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3 years ago

100 points! Formation of an Ion

2 answers:

8 0

$\huge\underline\pink {Answer}$

If every oxygen ion is combined with an aluminium ion has a charge of -2,the charge of each aluminum ion would be -3.

Uncharged Aluminum atom must need to lose it's electrons,in order to form the bond with oxygen which has vacant orbitals

ion

atom that has a positive or negative charge because it lost or gained one or more electrons

chemical bond

the attractive force that holds atoms or ions together

ionic bond

a chemical bond in which one atom loses an electron and the other atom gains electrons to form ions

chemical formula

a combination of chemical symbols and numbers to represent a substance

covalent bond

bond formed by the sharing of electrons between atoms

kvasek [131]3 years ago

4 0

What you have to realize is who is creating the charge and who is receiving it.

Oxygen and Aluminum are in their ionic form. Ions are produced by electron movement.

Aluminum gives away electrons (3 of them).

Oxygen takes on electrons (2 of them).

The 2 and the 3 in Al2O3 tells you how many electrons were given up and how they were taken on.

The 2 and the 3 in Al2O3 also tells you (indirectly) what the lowest common denominator is.

So once you understand the above 2 sentences you will understand the answer to your questions.

The lowest common multiple of 3 and 2 is six.

If the Aluminum gives up 3 electrons per atom, then to get to 6, you will have to have two aluminums.

If oxygen takes on 2 electrons then to get to six. There must be 3 oxygens present. Look at the formula

Al2 there are two aluminums

O3 there are three oxygens. That's why the formula is written the way it is.

==============

Aluminum is never seen alone because it is very eager to give up it's 3 electrons. It will combine with almost anything that is willing to receive those electrons.

Compare it to gold. Gold will combine with almost nothing ionically. Separating gold from a none metal won't happen normally, and refining to get rid of that problem just won't occur.

You might be interested in

Hess’s law

Delvig [45]

From the statement of Hess' law, the enthalpy of the reaction A---> C is +90 kJ

Hess' law of constant heat summation states that for a multistep reaction, the standard enthalpy of reaction is always constant and is independent of the pathway or intermediate routes taken.

From Hess' law, the enthalpy change for the reaction A ----> C is calculated as follows:

A---> C = A ---> B + B ---> C

ΔH of A---> C = 30 kJ + 60 kJ

ΔH = 90 kJ

Therefore, the enthalpy of the reaction A---> C is +90 kJ

The above reaction A---> C can be shown in the enthalpy diagram below:

A -------------------> C (ΔH = +90 kJ)

\ /

\ / (ΔH = +60 kJ)

(ΔH = +30 J) \ /

> B

Learn more about enthalpy and Hess law at: brainly.com/question/9328637

8 0

2 years ago

Calculate the number of moles present in 9.50g of co2

pshichka [43]

Answer:

0.4318 mol

Explanation:

Moles= mass/ mollar mass

Moles= 9.50/22

Moles= 0.4318 mol

3 0

3 years ago

PLEASE HELP!! (I will mark brainiest) (REAL ANSWERS ONLY PLEASE!)

Inessa [10]

The phosphate group of one nucleotide bonds covalently with the sugar molecule of the next nucleotide, and so on, forming a long polymer of nucleotide monomers. The sugar–phosphate groups line up in a “backbone” for each single strand of DNA, and the nucleotide bases stick out from this backbone. The carbon atoms of the five-carbon sugar are numbered clockwise from the oxygen as 1′, 2′, 3′, 4′, and 5′ (1′ is read as “one prime”). The phosphate group is attached to the 5′ carbon of one nucleotide and the 3′ carbon of the next nucleotide. In its natural state, each DNA molecule is actually composed of two single strands held together along their length with hydrogen bonds between the bases.

8 0

3 years ago

Read 2 more answers

what would be the ph of an aqueous solution of sulphuric acid which is 5×10^_5 mol l^_1 in concentration

guajiro [1.7K]

Answer:

the ph of an aqueous solution of sulphuric acid which is 5*10^5 mol in concentration is basic in nature

6 0

2 years ago

Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)

saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if $\rm O_2$ is removed from the mixture. The reason is that collisions between $\rm SO_2$ molecules and $\rm O_2\!$ molecules would become less frequent.

The reaction would not be at equilibrium for a while after $\rm O_2$ was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of $\rm SO_3\, (g)$ to the system, the backward reaction would have broken down the same amount of $\rm SO_3\, (g)\!$ . So is the case for $\rm SO_2\, (g)$ and $\rm O_2\, (g)$ .

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that $\rm SO_2\, (g)$ and $\rm O_2\, (g)$ molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield $\rm SO_3\, (g)$ , only a fraction of the reactions will be fruitful.

Assume that $\rm O_2\, (g)$ molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer $\!\rm O_2\, (g)$ molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between $\rm SO_2\, (g)$ and $\rm O_2\, (g)\!$ molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after $\rm O_2\, (g)$ was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more $\rm SO_3\, (g)$ gets converted to $\rm SO_2\, (g)$ and $\rm O_2\, (g)$ , the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0

2 years ago