0.13 = \frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}

Where p(SO₃) is the partial pressure of SO₃, p(SO₂) is the partial pressure of SO₂ and p(O₂) is the partial pressure of O₂.

  • With 2.00 mol SO₂ and 2.00 mol O₂ if there was a 100% yield of SO₃, then 2 moles of SO₃ would be produced and 1.00 mol of O₂ would remain.
  • With a 71.0% yield, there are only 2*0.71 = 1.42 mol SO₃, the moles of SO₂ that didn't react would be 2 - 1.42 = 0.58; and the moles of O₂ that didn't react would be 2 - 1.42/2 = 1.29.

The total number of moles is 1.42 + 0.58 + 1.29 = 3.29. With that value we can calculate the molar fraction (X) of each component:

  • XSO₂ = 0.58/3.29 = 0.176
  • XO₂ = 1.29/3.29 = 0.392
  • XSO₃ = 1.42/3.29 = 0.432

The partial pressure of each gas is equal to the total pressure (PT) multiplied by the molar fraction of each component.

  • p(SO₂) = 0.176 * PT
  • p(O₂) = 0.392 * PT
  • p(SO₃) = 0.432 * PT

Rewriting KP and solving for PT:

\frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}=0.13\\\frac{(0.432*P_{T})^{2}}{(0.176*P_{T})^{2}(0.392*P_{T})} =0.13\\\frac{0.1866*P_{T}^{2}}{0.0121*P_{T}^{3}} =0.13\\\frac{15.369}{P_{T}}=0.13\\P_{T}=118.22 atm

5 0
3 years ago
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