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Anastasy [175]
4 years ago
5

What is the name of this organic molecule

Chemistry
2 answers:
Alex_Xolod [135]4 years ago
8 0

Answer:

C2H4 Ethylene

Explanation:

can u mark me brainliest pls?

Have a great day! ^_^

never [62]4 years ago
4 0

Answer:

C2H4 Ethylene

Explanation:

Hope it helps! :)

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6 0
3 years ago
Compound X has the formula C8H14. X reacts with one molar equivalent of hydrogen in the presence of a palladium catalyst to form
REY [17]

Answer:

1-Ethyl-3-methylidenecyclopentane  

Step-by-step explanation:

Formula = C₈H₁₄. An alkane has formula C₈H₁₈. ∴ X contains 2 double bonds, 2 rings, or 1 ring and 1 double bond.

X absorbs only 1 mol of hydrogen. ∴ X contains 1 ring and 1 double bond.

Hydrogenation gives 1-ethyl-3-methylcyclopentane.

Ozonolysis gives formaldehyde, so X must contain a =CH₂ group.  

Hydrogenation of X converted the =CH₂ to -CH₃.

X is 1-ethyl-3-methylidenecyclopentane.

You can see the reactions in the image below.

3 0
3 years ago
I need help with this packet
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Q1. An inorganic compound is a compound where the main constituent or substance is not that of Carbon but predominantly other elements, such as I, N etc. An organic compound is one where the main substituent or main element, the element found in much greater amounts would be Carbon.

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B. Oxygen atom.
5 0
3 years ago
calculate the volume of carbon dioxide of room temperature and pressure obtained from 30 grams of glucose​
ivanzaharov [21]

Answer:

A

Explanation:

4 0
3 years ago
An unknown compound, X is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 an
Westkost [7]

Answer:

7.3

Explanation:

By Henderson Hasselbalch equation we can calculate the pH or the pOH of a solution by its pKa. Remember that pH = -log[H^{+}], and pKa = -logKa. Ka is the equilibrium constant of the acid.

Henderson Hasselbalch equation :

pH = pKa - log \frac{[HA]}{[A^{-}]}

Where [HA] is the concentration of the acid, and [A^{-}] is the concentration of the anion which forms the acid.

So, acid X, has two ionic forms, the carboxyl group and the other one. First, we have 0.1 mol/L of the acid, in 100 mL, so the number of moles of X

n1 = (0.1 mol/L)x(0.1 L) = 0.01 mol

When it dissociates, it forms 0.005 mol of the carboxyl group and 0.005 mol of the other group. Assuming same  stoichiometry.

Adding NaOH, with 0.1 mol/L and 75 mL, the number of moles of OH^- will be

n2 = (0.1 mol/L)x(0.075 L) = 0.0075 mol

So, the 0.0075 mol of OH^- reacts with 0.005 mol of carboxyl, remaining 0.0025 mol of OH^-, which will react with the 0.005 mol of the other group. So, it will remain 0.0025 mol of the other group.

The final volume of the solution will be 175 mL, but both concentrations (the acid form and ionic form) have the same volume, so we can use the number of mol in the equation.

Note that, the number of moles of the acid form is still 0.01 mol because it doesn't react!

So,

6.72 = pKa - log \frac{0.01}{0.0025}

6.72 = pKa - log 4

pKa - log4 = 6.72

pKa = 6.72 + log4

pKa = 6.72 + 0.6

pKa = 7.3

8 0
4 years ago
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