1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
AfilCa [17]
1 year ago
9

When 20 ml of 0.1 M HCl is mixed with 20

Chemistry
1 answer:
Bas_tet [7]1 year ago
4 0

Answer:

it is b i took the test

Explanation:

You might be interested in
Use the drop-down menus to select the correct name for each of the organic compounds.
Molodets [167]

Answer:

Here's what I get  

Explanation:

CH₃CH₂CH₂CH₂CH₂CH₃ —  hexane

CH₂=CHCH₂CH₂CH₂CH₃ —  hex-1-ene is the preferred IUPAC name (PIN). 1-Hexene is accepted

CH₃C≡CCH₃ —  but-2-yne (PIN); 2-butyne is accepted

CH₃CH(CH₃)CH₂CH₂CH₃ — 2-methylpentane

CH₃CH₂CHCICH₂CH₃ — 3-chloropentane

5 0
3 years ago
Read 2 more answers
Helium, neon, and argon are examples of
iogann1982 [59]
C) Noble gases
The six noble gases are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn). Their atomic numbers are, respectively, 2, 10, 18, 36, 54, and 86.
4 0
3 years ago
Read 2 more answers
What is needed to release the energy in a bond?
IRISSAK [1]
Answer:

More Energy

Explanation:

Energy is required to break bonds
3 0
3 years ago
Read 2 more answers
Draw all four products obtained when 2-ethyl-3-methyl-1,3-cyclohexadiene is treated with HBr at room temperature and show the me
LenKa [72]

Answer:

See explanation below

Explanation:

In this case we have reaction of addition. In this case a diene reacting with an acid as HBr. This reaction is known as Hydrohalogenation, and, as we have a diene, this kind of reaction can be done as 1,4 addition. Which means that the reaction will be undergoing with an adition in the carbon 1, and carbon 4.

At room temperature we can expect that this reaction can be done in thermodynamic conditions, Now, as the problem states that is forming 4 products, we can expect products of a 1,2 addition too. This product can be formed if the reaction is taking place in the most stable carbocation, and then, by resonance, we can expect the 1,4 product too.

Now, the HBr can be attacked by the double bond of the first position, giving two possible products or by the double bond of the third position giving the other two products. These products are all possible, obviously the most stable will be the major of all of them, but the other three are perfectly possible. One product is formed without doing much, and the other by resonance. Same happens with the other double bond.

In the picture below, you have the mechanism for all the 4 products.

Hope this helps

5 0
3 years ago
Translate the words into formulas, predict the product, & balance the equations. Include states of matter.
Andreyy89
Hihihihihihihihihihihihihi
Hibiscus I
Hi yes



8 0
3 years ago
Other questions:
  • What is the relationship between the improved microscopes and the discoveries made about cells?
    14·1 answer
  • A compound of low solubility
    10·1 answer
  • Solve for t In(A/B) -kt where B-1.65 x 102 M A 1.00 x 10 M. k-4.80 x 104s* and t=
    13·1 answer
  • Please answer this question ASAP. This is on a homework assignment that is due tonight!! Thank you in advance! A voltaic cell em
    14·2 answers
  • Determine the pH of the following base solutions. (Assume that all solutions are at 25°C and the ion-product constant of water,
    6·1 answer
  • Identify and name the functional group present<br> CH4
    8·1 answer
  • Calculate the value of the equilibrium constant for the following system, where the concentrations of reactants and products are
    6·1 answer
  • Manganese (Mn) has many possible oxidation numbers (+2 and +4 are just two of these). Write the correct formulas and names for t
    14·1 answer
  • In which series is the order of classification correct
    7·1 answer
  • The decrease in the water table due to overuse of water.
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!