Ask question

Ask question

All categories

4 years ago

8

In the Gabriel synthesis of primary amines, N-potassiophthalimide is used as a source of the nitrogen atom. Complete the synthes

is reactions by drawing the missing structures. Ignore inorganic counterions.

1 answer:

Amiraneli [1.4K]4 years ago

4 0

Reaction is attached:

You might be interested in

Consider the following equilibrium reaction having gaseous reactants and products. 4HCl + O2 ⇌ 2H2O + Cl2 Which of the following

gulaghasi [49]

<h3><u>Answer</u>;</h3>

B.The rate of forward reaction increases.

<h3><u>Explanation;</u></h3>

Le Chatelier's principle states that changing a factor such as concentration, temperature, or pressure of a reaction at equilibrium will cause the reaction to shift in the direction that counteracts the effect of that change.
<em><u>Therefore, when reactants are added to a reaction at equilibrium shift when more reactants are added then the reaction shifts to the right to make more products.</u></em>

3 0

4 years ago

HCHO2 + NaOH = NaCHO2 + H2O complete ionic

Llana [10]

Answer:

HCHO₂(aq) + Na⁺(aq) + OH⁻(aq) ⟶ Na⁺(aq) + CHO₂⁻(aq) + H₂O(ℓ)

Explanation:

An ionic equation uses the symbols (aq) [aqueous] to indicate molecules and ions that are soluble in water, (s) [solid] to indicate insoluble solids, and (ℓ) to indicate substances (usually water) in the liquid state.

In this reaction, aqueous sodium hydroxide reacts with aqueous formic acid to form sodium formate and water.

1. Molecular equation

HCHO₂(aq) + NaOH(aq) ⟶ NaCHO₂(aq) + H₂O(ℓ)

2. Ionic equation

Sodium hydroxide and sodium formate are soluble ionic compounds, so we write them as hydrated ions.

HCHO₂(aq) + Na⁺(aq) + OH⁻(aq) ⟶ Na⁺(aq) + CHO₂⁻(aq) + H₂O(ℓ)

4 0

3 years ago

A solution of nitrous acid and potassium nitrite acts as a buffer due to reactions that occur within the solution when a strong

Ghella [55]

Answer:

a. NO₂⁻ + H⁺ → HNO₂

b. HNO₂ + OH⁻ → NO₂⁻ + H₂O

Explanation:

A buffer is defined as an aqueous mixture of a weak acid and its conjugate base or vice versa.

The buffer of the problem is HNO₂/NO₂⁻ <em>where nitrous acid is the weak acid and NO₂⁻ is its conjugate base.</em>

a. When a acid is added to a buffer as the buffer of the problem, the conjugate base will react with the acid, to produce the weak acid, thus:

NO₂⁻ + HCl → HNO₂ + Cl⁻

Ionic equation is:

NO₂⁻ + H⁺ + Cl⁻ → HNO₂ + Cl⁻

In the net ionic equation, you avoid the ions that don't react, that is:

<h3>NO₂⁻ + H⁺ → HNO₂</h3>

b. In the same way, the weak acid will react with the strong acid producing water and the conjugate base, thus:

HNO₂ + NaOH → NO₂⁻ + H₂O + Na⁺

The ionic equation is:

HNO₂ + Na⁺ + OH⁻ → NO₂⁻ + H₂O + Na⁺

And the net ionic equation is:

<h3>HNO₂ + OH⁻ → NO₂⁻ + H₂O</h3>

5 0

3 years ago

The acid-base reaction between HCl and NaOH produces a solution with a pH of 7 at the equivalence point (NaCl H2O). Why does an

Bess [88]

Answer:

Answer is in the explanation.

Explanation:

Before the equivalence point, the pH of the solution of HCl that is titrated with NaOH has a pH <<< 7. When you are adding more NaOH nearing, thus, to the equivalence point the change in pH occurs quickly, and, with 1 drop of excess of NaOH after equivalence, the pH of the solution change to a pH >>> 7

That means the volume added at pH 5 or pH 9 is, almost, the same doing the indicator work just as well as an indicator with change color at pH 7

5 0

3 years ago

What is the volume occupied by 3.00 mol of hydrogen gas at STP? *

xxTIMURxx [149]

Answer: V=67.2 L

Explanation:

For this problem we will need to use the Ideal Gas Law.

Ideal Gas Law: PV=nRT

P=1.00 atm (STP)

V=?

n=3.00 mol

R=0.08206Latm/Kmol

T=273.15 K (STP)

To find V, we would manipulate the equation to V=nRT/P

$V=\frac{(3.00 mol)(0.08206Latm/Kmol)(273.15K)}{1.00 atm}$

$V=67.244 L$

With significan figures, our answer is V=67.2 L.

6 0

3 years ago