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4 years ago

Write the empirical formula for at least four ionic compounds that could be formed from the following ions: Fe2+,Cr4+,Cl-,O2-

Chemistry

1 answer:

brilliants [131]4 years ago

4 0

Answer: Four ionic compounds formed will be $FeCl_2,FeO,CrCl_4\text{ and }CrO_2$

Explanation:

Ionic compound is formed by the complete transfer of electrons from 1 atom to another atom. The cation is formed by the loss of electrons by metals and anions are formed by gain of electrons by non metals.

We are given:

Two cations having formulas $Fe^{2+}\text{ and } Cr^{4+}$

Two anions having formulas $Cl^-\text{ and }O^{2-}$

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

The four ionic compounds are:

When iron ion and chloride ions combine, it results in the formation of $FeCl_2$ compound. This is named as iron (II) chloride.
When iron ion and oxide ions combine, it results in the formation of $FeO$ compound. This is named as iron (II) oxide.
When chromium ion and chloride ions combine, it results in the formation of $CrCl_4$ compound. This is named as chromium (IV) chloride.
When chromium ion and oxide ions combine, it results in the formation of $Cr_2O_4=CrO_2$ compound. This is named as chromium (IV) oxide.

Hence, the four ionic compounds formed will be $FeCl_2,FeO,CrCl_4\text{ and }CrO_2$

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Hiii pls help me to write out the ionic equation

emmasim [6.3K]

Answer:

STEP I

This is the balanced equation for the given reaction:-

$2KOH_{(aq)} + H_2SO_4{}_{(aq)} \rightarrow K_2SO_4{}_{(aq)} + 2H_2O_{(l)}$

STEP II

The compounds marked with (aq) are soluble ionic compounds. They must be

broken into their respective ions.

see, in the equation KOH, H2SO4, and K2SO4 are marked with (aq).

On breaking them into their respective ions :-

2KOH -> 2K+ + 2OH-
H2SO4 -> 2H+ + (SO4)2-
K2SO4 -> 2K+ + (SO4)2-

STEP III

Rewriting these in the form of equation

$\underline{\pmb{2K^+} }+ 2OH^- + 2H^+ + \pmb{\underline{{SO_4{}^{2-}}} \: \rightarrow \: \underline{\pmb{2K^+}}} + \underline{\pmb{SO_4{}^{2-}}} + 2H_2O$

STEP IV

Canceling spectator ions, the ions that appear the same on either side of the equation

(note: in the above step the ions in bold have gotten canceled.)

$\boxed{ \mathfrak{ \red{ 2OH^-{}_{(aq)} + 2H^+{(aq.)} \rightarrow H_2O{}_{(l)}}}}$

This is the net ionic equation.

____________________________

$\\$

$\mathfrak{\underline{\green{ Why\: KOH \:has\: been\: taken\: as\: aqueous ?}}}$

KOH has been taken as aqueous because the question informs us that we have a solution of KOH. by solution it means that KOH has been dissolved in water before use.

[Alkali metal hydroxides are the only halides soluble in water ]

4 0

3 years ago

Draw the resonance structures for the conjugate base of Phenol (C6H6O). In one sentence, explain why phenol (C6H6O) has a pKa of

notsponge [240]

Answer:

The four resonance structures of the phenoxide ion are shown in the image attached

The conjugate base of cyclohexanol has only one resonance contributor, while

the conjugate base of phenol has four resonance contributors.

Explanation:

In organic chemistry, it is known that structures are more stable if they possess more resonance contributors. The greater the number of contributing canonical structures, the more stable the organic specie. Since the phenoxide ion has four contributing canonical structures, it is quite much more stable than cyclohexanol having only one contributing structure to its conjugate base. Hence the PKa(acid dissociation constant) of phenol is lesser than that of cyclohexanol. The conjugate base of phenol is stabilized by resonance.

6 0

3 years ago

Molecule A is: A. Isocitrate B. Pyruvate C. Alpha-ketoglutarate D. Oxaloacete E. None of the above

hodyreva [135]

Answer:... I'm sorry.. I think you're missing something

Explanation:

3 0

3 years ago

Noble gas compounds like KrF, XeCl, and XeBr are used in excimer lasers. Draw an approximate molecular orbital diagram appropria

Aneli [31]

Answer:

Here's what I get.

Explanation:

The MO diagrams of KrBr, XeCl, and XeBr are shown below.

They are similar, except for the numbering of the valence shell orbitals.

Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.

However, the MO diagrams are approximately correct.

The ground state electron configuration of KrF is

$(1\sigma_{g})^{2}\, (1\sigma_{u}^{*})^{2} \, (2\sigma_{g})^{2} \, (2\sigma_{u}^{*})^{2} \, (3\sigma_{g})^{2} \, (1\pi_{u})^{4} \, (1\pi_{g}^{*})^{4} \, (3\sigma_{g}^{*})^{1}$

KrF⁺ will have one less electron than KrF.

You remove the antibonding electron from the highest energy orbital, so the bond order increases.

The KrF bond will be stronger.

6 0

4 years ago

Oh no... not again... Prof. Vitarelli spots Sybil running down the hall... yelling something... something about her tea cups...

Y_Kistochka [10]

Answer:

The reducing agent is Zn.

Explanation:

Let's consider the reaction between zinc and hydrochloric acid.

Zn(s) + 2 HCl(aq) ⇄ ZnCl₂(aq) + H₂(g)

This is a redox reaction, which can be divided in 2 half-reactions: reduction and oxidation.

In the reduction, H⁺ gains electrons and it is considered the oxidizing agent.

2H⁺ + 2 e⁻ ⇒ H₂

In the oxidation, Zn loses electrons and it is considered the reducing agent.

Zn ⇒ Zn²⁺ + 2 e⁻

6 0

3 years ago