The molecular formula C8H8 might represent a cyclic compound where carbon atoms are linked in an 8-membered ring, with each carbon atom being bonded to two other carbon atoms and a hydrogen atom, maintaining the tetravalency of carbon.
Explanation:The compound you mentioned, C8H8, does not fit with the typical hydrocarbon pattern which typically results in hydrocarbons like alkanes having a general formula of CnH2n+2. However, since the question stated clearly that there are no double or triple bonds and all carbon atoms are identical, there's a possibility that the compound is a cyclic structure. Let's consider a structure where carbon atoms are linked in a cyclic form, forming an 8-membered ring. Each carbon atom in this ring will form two bonds with two other carbon atoms and the remaining two bonds will be with hydrogen atoms. In this way, each carbon atom will maintain its tetravalency (4 bonds per carbon atom), and there will be 8 hydrogen atoms, exactly fitting the given formula C8H8. To draw the Lewis structure, arrange 8 carbon atoms in a ring and attach a hydrogen atom to each carbon atom. Remember, each line represents a single bond between atoms.
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The request for a Lewis structure of C8H8 without double or triple bonds and with all carbon and hydrogen atoms being chemically identical suggests an unusual compound that contradicts common organic chemistry principles, hence a conventional structure cannot be provided.
Explanation:The student is asking for a Lewis structure of a compound with the formula C8H8 that does not contain any double or triple bonds. Given that all carbon and hydrogen atoms in this compound are chemically identical, this suggests a cyclic structure where each carbon atom is bound to two other carbons and a single hydrogen atom, forming a ring. The best structure fitting this description would be cyclooctatetraene, although it has alternating single and double bonds, which contradicts the no double or triple bonds condition.
However, considering the compound has no double or triple bonds, the compound would be highly unusual and not in line with standard organic chemistry principles. All carbon atoms making four bonds is a fundamental concept in organic chemistry, which leads to the conclusion that either the information given in the question is incorrect or it is a highly unconventional compound not typically covered in standard chemistry curricula. Therefore, without double or triple bonds, the compound's structure cannot be satisfactorily drawn while adhering to typical valency rules.
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Which statement is true?
A) Unicellular organisms cannot obtain energy or reproduce, but they can maintain structure (homeostasis).
B) Unicellular organisms don't have to do anything - they are only just one cell!
C) Unicellular organisms have to meet the 3 main challenges of life. They have to obtain energy, reproduce, and maintain structure (homeostasis).
D) Unicellular organisms are not living.
Final answer:
The true statement about unicellular organisms is that they must obtain energy, reproduce, and maintain homeostasis, despite being single-celled.
Explanation:
The statement that is true about unicellular organisms is: C) Unicellular organisms have to meet the 3 main challenges of life. They have to obtain energy, reproduce, and maintain structure (homeostasis).
All living organisms, including unicellular ones, need to carry out several basic functions to sustain life. These functions include obtaining energy to power their biological processes, growing and reproducing to ensure the survival of their species, and maintaining a stable internal environment through the process of homeostasis. Even though they are made of only one cell, unicellular organisms are complex and very much alive. They are capable of responding to their environment and have complex chemistry. Additionally, these organisms must exchange matter with their surroundings in order to grow, reproduce, and maintain their organization.
sodium chloride (NaCl)
silver (Ag)
nitrogen
propane
Which ones likely property is malleable?
Silver (Ag) is the substance most likely to be malleable among the listed materials because it is a metal with characteristics of ductility and malleability, unlike the brittle ionic compound sodium chloride or the gaseous nitrogen and propane. Hence, correct option B.
Among sodium chloride (NaCl), silver (Ag), nitrogen, and propane, the substance that is likely malleable is silver (Ag). Malleability is a property characteristic of metals that allows them to be hammered or rolled into thin sheets without breaking. Sodium chloride, being an ionic compound, is hard, brittle, and not malleable. Nitrogen is a diatomic gas at room temperature, and propane is a hydrocarbon gas, neither of which has malleable properties. Silver, however, is a metal known for its excellent malleability and ductility. It is soft enough to be worked into various shapes and maintains its structural integrity when manipulated.
What process is primarily responsible for the formation of precipitation in warm clouds?
A chemical engineer has determined by measurements that there are 69.0 moles of hydrogen in a sample of methyl tert-butyl ether. how many moles of oxygen are in the sample? g
In a methyl tert-butyl ether sample that contains 69.0 moles of hydrogen, there are 5.75 moles of oxygen.
Methyl tert-butyl ether is an organic compound that can be represented through the semi condensed formula (CH₃)₃COCH₃ or through the condensed formula C₅H₁₂O. As we can see, in methyl tert-butyl ether the molar ratio of H to O is 12:1. The number of moles of oxygen in a sample that contains 69.0 moles of H are:
[tex]69.0 mol H \times \frac{1mol O}{12 mol H} = 5.75 mol O[/tex]
In a methyl tert-butyl ether sample that contains 69.0 moles of hydrogen, there are 5.75 moles of oxygen.
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29 Points!!
What is NOT true of mitosis
It is the division of the genetic material of a dividing eukaryotic cell into two parts.
Each daughter cell receives a full set of chromosomes identical to the parent cell.
It is a two-step process involving the division of the nucleus and then cytokinesis.
It is a process that takes an average of eight minutes for any cell to complete the process.
Calculate the mass of 1.23×1024 helium atoms.
6.023*1023 He atoms=1 mole He
1.23×1024 helium atoms=1.23×1024/6.023*1023 =2.042 mole He
one mole He=4.0 gm
2.042 mole He=2.042*4.0=8.168 gm
answer in grams of Helium is 8.168 gm
Tris {(hoch2)3cnh2} is one of the most common buffers used in biochemistry. a solution is prepared by adding enough tris and 12 m hcl(aq) to give 1.00 l of solution with [tris] = 0.30 m and [trish+] = 0.60 m. what is the ph of this buffered system if the pkb is 5.92?
Given that,
The concentration of TRIS = 0.30 M
The concentration of TRIS+ = 0.60 M
Kb = 1.2 x 10^-6
pKb = -log Kb = - log (1.2 x 10^-6) = 5.920
Now, by using the Hendersonn equation,
pH = pKa + log TRIS+/TRIS = 5.920 + log (0.60/0.30) = 6.221
pOH=14-pH=14-6.221 = 7.779
Given:
Buffer system : Tris/TrisH+
[Tris] = 0.30 M
[TrisH+] = 0.60 M
pKb = 5.92
To determine:
pH of the buffer
Explanation:
The pH of a buffer can be obtain using the Henderson-Hasselbalch equation:
pH = pKa + log[Base]/[Acid]
In this case the conjugate base = [Tris]
Acid = [TrisH+]
Now, pKa = 14-pKa = 14-5.92 = 8.08
pH = 8.08 + log[0.30]/[0.60] = 7.778
Ans: pH of the buffer = 7.78
In the molecule on the left, areas that have a partial negative charge are pink and areas that have a partial positive charge are blue. The molecule on the right is nonpolar.
What will most likely happen when these molecules get closer?
The molecules will repel each other because of a dipole-dipole interaction.
The molecules will attract each other because of a dipole-dipole interaction.
An induced dipole will be produced in the molecule on the right.
An induced dipole will be produced in the molecule on the left.
Planck's equation E = h × v is used to determine which of the following?
Question options:
the energy released as an electron moves to a higher orbital
the number of lines on the atomic spectrum for an atom
the energy released as an electron moves to a lower orbital
The Planck's equation is used to determine the energy released as an electron moves to a lower orbital.
According to the Bohr's model of the atom, energy is released when an electron moves from a higher to a lower energy level.
The magnitude of this energy released is obtained by the Plank's equation; E = h × v.
This energy often appears as a photon of light.
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Why does a thermometer need to be submerged in a ice water slush instead of only ice?
The tendency of a moving object to continue moving in a straight line or a stationary object to remain in place is called
a.
orbital speed.
b.
mass.
c.
inertia.
d.
gravity.
Answer:
c. inertia.
Explanation:
According to Newton’s First Law of Motion or the Law of Inertia a body at rest continues to be at rest unless acted upon by some force. A body in uniform motion continues to move in uniform motion unless acted upon by some external force. The external motion may change the uniform motion of the object.
Suppose that the ion is excited by light, so that an electron moves from a lower-energy to a higher-energy molecular orbital. would you expect the excited-state h−2 ion to be stable?
The stability of an excited-state ion, in this case, the H⁻₂ ion (diatomic hydrogen anion), depends on several factors, including the energy of the excited state, the electronic configuration, and the propensity for the ion to release energy and return to a lower energy state.
What is expectedIn general, when an electron is excited to a higher-energy molecular orbital, the resulting ion is not stable in the long term. Excited states are typically higher in energy and have higher potential energy compared to the ground state. As a result, the excited-state ion is often in an energetically unfavorable condition.
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The excited-state H₂²- ion is unlikely to be stable because its ground state already has a bond order of zero, indicating instability, and excitation to an antibonding orbital further destabilizes it.
The stability of the H₂²- ion in an excited state can be assessed using a molecular orbital energy-level diagram. When an electron in this ion is excited to a higher energy level, such as from a bonding to an antibonding orbital due to the absorption of light, the bond order decreases, affecting stability. In the case of H₂²-, the bond order in the ground state is already zero, indicating that the ion is not stable. Exciting an electron to an antibonding orbital would not change the bond order but would further destabilize any remnant bonding interactions. Thus, it would be unlikely for the excited-state H₂²- ion to be stable.
According to the equation for this chemical reaction, what does the 2 in front of 2Ag tell us?
A) Two atoms of silver are lost in this reaction.
B) Two atoms of silver are produced by this reaction.
C) Two atoms of silver are needed to complete the reaction.
D) Two atoms of silver are gained to produce silver sulfide.
the right answer is "Two atoms of silver are needed to complete the reaction." because two atoms of silver are needed to complete the reaction. The silver atoms are used to produce silver sulfide. There must be two atoms of silver on each side of the arrow to balance the equation. Atoms are never lost or gained in a chemical reaction.
the 2 in front of 2Ag in the chemical equation indicates that two atoms of silver are produced by this reaction. Therefore, B) Two atoms of silver are produced by this reaction.
The number in front of a chemical formula in a balanced chemical equation represents the coefficient, which indicates the ratio of moles of each substance involved in the reaction.
In this case, the 2 in front of 2Ag tells us that two moles of silver (Ag) are involved in the reaction.
Given the options provided:
B) Two atoms of silver are produced by this reaction.
Therefore, the 2 in front of 2Ag in the chemical equation indicates that two atoms of silver are produced by this reaction.
In the sn2 experiment, what was the purpose of washing the distilled product with 5% naoh
The purpose of washing the product with NaOH is simply to neutralize any acid which remained or leaked after the 1st initial separation. The NaOH base reacts with the acid to form neutralization reaction products which are soluble in water.
The distilled product in an SN2 experiment is washed with 5% NaO for the purpose of extracting and neutralizing acidic impurities to produce a pure product. This relates to the use of NaOH during titration experiments where it is used to neutralize an acid and find its acidity.
Explanation:In the SN2 experiment, the distilled product was washed with 5% NaOH for the purpose of neutralization and extraction of acidic impurities. During the reaction performed, there might be some acidic byproducts which can affect the final result or the purity of the product. The washing step with sodium hydroxide, which is a base, neutralizes them and helps to obtain a pure product. This is similar to the provided titration examples where an acid is neutralized by a base. For example, in titration, HCl is neutralized by NaOH.A closely related real-world application of this concept is the use of NaOH (sodium hydroxide) during titration experiments. Here, NaOH is used to determine the acidity of an unknown solution. As indicated in the information, different solutions and volumes can be used to reach the so-called 'equivalence point' or the point at which the acid is fully neutralized by the base.
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Iven the parent compound, draw the mass spectrum fragment that is observed at m/z 77. include any hydrogen atoms and the charge.
The first step is to determine each compound's total molar mass by adding up all of its constituent parts.
What are compounds?Compounds are defined as anything made up of similar molecules with atoms from two or more different chemical elements. Chemical connections that are challenging to break are created when the elements interact with one another.
Chlorobenzene, the chemical on the left, has a mass of 112 and is composed of 5 carbons (12 g/mol), 1 chlorine (35 g/mol), and 5 hydrogens (1 g/mol). The right-hand chemical (2-chloropentane), which consists of 3 carbons, 1 chlorine, and 7 hydrogens, has a mass of 78. The mass loss for chlorobenzene is 112 - 77 = 35. The mass loss for 2-chloropropane is 78 - 77 = 1. The loss of a hydrogen ion results in a loss of one unit. This is typical and is referred to as the "M-1 peak." Therefore, the dehydrogenated molecular ion is represented by the 77 m/z fragment.
Thus, the first step is to determine each compound's total molar mass by adding up all of its constituent parts. '
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Calculate the number of moles of aluminum, sulfur, and oxygen atoms in 9.00 moles of aluminum sulfate, al2(so4)3. express the number of moles of al, s, and o atoms numerically, separated by commas.
The number of moles of aluminum, sulfur, and oxygen in 9.00 moles of aluminum sulfate are 18.00 moles of Al, 27.00 moles of S, and 108.00 moles of O, respectively.
To calculate the number of moles of aluminum (Al), sulfur (S), and oxygen (O) atoms in 9.00 moles of aluminum sulfate,
Al₂(SO₄)₃, we need to consider the stoichiometry of the compound. Aluminum sulfate contains 2 atoms of aluminum, 3 atoms of sulfur, and 12 atoms of oxygen per formula unit.
For aluminum (Al):
2 moles of Al are in 1 mole of Al₂(SO₄)₃, so in 9.00 moles of Al₂(SO₄)₃, there are 2 x 9.00 moles = 18.00 moles of Al.
For sulfur (S):
3 moles of S are in 1 mole of Al₂(SO₄)₃, so in 9.00 moles of Al₂(SO₄)₃, there are 3 x 9.00 moles = 27.00 moles of S.
For oxygen (O):
12 moles of O are in 1 mole of Al₂(SO₄)₃, so in 9.00 moles of Al₂(SO₄)₃, there are 12 x 9.00 moles = 108.00 moles of O.
Therefore, the number of moles of Al, S, and O atoms in 9.00 moles of Al₂(SO₄)₃ are 18.00, 27.00, 108.00 respectively, separated by commas.
Assign an oxidation number to each element in the reaction.
CO(g) + 2H2(g) → CH3OH(g)
In CO, the oxidation number of C is _, and that of O is _.
In H2, the oxidation number of H is _.
In CH3OH, the oxidation number of C is _, that of O is _, and that of H is _.
In CO, the oxidation number of C is +2 and that of O is -2. In H2, the oxidation number of H is 0. In CH3OH, the oxidation number of C is -2, that of O is -2, and that of H is +1.
Explanation:To assign an oxidation number to each element in the reaction of CO(g) + 2H2(g) → CH3OH(g), we follow the standard rules for oxidation states.
In CO, carbon is more electropositive compared to oxygen, so while oxygen typically has an oxidation number of -2, carbon must balance this with a +2.
In H2, as an elemental form, hydrogen has an oxidation number of 0.
In CH3OH (methanol), the carbon atom is bonded to one oxygen atom and three hydrogen atoms. Following the rules, oxygen has an oxidation number of -2, and each hydrogen atom has an oxidation number of +1. Since carbon is bonded to four hydrogen atoms and one oxygen atom, the total for hydrogen is +3 (3*+1) and for oxygen is -2. To balance the charges, carbon must have an oxidation number of -2 in CH3OH.
The oxidation numbers in CO are +2 for carbon and -2 for oxygen. Hydrogen in H2 has an oxidation number of 0. In CH3OH, carbon's oxidation number is -2, oxygen's is -2, and hydrogen's is +1 for each atom.
Explanation:The oxidation numbers for the elements in the given compounds and molecule can be assigned following certain rules. Let's go through each substance in the reaction CO(g) + 2H2(g) → CH3OH(g).
In CO, carbon (C) is more electropositive than oxygen (O), so it has a lower oxidation number. By default, oxygen has a -2 oxidation state in most compounds (except peroxides and other special cases). Since the molecule is neutral, C must have an oxidation state of +2 to balance oxygen's -2.In H2, hydrogen is in its elemental form, so its oxidation number is 0.In CH3OH (methanol), for oxygen, the oxidation state is typically -2. Each hydrogen has an oxidation state of +1. There are four hydrogens totaling +4 and one oxygen at -2, thus carbon must have an oxidation state of -2 to balance.Therefore, in CO, the oxidation number of C is +2, and that of O is -2. In H2, the oxidation number of H is 0. In CH3OH, the oxidation number of C is -2, that of O is -2, and that of H is +1.
How many significant figures are in the number 2903?
Which of the following elements would be the most different from magnesium (Mg)?
A. Na
B. Ca
C. Be
D. Sr
The answer is A. Na (apex answer)
Pure acetic acid (hc2h3o2) is a liquid and is known as glacial acetic acid. calculate the molarity of a solution prepared by dissolving 50.00 ml of glacial acetic acid at 25 °c in sufficient water to give 500.0 ml of solution. the density of glacial acetic acid at 25 °c is 1.05 g/ml
The molarity of a solution prepared is 1.735 M
From the question,
We are to calculate the molarity of a solution prepared by dissolving 50.00 ml of glacial acetic acid in sufficient water to give 500.0 ml of solution.
First, we will determine the mass of glacial acetic acid present in the 50.00 mL solution
From the question,
The density of glacial acetic acid at 25 °c is 1.05 g/ml
This means, there are 1.05 g of glacial acetic acid in 1 mL of the solution
If, 1mL of the solution contains 1.05 g of glacial acetic acid
Then, 50 mL of the solution will contain 50 × 1.05 g of glacial acetic acid
50 × 1.05 g = 52.5 g
∴ The mass of glacial acetic acid in the 50 mL solution is 52.5 g
Now, we will determine the number of moles of the glacial acetic acid
From the formula
[tex]Number \ of\ moles =\frac{Mass }{Molar\ mass}[/tex]
Mass of glacial acetic acid = 52.5 g
Molar mass of glacial acetic acid = 60.052 g/mol
∴ Number of moles of glacial acetic acid = [tex]\frac{52.5}{60.052 }[/tex]
Number of moles of glacial acetic acid = 0.86748 moles
Now, for the molarity (that is, concentration) of the solution
From the formula
Number of moles = Concentration × Volume
Then,
[tex]Concentration = \frac{Number \ of \ moles}{Volume}[/tex]
Number of moles of glacial acetic acid = 0.86748 moles
Volume of the final solution = 500.0 mL = 0.5 L
∴ Concentration of the solution = [tex]\frac{0.86748}{0.5}[/tex]
Concentration of the solution = 1.735 M
Hence, the molarity of a solution prepared is 1.735 M
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why is the chemical energy of a battery potential energy and not kinetic energy
A battery stores chemical potential energy, which is converted into electric energy when it is connected to a circuit. The energy is then used to power the circuit and is eventually transformed into other forms of energy.
Explanation:A battery stores chemical potential energy. When it is connected in a circuit, a chemical reaction takes place inside the battery which converts chemical potential energy to electric energy. This electric energy powers the electrons to move through the circuit, and the energy is converted into other forms like heat and light by the circuit elements. The battery goes flat when all its chemical potential energy has been converted into other forms of energy.
A circuit contains two devices that are connected in parallel. If the resistance of one of these devices is 12 ohms and the resistance of the other device is 4 ohms, the total resistance of the two devices is
A. 0.333 ohms.
B. 16 ohms.
C. 3 ohms.
D. 0.0625 ohms
Answer: The correct option is 3 ohms.
Solution:
When resistors are connected in parallel the total resistance is given by :
[tex]\frac{1}{R_{total}}=\sum_{i=1}^n\frac{1}{R_1}+\frac{1}{R_2}+...\frac{1}{R_n}[/tex]
[tex]R_1=12 ohm[/tex]
[tex]R_2=4 ohm[/tex]
[tex]\frac{1}{R_{total}}=\frac{1}{12}+\frac{1}{4}=3 ohm[/tex]
The total resistance of the two devices is 3 ohm .
Consider that cuo(s) + h2so4(aq) reacts to give cuso4(aq) and cuo(s) + k2so4does not give a product.
Answer : 1) CuO is a basic salt and therefore reacts with sulfuric acid to give copper (II) sulfate and water.
2) CuO is very stable and therefore does not dissociate in water and therefore cannot react with the sulfate ion in potassium sulfate.
In first case, [tex]CuO_{(s)} + H_{2}SO_{4}_{(aq)} ----> CuSO_{4}_{(s)} + H_{2}O_{(g)}[/tex]
Wherein when CuO was made to react with potassium sulphate it didn't had any product because CuO was not ionised by water in presence of potassium sulphate.
The lattice energy of CuO (s) is very high, so it does not dissolve in water to give its ions. But it is a Bronsted–Lowry base thus it can react with the Bronsted–Lowry acid such as sulphuric acid but not with Bronsted–Lowry base [tex]{{\mathbf{K}}_{\mathbf{2}}}{\mathbf{S}}{{\mathbf{O}}_{\mathbf{4}}}[/tex].
Further Explanation:
The definition of acids and bases can be expressed in many ways based on different theories, which are as follows:
• According to Arrhenius theory, acid is defined as the one which produces hydrogen ions in a solution, while the base is defined as the one which produces hydroxide ions in a solution.
• According to Bronsted–Lowry theory, the acid in the reaction donates a proton while a base is one that accepts a proton.
• According to Lewis theory, the acid in the reaction accepts a pair of electrons while a base donates a pair of electrons.
Lattice energy is termed as the amount of energy released when the ions that exist in gaseous state combine to form compound. The lattice energy of a compound is inversely related to the size of the ions present in it.
The size of [tex]{\text{C}}{{\text{u}}^{2+}}[/tex]and [tex]{{\text{O}}^{2-}}[/tex]is small and therefore lattice energy of CuO(s) is very high. Thus it does not dissolve in water to give its ions. But since it is a Bronsted–Lowry base thus, it can accept the hydrogen ions from sulphuric acid and form [tex]{\text{CuS}}{{\text{O}}_{\text{4}}}[/tex]and [tex]{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}[/tex].
Therefore, the complete reaction is,
[tex]{\text{CuO}}\left(s\right)+{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\left({aq}\right)\to{\text{CuS}}{{\text{O}}_{\text{4}}}\left({aq}\right)+{{\text{H}}_{\text{2}}}{\text{O}}\left(l\right)[/tex]
But [tex]{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\left({aq}\right)[/tex] is not a Bronsted–Lowry acid. Therefore CuO (s) can not react with [tex]{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\left({aq}\right)[/tex]in a way as it does with the sulphuric acid.
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Answer details:
Grade: Senior school
Subject: Chemistry
Chapter: Acids and bases
Keywords: Acids, bases, lattice energy, CuO, k2so4, Bronsted–Lowry theory, proton acceptor, proton donor, h2so4, k2so4.
How much heat is released if 7.15 g cao(s) is added to 152 g of h2o(l)? cao(s) + h2o(l) → ca(oh)2(s) δh°rxn = –64.8 kj/mol?
The heat released when 7.15 g CaO(s) is added to 152 g H2O(l) is -8.28 kJ.
Explanation:According to the given equation and the enthalpy change for the reaction (ΔHrxn), we can calculate the amount of heat released when 7.15 g of CaO(s) is added to 152 g of H2O(l). First, we need to convert the masses of CaO and H2O to moles. The molar mass of CaO is 56.08 g/mol and the molar mass of H2O is 18.02 g/mol. So, the number of moles of CaO is 7.15 g / 56.08 g/mol = 0.1275 mol and the number of moles of H2O is 152 g / 18.02 g/mol = 8.4417 mol.
Next, we can use the stoichiometric coefficients of the balanced equation to determine the moles of products formed. From the equation, we see that 1 mole of CaO reacts to form 1 mole of Ca(OH)2. So, the moles of Ca(OH)2 formed is also 0.1275 mol.
Finally, we can use the enthalpy change value (ΔHrxn) to calculate the heat released. Since the reaction releases -64.8 kJ/mol, we can multiply this value by the number of moles of Ca(OH)2 formed (0.1275 mol) to get the heat released: -64.8 kJ/mol * 0.1275 mol = -8.28 kJ.
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Complete the balanced overall ionic equation for sodium iodide dissolving in water.
Answer: NaI (aq) → Na⁺ (aq) + I⁻(aq)
Explanation:
1) The formula of sodium iodide is NaI
2) Due to the great electronegativities of sodium (Na) and iodine (I), sodium iodide is a ionic compound.
3) Ionic compounds dissociate in water to produce the corresponding ions. In this case, since I has the greatest electronegativity, the ions are Na⁺ and I⁻ .
4) When you write the ionic equation, you have to take into account, the number of ions per unit formula and the phases.
So, in this case, the species are in water solutions, this is aqueous solutions. Then you use the symbol aq to indicate the aqueous phase.
The result is the ionic equation: NaI (aq) → Na⁺ (aq) + I⁻(aq)
The overall balanced ionic equation for sodium iodide dissolved in water is
[tex]\boxed{{{\text{H}}_2}{\text{O}}\left( l \right)\to{{\text{H}}^ + }\left({aq}\right) + {\text{O}}{{\text{H}}^ - }\left({aq}\right)}[/tex]
Further Explanation:
The three types of equations that are used to represent the chemical reaction are as follows:
1. Molecular equation
2. Total ionic equation
3. Net ionic equation
The reactants and products remain in undissociated form in molecular equation. In the case of total ionic equation, all the ions that are dissociated and present in the reaction mixture are represented while in the case of net or overall ionic equation only the useful ions that participate in the reaction are represented.
The steps to write the overall ionic reaction are as follows:
Step 1. Write the molecular equation for the reaction with the phases in the bracket.
In the reaction, NaI reacts with [tex]{{\text{H}}_2}{\text{O}}[/tex] to form NaOH and HI. The balanced molecular equation of the reaction is as follows:
[tex]{\text{NaI}}\left({aq}\right)+{{\text{H}}_2}{\text{O}}\left( l \right)\to{\text{NaOH}}\left( {aq} \right)+{\text{HI}}\left( {aq}\right)[/tex]
Step2. Dissociate all the compounds with the aqueous phase to write the total ionic equation. The compounds with solid and liquid phase remain same. The total ionic equation is as follows:
[tex]{\text{N}}{{\text{a}}^ + }\left({aq} \right) + {{\text{I}}^ - }\left({aq} \right) + {{\text{H}}_2}{\text{O}}\left( l \right) \to{\text{N}}{{\text{a}}^ + }\left({aq} \right)+{\text{O}}{{\text{H}}^ - }\left( {aq}\right)+{{\text{H}}^ + }\left( {aq} \right)+{{\text{I}}^ - }\left( {aq} \right)[/tex]
Step3. The common ions on both the sides of the reaction get cancelled out to get the overall ionic equation.
[tex]\boxed{{\text{N}}{{\text{a}}^ + }\left( {aq} \right)}+\boxed{{{\text{I}}^ - }\left( {aq}\right)} + {{\text{H}}_2}{\text{O}}\left( l \right) \to\boxed{{\text{N}}{{\text{a}}^ + }\left( {aq}\right)}+{\text{O}}{{\text{H}}^ - }\left( {aq} \right)+{{\text{H}}^ + }\left( {aq}\right) + \boxed{{{\text{I}}^ - }\left( {aq}\right)}[/tex]
Therefore, the overall ionic equation obtained is as follows:
[tex]{{\text{H}}_2}{\text{O}}\left( l \right) \to{{\text{H}}^ + }\left({aq} \right)+{\text{O}}{{\text{H}}^ - }\left({aq} \right)[/tex]
Learn more:
1. Balanced chemical equation: https://brainly.com/question/1405182
2. Oxidation and reduction reaction: https://brainly.com/question/2973661
Answer details:
Grade: High School
Subject: Chemistry
Chapter: Chemical reaction and equation
Keywords: overall ionic equation, NaI, NaOH, H2O, HI, H+, I-, aqueous phase, dissociate, molecular equation, reaction, sodium iodide, water.
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