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Cerrena [4.2K]
3 years ago
7

Write the formulas of the three singly chlorinated isomers formed when 2,2-dimethylbutane reacts with Cl2 in the presence of lig

ht.

Chemistry
1 answer:
Tems11 [23]3 years ago
3 0

The molecule with same molecular formula but different arrangement of atoms is said to be an isomer.

When 2,2-dimethylbutane reacts with chlorine in the presence of light gives three isomers that is CH_{3}C(CH_{3})_{2}CHClCH_{3} (3-chloro-2,2-dimethylbutane), ClCH_{2}C(CH_{3})_{2}CH_{2}CH_{3} (1-chloro-2,2-dimethylbutane) and ClCH_{2}CH_{2}C(CH_{3})_{2}CH_{3} (1-chloro-3,3-dimethylbutane).

In above case, the molecular formula of all isomers are same i.e.C_{6}H_{13}Cl but chlorine is arranged in different positions of carbon. Thus, results isomers.

The reaction is shown in the image.


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A titration of 25.0 mL of a solution of the weak base aniline, C6H5NH2, requires 25.67 mL of 0.175 M HCl to reach the equivalenc
Lorico [155]

Answer:

a. 0.180M of C₆H₅NH₂

b. 0.0887M C₆H₅NH₃⁺

c. pH = 2.83

Explanation:

a. Based in the chemical equation:

C₆H₅NH₂(aq) + HCl(aq) → C₆H₅NH₃⁺(aq) + Cl⁻(aq)

<em>1 mole of aniline reacts per mole of HCl</em>

Moles required to reach equivalence point are:

Moles HCl = 0.02567L ₓ (0.175mol / L) = 4.492x10⁻³ moles HCl = moles C₆H₅NH₂

As the original solution had a volume of 25.0mL = 0.0250L:

4.492x10⁻³ moles C₆H₅NH₂ / 0.0250L = 0.180M of C₆H₅NH₂

b. At equivalence point, moles of C₆H₅NH₃⁺ are equal to initial moles of C₆H₅NH₂, that is 4.492x10⁻³ moles

But now, volume is 25.0mL + 25.67mL = 50.67mL = 0.05067L. Thus, molar concentration of C₆H₅NH₃⁺ is:

[C₆H₅NH₃⁺] = 4.492x10⁻³ moles / 0.05067L = 0.0887M C₆H₅NH₃⁺

c. At equivalence point you have just 0.0887M C₆H₅NH₃⁺ in solution. C₆H₅NH₃⁺ has as equilibrium in water:

C₆H₅NH₃⁺(aq) + H₂O(l) → C₆H₅NH₂ + H₃O⁺

Where Ka = Kw / Kb = 1x10⁻¹⁴ / 4.0x10⁻¹⁰ =

<em>2.5x10⁻⁵ = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺]</em>

When the system reaches equilibrium, molar concentrations are:

[C₆H₅NH₃⁺] = 0.0887M - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in Ka formula:

2.5x10⁻⁵ = [X] [X] / [0.0887M - X]

2.2175x10⁻⁶ - 2.5x10⁻⁵X = X²

0 = X² + 2.5x10⁻⁵X - 2.2175x10⁻⁶

Solving for X:

X = -0.0015 → False solution. There is no negative concentrations.

X = 0.001477 → Right solution.

As [H₃O⁺] = X, [H₃O⁺] = 0.001477

Knowing pH = -log [H₃O⁺]

pH = -log 0.001477

<h3>pH = 2.83</h3>

7 0
3 years ago
The Dew Point... a. is the same as temperature when the relative humidity is 100%. b. can only be given in Kelvin degrees. c. is
telo118 [61]

Answer:

Option (A)

Explanation:

The dew point refers to the temperature at which the amount of water vapor present in the air is so high that the relative humidity becomes 100%, and with the increasing rate of cooling, the condensation process takes place and dew is formed.

So for dew point to occur, the air temperature must reach a condition where the air is fully saturated or the relative humidity is 100%.

Thus, the correct answer is option (A).

6 0
3 years ago
HURRY
Ugo [173]

Answer:

11.647g of water may be produced.

Explanation:

chemistry of the reaction:

2O2 + 2H2 ----> 2H2O + O2

64g + 4g yields 36g of water,

20g + 2g yields 11.647058824g of water

5 0
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Consider the reaction 2Al + 6HBr → 2AlBr3 + 3H2. If 8 moles of Al react with 8 moles of HBr, what is the limiting reactant?
TiliK225 [7]

Answer:- HBr is limiting reactant.

Solution:- The given balanced equation is:

2Al+6HBr\rightarrow 2AlBr_3+3H_2

From this equation, There is 2:6 mol or 1:3 mol ratio between Al and HBr. Since we have 8 moles of each, HBr is the limiting reactant as we need 3 moles of HBr for each mol of Al.

The calculations could be shown as:

8molAl(\frac{6molHBr}{2molAl})

= 24 mol HBr

From calculations, 24 moles of HBr are required to react completely with 8 moles of Al but only 8 moles of it are available. It clearly indicates, HBr is limiting reactant.

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3 years ago
True or false. adding more neutrons to a neutrally charged atom will not change the charge
san4es73 [151]
The statement is true.
6 0
3 years ago
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