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Orlov [11]
3 years ago
14

What would you expect to observe when Br2 reacts with 2-butyne?

Chemistry
1 answer:
Nookie1986 [14]3 years ago
6 0

Answer:

I expect to observe a change in colour from reddish brown to a colourless solution

Explanation:

Bromine (Br2) attacks the electron rich carbon-carbon triple bond in but-2-yne, an alkyne to form an initial product 2, 3 dibromobut-2-ene; which reacts with excess bromine to form a final product 2,2,3,3 -tetrabromobutane.

The reaction occurs in two steps. On approaching but-2-yne, bromine molecule becomes polarised forming an induced dipole containing a bromonium ion.

Br - Br → Br+ - Br-

The bromonium ion (Br+) formed then attacks the carbon - carbon triple bond to form the initial product

2,3- dibromobut-2-ene

CH2-C≡C-CH2 + Br+ →

CH2 - CBr =CBr-CH2

(2,3- dibromobut-2-ene)

Which in the presence of excess bromine gives the final product

2,2,3,3 - tetrabromobutane.

CH2 - CBr =CBr-CH2 + Br2 →

CH3 -CBr2-CBr2 - CH3

2,2,3,3 - tetrabromobutane.

A visible change in colour from the reddish-brown colour of Bromine to a colourless solution is observed during the reaction.

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I NEED HELP ASAP! WILL MARK AS BRAINLIEST
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Answer:

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Explanation:

There are a couple of ways to prepare a buffer solution of a specific pH. In the first method, prepare a solution with an acid and its conjugate base by dissolving the acid form of the buffer in about 60% of the volume of water required to obtain the final solution volume

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3 years ago
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A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the
baherus [9]

Answer: 1. 9.08\times 10^{-6} moles

2. 90 mg

Explanation:

O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 4.54 \times 10^{-6} moles of ozone is removed by =\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6} moles of sodium iodide.

Thus 9.08\times 10^{-6} moles of sodium iodide are needed to remove 4.54\times 10^{-6} moles of O_3

2. \text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by =\frac{2}{1}\times 0.0003=0.0006 moles of sodium iodide.

Mass of sodium iodide= moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg    (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of O_3.

3 0
3 years ago
Balance the following redox reaction occurring in an acidic solution. The coefficient of Mn2+(aq) is given. Enter the coefficien
Vadim26 [7]

Answer:

_5_ AsO2−(aq) + 3 Mn2+(aq) + _2_ H2O(l) → _5_ As(s) + _3_ MnO4−(aq) + _4_ H+(aq)

Explanation:

Step 1:

The unbalanced equation:

AsO2−(aq) + 3 Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

Step 2:

Balancing the equation.

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

The above equation can be balanced as follow:

There are 3 atoms of Mn on the left side of the equation and 1 atom on the right side. It can be balance by putting 3 in front of MnO4− as shown below:

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 12 atoms of O on the right side and a total of 3 atoms on the left side. It can be balance by putting 5 in front of AsO2− and 2 in front of H2O as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 4 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 4 in front of H+ as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + 4H+(aq)

There are 5 atoms of As on the left side and 1 atom on the right side. It can be balance by putting 5 in front of As as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → 5As(s) + 3MnO4−(aq) + 4H+(aq)

Now the equation is balanced

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3 years ago
5 The apparatus shown below was set up. Give explanations for the following observations.
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Answer:

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Explanation:

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