Alkali metals nitrates on heating give the corresponding __ and oxygen.

The activation energy for this reaction is - 55.5 kJ/mol

Calculation,

Given data,

First temperature [tex]T_{1}[/tex] =  302°C = 302+273 = 575 K

Second temperature [tex]T_{2}[/tex] =  508°C =  508+273 = 781 K

rate constant at  302°C = 2. 45 × [tex]10^{-4}[/tex] lit/mol s

rate constant at 508°C =  0. 0965 lit/mol s.

Value of universal gas constant = 8.3145 J/k mol.

Apply Arrhenius equation,

㏒[tex]K_{1} /K_{2}[/tex] = [tex]E_{a}/2.203R[/tex] [1/ 575 K - 1/781 K]

㏒2. 45×[tex]10^{-4}[/tex] lit/mol s/0.0965 lit/mol s = [tex]E_{a}[/tex] /2.303×8.3145 J/k mol[781-575/575K×781 K]

[tex]E_{a}[/tex] = - 55.5 kJ/mol

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According to the IUPAC convention alkyl substituents on a hydrocarbon chain should be listed in alphabetical without considering prefixes order.

IUPAC convention of organic chemistry is a method of naming organic chemical compounds as recommended by the International Union of Pure and Applied Chemistry (IUPAC).

The name of the compound is written out with the substituents in alphabetical order followed by the base name (derived from the number of carbons in the parent chain).

Types of IUPAC  Nomenclature of a few important aliphatic compounds:

Example :

Ethane, which has 2 carbon atoms and 6 hydrogen atoms, with the molecular formula of = C₂H₆

Formation of alkyl group:

Methane (CH₄)  Remove 1 hydrogen (H) convert to methyl (H₃-C-)

Example: Propyl (-CH₃ - CH₂ - CH₂ - )

According to the IUPAC convention alkyl substituents on a hydrocarbon chain should be listed in alphabetical without considering prefixes order.

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The element "X" is "O" (oxygen).

Given,

Chemical formula = Na₂CX₃

Formula mass = 106 amu

Molar mass of Na = 23 amu

Molar mass of C = 12 amu

To find,

Element X =?

We will equate the equation as follows,

2(23) + 12 + 3(y) = 106

46 + 12 + 3y =106

58 + 3y = 106

3y = 106 - 58

3y = 48

y = 48/3

y = 16

We know that Oxygen has molecular mass of 16. Therefore the element "X" is "O".

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Answer:

0.453 moles

Explanation:

The Ideal Gas Law equation looks like this:

PV = nRT

In this formula,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas Constant (0.08206 atm*L/mol*K)

-----> T = temperature (K)

To find the amount of moles of gas, you can plug the given values into the equation and then simplify.

P = 2.7 atm                                 R = 0.08206 atm*L/mol*K

V = 4.2 L                                     T = 305 K

n = ? moles

PV = nRT

(2.7 atm)(4.2 L) = n(0.08206 atm*L/mol*K)(305 K)

11.34 = n(25.0283)

0.453 = n

Answer:

136 g Al₂O₃

Explanation:

Assuming you do not need to find the limiting reactant, to find the mass of Al₂O₃, you need to (1) convert grams O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles Al₂O₃ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles Al₂O₃ to grams Al₂O₃ (via molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to match the sig figs of the given value (64.0 g).

Molar Mass (O₂): 32 g/mol

Molar Mass (Al₂O₃): 102 g/mol

4 Al + 3 O₂ -----> 2 Al₂O₃

64.0 g O₂           1 mole           2 moles Al₂O₃            102 g
-----------------  x  --------------  x  ------------------------  x   -------------  =  136 g Al₂O₃
                             32 g               3 moles O₂             1 mole

Answer:

Explanation:

Based of the fact that you were given 2 masses I would assume this to be a limiting reagent question. However amount on the left side both equal 2. Ignoring limiting reagents and focusing on just O2 the steps would be:

1. Make sure the equation is balanced ( already given)

2- Use given values to find the mols of O2 (mass/molar mass)

3. Mols are conserved but due to the coefficients the molar value from O2 must be divided by three and multiplied by 2 to ensure proper ratios

4. Using that amount the mass can derived using amount/molar mass

5. Use proper significant digits and units(3 in this case)

The binding energy of the electron is 1.237 ×10^(-16) kj/ mol.

The given parameter

wavelength of photon,

                       λ = 0.999nm

                          = 0.999x10⁻⁹ m

K.E of emitted photon,

                    K.E = 940 ev

The binding energy of electron is calculated as follows:

From Einstein's mass defect equation=

                    ΔE = Δmc²

Also, from Einstein's photo electric equation=

                       E = Ф + K.E

Where;

          Ф is binding energy of electron on metal surface.

The energy of on mole of electron, the emitted is calculated as:

                       [tex]E = hf =h \frac{c}{λ} = (6.626×10^{-34} ) × \frac{{3×10^{8} }}{0.999×10^{-9} } \\\\E= 1.989×10^{-16}[/tex]

The K.E of emitted electron in Joules is

                 [tex]K.E = 940 × \frac{1}{2} ×1.602 ×\ 10^{-19} J\\\\ = 0.752 × 10^{-16} J[/tex]

The binding energy is electron is calculated as:

               [tex]Ф = E - K.E\\= (1.989 - 0.752) ×10^{-16}\\ =1.237 × 10^{-16} J[/tex]

Question: An X-ray photon of wavelength 0.999nm strikes a surface. The emitted electron has a kinetic energy of 940 eV. What isthe binding energy of the electron in kJ/mol? (KE=1/2mv2; 1 electron volt (eV) = 1.602 x10-19J)

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Answer:

3.3 x 10⁻¹⁷ Hz    

Explanation:

To find the frequency, you can use the following equation:

E = h / f

In this equation,

-----> E = energy (J)

-----> h = Planck's Constant (6.626 x 10⁻³⁴ J*s)

-----> f = frequency (Hz)

You can plug the given values into the equation and simplify to find the frequency. This equation will require a little bit of rearranging.

E = h / f                                                        <----- Given equation

(2.0 x 10⁻¹⁷ J) = (6.626 x 10⁻³⁴ J*s) / f         <----- Insert values

(2.0 x 10⁻¹⁷ J) x f = (6.626 x 10⁻³⁴ J*s)        <----- Multiply both sides by f

f = 3.3 x 10⁻¹⁷ Hz                                        <----- Divide both sides by 2.0 x 10⁻¹⁷

The frequency of an x-ray wave with an energy of 2.0 x 10^-17 J is 3.3 x 10⁻¹⁷ Hz.

The term frequency is defined as the number of waves that pass a fixed point in unit time. Frequency is measured in hertz which is equal to one event per second.

Frequency also describes the number of cycles undergoes during one unit of time by a body in periodic motion.

Calculating the frequency, you can use the following equation:

E = h / f

Where,

E = energy (J)

h = Planck's Constant (6.626 x 10⁻³⁴ J*s)

f = frequency (Hz)

Insert his values in the given equation

(2.0 x 10⁻¹⁷ J) = (6.626 x 10⁻³⁴ J × s) / f        

(2.0 x 10⁻¹⁷ J) x f = (6.626 x 10⁻³⁴ J × s)        

f = 3.3 x 10⁻¹⁷ Hz                

Thus, The frequency of an x-ray wave with an energy of 2.0 x 10^-17 J is 3.3 x 10⁻¹⁷ Hz.

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The hybridization of the central atom in SCl₂ is sp³.

Lewis structure is also known as Lewis dot structure which simply represent the valence electron of an atom.

Hybridization is defined as the mixing of two atomic orbitals belongs to same atom but having entirely different shapes , energies which produces a new hybridized orbitals.

Now first write the electronic configuration of central atom.

The central atom in SCl₂ is S (Sulphur).

Electronic configuration of S: [Ne] 3s² 3p⁴

One 3s and three 3p undergo hybridization to form sp³ orbitals. Here each atom forms bond with one sp³ hybrid orbital.

Thus from the above conclusion we can say that The hybridization of the central atom in SCl₂ is sp³.

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Rutherford's scattering experiment demonstrated that the mass of an atom was uniformly distributed throughout its volume this statement is incorrect.

Hence, Option B is correct answer.  

Rutherford's Alpha Scattering Experiment demonstrated that Rutherford bombarded most of the alpha particles passed straight through the gold foil. Rutherford observed that most part of the atom is empty.

The mass of an atom was uniformly distributed throughout its volume this experiment was given by Thomson Model.

Thus from the above conclusion we can say that Rutherford's scattering experiment demonstrated that the mass of an atom was uniformly distributed throughout its volume this statement is incorrect.

Hence, Option B is correct answer.  

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The option that is most likely to be true regarding L76, L77, and L79  is that option B: The assembly would incur an entropic penalty if they occupied a solvent-exposed site.

The entropic penalty in regards to ordered water is known to be one that tells or account  for any form of  weaker binding of the antibiotic novobiocin to what we call resistant mutant of DNA gyrase.

Note that in regards to the scenario above, the Entropic penalty is seen as the thermodynamically disfavored needs that is required in forming a cage of polar solvent molecules that is known to be seen around surface that has exposed hydrophobic potion of a molecule.

Hence, The option that is most likely to be true regarding L76, L77, and L79  is that option B: The assembly would incur an entropic penalty if they occupied a solvent-exposed site.

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See full question below

When Bs1A is assembled as an octamer, what is most likely to be true regarding L76, L77, and L79?

A. They are oriented toward the solvent-exposed exterior of the protein assembly.

B. The assembly would incur an entropic penalty if they occupied a solvent-exposed site.

C. Unlike in a monomer, they are not situated within the hydrophobic cap.

D. Their physiochemical properties are not substantially dependent on their hydrophobicity

Yes, we can use this balance to  estimate the mass to the nearest milligram.

The weighing balance is the kind of scale that can be used to measure the mass of a body. We know that mass is the quantity of matter that is contained in a  body and the the SI unit of mass is the kilogram.

Looking at the balance as it is graduated in milligrams, we know that measurement to the nearest gram means measurement to the nearest whole number so we can say that the answer is yes. We can use this balance to  estimate the mass to the nearest milligram.

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The consideration element is the correct answer.

The exchange of the value is called consideration. The exchange of money is known as value for anything such as services/property, exchange of property, or services.

There are many types of considerations that are given below:

The four important elements of a contract are given below:

If any of the given elements is missing then the contract is voided, and the parties released from the agreement.

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The oxygen consumption for a burmese python at 20 contractions per minute 5,30,35,10,15,20,25 contractions per minute.

The Eastern Hemisphere's tropical and subtropical regions are home to the genus Python, a family of constricting snakes. For non-venomous flecked snakes, François Marie Daudin suggested the name "Python" in 1803. There are now 10 recognized python species. A new species has been established, and three previously recognized python subspecies have been elevated.

Ythons lack venom, and colubrids (rear-fanged snakes) either lack venom entirely or have a weak venom. It is important to take venomous elapid (front-fanged snake) bites seriously and treat them accordingly. Massive snakes called pythons have a brutal killing style. These snakes, like other strong constrictors like anacondas, will capture their victims with their mouths and wrap around them.

What was the oxygen consumption for a burmese python at 20 contractions per minute?

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The pH of a buffer solution with acid that (PKA 6. 1) is exactly half as concentrated as its conjugate base is 6.4.

A buffer solution is a solution that has a maintained pH, not basic or not acidic. Its pH changes when acid or base is added to the solution.

We had to figure out the acid's concentration, which is exactly half that of its potential base.

We know that pH = pH_log

We have less than 6.1 pH so this is a conjugated base.

This will equal to 6.1 + log2 = 6.4

Thus, the pH of a buffer solution with acid is 6.4.

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Answer:

0.0423 L

Explanation:

Molarity can be represented by the following ratio:

Molarity = moles / volume (L)

You can plug the given values into the equation and simplify to find the volume.

Molarity = moles / volume

0.730 M = 0.0309 moles / volume

(0.730 M) x volume = 0.0309 moles

volume = 0.0423 L

When hexan-1-ol is treated with conc. H₂SO₄ at moderate temperatures, Di-n-hexyl ether is formed via SN₂ mechanism.

SN₂ Mechanism is mechanism involving the factors of steric hindrance following the order of 1° > 2° > 3°. It is a Nucleophilic substitution reaction.

Williamson Ether Synthesis is the synthesis of ether through SN₂ mechanism in which an ether is treated with conc. H₂SO₄ at moderate temperature.

When hexan-1-ol is treated with conc. H₂SO₄ at moderate temperatures, Di-n-hexyl ether is formed as below:

CH₃CH₂CH₂CH₂CH₂CH₂OH +Conc. H₂SO₄-->(CH₃CH₂CH₂CH₂CH₂CH₂)₂0

Hexan-1-ol                                                                 Di-n-hexyl ether

Williamson Ether Synthesis is a reaction involving a alcohol that is deprotonated to form an ether.

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It is observed that when magnesium reacts with hydrochloric acid, it produces visible bubbles of hydrogen gas.

A vigorous reaction will occur giving rise of heat as it is an exothermic reaction. If heat is applied then it should be more vigorous.

Reaction is as follows:

Mg(s) + HCl (aq) --> MgCl2 (aq) + H2(g)

Magnesium reacts easily with HCl to produce H2 gas and magnesium ions, Mg2+, and heat. The reaction is exothermic, so it heats up quickly.

Mg(s) + HCl (aq) --> MgCl2(aq) + H2(g)

The net ionic equation :

Mg(s) + 2H+ --> Mg2 + H2(g)

If water is removed from the solution then white crystals of Mgcl2 is obtained.

Or in simple words,

2Mg + 2HCl -> 2Mg+ + 2Cl- + H2(gas)

The magnesium is attacked by the hydrochloric acid resulting in the magnesium dissolving into the solution resulting in a solution of magnesium chloride in hydrochloric acid and the production of hydrogen gas.

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91.4 grams

91.4 grams of kcl in grams must be added to 500 ml of a 0. 15 m kcl solution to produce a 0. 40 m solution.

C = mol/volume

2.45M=mol/0.5L

2.45M⋅0.5L = mol

mol = 1.225

Convert no. of moles to grams using the atomic mass of K + Cl

1.225mol * [tex]\frac{39.1+35.5}{mol}[/tex]

mol=1.225

=1.225 mol . [tex]\frac{74.6g}{mol}[/tex]

=1.225 . 74.6

=91.4g

therefore, 91.4 grams of kcl in grams must be added to 500 ml of a 0. 15 m kcl solution to produce a 0. 40 m solution.

What is 1 molar solution?

In order to create a 1 molar (M) solution, 1.0 Gram Molecular Weight of the chemical must be dissolved in 1 liter of water.

58.44 g make up a 1M solution of NaCl.

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The balanced chemical equation for the standard formation reaction of liquid acetic acid is given as ,

[tex]2C(gr) +2H_{2} (g) +O_{2} (g)[/tex]  → [tex]CH_{3} COOH(l)[/tex]

The reaction that form the products from their elements in their standard state is called formation of reaction .The acetic acid consist C , H , and O , So, determine their standard state . Carbon is graphite at 25°C and 1 atm , whereas hydrogen and oxygen are diatomic gases . Hence , we start with unbalanced reaction.

[tex]C(gr) +H_{2} (g) +O_{2} (g)[/tex] → [tex]CH_{3} COOH(l)[/tex]

The balanced chemical equation for the standard formation reaction of liquid acetic acid as,

[tex]2C(gr) +2H_{2} (g) +O_{2} (g)[/tex]  → [tex]CH_{3} COOH(l)[/tex]

The combustion of liquid acetic acid is given as,

[tex]CH_{3} COOH(l) + 2O(g)[/tex] → [tex]2CO_{2}((g) +2H_{2} O(l)[/tex]                    ΔH =-873

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Answer:

102 g Al₂O₃

Explanation:

The theoretical yield is the mass of product calculated via the molar masses and balanced chemical equation.

The limiting reagent is the reactant that is completely reacted before the other reactant(s) are used up. Since Al produces the smaller amount of product, it appears that Al is the limiting reactant. You can only make as much product as the limiting reactant allows. As such, the theoretical yield is 102 grams Al₂O₃.

Number of moles of KOH present in solution is 3.75.

The term "mole" refers to a mole's atomic mass, which is expressed in grams. A mole is an element's gram-atomic mass, or mass.

The mole idea equates the mass of a single atom or molecule (measured in a.m.u.) to the mass (measured in grams) of a large collection of equivalent molecules.

To calculate moles of potassium hydroxide in 750 ml of 5.00 m of KOH solution -

Mathematically molarity  is expressed as,

Molarity =no. of molecules of solute/volume of solution

Given: Molarity of solution = 5.00 M

Volume of solution = 750 ml = 0.750 l

∴ 5 = no. of moles/0.750

∴ number of moles = 3.75

Hence, number of moles of KOH present in solution is 3.75.

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The boiling point of a solution consisting of cyclohexane (c6h12) dissolved in benzene if the freezing point of this solution is 0. 0°c is 80.9°C.

When a solute and a solvent are combined to form a liquid homogeneous mixture or solution, the properties that result are different from those of the pure substances. As a result, we can find a freezing point depression, boiling point elevation, and a few other cooperative properties, but these two are the focus of this article.

Taking into account that the freezing point depression constant solely depends on the solvent (cyclohexane), we can begin by utilizing the freezing point depression to determine the molality of the solution:

[tex]T_{fsolution} - T_{fsolvent} = -m * {K_{fsolvent}[/tex]

0. 0°c - 5.5°c = -m * 20.0 °c

[tex]m = \frac{-5.5}{- 20.0 }[/tex]

m = 0.275 m

The boiling point elevation formula is then used to get the solution's boiling point, which is once more dependent on the solvent's boiling point elevation constant:

[tex]T_{b, solution} - T_{b, solute} = m * Kb_{solvent} \\\\T_{b, solution} - 80.1 = 0.275m + 2.79\\\\T_{b, solution} = 80.9[/tex]

Therefore, The boiling point of a solution is 80.9°C.

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Carbon atoms present in 2. 00 g of butane is 8.28 × 10^22.

Molecular weight of C4H10 is = 58.1 grams

Moles of butane = given mass/ molar mass

= 2/58.1

= 0.0344 mol

Multiply the number of moles by Avogadro's number. This will let you obtain the number of butane molecules:

0.0344 x 6.023 x 10^{23} molecules of butane = 2.07 × 10^22 molecules

of butane

Now multiply this number by four (due to four carbon atoms per butane molecule) to obtain the answer:

so, no. of carbon atoms = 4 x 2.07x 10^22 = 8.28x 10^22 atoms.

Thus the no. of carbon atom is 8.28×10^22.

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The major product of reaction between 1-butanol and Na2Cr2O7 is butanoic acid.

When a primary alcohol like 1-butanol (OH is bonded to a primary carbon) is begin to oxidize in the presence of strong oxidizing reagent such as sodium dichromate (Na2Cr2O7) and H2SO4, sulfuric acid, the stepwise oxidation take place as above firstly to the corresponding aldehyde which undergoes further oxidation to the corresponding carboxylic acid.

You can find that the formed aldehyde after first oxidation is butanal and the only organic product, due to the strong oxidizing reagent is butanoic acid.

Thus, the major product formed is butanoic acid.

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There are 372 grams of calcium phosphate are theoretically produced if we start with 3. 40 moles of calcium nitrate and 2. 40 moles of lithium phosphate .

Calculation,

The reactant or reagent that produces the least moles of the products is called limiting reagents. When  limiting reagents used up , the reaction stops.

The balanced equation is given as,

[tex]3Ca(NO_{3} )_{2} + 2Li_{3} ( PO_{4} )[/tex] → [tex]3LiNO_{3} + Ca_{3}( PO_{4} )_{2}[/tex]

Multiply the moles of each reactant by the mole ratio between it and calcium phosphate in the balanced equation . so that the moles of the reactant cancel , leaving moles of calcium phosphate.

3.4 mol of calcium nitrate × 1 mol calcium phosphate / 3 mol calcium nitrate = 1.13 mol calcium phosphate

2.4 mol of  lithium phosphate× 1 mol calcium phosphate / 2 mol  lithium phosphate = 1.02 mol calcium phosphate

So, calcium nitrate  is limiting reactant .

Calculation of mass of 1.02 mol calcium phosphate.

Multiply the moles of  calcium phosphate by its molar mass.

molar mass of  calcium phosphate = 3×40.078 g/mol calcium ion+2×30.9 g/mol phosphorus + 8×15.99 g/mol calcium phosphate = 310.178 g/mol calcium phosphate

1.20 mol calcium phosphate × 310.178 g/mol = 372 gram

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Oxygen is the crucial gas required for an organism to carry out the cellular process. The seal can dive for 4 minutes before it runs out of oxygen. Thus, option a is correct.

Oxygen is the most vital element of the organism that is required by almost all cellular and metabolic activities. It is inhaled by the respiratory system and passed to cells and tissues through the circulatory system.

Given,

Mass of seal = 100 kg = 100000 gm

Rate of oxygen consumption (r) = 1 ml/ (gh)

The volume of the stored O₂ = 7000 mL

The time (t) is calculated as,

t = V ÷ mr

Substituting the values above,

t = 7000 ÷ 100000 × 1

= 0.07 hour

In minutes it will be given as,

0.07 × 60 = 4.2 minutes

Therefore, option a. 4 minutes is the time till the seal can remain underwater without oxygen.

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Your question is incomplete, but most probably your full question was, The oxygen consumption rate of a 100 kg seal is 1 ml/(g h). assuming that it has oxygen stores of 7 liters, how many minutes can it dive before running completely out of oxygen?

Answer:

Reactants: Fe₂O₃ and C

Products: Fe and CO₂

Explanation:

The reactants are the molecules that undergo changes in a reaction.

The products are the molecules which are made during a reaction.

The reactants are on the left side of the arrow and the products are on the right side of the arrow. Therefore.....

The chemical reaction:

2 Fe₂O₃ + 3 C -----> 4 Fe + 3 CO₃

Reactants: Fe₂O₃ and C

Products: Fe and CO₂

Answer: All have something wrong with them!!!

Explanation: #1: He is using chemicals without using any goggles or protection

#2: She is unsupervised while around fire as well as no protection

#3:

#4: There are many flammable things around the fire especially since its paper it could start a major fire

#5: Again has no protection (goggles. gloves, e.t.c.) and putting chemicals close to your face is just a Don't in science

#6: They are "horseplaying" and also someone else is doing all the work

Answer:

CrO₂ --------------------> Cr⁴⁺ and O²⁻

VCO₃ -------------------> V²⁺ and CO₃²⁻

Cr₂(SO₄)₃ -------------> Cr³⁺ and SO₄²⁻

(NH₄)₂S ----------------> NH₄⁺ and S²⁻

Explanation:

Within ionic compounds, the cation is listed first, followed by the anion. Some of the ions are polyatomic, meaning they are covalently bonded to other elements. Polyatomic ions always have a specific charge.

All of these ionic compounds have an overall charge of 0. As such, the charges of the cations and anions must balance out. In order to do so, there are some compounds which have more than one atom of each ion.

2.) CrO₂

------> Oxygen (O) always forms the anion, O²⁻.

------> Therefore, if there are 2 oxygen anions, the chromium (Cr) must have the cationic form of Cr⁴⁺.

------> +4 + (-2) + (-2) = 0

3.) VCO₃

------> Carbonate (CO₃), a polyatomic ion, always has the state CO₃²⁻.

------> If there is only one atom of each ion, the charges must perfectly balance, making vanadium (V) be the cation V²⁺.

------> +2 + (-2) = 0

4.) Cr₂(SO₄)₃

------> Sulfate (SO₄), a polyatomic ion, always has the state SO₄²⁻.

-------> The only way the charges could balance out is if the chromium (Cr) is in the cationic form Cr³⁺.

------> +3 + 3 + (-2) + (-2) + (-2) = 0

5.) (NH₄)₂S

------> Ammonium (NH₄), a polyatomic ion, always has the state NH₄⁺.

------> Sulfur (S) always forms the anion S²⁻.

------> +1 + 1 + (-2) = 0