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Lemur [1.5K]
3 years ago
12

Structural formula for alkene with double bond st carbon 2 that shows no trans -cis isomerism C6H12​

Chemistry
1 answer:
svetlana [45]3 years ago
5 0

Answer:

Explanation:

Read up on this:

https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_The_Basics_of_GOB_Chemistry_(Ball_et_al.)/13%3A_Unsaturated_and_Aromatic_Hydrocarbons/13.02%3A_Cis-Trans_Isomers_(Geometric_Isomers)

I think the answer is going to structure of 2-methyl-2-pentene.

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In order for the molecule to change phase from liquid to gas and evaporates, it needs to overcome the force from other molecules around it. as the force bigger evaporation gets harder. so e has the highest force and higher boiling point.
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99% of matter that makes up living organisms consists of how many elements?
DiKsa [7]

Answer:

h2o or air

Explanation:

6 0
3 years ago
It has been proven that North America is moving west about 2 cm per year. How many kilometers wouldn’t move in 5000 years
Viefleur [7K]
0.1 km

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7 0
3 years ago
Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid
Phoenix [80]

Answer:

pH = 8.0

Explanation:

First, we have to calculate the moles of NaOH.

35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol

Let's consider the balanced equation.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

The concentration of C₂H₃O₃Na is:

\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M

C₂H₃O₃Na dissociates according to the following equation:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:

pKa + pKb = 14

pKb = 14 -3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can calculate [OH⁻] using the following expression:

[OH⁻] = √(Kb.Cb)               <em>where Cb is the initial concentration of the base</em>

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

Now, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0

7 0
2 years ago
How many moles of NH3 can be produced from 30.0 mol of H2 and excess N2?
VMariaS [17]
20 mol of NH, can be produce from 30 mol o H2
6 0
3 years ago
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