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emmainna [20.7K]
3 years ago
14

(f) Which of the following alkenes is the major product when 2-bromo-2-methylpentane is treated with sodium ethoxide in ethanol?

Chemistry
2 answers:
belka [17]3 years ago
8 0

Answer:

D) 2-methylpent-2-ene

Explanation:

This is an elimination reaction of Halogenoalkane. 2-bromo-2-methylpentane when is heated with NaOH or NaOC2O5( sodium ethoxide) in ethanol will form alkene rather than alcohol.

2-methylpent-1-ene is minor product since double bond form with secondary Carbon rather than primary Carbon.

serious [3.7K]3 years ago
4 0

Answer:

The major product when 2-bromo-2-methylpentane is treated with sodium ethoxide in ethanol is: 2-methylpent-2-ene

Explanation:

The elimination reaction generates the most substituted alkene (Zaitsev's rule)

Hence, the major product when 2-bromo-2-methylpentane is treated with sodium ethoxide in ethanol is 2-methylpent-2-ene.

In the attached picture, you can see that the 2-methylpent-1-ene is less substituted

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If 125g of KClO3 is heated, what is the total mass of the products?​
Andrej [43]

Given parameters:

Mass of KClO₃  = 125g

Unknown:

Total mass of the products = ?

When  KClO₃ is heated, it thermally decomposes to KCl and O₂ according to the chemical equation below;

               2KClO₃  →  2KCl + 3O₂

All chemical equations obeys the law of conservation of matter and with this regard, we know that the amount of reactants used is the same as that of the product.

The total mass of the products must give us 125g according to this law of conservation of matter.

Now to find the masses of each product,

  1. Find the number of moles of the given reactant:

     Number of moles  = \frac{mass}{molar mass}

  molar mass of  KClO₃  = 39 + 35.5 + 3(16)  = 122.5g/mol

    So number of moles of KClO₃ = \frac{125}{122.5}  = 1.02moles

    2. Now, using this number of moles, find the number of moles of the products using this value;

   2 moles of KClO₃ produced 2 moles of KCl

  1.02 moles of KClO₃ will also produce 1.02moles of KCl

   2 moles of KClO₃ produced 3 moles of O₂

   1.02 moles of KClO₃ will produce   \frac{1.02 x 3} {2} mole = 1.53 moles of O₂

   3. Now find the masses of each product;

Mass  = number of moles x molar mass

  molar mass of KCl  = 39 + 35.5 = 74.5g/mol

  molar mass of O₂  = 16 x 2  = 32g/mol

  Mass of KCl  = 74.5 x 1.02  = 75.99g

  Mass of O₂  = 32 x 1.53 = 48.96g

Total mass of products = mass of KCl + Mass of O₂ = 75.99g + 48.96g

                                        = 124.95g

This value is approximately the same as that of mass of  KClO₃

 

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What is the total amount of heat released when 94.0 g water at 80.0 °C cools to form ice at −30.0 °C?
Wewaii [24]

Answer:

The total amount of heat released  is  68.7 kJ

Explanation:

Given that:

mass of water = 94.0 g

moles of water = 94 / 18.02 = 5.216

80⁰C   ------>  0⁰C  -------->    -30⁰C

Q1 = m Cp dT

      = 94 x 4.184 x (0 - 80)

     = -31463.68 J

     = -31.43 kJ

Q2 = 6.01 x 10^3 x 5.216

    = - 31348.16 J

   = -31.35 kJ

Q3 = - 94 x 2.09 x 30

    = - 5893.8 J

   = -5.894 kJ

Total heat = Q1 + Q2 + Q3  = -31.43 kJ  + (-31.35 kJ  ) + (-5.894 kJ )  = -68.7 kJ

Total heat released = -68.7 kJ

Note that the "negative sign" simply indicates heat released, therefore no need to put it in the answer.

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3 years ago
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