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JulijaS [17]
3 years ago
10

1-Butanol, 1-butene, and di-n-butyl ether are all removed from the product by washing with concentrated sulfuric acid. Explain w

hy these are removed and why 1-bromobutane is not removed.
Chemistry
1 answer:
Alecsey [184]3 years ago
3 0

Answer:

1-bromobutane, unlike others, will not be removed because it is an <em>alkane</em> and inert to concentrated sulfuric acid.

Explanation:

<em>Alkanes</em> are saturated organic compounds, primarily made up of hydrogen and carbon atoms. Saturation implies that they posses no double or triple bonds, therefore making them stable and unresponsive to addition reactions and also, concentrated sulfuric acid.

1-Butanol is a primary alcohol. Primary alcohols will react with acids to produce alkyl halides. 1-butene, as an alkene, will react with concentrated sulfuric acid to produce alkyl hyrgensulfate, while di-n-butyl will produce an ether when it reacts with concentrated sulfuric acid.

Therefore, these organic compounds will be washed and converted to other compounds by concentrated sulfuric acid except 1-bromobutane.

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How does the law of conservation of matter apply to chemical equations?
Tasya [4]

Answer: option D.

The total number of atoms of each element on both sides of the

equation must be the same.

Explanation:

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4 HF(g)+SiO2(s) → SiF4(9)+2 H2O(9) <br> Is the Si oxidized or reduced?
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Si is reduced since it loses the oxygen atom

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3 years ago
The molar solubility of magnesium carbonate is 1.8 × 10–4 mol/l. What is ksp for this compound?
Serhud [2]

Answer:

3.2 × 10⁻⁸

Explanation:

Let's consider the solution of magnesium carbonate.

MgCO₃ ⇄ Mg²⁺(aq) + CO₃²⁻(aq)

We can relate the molar solubility (S) with the solubility product (Ksp) using an ICE chart.

         MgCO₃ ⇄ Mg²⁺(aq) + CO₃²⁻(aq)

I                             0                0

C                          +S              +S

E                            S                S

The Ksp is:

Ksp = [Mg²⁺] × [CO₃²⁻] = S × S = S² = (1.8 × 10⁻⁴)² = 3.2 × 10⁻⁸

4 0
3 years ago
Read 2 more answers
What is the mass (grams) of salt in 10.0 m' of ocean water? ball park-4x10's (1.000 molsalt -58.44 g salt, 1.0 L ocean water -0.
koban [17]

Answer:

3.5 × 10⁵ g of salt

Explanation:

<em>What is the mass (grams) of salt in 10.0 m³ of ocean water?</em>

We have this data:

  • 1.000 mol salt is equal to 58.44 g salt
  • 1.0 L of ocean water contains 0.60 mol of salt

We will need the following relations:

  • 1 L = 1dm³
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We can use proportions:

10.0m^{3} .\frac{10^{3}dm^{3}  }{1m^{3} } .\frac{1L}{1dm^{3} } .\frac{0.60molSalt}{1.0L} .\frac{58.44gSalt}{1molSalt} =3.5 \times 10^{5} gSalt

8 0
3 years ago
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