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Shtirlitz [24]
3 years ago
7

Using the reagents below, list in order (by letter, no period) those necessary to transform 1-chlorobutane into 1-butyne. Note:

Not all entry-fields provided may be needed. Choose "na" in any space where you have no reagent .
a. H2, Pt:
b. H2, Lindlar's Catalyst
c. H2, Ni2B{P-2)
d. Br2, CCI4
e. NaNH2, Mineral oil, heat
f. NH4CI G.
g. t -BuOK, t -BuOH, heat
h. CH3 I
i. CH3CH2 I
Chemistry
1 answer:
Andru [333]3 years ago
3 0

Answer:

gde

Explanation:

We are attempting to synthesize 1-butyne from 1-chlorobutane. Since 1-chlorobutane is a primary alkyl halide, 1-butene is formed when 1-chlorobutane is reacted with a bulky base such as t -BuOK or t -BuOH in presence of strong heat. This is an E2 reaction.

Secondly, the 1-butene is reacted with bromine in carbon tetrachloride. The vicinal dihalide (1,2-dibromobutane) is formed. This can now undergo further elimination reactions in the presence of sodamide and strong heat to yield 1-butyne which is the desired product. These reactions involve the elimination of the first HBr molecule to give an alkenyl bromide. A second elimination now gives the terminal alkyne.

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Paraphin [41]

Answer:

because it will control the birth of a child which can happen alot being overpopulated in the country or world. It also can mange their life and create a small and happy and healthy family.Also the child can get more love and support from their parents if they have less children and it wont be a taruma for the couple as well.  because growing a child can be expensive time costly and really hard.  

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Explanation:

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8 0
3 years ago
I need help on this question
uranmaximum [27]

Answer:

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Explanation:

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6 0
3 years ago
Read 2 more answers
Question 2
Alenkinab [10]

Answer:

1.53 atm

Explanation:

From the question given above, the following data were obtained:

Volume = constant

Initial pressure (P₁) = stp = 1 atm

Initial temperature (T₁) = 273 K

Final temperature (T₂) = 144 °C = 144 °C + 273 = 417 K

Final pressure (P₂) =?

Since the volume is constant, the final pressure can be obtained as follow:

P₁ / T₁ = P₂ / T₂

1 / 273 = P₂ / 417

Cross multiply

273 × P₂ = 417

Divide both side by 273

P₂ = 417 / 273

P₂ = 1.53 atm

Therefore, the final pressure (i.e the pressure inside the hot water bottle) is 1.53 atm.

8 0
3 years ago
1. Complete the following
Marat540 [252]

Answer:

\large \boxed{\text{0.603 mol}}

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of O₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:        180.16

           C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g:        18.1

(a) Moles of C₆H₁₂O₆

\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{18.1 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1005 mol C$_{6}$H$_{12}$O}_{6}

b) Moles of O₂

\text{Moles of O}_{2} =\text{0.1005 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol O}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.603 mol O}_{2}\\\\\text{The reaction requires $\large \boxed{\textbf{0.603 mol}}$ of oxygen}  

3 0
3 years ago
Identify the following neutral elements by their condensed electron configurations. (Enter the chemical symbol or name of the el
Xelga [282]

Answer: [Kr] 5s1 4d10 silver

[Xe] 6s2 4f14 5d7 Iridium

[Ar] 4s2 3d10 4p4 selenium

[Kr] 5s1 4d4 niobium

[Xe] 6s1 4f14 5d10 gold

[Kr] 5s2 4d10 5p4 tellurium

Explanation:

Selenium and tellurium will have the same properties because they both belong to group 16 in the periodic table.

Silver and gold will have the same properties because they both belong the group 11 in the periodic table.

Elements having the same outermost shell configuration show similar properties since chemical reactions occur by loss or gain of electrons. This loss or gain is determined by the number of electrons occupying the outermost shell. Elements in the same group have the same outermost shell electron configuration and similar properties.

5 0
3 years ago
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