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gregori [183]
3 years ago
7

One method for preparing a nitrile is the dehydration of a primary amide.

Chemistry
1 answer:
Olenka [21]3 years ago
4 0
True because a dehydration reaction is a conversion that involves the loss of water from the reacting molecule or ion. Dehydration reactions are common processes, the reverse of a hydration reaction. Common dehydrating agents used in organic synthesis include sulfuric acid and alumina. Often dehydration reactions are effected with heating.
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The man who developed the concept that light is emitted and absorbed in bundles or packets was:
ludmilkaskok [199]
Planck suggested that light/energy was absorbed/released in certain amounts, called quanta.
3 0
3 years ago
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In the following reaction, how many liters of o2 will produce 43.62 liters of co2 at stp? c3h8 5 o2 3 co2 4 h2o
antoniya [11.8K]
The balanced chemical reaction:

C3H8 + 5O2 = 3CO2 + 4H2O

We are given the amount of the carbon dioxide to be produced. This will be the starting point of our calculations.

<span>43.62 L CO2 ( 1 mol CO2 / 22.4 L CO2 ) (5 mol O2 / 3 mol CO2 ) ( 22.4 L O2 / 1 mol O2) = 72.7 L O2</span>



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4 years ago
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Porque el agua reduce la temperatura alta?
Naya [18.7K]
Porque la temperatura de la agua es neutral
8 0
3 years ago
Given 700 ml of oxygen at 7 ºC and 80.0 cm Hg pressure, what volume does it take at 27 ºC and 50.0 cm Hg pressure?
katen-ka-za [31]

Answer:

I think that it might be 2.7

Explanation:

7 0
3 years ago
50 ml of 0.2 M acetic acid is shaken with 10 g activated charcoal. The concentration of acetic acid is reduced to 0.5 times the
Olegator [25]

Answer:

0.03 g_{acid}/g_{charcoal}

Explanation:

The amount adsorbed (solute) is the acetic acid, and the adsorbent is the activated charcoal. The mass of the adsorbent is 10 g.

So, we need to calculate the mass of the acetic acid as follows:

m = n*M = C*V*M

Where:

n: is the number of moles = C*V

M: is the molecular mass =  60.052 g/mol

C: is the final concentration of the acid = 0.5*0.2 mol/L = 0.10 mol/L

V: is the volume = 50 ml = 0.050 L

m_{acid} = C*V*M = 0.10 mol/L*0.050 L*60.052 g/mol = 0.30 g

Now, the amount of solute adsorbed per gram of the adsorbent is:

\frac{m_{acid}}{m_{charcoal}} = \frac{0.30 g}{10 g} = 0.03 g_{acid}/g_{charcoal}

Therefore, the amount of solute adsorbed per gram of the adsorbent is 0.03 g/g.

I hope it helps you!

3 0
3 years ago
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