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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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A drop of vinegar ( a weak acid ) is placed on a sample of each of the materials below. Which will show the most active reaction
professor190 [17]

Answer: Option (B) is the correct answer.

Explanation:

Marble is also known as calcium carbonate and its chemical formula is CaCO_{3}.

When it combines with vinegar then it results in the formation of calcium acetate which is a water soluble compound along with the release of carbon dioxide which occurs in the form of bubbles and water.

The reaction is as follows.

    CaCO_{3} + CH_{3}COOH \rightarrow Ca(CH_{3}COOH)_{2} + H_{2}O + CO_{2}

Whereas vinegar being a weak acid will not react with glass, copper and steel.

8 0
3 years ago
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A substance with a pH of 4.0?
IrinaVladis [17]
Tomato juice or acid rain
7 0
3 years ago
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Are alchemy and chemistry the same
Anit [1.1K]

Are alchemy and chemistry the same: No

<u>Explanation:</u>

Alchemy and chemistry differ from each other. Their main difference lies in the concepts that they are based on. Alchemy has the base of viewing the reality of nature supernaturally. Chemistry has a base of viewing the reality in a natural way.

Protoscience is the name given for Alchemy. This is because that alchemy usually focuses on the traditional things like procedures and  terminologies. Chemistry always focuses on the matter and their transformations. The matter and its transformations are dealt with chemistry.

5 0
3 years ago
131i has a half-life of 8.04 days. assuming you start with a 1.53 mg sample of 131i, how many mg will remain after 13.0 days ___
inn [45]
For this problem we can use half-life formula and radioactive decay formula.

Half-life formula,
t1/2 = ln 2 / λ

where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days

Hence,         
8.04 days    = ln 2 / λ                         
λ   = ln 2 / 8.04 days

Radioactive decay law,
Nt = No e∧(-λt)

where, Nt is amount of compound at t time, No is amount of compound at  t = 0 time, t is time taken to decay and λ is radioactive decay constant.

Nt = ?
No = 1.53 mg
λ   = ln 2 / 8.04 days = 0.693 / 8.04 days
t    = 13.0 days 

By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg

Hence, mass of remaining sample after 13.0 days = 0.499 mg

The answer is "e"

8 0
3 years ago
Calculate the fractional saturation for hemoglobin when the partial pressure of oxygen is 40 mm Hg. Assume hemoglobin is 50%% sa
kumpel [21]

Answer:

The fractional saturation for hemoglobin is 0.86

Explanation:

The fractional saturation for hemoglobin can be calculated using the formula

Y_{O_{2} } = \frac{(P_{O_{2} })^{h}  } {(P_{50})^{h}  + (P_{O_{2} })^{h}   }

Where Y_{O_{2} } \\ is the fractional oxygen saturation

{P_{O_{2} } is the partial pressure of oxygen

P_{50} is the partial pressure when 50% hemoglobin is saturated with oxygen

and h is the Hill coefficient

From the question,

{P_{O_{2} } = 40 mm Hg

P_{50} = 22 mm Hg

h = 3

Putting these values into the equation, we get

Y_{O_{2} } = \frac{(P_{O_{2} })^{h}  } {(P_{50})^{h}  + (P_{O_{2} })^{h}   }

Y_{O_{2} } = \frac{40^{3} }{22^{3} + 40^{3}  }

Y_{O_{2} } = \frac{64000 }{10648 + 64000  }

Y_{O_{2} } = \frac{64000 }{74648 }

Y_{O_{2} } = 0.86

Hence, the fractional saturation for hemoglobin is 0.86.

4 0
3 years ago
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