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3 years ago

Draw the ketone produced from the oxidation of 2-pentanol.

1 answer:

8 0

When 2-pentanol gets oxidized you end up with 2-pentanone. In this case, since we are oxidizing a secondary alcohol you end up with a ketone. They cannot be oxidized further than that. If it was a primary alcohol you would end up with an aldehyde.

You can find the mechanism as well the molecule formula of both 2pentanol and 2pentanone attached.

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9966 [12]

Answer: c
that is the answer

5 0

2 years ago

What happens to the reaction c2h6 + 137 kj c2h4+h2 when the temperature of the reaction is increased?

Softa [21]

Answer is: at higher temperatures reaction will go to the right (forward), more products (C₂H₄ and H₂) will be produce, because this is endothermic reaction (ΔH is positive, energy is consumed) and according Le Chatelier's principle heat is included as a reactant. .

4 0

3 years ago

Calculate the number of formula units in 2.50 mol NaNO

marta [7]

The number of formula units in 2.50 mol of the compound is 15.1 * 10^23.

The question is unclear whether NaNO2 or NaNO3 is implied. However,in either case, the solution applies equally.

6.02 * 10^23 formula units of the compound are contained in 1 mole

x formula units are contained in 2.5 moles of the compound

x = 6.02 * 10^23 formula units * 2.5 moles/ 1 mole

x = 15.1 * 10^23 formula units of the compound.

Learn more; brainly.com/question/9743981

3 0

3 years ago

Which of the following possess the greatest concentration of hydroxide ions?

jek_recluse [69]

Answer : The correct option is (d) a solution of 0.10 M NaOH

Explanation :

(a) a solution of pH 3.0

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$11=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-11}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-11}M$

(b) a solution of 0.10 M $NH_3$

As we know that 1 mole of $NH_3$ is a weak base. So, in a solution it will not dissociates completely.

So, the $OH^-$ concentration will be less than 0.10 M

(c) a solution with a pOH of 12.

We have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$12=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-12}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-12}M$

(d) a solution of 0.10 M NaOH

As we know that $NaOH$ is a strong base. So, it dissociates to give $Na^+$ ion and $OH^-$ ion.

So, 0.10 M of $NaOH$ in a solution dissociates to give 0.10 M of $Na^+$ ion and 0.10 M of $OH^-$ ion.

Thus, the $OH^-$ concentration is, 0.10 M

(e) a $1\times 10^{-4}M$ solution of $HNO_2$

As we know that 1 mole of $HNO_2$ in a solution dissociates to give 1 mole of $H^+$ ion and 1 mole of $NO_2^-$ ion.

So, $1\times 10^{-4}M$ of $HNO_2$ in a solution dissociates to give $1\times 10^{-4}M$ of $H^+$ ion and $1\times 10^{-4}M$ of $NO_2^-$ ion.

The concentration of $H^+$ ion is $1\times 10^{-4}M$

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (1.0\times 10^{-4})$

$pH=4$

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$10=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-10}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-10}M$

From this we conclude that, a solution of 0.10 M NaOH possess the greatest concentration of hydroxide ions.

Hence, the correct option is (d)

3 0

3 years ago

How is Hydrogen in heavy water different from hydrogen in normal water

soldier1979 [14.2K]

Answer: An oxygen atom in heavy water has an extra neutron. A hydrogen atom in heavy water has an extra proton.

Explanation:

7 0

3 years ago