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user100 [1]
3 years ago
13

The amide formed in the reaction of 3-ethylbenzoic acid and dimethylamine is

Chemistry
1 answer:
crimeas [40]3 years ago
4 0

Answer: -

3-Ethyl-N,N-dimethylbenzamide.

Explanation: -

In the first step the nitrogen lone pair attacks the carbonyl carbon.

Then a OH becomes water molecule by protonation and leaves.

Final step involves proton loss to form amide.

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If i mix ammonia and bleach what do i get.
Harlamova29_29 [7]

Answer:

Death

Explanation:

Just kidding but its a toxic gas that will lead to nausea and coughing.

5 0
3 years ago
What is the percent by mass of carbon in c10 h14 n2?
bonufazy [111]
Add all weighs together
calculate carbons weigh
(carb weigh/total)×100
3 0
3 years ago
Can you pls tell me the net ironic equation of H₂²⁺(aq) + O₂²⁻(aq) + Mg²⁺(aq) + SO₃²⁻(aq) → Mg²⁺(aq)+SO²⁻₄(aq) + H₂O(l)
almond37 [142]

Answer:

H₂²⁺(aq) + O₂²⁻(aq) + SO₃²⁻(aq) → SO²⁻₄(aq) + H₂O(l)

Explanation:

H₂²⁺(aq) + O₂²⁻(aq) + Mg²⁺(aq) + SO₃²⁻(aq) → Mg²⁺(aq) + SO²⁻₄(aq) + H₂O(l)

A careful observation of the equation above, shows that the equation is already balanced.

To obtain the net ionic equation, we simply cancel Mg²⁺ from both side of the equation as shown below:

H₂²⁺(aq) + O₂²⁻(aq) + SO₃²⁻(aq) → SO²⁻₄(aq) + H₂O(l)

5 0
3 years ago
According to the periodic table, which two elements are in the same row? A. copper and gold B. sodium and sulfur C. lithium and
Korolek [52]
I think the answer is B. Correct me if i'm wrong :)
4 0
3 years ago
A gas sample has an initial volume of 63.2 mL, an initial temperature of 42.0 ?C, and an initial pressure of 751 mmHg. The volum
exis [7]

Answer:

Lets Write Down the Given Initial Conditions.

P_1 = 751mmHg           P_2 = --

V_1 = 63.2ml                V_2 = 47.6ml

T_1 = 42C                     T_2 = 77

In Order to Solve for the Unknown: P_2

we must use the Ideal Gas Law to Solve for the Second Unknown pressure:

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

Then Rearrange this equation in a form where P2 can be solved from:

P_2 = \frac{P_1V_1T_2}{T_1V_2}

Then Insert the Values from above to solve:

P_2 = \frac{(751 mmHg)(63.2ml)(77C)}{(47.6ml)(42C)}

The Answer is : 1830 mmHg considering sig figs

4 0
3 years ago
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