Write for empiricalformula
Fe2+, cr4+, cl-, O2-

Answers

Answer 1
Answer:

Answer:

{eq}Fe^{2+} {/eq} and {eq}I^- {/eq} forms {eq}FeI_2 {/eq}

{eq}Fe^{2+} {/eq} and {eq}S^{2-} {/eq} forms {eq}FeS {/eq}

{eq}Cr^{4+} {/eq} and...

Explanation:

Empirical formula:

The empirical formula gives the simple ratio of the different types of atoms in a compound. It is different from the molecular formula, which gives the exact number of each type of atom in a compound.

Answer 2
Answer:

Final answer:

In basic chemistry, an empirical formula represents the simplest ratio of atoms in a compound. For example, the empirical formula for a compound formed by Fe2+ (Iron II) and O2- (Oxide) would be FeO. In this question, a compound composed of all these ions (Fe2+, Cr4+, Cl-, O2-) is unusual and there's insufficient information to determine a reasonable structure.

Explanation:

The question is asking for an empirical formula, which is a formula that gives the simplest whole number ratio of atoms of each element in a compound. The formula you're asked to provide involves the ions Fe2+ (Iron II), Cr4+ (Chromium IV), Cl- (Chloride), and O2- (Oxide).

Creating an empirical formula is a matter of balancing out the charges in order to get a neutral compound. For instance, if you wanted to combine Fe2+ and O2-, the empirical formula would be FeO because one Fe2+ ion would balance out one O2- ion to make an electrically neutral compound.

It's important to remember that the charge value of the ion helps you determine the necessary ratio to achieve neutrality. In essence, we need the amount of positive charge to equal the amount of negative charge in the empirical formula.

For a compound involving all these ions, unfortunately, it's not common or reasonable to have a compound with four different ions. Iron, chromium, and oxygen are transition metals that could form complex ions, but we do not have enough information in this question to determine the structure.

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Related Questions

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Which of the following required Bohr's model of the atom to need modification ? A. Energies of electrons are quantized. B. Quantized electron energies are responsible for emission spectra lines. C. An electron's energy increases the farther it moves from the nucleus. D. Electrons do not follow circular orbits around the nucleus..

Answers

Answer:

Electrons do not follow circular orbits around the nucleus

Explanation:

Bohr's model of the atom is a combination of elements of quantum theory and classical physics in approaching the problem of the hydrogen atom. According to Neils Bohr, stationary states exist in which the energy of the electron is constant. These stationary states were referred to as circular orbits which encompasses the nucleus of the atom. Each orbit is characterized by a principal quantum number (n). Energy is absorbed or emitted when an electron transits between stationary states in the atom.

Sommerfeld improved on Bohr's proposal by postulating that instead of considering the electron in circular orbits, electrons actually orbited around the nucleus in elliptical orbits, this became a significant improvement on Bohr's model of the atom until the wave mechanical model of Erwin Schrödinger was proposed.

Answer:

Electrons do not follow circular orbits around the nucleus

Explanation:

The location oon the surface of the Earth directly above the focus of an earthquakeis called the

A.focus point

B.seismic wave

C.epicenter

Answers

Answer:

I think its epicenter

Explanation:

The epicenter is the point on the earth's surface vertically above the hypocenter, point in the crust where a seismic rupture begins.

The chemical reaction for the formation of syngas is: CH4 + H2O -> CO + 3 H2 What is the rate for the formation of hydrogen, if the rate of the formation of carbon monoxide is 0.35 M/s ? g

Answers

Answer :  The rate for the formation of hydrogen is, 1.05 M/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-(1)/(a)(d[A])/(dt)

\text{Rate of disappearance of B}=-(1)/(b)(d[B])/(dt)

\text{Rate of formation of C}=+(1)/(c)(d[C])/(dt)

\text{Rate of formation of D}=+(1)/(d)(d[D])/(dt)

Rate=-(1)/(a)(d[A])/(dt)=-(1)/(b)(d[B])/(dt)=+(1)/(c)(d[C])/(dt)=+(1)/(d)(d[D])/(dt)

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

CH_4+H_2O\rightarrow CO+3H_2

The expression for rate of reaction :

\text{Rate of disappearance of }CH_4=-(d[CH_4])/(dt)

\text{Rate of disappearance of }H_2O=-(d[H_2O])/(dt)

\text{Rate of formation of }CO=+(d[CO])/(dt)

\text{Rate of formation of }H_2=+(1)/(3)(d[H_2])/(dt)

The rate of reaction expression is:

\text{Rate of reaction}=-(d[CH_4])/(dt)=-(d[H_2O])/(dt)=+(d[CO])/(dt)=+(1)/(3)(d[H_2])/(dt)

As we are given that:

+(d[CO])/(dt)=0.35M/s

Now we to determine the rate for the formation of hydrogen.

+(1)/(3)(d[H_2])/(dt)=+(d[CO])/(dt)

+(1)/(3)(d[H_2])/(dt)=0.35M/s

(d[H_2])/(dt)=3* 0.35M/s

(d[H_2])/(dt)=1.05M/s

Thus, the rate for the formation of hydrogen is, 1.05 M/s

Draw 1,2,3,4,5,6-hexachlorocyclohexane with :a. all the chloro groups in axial positions.
b. all the chloro groups in equatorial positions.

Answers

Answer:

This is required answer.

Explanation:

Given that,

1,2,3,4,5,6-hexachlorocyclohexane

(a). We need to draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in axial positions

Using given data

We draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in axial positions.

When we say that all the chloro groups in axial position that means axial bonds are vertical.

(b). We need to draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in equatorial positions

Using given data

We draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in  equatorial positions.

When we say that all the chloro groups in equatorial position that means axial bonds are horizontal.

Hence, This is required answer.

When you convert feet to inches, how do you decide which portion of the conversion factor should be in the numerator and which in the denominator?

Answers

Answer : The conversion used is, 1\text{ feet}=12\text{ inches}

Explanation :

The conversion used :

1\text{ feet}=12\text{ inches}

For example : To convert 10 feet into inches.

1\text{ feet}=12\text{ inches}

10\text{ feet}=10\text{ feet}* \frac{12\text{ inches}}{1\text{ feet}}

                    =120\text{ inches}

From this we conclude that, the conversion factor used in the numerator and denominator for 1 feet should be, \frac{12\text{ inches}}{1\text{ feet}}

Final answer:

When converting from feet to inches, use the conversion factor 12 inches/1 foot. This way, the unit 'foot' cancels out and leaves 'inches', performing the conversion. An example is converting 5 feet to inches is 5 times 12, which equals to 60 inches.

Explanation:

When you're converting units, such as feet to inches, the main guideline is to set up your conversion factor in a way that cancels out the unit you want to convert from and leaves you with the unit you want to convert to. In this case, since you're converting from feet to inches, you'll be using the fact that there are 12 inches in 1 foot as your conversion factor. Therefore, when converting, the conversion factor should be set up as 12 inches/1 foot. This in essence means you're multiplying by the number 1, which doesn't change the value, just the units.

For instance, if you have 5 feet and you want to convert this to inches, you'll set your conversion factor up as 12 inches/1 foot, with inches in the numerator and feet in the denominator, to cancel out feet. Multiply 5 feet by the conversion factor (12 inches/1 foot), the result would be 60 inches. Here, the 'feet' units cancel leaving the answer in inches, completing the conversion.

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1. What is the angular distance north or south from the earth's equator beginning at0° at the equator and ending at 90° at either pole?
a. Weather
b. Altitude
c. Latitude
d. Climate

Answers

Answer:

c. Latitude

Explanation:

The angular distance north or south from the earth's equator beginning at 0° at the equator and ending at 90° of either pole is the latitude.

The equator is a line of latitude that divided the earth into two hemispheres.

Only the equator is a great circle as a line of latitude. Others are small circles.

  • Weather is the atmospheric condition of a place over a short period of time
  • Climate is the average weather condition at a place over a long period of time.
  • Altitude of a place is its elevation above a reference plane.