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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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A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical formula? The second step is to calcul
Free_Kalibri [48]
Lets take 100 g of this compound,
so it is going to be 2.00 g H, 32.7 g S and 65.3 g O.

2.00 g H *1 mol H/1.01 g H ≈ 1.98 mol H
32.7 g S *1 mol S/ 32.1 g S ≈ 1.02 mol S
65.3 g O * 1 mol O/16.0 g O ≈ 4.08 mol O

1.98 mol H : 1.02 mol S : 4.08 mol O = 2 mol H : 1 mol S : 4 mol O

Empirical formula
H2SO4
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3 years ago
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17 points Pls helpppppopppp
pentagon [3]

Answer:

Lipids

Explanation:

Lipids perform many different functions in a cell. Cells store energy for long-term use in the form of lipids called fats. Lipids also provide insulated from the environment for plants and animals

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1. Draw a wave with a longer wavelength and the same amplitude as the wave shown. Then draw a second wave with a larger amplitud
velikii [3]

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Sound traveling

Explanation:

7 0
2 years ago
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4 examples of Substances that sublime?
AVprozaik [17]

1. Dry Ice (solid carbon dioxide)

2. Iodine

3. Arsenic

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3 0
3 years ago
How many moles of na2co3 are necessary to reach stoichiometric quantities with cacl2
lbvjy [14]

0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary  to reach stoichiometric quantities with cacl2.

<h3>Explanation:</h3>

Based on the reaction

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

1 mole of CaCl₂ reacts per mole of Na₂CO₃

we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g

  • We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
  • These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
  • Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:

<h3>Moles CaCl₂.2H₂O:</h3>

1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂

Moles Na₂CO₃:

0.0102 moles Na₂CO₃

Mass Na₂CO₃:

0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present

Therefore, we can conclude that 0.0102 moles Na₂CO₃  is necessary.to reach stoichiometric quantities with cacl2.

To learn more about stoichiometric quantities visit:

<h3>brainly.com/question/28174111</h3>

#SPJ4

7 0
2 years ago
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