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2 years ago

Compare Ion and Radical Atom and molecule Organic and inorganic compounds

Chemistry

2 answers:

horrorfan [7]2 years ago

6 0

Answer:

'An ion has a non-zero electric charge. A radical has an atom with unfilled electron shells and so is very reactive, but is electrically neutral.'

'Atoms are single neutral particles. Molecules are neutral particles made of two or more atoms bonded together.'

'The primary difference that lies between these organic compounds and inorganic compounds is that organic compounds always have a carbon atom while most of the inorganic compounds do not contain the carbon atom in them.'

goldenfox [79]2 years ago

5 0

Organic and inorganic compounds:

Similarities:::---

1) In both organic and inorganic compound, the elements joining in making compound complete their octet.

2) Organic compound shows isomerism (R/S , E/Z , cis/trans, enantiomers, diastereomers, etc), just as inorganic compound do (Δ/Λ, enantiomers, R/S enantiomers, cis/trans, fac/mer isomers, etc).

3) Organic compound and inorganic compounds can be very active in chemical reactions, e.g. organolithium reagents could be flammable or even pyrophoric (generally stored under 10°C), while inorganic reagents like in this paper are quite rapid...

Differences:::---

1) Organic compound are very long in size so they can make polymer but inorganic compound are not very long but its structure might me complex.

2) The boiling,melting point of organic compound is lass than inorganic compound.

3) Organic compound always contains carbon but inorganic compounds might not.

4) Organic compound

_____________________________

Ion and Radical Atoms:

In some sense they are a bit similar. The main difference is that a neutral radical has no charge imbalance between the protons and electrons, but the cation or anion does.

Ions are written with an explicit charge because of that charge imbalance, but radicals may or may not have an imbalance of charge. Just because a molecule is a radical doesn't mean it's neutral.

For example, if you shoot 2-pentanone with an electron beam for Electron "Impact" Mass Spectroscopy, where you essentially study molecules whose structures break into smaller pieces as a result of interacting with stray electrons to identify the molecules, you get a cationic radical (middle, or far right of the following diagram).

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A 42.0g sample of compound containing only C and H was analyzed. The results showed that the sample contained 36.0g of C and 6.0

user100 [1]

Answer:

What is the empirical formula of the compound?

Explanation:

When the relative masses of elements in a hydrocarbon are given, it is possible to use this information to obtain the empirical formula by dividing the given masses of each element by the relative atomic masses of the element. The lowest ratio is now used to divide through to obtain the empirical formula of the compound.

The empirical formula only shows that ratio of atoms of each element present in the compound. From the information provided, the empirical formula of the compound is CH2. Hence the answer.

5 0

3 years ago

Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves

hjlf

This is an incomplete question, here is a complete question.

Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves a two-step process. In the first step, manganese (II) carbonate and oxygen react to form manganese (IV) oxide and carbon dioxide:

$2MnCO_3(s)+O_2(g)\rightarrow 2MnO_2(s)+2CO_2(g)$

In the second step, manganese (IV) oxide and aluminum react to form manganese and aluminum oxide:

$3MnO_2(s)+4Al(s)\rightarrow 3Mn(s)+2Al_2O_3(s)$

Write the net chemical equation for the production of manganese from manganese (II) carbonate, oxygen and aluminum. Be sure your equation is balanced.

Answer : The net chemical equation for the production of manganese is:

$6MnCO_3(s)+3O_2(g)+8Al(s)\rightarrow 6CO_2(g)+6Mn(s)+4Al_2O_3(s)$

Explanation :

The given two chemical reactions are:

(1) $2MnCO_3(s)+O_2(g)\rightarrow 2MnO_2(s)+2CO_2(g)$

(2) $3MnO_2(s)+4Al(s)\rightarrow 3Mn(s)+2Al_2O_3(s)$

First we are multiplying reaction 1 by 3, and reaction 2 by 2, we get:

(1)

(2) $6MnO_2(s)+8Al(s)\rightarrow 6Mn(s)+4Al_2O_3(s)$

Now we are adding both the reactions, we get the overall chemical reaction.

$6MnCO_3(s)+3O_2(g)+6MnO_2(s)+8Al(s)\rightarrow 6MnO_2(s)+6CO_2(g)+6Mn(s)+4Al_2O_3(s)$

The $MnO_2$ is common on both side, by cancelling it, we get:

The net chemical equation for the production of manganese is:

$6MnCO_3(s)+3O_2(g)+8Al(s)\rightarrow 6CO_2(g)+6Mn(s)+4Al_2O_3(s)$

5 0

3 years ago

One-step constrained-volume synthesis of silver decorated polymer colloids with antimicrobial and sensing properties,

Anarel [89]

It would be the thurd

5 0

3 years ago

When a solute is added to a solution why does it remain homogeneous?

Olenka [21]

<h3>Answer:</h3>

When a solute is added to a solution, it remains homogeneous because the solute is soluble in given solvent.

<h3>Explanation:</h3>

Homogeneous mixtures, also called true solutions are those mixtures in which the components proportions are same throughout in any given sample. For example, the mixture of table salt (NaCl) and water. When the solution is unsaturated and further NaCl is added to it, it will dissolve the NaCl because the saturation point is still not reached. Remember, as "<em>Like Dissolves Like</em>" NaCl being polar in nature will interact with water molecules and will dissociate into Na⁺ and Cl⁻ ions surrounded by δ- O and δ+ H atoms of water molecules.

<h3>Conclusion:</h3>

In order to form a Homogeneous mixture the solution must be unsaturated, solvent must have affinity for incoming solute particles and the size of solute should be equal to 1 Â (Angstrom).

8 0

3 years ago

Read 2 more answers

What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m

djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL (1cm³ = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³ mol / 0.4327 L = 0.0075 M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x 432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

8 0

3 years ago