The percent ionization of a 0.350 M HC,H,O2 solution. Ka(HC2H302)

What is a particle diagram to illustrate the composition of air

navik [9.2K]

A diagram of the composition of air would be the air's chemical formula and it's structure.

4 0

3 years ago

A piece of an unknown substance weighing 124.0 grams is heated in boiling water to 100.0oc. when the substance is placed in a ca

MakcuM [25]

The heat cause 300g water temperature increase from 20 to 26 celcius. The heat transferred would be: 300g * (26 °C -20 °C) *4.2 joule/gram °C= 7560J

The unknown substance is added to the water, so its final temperature should be the same as the water. The calculation would be:

7560J= 124g * (100-26)* specific heat

specific heat= 7560J / 124g / 74 °C= 0.8238 J/gram °C

3 0

4 years ago

Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of

harkovskaia [24]

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g) (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

$n = \frac{m}{M}$

Where:

m: is the mass

M: is the molar mass

For hydrogen we have:

$n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles$

And for nitrogen:

$n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles$

We can see in reaction (1) that 3 moles of hydrogen react with 1 mol of nitrogen, so the number of hydrogen moles needed to react nitrogen is:

$n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles$

Since we have 3.47 moles of hydrogen and we need 7.50 moles to react with all the mass of nitrogen, the limiting reactant is hydrogen.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that 3 moles of hydrogen produces 2 moles of ammonia, so:

$n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles$

Hence, hydrogen would produce 2.31 moles of ammonia.

Therefore, hydrogen is the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

Find more about limiting reactants here:

brainly.com/question/2948214?referrer=searchResults

I hope it helps you!

6 0

3 years ago

Identify the most acidic hydrogens in each of the following molecules. Give the structure of the enolate ion arising from deprot

san4es73 [151]

Answer:

See explanation below (Brainlist please)

Explanation:

First of all, we need to understand what is an acidic hydrogen.

An acidic hydrogen, is the atom of hydrogen which is more propense to undergo an acid base reaction, and form a stable ion or molecule in the process.

In other words, is the hydrogen that is more vulnerable to get substracted in an acid base reaction to form another compound.

Knowing this information, gives us an idea of how a molecule can be formed and which kind of compound is formed.

Now, in this question, we have 5 molecules. Each of them is either a ketone or aldehyde, so this mean that we have the carbonile group (C = O), which means that is easier to identify the acidic hydrogen. This is because the Carbonile group is an attractor group, so, it will attract the charges by inductive effect (in some cases by resonance), and the molecule is more stable.

This can be shown by drawing the enolate ion that is formed once the molecule undergo the acid base reaction. As it's an enolate form that we are looking for, then it means that the ketone or aldehyde is undergoing an electrofilic attack with a base. This base will substract the most acidic hydrogen to form a better and stable enolate. The acidic hydrogen and the enolate form can be seen in the attached picture.

a) In the case of acetaldehyde, the most acidic will be the hydrogen of carbon 2, because the hydrogen from the carbonile, once it's substracted, the charge of the carbon cannot be stabilized by resonance. Carbon 2 hydrogens, can do this job easily.

b) Propanal happens something similar to acetaldehyde, the terminal hydrogen cannot be substracted, and carbon 3, once the hydrogen is gone, the negative charge cannot be stabilized by resonance, so hydrogens of carbon 2 can do this.

c) in the case of acetone, is easier to look because we only have the C = O between two methyl group, so you can use either carbon 1 or 3 to do the job.

d) 4 heptanone the most acidic hydrogen would be carbon 3 or 5, because they are closer to the C=O and the ion can be stabilized by resonance.

e) Finally in ciclopentanone, the most acidic hydrogen would be carbon 2 or 5.

See picture for a better understanding.

Hope it helps.

7 0

3 years ago

10 + 32 -6=any one?

Kryger [21]

Answer:

Your answer will be 36

Explanation:

10+32=42

42-6=36

3 0

3 years ago

The percent ionization of a 0.350 M HC,H,O2 solution. Ka(HC2H302)​

The percent ionization of a 0.350 M HC,H,O2 solution. Ka(HC2H302)