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3 years ago

What experimental evidence can you provide that the product isolated is 1-bromobutane?

2 answers:

5 0

The experimental evidence that you could provide that <span> the product isolated is 1-bromobutane would be the changes that happens that are observable by the naked eye. Hope this helps. Have a nice day. Feel free to ask more questions.</span>

shepuryov [24]3 years ago

5 0

Answer:

Boiling point and IR Spectra.

Explanation:

The only thing that can be determined to know that the product formed is 1-Bromobutane are two physical methods. The first one is isolation of the product and testing for boiling point because it is an organic compound. If the observed boiling range is 95 to 97 degree Celsius which is very close to published boiling range (100 to 104 degree Celsius), the product formed is 1-bromobutane.

A more readible method is taking the IR spectra of the compound. The characteristic peaks of the compound will be at 2960.20/cm which is relatively close to 2962/cm.

It all depends what reactants were used for forming 1-Bromobutane as there are number of reactants which can be used to form it.

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What is the pH of a solution with an [H+] of (a) 5.4 x 10-10, (b) 4.3 x 10-5, (c) 5.4 x 10-7?

IgorLugansk [536]

Answer:

a. 9.2

b. 4.4

c. 6.3

Explanation:

In order to calculate the pH of each solution, we will use the definition of pH.

pH = -log [H⁺]

(a) [H⁺] = 5.4 × 10⁻¹⁰ M

pH = -log [H⁺] = -log 5.4 × 10⁻¹⁰ = 9.2

Since pH > 7, the solution is basic.

(b) [H⁺] = 4.3 × 10⁻⁵ M

pH = -log [H⁺] = -log 4.3 × 10⁻⁵ = 4.4

Since pH < 7, the solution is acid.

(c) [H⁺] = 5.4 × 10⁻⁷ M

pH = -log [H⁺] = -log 5.4 × 10⁻⁷ = 6.3

Since pH < 7, the solution is acid.

3 0

2 years ago

A sealed reaction vessel initially contains 1.113×10-2 moles of water vapor and 1.490×10-2 moles of CO(g).

Molodets [167]

Answer:

Kc = $\frac{[8.326x10-3]^{1} }{[1.113x10-2]^{1}[1.490x10-2]^{1} }$

Kc = 50.2059

Explanation:

1. Balance the equation

2. Use the Kc formula

Remember that pure substances, like H2 are not included on the Kc formula

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3 years ago

In a constant‑pressure calorimeter, 70.0 mL of 0.350 M Ba(OH)2 was added to 70.0 mL of 0.700 M HCl. The reaction caused the temp

asambeis [7]

Answer:

ΔH = 57 Kj/mole H₂O

Explanation:

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3 years ago

Read 2 more answers

Which of the following pairs lists a substance that can neutralize H2SO4 and the salt that would be produced from the reaction?

Bogdan [553]

The second option only.

LiOH, Li₂SO₄.

<h3>Explanation</h3>

A base neutralizes an acid when the two reacts to produce water and a salt.

Sulfuric acid H₂SO₄ is the acid here. There are more than one classes of bases that can neutralize H₂SO₄. Among the options, there are:

Metal hydroxides

Ca(OH)₂ and
LiOH.

Metal hydroxides react with sulfuric acid to produce water and the sulfate salt of the metal.

$\text{Ca}(\text{OH})_{\bf 2}+\text{H}_2\text{SO}_4 \to \textbf{Ca}\textbf{SO}_{\bf 4} +{\bf 2}\;\text{H}_2\text{O}$ .

The formula for calcium sulfate $\text{CaSO}_4$ in option A is spelled incorrectly. Why? The charge on each calcium $\text{Ca}^{2+}$ is +2. The charge on each sulfate ion ${\text{SO}_4}^{2-}$ is -2. Unlike $\text{Li}^{+}$ ions, it takes only one $\text{Ca}^{2+}$ ion to balance the charge on each ${\text{SO}_4}^{2-}$ ion. As a result, $\text{Ca}^{2+}$ and ${\text{SO}_4}^{2-}$ ions in calcium sulfate exist on a 1:1 ratio.

$2\;\text{LiOH} +\text{H}_2\text{SO}_4 \to \text{Li}_2\text{SO}_4 + 2\;\text{H}_2\text{O}$ .

Ammonia, NH₃

Ammonia NH₃ can also act as a base and neutralize acids. NH₃ exists as NH₄OH in water:

$\text{NH}_3 + \text{H}_2\text{O} \to \textbf{NH}_{\bf 4}\text{OH}$ .

The ion ${\text{NH}_4}^{+}$ acts like a metal cation. Similarly to the metal hydroxides, NH₃ (or NH₄OH) neutralizes H₂SO₄ to produce water and a salt:

$2\;\textbf{NH}_{\bf 4}\text{OH}+ \text{H}_2\text{SO}_4 \to (\textbf{NH}_{\bf 4})_2\text{SO}_4+2\;\text{H}_2\text{O}$ .

The formula of the salt (NH₄)₂SO₄ in the fourth option spelled the ammonium ion incorrectly.

As part of the salt (NH₄)₂SO₄, the ammonium ion NH₄⁺ is one of the products of this reaction and can't neutralize H₂SO₄ any further.

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What do liver cells do?

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