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qaws [65]
3 years ago
8

Draw the major organic product of the reaction. Indicate the stereochemistry via wedge/dash bonds, including explicit H and D at

oms, at the stereogenic center. Omit byproducts such as salts or methanol

Chemistry
1 answer:
Nat2105 [25]3 years ago
4 0

Answer:

In the reaction ocurrs a nucleophilic substitution of a primary alkyl bromide, where the nucleophile is CH3O-.

Explanation:

Nucleophilic substitution is a type of substitution reaction in which a nucleophile replaces an atom or group in an electrophilic position of a molecule, called a leaving group.

It is a type of fundamental reaction in organic chemistry, where the reaction occurs on an electrophilic carbon. Although nucleophilic substitution reactions can also take place on covalent inorganic compounds.

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VMariaS [17]

Answer:

magnitude and direction

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what is the specific latent heat of fusion for a substance that takes 550 kj to melt 14 kg at 262 k? brainly answers
Ket [755]

q = mL

L = q/m = 550 kJ/14 kg = 39 kJ/kg

3 0
3 years ago
What mass of N2O5 will result from the reaction of 6.0 mol of NO2 if there is a 61.1% yield in the reaction
Nata [24]

Answer:

2.0 × 10² g

Explanation:

Step 1: Write the balanced equation

2 NO₂ + 0.5 O₂ ⇒ N₂O₅

Step 2: Calculate the theoretical yield, in moles, of N₂O₅

The molar ratio of NO₂ to N₂O₅ is 2:1.

6.0 mol NO₂ × 1 mol N₂O₅/2 mol NO₂ = 3.0 mol N₂O₅

Step 3: Calculate the theoretical yield, in grams, of N₂O₅

The molar mass of N₂O₅ is 108.01 g/mol.

3.0 mol × 108.01 g/mol = 3.2 × 10² g

Step 4: Calculate the real yield, in grams, of N₂O₅

real yield = theoretical yield × percent yield

real yield = 3.2 × 10² g × 61.1% = 2.0 × 10² g

6 0
3 years ago
Which statement best describes what occurs during a chemical reaction?
In-s [12.5K]

Answer:

bonds within the nucleus of reactants atoms are broken and rearranged to form new product atoms

3 0
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The molar heat of the vaporization of ammonia i 23. 3 kJ/mol. What i the molar
dimaraw [331]

The molar heat of vaporization of ammonia is 23.3 kJ/mol. The molar heat of condensation of ammonia is - 23.3 kJ/mol.

The molar heat of condensation is the opposite of the molar heat of vaporization. The molar heat of vaporization of ammonia is given :

ΔH evaporation = - ΔH condensation

Therefore the molar heat of  condensation of ammonia is given by:

ΔH condensation = - 23.3 kJ / mol

That's right. The molar heat  of vaporization of ammonia is 23.3 kJ/mol. The molar heat of condesation of ammonia is - 23.3 kJ/mol.

Learn more about heat of condensation at

brainly.com/question/14380051

#SPJ4

3 0
1 year ago
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