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maria [59]
3 years ago
14

t the R group in the reagent over the arrow be isopropyl. (i.e. The reagent is LiN[CH(CH3)2]2.) Now, fill in the blanks. This re

action will follow ______ rule and will form the _________ substituted alkene. a. Zaitzev's; more b. Zaitzev's; less c. Hofmann's; less d. Hofmann's; more
Chemistry
2 answers:
vladimir2022 [97]3 years ago
6 0

Answer:. A

Explanation:

T the R group in the reagent over the arrow be isopropyl. (i.e. The reagent is LiN[CH(CH3)2]2.) Now, fill in the blanks. This reaction will follow Zaitzev'srule and will form the more substituted alkene.

Tanzania [10]3 years ago
3 0

Answer:c. Hofmann's; less

Explanation:

The Hofmann elimination process, named after its discoverer, the German chemist August Wilhelm Von Hofmann rule states that the major alkene product is the <u>least substituted </u>and least stable product when it comes to asymmetrical amines.  The Hofmann elimination can be illustrated in t e formation of “Hoffman” products in elimination reactions  using a less common, LDA LiN[CH(CH3)2]2  Lithium Di-isopropyl Amide (LDA) usually with base  potassium t-butoxide  which will give rise to a less substituted alkene.

This reaction will follow Hofmann's_____ rule and will form the ___less______ substituted alkene.

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773.33 degrees F
F=Fahrenheit
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F=K x 9/5 - 459.67
773.33 = 685 x 9/5 - 459.67

8 0
3 years ago
Balance the equation 2Fe + 3Cl2 → 2FeCl3
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Answer:

6 Cl0 + 6 e- → 6 Cl-I (reduction)

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2 years ago
An atom of argon in the ground state tends not to bond with an atom of a different element because the argon atom has (a)a total
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The answer is <span>a total of eight valence electrons
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5 0
3 years ago
Read 2 more answers
A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the
Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, C_6H_4N_2O_4

Explanation :

First we have to calculate the mass of benzene.

\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}

\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g  = 0.04395 kg

Formula used :  

\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}

where,

\Delta T_f = change in freezing point  = 5.53-1.37=4.16^oC

\Delta T_s = freezing point of solution

\Delta T^o = freezing point of benzene

Molal-freezing-point-depression constant (K_f) for benzene = 5.12^oC/m

m = molality

Now put all the given values in this formula, we get

4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}

\text{Molar mass of unknown compound}=180.6g/mol

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles

Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles

Moles of N = \frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles

Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{3.575}{1.193}=2.99\approx 3

For H = \frac{2.4}{1.193}=2.01\approx 2

For N = \frac{1.193}{1.193}=1

For O = \frac{2.381}{1.193}=1.99\approx 2

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = C_3H_2N_1O_2

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}

n=\frac{180.6}{84}=2

Molecular formula = (C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4

Therefore, the molecular of the compound is, C_6H_4N_2O_4

3 0
3 years ago
A perfect cube of aluminum metal was found to weigh 20.00 g. The density of aluminum is 2.7 g/ml. What are the dimensions of the
denis23 [38]

Answer:

             Height  = 1.9493 cm

             Width =  1.9493 cm

             Depth  =  1.9493 cm

Solution:

Data Given:

                  Mass  =  20 g

                  Density  =  2.7 g/mL

Step 1: Calculate the Volume,

As,

                                        Density  =  Mass ÷ Volume

Or,

                                        Volume  =  Mass ÷ Density

Putting values,

                                        Volume  =  20 g ÷ 2.7 g/mL

                                        Volume  =  7.407 mL or 7.407 cm³

Step 2: Calculate Dimensions of the Cube:

As we know,

                                        Volume  =  length × width × depth

So, we will take the cube root of 7.407 cm³ which is 1.9493 cm.

Hence,

                                        Volume  =  1.9493 cm × 1.9493 cm × 1.9493 cm

                                        Volume  =  7.407 cm³

4 0
3 years ago
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