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allsm [11]
1 year ago
7

given your proposed electron pushing mechanism, list as many factors that could play a role in the success or failure of the rea

ction:
Chemistry
1 answer:
miskamm [114]1 year ago
3 0

The factors that play the role in success is solvent, concentration of nucleophile and the substrate.

What is S_{N}2 reaction ?

When sp^{3} hybridized  carbon containing a leaving group and nucleophile attack from the remote side of the leaving group then the reaction is known as S_{N}2 reaction.

Factors depending on S_{N}2 reaction:

  1. Solvent: Polar aprotic solvents are very useful for this reaction as they increase the intermediate concentration and also solvate the nucleophile.
  2. Nucleophile: More strong the nucleophile  more is the rate of reaction.
  3. Substrate:1°  substrate react faster than 2° than 3°

Read more about S_{N}2 reaction:

brainly.com/question/10143438

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How many moles of ba(oh)2 are in 2400 grams of ba(oh)2?
Lerok [7]

Answer: 14.01 moles

To solve this question, you need to determine the molecular mass of the Ba(OH)2. The molecular mass would be: 137.3 + 2(16+1)= 171.3g/mol.

If the total weight of Ba(OH)2 is 2400g and its molecular mass is 171.3g/mol, the number of moles would be: 2400g/ (171.3g/mol)=14.01 moles

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Measure the level of lead in blood
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It's measured in micrograms per deciliter. The amount is 5 micrograms.
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3 years ago
Complete and balance the following equation:
sertanlavr [38]
You need to add the last substance to the products side(the right sode of the arrow). You have hydrogen and oxygen - water.

You get: BrO3 + N2H4 -> Br2 + N2 + H2O

# of Br: 1x1 = 1 # of Br: 2x1 = 2
O: 3x1 = 3 O: 1x1 = 1
N: 2x1 = 2 B N: 2x1 = 2
H: 4x1 = 4. H: 2x1 = 2

Br:
Multiply the reactant (left) side by 2 to balance.

O:
You've just multiplied the reactant oxygen by 2 so now the reactant side equals 6. Multiply the product (right) side by six as well.

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The product side is now equal to 12. Multiply the reactant side by 3 to balance.

N:
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You end up with the complete and balanced equation:

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4 0
4 years ago
Please help me find the final answer
dlinn [17]

Answer:

5. The mass of Na₂CO₃, that will produce 5 g of CO₂ is approximately 12.04 grams of Na₂CO₃

6. The mass of nitrogen gas (N₂) that will react completely with 150 g of hydrogen (H₂) in the production of NH₃ is 693.\overline{3168} grams of N₂

Explanation:

5. The given equation for the formation of carbon dioxide (CO₂) from sodium bicarbonate (Na₂CO₃)  is presented as follows;

(Na₂CO₃) + 2HCl → 2NaCl + CO₂ + H₂O

One mole (105.99 g) of Na₂CO₃ produces 1 mole (44.01 g) of CO₂

The mass, 'x' g of Na₂CO₃, that will produce 5 g of CO₂ is given by the law of definite proportions as follows;

\dfrac{x \ g}{105.99 \ g} = \dfrac{5 \ g}{44.01 \ g}

\therefore {x \ g} = \dfrac{5 \ g}{44.01 \ g} \times 105.99 \ g \approx 12.04 \ g

The mass of Na₂CO₃, that will produce 5 g of CO₂, x ≈ 12.04 g

6. The chemical equation for the reaction is presented as follows;

N₂ + 3H₂ → 2NH₃

Therefore, one mole (28.01 g) of nitrogen gas, (N₂), reacts with three moles (3 × 2.02 g) of hydrogen gas (H₂) to produce 2 moles of ammonia (NH₃)

The mass 'x' grams of nitrogen gas (N₂) that will react completely with150 g of hydrogen (H₂) in the production of NH₃ is given as follows;

\dfrac{x \ g}{28 .01 \ g} = \dfrac{150 \ g}{3 \times 2.02 \ g} = \dfrac{150 \ g}{6.06 \ g}

\therefore \ x \ g= \dfrac{150 \ g}{6.06 \ g} \times 28.01 \ g = 693.\overline {3168} \ g

The mass of nitrogen gas (N₂) that will react completely with 150 g of hydrogen (H₂) in the production of NH₃, x = 693.\overline{3168} grams

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