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Mandarinka [93]
3 years ago
8

This molecule can be synthesized from an alkyne anion and an alkyl bromide. However, there are two ways in which this molecule c

an be formed. One way uses a higher molecular weight alkyne anion (Part 1) and the other uses a lower molecular weight anion (Part 2). Draw the two versions in the boxes below. Omit spectator ions.

Chemistry
1 answer:
kozerog [31]3 years ago
4 0

Answer:

See figures

Explanation:

In this case, we will have different options around the<u> triple bond</u>. On part 1 we have to use the <u>bigger anion</u>, so we have to use the anion in which the <u>benzene ring is included </u>therefore the <u>ethyl</u> would be the <u>halide</u> (see figure 1).

On part 2 we have to use the <u>smaller anion</u>, so the <u>ethyl group</u> would be the anion and the other part of the molecule would be the <u>alkyl halide</u> (see figure 2).

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Written in this form, carbon-13, what does the 13 represent and what is the atomic number?
tigry1 [53]
Explanation:
Carbon has 3 isotopes. Carbon-12, Carbon-13, and Carbon-14
All of them have an atomic number of 6, which means all of them have 6 protons (if they had different numbers of protons, they'd be different elements).
Since they all have 6 protons, in their elemental forms, they all have 6 electrons, too.
Carbon-12:
Atomic number: 6
Mass number: 12
protons: 6
electrons: 6
neutrons: 6
Carbon-13:
Atomic number: 6
Mass number: 13
protons: 6
electrons: 6
neutrons: 7
Carbon-14:
Atomic number: 6
Mass number: 14
protons: 6
electrons: 6
neutrons: 8
8 0
3 years ago
On.<br> pe. Where is the Hubble telescope located?
Lubov Fominskaja [6]

Answer:

space

Explanation:

6 0
3 years ago
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Give the oxidation state of the metal species in each complex. ru(cn)(co)4 -
kogti [31]
The given complex ion is as follow,

                                              [Ru (CN) (CO)₄]⁻

Where;
            [ ]  =  Coordination Sphere

            Ru  =  Central Metal Atom  =  <span>Ruthenium

            CN  =  Cyanide Ligand

            CO  =  Carbonyl Ligand

The charge on Ru is calculated as follow,

                               Ru + (CN) + (CO)</span>₄  =  -1
Where;
            -1  =  overall charge on sphere

             0  =  Charge on neutral CO

            -1  =  Charge on CN

So, Putting values,


                               Ru + (-1) + (0)₄  =  -1

                               Ru - 1 + 0  =  -1

                               Ru - 1  =  -1

                               Ru  =  -1 + 1

                               Ru  =  0
Result:
          <span>Oxidation state of the metal species in each complex [Ru(CN)(CO)</span>₄]⁻ is zero.
3 0
4 years ago
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Consider the molecule BF3. (a) What is the electron config- uration of an isolated B atom? (b) What is the electron con- figurat
anzhelika [568]

For the given molecule, we are asked to give-

  1. The electron configuration of an isolated B atom
  2. The electron configuration of an isolated F atom
  3. Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride
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as atomic number of B is 5

electronic configuration will be [He] 2s² 2p¹

  • The electron configuration of an isolated F atom:

as atomic number of F is 9

electronic configuration will be  [He] 2s² 2p5

  • Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride will be sp2.

as the one s and two of p orbital from the valance shell will hybridised to make 3 hybrid orbital of B resulting in 3 B-F bonds.

  • valence orbitals, if any, remain unhybridized on the B atom will be 1

To know more about hybrisisation:

brainly.com/question/23038117

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What is the name of the molecule shown below?
xxTIMURxx [149]

Answer:

B. 1-heptene

Explanation:

The 1-HEPTENE molecule contains a total of 20 bond(s) There are 6 non-H bond(s), 1 multiple bond(s), 4 rotatable bond(s) and 1 double bond(s).

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