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amid [387]
3 years ago
5

A chemist combines ethane and chlorine with the goal of producing chloroethane which other product could be reasonably expected

to form
Chemistry
2 answers:
lesya692 [45]3 years ago
5 0
<span>So when the chemist combines Ethane (CH3CH3) and Chlorine (Cl2) with the intention of producing Chloroethane (CH3CH2Cl), the other product that's formed in this reaction is 1,2-dichloroethane (ClCH2CH2Cl) also called as Ethylene dichloride with molecular weight of 98.954 g/mol. This is a colorless oily flammable substance that weighs heaver when vaporized.</span>
morpeh [17]3 years ago
3 0

Answer: other products are 1,2-dichloroethane, 1,1-dichloroethane and 1,1,1-trichloro ethane formed.

Explanation: since in chlorination of Alkane is an example for free-radical Substitution and due to more reactivity of free-radical excess chlorination (poly chlorination) Occurs.

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what is the concentration of an NaOH solution that requires 50 mL of a 1.25 M H2SO4 solution to neutralize 78.0 ml of NaOH​
Dimas [21]

Answer:

  • The answer is the concentration of an NaOH = 1.6 M

Explanation:

The most common way to solve this kind of problem is to use the formula  

  • C₁ * V₁ = C₂ * V₂

In your problem,

For NaOH

C₁ =??     v₁= 78.0 mL = 0.078 L

For H₂SO₄

C₁ =1.25 M     v₁= 50.0 mL = 0.05 L

but you must note that for the reaction of NaOH with H₂SO₄

2 mol of NaOH raect with 1 mol H₂SO₄

So, by applying in above formula

  • C₁ * V₁ = 2 * C₂ * V₂
  • (C₁ * 0.078 L) = (2*  1.25 M * 0.05 L)
  • C₁ = (2*  1.25 M * 0.05 L) / (0.078 L) = 1.6 M  

<u>So, the answer is the concentration of an NaOH = 1.6 M</u>

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