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Elden [556K]
3 years ago
9

Which one of the following reactions would produce t-butyl methyl ether in high yield?

Chemistry
1 answer:
swat323 years ago
7 0

Answer:

D) sodium t-butoxide + bromomethane

Explanation:

The alkoxide ion is a strong nucleophile, that unlike alcohols, will react with primary alkyl halides to form ether. This general reaction is known as <em>the Williamson synthesis</em>, and is a SN₂ displacement. The alkyl halide must be primary so the back side attack is not hindered, and the alkoxide ion must be formed with the most hindered group.

The mechanism can be seen in the attachment.

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What is the "reproductive system"
bija089 [108]

Answer: The reproductive system of an organism, also known as the genital system, is the biological system made up of all the anatomical organs involved in sexual reproduction

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6 0
3 years ago
Read 2 more answers
2. Suppose that 21.37 mL of NaOH is needed to titrate 10.00 mL of 0.1450 M H2SO4 solution.
AysviL [449]

Answer:

0.1357 M

Explanation:

(a) The balanced reaction is shown below as:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O

(b) Moles of H_2SO_4 can be calculated as:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For H_2SO_4 :

Molarity = 0.1450 M

Volume = 10.00 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 10×10⁻³ L

Thus, moles of H_2SO_4 :

Moles=0.1450 \times {10\times 10^{-3}}\ moles

Moles of H_2SO_4  = 0.00145 moles

From the reaction,

1 mole of H_2SO_4 react with 2 moles of NaOH

0.00145 mole of H_2SO_4 react with 2*0.00145 mole of NaOH

Moles of NaOH = 0.0029 moles

Volume = 21.37 mL = 21.37×10⁻³ L

Molarity = Moles / Volume = 0.0029 /  21.37×10⁻³  M = 0.1357 M

7 0
3 years ago
Most particles would travel____from their source to a screen that lit up<br> when struck.
quester [9]

Answer: in a straight path!

Explanation: hope this helps

5 0
3 years ago
Sodium electron configuration
Alex73 [517]

Full:

1s² 2s² 2p⁶ 3s¹

Abbreviated:

[Ne] 3s¹

4 0
3 years ago
Hydrocarbons, compounds containing only carbon and hydrogen, are important in fuels. The heat of combustion of cyclobutane, C4H8
Gennadij [26K]

Answer:

1. C4H8 + 6O2  -----> 4CO2 + 4H20

2. 3836.77 kcal

Explanation:

1. Balanced equation for the complete combustion of cyclobutane:

C4H8 + 6O2  -----> 4CO2 + 4H20

2. Heat of combustion of cyclobutane = 650.3 kcal/mol

    Molecular weight of cyclobutane, C4H8 = 56.1 g/mol

   Mole of C4H8 : mass of cyclobutane/Molecular weight of cyclobutane

  Mole of C4H8 = 331/56.1 = 5.9 mol

Energy released during combustion =  5.9 mol × 650.3 kcal/mol = 3836.77kcal

Therefore the energythat  is released during the complete combustion of 331 grams of cyclobutane is 3836.77kcal

8 0
3 years ago
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