Chemistry Regents Exam January 2026 Paper with Step-by-Step Solutions
Access the official January 2026 Chemistry Regents Exam Paper and prepare effectively with detailed explanations, expert guidance, and step-by-step solutions. This page helps students review the complete exam, understand the structure of the Regents Chemistry test, and improve their exam performance through clear conceptual explanations.
January 2026 Chemistry Regents Exam β Overview
The Chemistry Regents Exam January 2026 evaluates students on major chemistry concepts including atomic structure, chemical bonding, stoichiometry, thermodynamics, acid-base chemistry, and equilibrium. The Regents exam is a key assessment for high school students studying chemistry in New York State.
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- Access the official Chemistry Regents January 2026 exam paper
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Chemistry Regents January 2026 Exam Paper
Below is the official exam paper released by the New York State Education Department. You can scroll, zoom, and review all questions directly on this page.
Chemistry Regents Worked Solutions
Question No. 01 | Gold Foil Experiment
Step-by-Step Solution
-
Step 1 β Remember what the gold foil experiment was.
Rutherford shot alpha particles at a very thin sheet of gold foil. -
Step 2 β Observe what happened.
Most of the alpha particles passed straight through the foil, but a few were deflected sharply, and a very small number bounced back. -
Step 3 β Interpret those results.
If most particles passed through, then most of the atom must be empty space.
If a few particles were strongly deflected, then the positive charge and most of the mass must be concentrated in a very tiny part of the atom. -
Step 4 β State the direct conclusion.
That tiny central part is the nucleus.
So the experiment showed that atoms contain a small, dense nucleus. -
Step 5 β Eliminate the other choices.
Choice (1) is Dalton's old model.
Choice (2) is a fact about electrons, but not the direct conclusion of Rutherford's experiment.
Choice (4) relates to later atomic models involving energy levels.
Connect the scientist to the discovery:
- Dalton β atoms are indivisible spheres
- Thomson β electrons exist
- Rutherford β small, dense, positive nucleus
- Bohr β electrons in energy levels
Question No. 02 | Protons, Neutrons, and Electrons
Step-by-Step Solution
-
Step 1 β Focus on the word atom.
Since the question says atom and does not mention an ion, we assume the atom is neutral. -
Step 2 β Use the neutral atom rule.
In a neutral atom:
number of protons = number of electrons -
Step 3 β Substitute the given value.
The atom has 37 protons, so it must also have:
37 electrons -
Step 4 β Ignore neutrons for electron count.
The 49 neutrons affect the atom's mass, but they do not change the number of electrons in a neutral atom. -
Step 5 β Choose the matching answer.
The correct choice is (2) 37.
Use these quick rules:
- Protons = atomic number
- Neutral atom: protons = electrons
- Mass number: protons + neutrons
- Neutrons change mass, not charge
Question No. 03 | Mass of an Electron
Step-by-Step Solution
-
Step 1 β Compare the three main subatomic particles.
Protons and neutrons each have about 1 atomic mass unit.
Electrons have an extremely tiny mass compared with both. -
Step 2 β Recall the key chemistry fact.
The mass of an electron is about 1/1836 the mass of a proton. -
Step 3 β Apply that fact to the choices.
Since an electron is far lighter than a proton, it definitely does not have the same mass as a proton or neutron.
It also does not have more mass than a neutron. -
Step 4 β Select the best description.
The only correct statement is:
An electron has much less mass than a proton.
A fast memory trick:
- Proton β 1 amu
- Neutron β 1 amu
- Electron β almost 0 amu
Question No. 04 | Wave-Mechanical Model
Step-by-Step Solution
-
Step 1 β Identify the atomic model mentioned.
The question specifically says the wave-mechanical model, which is the modern model of the atom. -
Step 2 β Recall what this model says.
In the modern model, electrons are not traveling in fixed circular paths like in the Bohr model.
Instead, they are found in regions of highest probability. -
Step 3 β Name those regions.
Those regions are called orbitals. -
Step 4 β Eliminate the other answers.
Spectra are patterns of light.
Quanta refers to packets of energy.
Nuclei are the centers of atoms, not the regions where electrons are found.
Watch for these model keywords:
- Bohr model β fixed energy levels
- Wave-mechanical model β orbitals
- Ground state β lowest energy arrangement
Question No. 05 | Distinguishing Potassium and Rubidium
Step-by-Step Solution
-
Step 1 β Notice the phrase βat STP.β
STP means standard temperature and pressure. -
Step 2 β Eliminate temperature and pressure.
If both samples are at STP, then they are already at the same temperature and same pressure.
So choices (3) and (4) cannot be used to tell them apart. -
Step 3 β Compare the phase.
Potassium and rubidium are both metals, and at STP both are solids.
So phase is also the same, which eliminates choice (2). -
Step 4 β Find a physical property that is different.
Different elements can have different densities.
Since density is a characteristic physical property, it can distinguish potassium from rubidium. -
Step 5 β Select the correct choice.
The only property listed that can differentiate the two samples is density.
To identify a substance, look for a characteristic property:
- density
- melting point
- boiling point
- solubility
Question No. 06 | Chemical Property of Chlorine
Step-by-Step Solution
-
Step 1 β Know the difference between physical and chemical properties.
A physical property can be observed without changing the substance.
A chemical property describes how a substance reacts to form new substances. -
Step 2 β Examine each choice.
Atomic radius is a physical property.
Melting point is a physical property.
Color and phase at STP are physical properties. -
Step 3 β Identify the reactive statement.
The statement βchlorine reacts with sodium to form a saltβ describes chlorine undergoing a chemical change and making a new compound. -
Step 4 β Select the correct answer.
Because it describes reactivity, the correct choice is:
Chlorine reacts with sodium to form a salt.
Quick way to tell the difference:
- Physical property β observed without changing identity
- Chemical property β describes ability to react and form something new
Question No. 07 | Allotropes of Oxygen
Step-by-Step Solution
-
Step 1 β Notice that both substances are made of oxygen only.
O2 and O3 are both forms of the same element, oxygen. -
Step 2 β Identify the chemistry idea being tested.
Different forms of the same element are called allotropes.
Oxygen gas and ozone are allotropes of oxygen. -
Step 3 β Compare their formulas.
O2 has 2 oxygen atoms in each molecule.
O3 has 3 oxygen atoms in each molecule.
So they clearly have different molecular structures. -
Step 4 β Think about their properties.
Since their structures are different, their properties are also different.
For example, ozone is much more reactive than ordinary oxygen gas. -
Step 5 β Select the correct choice.
The correct answer is:
different molecular structures and different properties
When you see two forms of the same element, think allotropes.
- Same element
- Different molecular arrangement or structure
- Different physical and chemical properties
Question No. 08 | Definition of a Compound
Step-by-Step Solution
-
Step 1 β Focus on the phrase βchemically combined.β
When elements are chemically combined, they form a compound, not a mixture. -
Step 2 β Recall the difference between a mixture and a compound.
A mixture is a physical combination, so substances are not chemically bonded.
A compound is a chemical combination, so the elements are bonded together. -
Step 3 β Remember the law of definite proportions.
In a compound, the elements are always present in a fixed ratio.
Example: water is always H2O, not sometimes H3O or HO. -
Step 4 β Eliminate wrong choices.
Choices (1) and (2) are mixtures, so they are wrong.
Choice (4) says variable proportions, but compounds have fixed proportions. -
Step 5 β Choose the correct answer.
The correct phrase is:
a compound with a fixed proportion of elements
Use this quick comparison:
- Mixture β physically combined, variable composition
- Compound β chemically combined, fixed composition
Question No. 09 | Empirical Formula
Step-by-Step Solution
-
Step 1 β Recall the definition of empirical formula.
An empirical formula shows the simplest whole-number ratio of atoms in a compound. -
Step 2 β Check each formula to see whether it can be simplified.
We look at the subscripts and ask: can they all be divided by the same whole number? -
Step 3 β Test each choice.
N2O4 β divide by 2 β NO2 so it is not empirical.
N2O5 cannot be reduced, so it is empirical.
N2F2 β divide by 2 β NF so it is not empirical.
N2H4 β divide by 2 β NH2 so it is not empirical. -
Step 4 β Identify the only simplest formula.
The only formula already in simplest whole-number ratio is:
N2O5
To test for an empirical formula:
- Look at all subscripts
- Try dividing them by the same whole number
- If you can reduce them, it is not empirical
- If you cannot reduce them, it is empirical
Question No. 10 | Categories of Compounds
Step-by-Step Solution
-
Step 1 β Recall the main Regents categories of compounds.
Compounds are commonly classified as ionic or molecular (also called covalent). -
Step 2 β Check choice (1).
βCovalentβ and βmolecularβ do not represent two different categories here because they refer to the same general type of compound. -
Step 3 β Eliminate the non-category terms.
βElectricalβ is not a category of compounds, so choices (2) and (4) are wrong. -
Step 4 β Find the two real categories.
The two different categories are:
ionic and molecular
Remember this shortcut:
- Ionic compound β metal + nonmetal
- Molecular compound β nonmetal + nonmetal
- Covalent usually means molecular bonding
Question No. 11 | Multiple Covalent Bond
Step-by-Step Solution
-
Step 1 β Understand the phrase βmultiple covalent bond.β
A multiple covalent bond means either a double bond or a triple bond. -
Step 2 β Recall the common diatomic elements.
Bromine exists as Br2, hydrogen as H2, iodine as I2, and oxygen as O2. -
Step 3 β Compare their bonds.
Br2 has a single bond.
H2 has a single bond.
I2 has a single bond.
O2 has a double bond. -
Step 4 β Identify the molecule with a multiple bond.
Since O2 contains a double bond, oxygen is the correct answer.
Common elemental molecules to remember:
- H2, N2, O2, F2, Cl2, Br2, I2
- O2 has a double bond
- N2 has a triple bond
- H2, F2, Cl2, Br2, I2 have single bonds
Question No. 12 | Asymmetrical Molecule
Step-by-Step Solution
- Step 1 β Understand what asymmetrical means.
An asymmetrical molecule does not have the same arrangement all around. Its shape or atom placement is uneven. - Step 2 β Check each choice.
HCl has two different atoms joined together, so the molecule is not balanced equally on both sides.
CO2 is linear and symmetrical: O=C=O.
Cl2 has two identical atoms, so it is symmetrical.
CH4 is tetrahedral and symmetrical. - Step 3 β Choose the asymmetrical one.
The only asymmetrical molecule listed is HCl.
Symmetrical molecules usually have balanced shapes like:
- CO2 β linear and balanced
- CH4 β tetrahedral and balanced
- HCl β two different atoms, so asymmetrical
Question No. 13 | Gaining an Electron
Step-by-Step Solution
- Step 1 β Determine the charge change.
If an atom gains one electron, it now has more electrons than protons.
That makes it a negative ion, also called an anion. - Step 2 β Determine the radius change.
When an atom gains an electron, electron-electron repulsion increases.
This causes the electron cloud to spread out more. - Step 3 β State the result.
A negative ion forms, and its radius becomes larger than the original atom.
Quick ion-size rule:
- Lose electrons β positive ion β smaller radius
- Gain electrons β negative ion β larger radius
Question No. 14 | Bond Formation and Energy
Step-by-Step Solution
- Step 1 β Look at what the reaction shows.
Two separate nitrogen atoms combine to make N2. - Step 2 β Decide whether a bond is being broken or formed.
Since two atoms are joining together, a bond is being formed. - Step 3 β Recall the energy rule.
Bond breaking absorbs energy.
Bond forming releases energy. - Step 4 β Apply the rule.
Since a bond is formed in N2, energy is released.
Memorize this pair:
- Breaking bonds β absorbs energy
- Forming bonds β releases energy
Question No. 15 | Ionic and Covalent Bonds in One Compound
Step-by-Step Solution
- Step 1 β Know when both bond types appear.
A compound has both ionic and covalent bonds when it contains:
β’ a metal with a polyatomic ion, or
β’ ionic attraction between ions and covalent bonds inside the polyatomic ion. - Step 2 β Test each choice.
CaCl2 is ionic only.
KClO2 has K+ and ClO2-.
The bond between K+ and ClO2- is ionic, but inside ClO2- the atoms are covalently bonded.
Li2S is ionic only.
MnO2 is treated as an ionic compound in Regents-level classification. - Step 3 β Identify the correct compound.
The one containing both types of bonding is KClO2.
If you see a metal + polyatomic ion, think:
- Ionic bond between the metal and the ion
- Covalent bonds inside the polyatomic ion
Question No. 16 | Cannot Be Broken Down Chemically
Step-by-Step Solution
- Step 1 β Recall what can and cannot be broken down.
Elements cannot be broken down by chemical means.
Compounds can be broken down chemically into simpler substances. - Step 2 β Identify the element.
Fe is the symbol for iron, which is an element. - Step 3 β Check the others.
Fe2O3, FeSO4, and Fe2(SO4)3 are all compounds because they contain more than one element. - Step 4 β Choose the correct answer.
The only substance that cannot be chemically decomposed is Fe.
Quick rule:
- Element β cannot be broken down chemically
- Compound β can be broken down chemically
Question No. 17 | Kinetic Molecular Theory
Step-by-Step Solution
- Step 1 β Recall the ideal gas assumptions.
Gas particles are far apart, have negligible volume, and move continuously. - Step 2 β Focus on particle motion.
According to kinetic molecular theory, gas particles move in random, constant, straight-line motion until they collide. - Step 3 β Eliminate the wrong choices.
Collisions do transfer energy, so choice (1) is wrong.
Ideal gases do not have strong attractive forces, so choice (2) is wrong.
Particle volumes are tiny compared with the empty space between them, so choice (4) is wrong.
Kinetic molecular theory keywords:
- random motion
- constant motion
- straight-line motion
- elastic collisions
- very weak attractions
Question No. 18 | Collision Theory
Step-by-Step Solution
- Step 1 β Recall collision theory.
Not every collision causes a reaction. - Step 2 β Know the two requirements.
For a collision to be successful, particles must have:
β’ the correct orientation
β’ enough energy to overcome activation energy - Step 3 β Choose the correct term.
The missing word is energy.
Think: successful collision = right direction + enough energy.
Question No. 19 | Same Number of Atoms
Step-by-Step Solution
- Step 1 β Use Avogadroβs law.
At the same temperature and pressure, equal volumes of gases contain equal numbers of particles. - Step 2 β Notice these are noble gases.
Ne, Ar, He, Kr, and Xe are all monatomic gases.
That means each particle is one atom. - Step 3 β Match the volume.
Since 4.0 L of Ne has a certain number of atoms, another monatomic gas will have the same number of atoms only if it has the same volume at STP. - Step 4 β Choose the matching sample.
The only choice with the same volume is 4.0 L of Xe(g).
For noble gases:
- particles = atoms
- same T and P + same volume = same number of atoms
Question No. 20 | Exothermic Dissolving
Step-by-Step Solution
- Step 1 β Understand exothermic dissolving.
An exothermic process releases heat, so the heat of solution is negative. - Step 2 β Recall common Table I values.
NaOH(s) dissolves exothermically.
KNO3, NH4Cl, and NH4NO3 dissolve endothermically. - Step 3 β Choose the only exothermic compound.
The correct answer is NaOH(s).
Common endothermic dissolving substances to remember:
- KNO3
- NH4Cl
- NH4NO3
Question No. 21 | Heat of Reaction Formula
Step-by-Step Solution
- Step 1 β Recall the heat of reaction formula.
In Regents Chemistry:
Heat of reaction = PE of products β PE of reactants - Step 2 β Match the formula to the choices.
The correct expression is choice (4).
Always write:
- ΞH = PE(products) β PE(reactants)
Question No. 22 | How a Catalyst Works
Step-by-Step Solution
- Step 1 β Recall the job of a catalyst.
A catalyst speeds up a reaction without being used up. - Step 2 β Know how it does this.
A catalyst provides an alternate reaction pathway that has a lower activation energy. - Step 3 β Eliminate the wrong choices.
A catalyst does not provide energy itself, does not raise reactant PE, and does not increase activation energy.
Catalyst = alternate path + lower activation energy.
Question No. 23 | Greatest Entropy
Step-by-Step Solution
- Step 1 β Recall what entropy means.
Entropy is a measure of disorder or randomness. - Step 2 β Compare phases.
Entropy increases in this order:
solid < liquid < gas - Step 3 β Compare temperatures.
For the same substance, a higher temperature usually means higher entropy. - Step 4 β Pick the sample with the most disorder.
The gas sample at the higher temperature, Xe(g) at 165 K, has the greatest entropy.
Entropy trend:
- solid β lowest entropy
- liquid β medium entropy
- gas β highest entropy
- higher temperature β greater entropy
Question No. 24 | Natural Changes in Systems
Step-by-Step Solution
- Step 1 β Recall the natural tendency of systems.
In nature, systems tend to move toward lower energy and greater disorder. - Step 2 β Translate βgreater disorder.β
Greater disorder means higher entropy. - Step 3 β Choose the statement that matches.
The correct combination is:
lower energy and greater disorder
Nature tends to favor:
- lower energy
- greater disorder
- higher entropy
Question No. 25 | Homologous Series Formula
Step-by-Step Solution
-
Step 1 β Use the given general formula.
The homologous series is represented by:
CnH2n -
Step 2 β Test each choice.
For each compound, compare the number of hydrogens to 2n. -
Step 3 β Check the choices one by one.
C2H2 β if n = 2, then H should be 4, not 2 β
C3H6 β if n = 3, then H should be 6 β
C4H10 β if n = 4, then H should be 8, not 10 β
C5H8 β if n = 5, then H should be 10, not 8 β -
Step 4 β Choose the match.
The only compound that fits CnH2n is C3H6.
Common organic general formulas:
- Alkanes: CnH2n+2
- Alkenes / cycloalkanes: CnH2n
- Alkynes: CnH2nβ2
Question No. 26 | Reaction That Produces an Alcohol
Step-by-Step Solution
-
Step 1 β Recall how alcohol can be made in chemistry.
In Regents Chemistry, fermentation is the process in which sugars are converted into ethanol and carbon dioxide. -
Step 2 β Eliminate the other choices.
Combustion is burning.
Deposition is a phase change from gas to solid.
Sublimation is a phase change from solid to gas.
None of these produce alcohol. -
Step 3 β Identify the correct term.
The reaction that produces an alcohol is fermentation.
Remember this organic chemistry fact:
- Fermentation β produces ethanol
- Ethanol is an alcohol
Question No. 27 | Identifying an Electrolyte
Step-by-Step Solution
-
Step 1 β Recall the definition of an electrolyte.
An electrolyte is a substance that forms ions in water, allowing the solution to conduct electricity. -
Step 2 β Focus on the key idea.
The important point is not whether the solid substance conducts electricity by itself.
The question is whether the aqueous solution conducts electricity. -
Step 3 β Identify the correct statement.
The correct description is:
When the substance dissolves in water, the solution can conduct an electric current.
Think of an electrolyte this way:
- Dissolves in water
- Forms ions
- Solution conducts electricity
Question No. 28 | Arrhenius Acid
Step-by-Step Solution
-
Step 1 β Recall the Arrhenius acid definition.
An Arrhenius acid is a substance that increases the concentration of H+ ions in water. -
Step 2 β Check each choice.
NH3 is a base.
HNO3 is nitric acid, so it is an Arrhenius acid.
NaCl is a salt.
NaOH is a base. -
Step 3 β Select the acid.
The correct answer is HNO3.
Quick acid-base check:
- Arrhenius acid β produces H+ in water
- Arrhenius base β produces OHβ in water
Question No. 29 | Splitting a Heavy Nucleus
Step-by-Step Solution
-
Step 1 β Focus on the word βsplits.β
The question says a heavy nucleus splits into two lighter nuclei. -
Step 2 β Recall the nuclear definitions.
Fission means splitting a heavy nucleus.
Fusion means combining light nuclei. -
Step 3 β Eliminate non-nuclear processes.
Decomposition is a chemical reaction type.
Deposition is a phase change.
Neither fits the question. -
Step 4 β Choose the correct process.
The heavy nucleus splits during fission.
Nuclear memory trick:
- Fission β one big nucleus splits
- Fusion β small nuclei join together
Question No. 30 | Fusion in the Sun
Step-by-Step Solution
-
Step 1 β Recall what happens in nuclear fusion.
In the Sun, light nuclei join together to form heavier nuclei. -
Step 2 β Understand where the released energy comes from.
A small amount of mass is lost during fusion, and that mass is converted into energy. -
Step 3 β Use Einsteinβs idea.
This is described by the idea behind E = mcΒ², where mass can be converted into energy. -
Step 4 β Choose the correct answer.
The Sunβs fusion energy comes from converting matter to energy.
For nuclear reactions, remember:
- A little mass can turn into a lot of energy
- This is the idea behind E = mcΒ²
- Fusion in the Sun is a classic example of matter β energy
Question No. 31 | Lithium in an Excited State
Step-by-Step Solution
-
Step 1 β Find the number of electrons in lithium.
Lithium has atomic number 3, so a neutral lithium atom has 3 electrons. -
Step 2 β Recall the ground-state configuration.
In the ground state, lithium is:
2-1
That means 2 electrons in the first shell and 1 electron in the second shell. -
Step 3 β Understand what βexcited stateβ means.
An excited state happens when an electron moves to a higher energy level than usual. -
Step 4 β Check the choices.
Choice (1) 2-1 is the ground state, not excited.
Choice (2) 2-2 has 4 electrons, so it cannot be lithium.
Choice (3) 1-2-1 also has 4 electrons, so it cannot be lithium.
Choice (4) 1-1-1 has 3 electrons and shows one electron promoted to a higher shell. -
Step 5 β Choose the excited-state configuration.
The correct answer is 1-1-1.
For excited-state electron configurations:
- Keep the same total number of electrons
- Look for an electron moved to a higher shell
- Do not confuse ground state with excited state
Question No. 32 | Lewis Electron-Dot Diagram for Selenium
Step-by-Step Solution
-
Step 1 β Find selenium on the periodic table.
Selenium, Se, is in Group 16. -
Step 2 β Use the group number to find valence electrons.
Group 16 elements have 6 valence electrons. -
Step 3 β Recall what a Lewis dot diagram shows.
A Lewis electron-dot diagram shows only the valence electrons around the symbol. -
Step 4 β Look for the diagram with 6 dots.
Among the four diagrams, the correct one is the one with exactly 6 dots around Se. -
Step 5 β Select the correct choice.
The diagram that shows 6 valence electrons is choice (3).
For main-group elements:
- Group 1 β 1 valence electron
- Group 2 β 2 valence electrons
- Group 13β18 β 3 to 8 valence electrons
- Group 16 elements like Se always have 6 valence electrons
Question No. 33 | Oxide Formula XβO
Step-by-Step Solution
-
Step 1 β Recall the charge of oxide.
Oxide is O2β. -
Step 2 β Use the formula X2O.
Since there are 2 X atoms balancing one O2β, the two X ions together must have a total charge of +2. -
Step 3 β Find the charge of one X ion.
If 2 X ions add up to +2, then each X must be:
X+ -
Step 4 β Choose the element that forms a +1 ion.
Cs is an alkali metal in Group 1, so it forms Cs+.
When working with formulas:
- Oxide always has charge β2
- Use the subscripts to figure out the ion charges
- Group 1 metals usually form +1 ions
Question No. 34 | Naming PdO
Step-by-Step Solution
-
Step 1 β Identify the negative ion.
Oxygen in oxide form is O2β. -
Step 2 β Use the formula PdO.
Since there is one Pd and one O, palladium must balance the β2 charge of oxygen. -
Step 3 β Determine the palladium charge.
Palladium must be Pd2+. -
Step 4 β Name the compound.
A transition metal needs a Roman numeral to show its charge.
So PdO is named palladium (II) oxide.
For transition metals:
- Find the charge first
- Use a Roman numeral in the name
- Example: FeO = iron(II) oxide
Question No. 35 | Mass Conservation in a Reaction
Step-by-Step Solution
-
Step 1 β Use the law of conservation of mass.
Total mass of reactants = total mass of products. -
Step 2 β Add the masses of the reactants.
12 g CH4 + 48 g O2 = 60 g total -
Step 3 β Use the mass of one product.
The products are CO2 and H2O.
We are told that CO2 has a mass of 33 g. -
Step 4 β Solve for the water mass.
60 g = 33 g + mass of H2O
mass of H2O = 60 β 33 = 27 g
For mass problems:
- Total reactant mass = total product mass
- If one product mass is missing, subtract from the total
Question No. 36 | Gram-Formula Mass of PβOβ
Step-by-Step Solution
-
Step 1 β Find the atomic masses.
Phosphorus, P = 31
Oxygen, O = 16 -
Step 2 β Multiply by the subscripts.
For P2O5:
2(31) = 62
5(16) = 80 -
Step 3 β Add the totals.
62 + 80 = 142 -
Step 4 β State the gram-formula mass.
142 g/mol
To find gram-formula mass:
- Multiply each elementβs atomic mass by its subscript
- Add all the results together
Question No. 37 | Same Electron Configuration as Neon
Step-by-Step Solution
-
Step 1 β Find neonβs electron count.
Neon has atomic number 10, so a neutral neon atom has 10 electrons. -
Step 2 β Check each ion pair.
We need both ions to end up with 10 electrons. -
Step 3 β Test the correct pair.
Magnesium forms Mg2+.
Magnesium has 12 electrons as an atom, so Mg2+ has 10 electrons.
Oxygen forms O2β.
Oxygen has 8 electrons as an atom, so O2β has 10 electrons. -
Step 4 β Choose the pair that matches neon.
Both magnesium ion and oxide ion have the same electron configuration as neon.
Noble gas configurations are very common in ions:
- Metals lose electrons to match a noble gas
- Nonmetals gain electrons to match a noble gas
- Neon configuration = 10 electrons
Question No. 38 | Particle Model of Fluorine at STP
Step-by-Step Solution
-
Step 1 β Identify fluorineβs form at STP.
Fluorine exists as F2(g). -
Step 2 β Understand what that means.
Diatomic means each particle has 2 atoms.
Since both atoms are fluorine, the two atoms in each molecule must be the same kind. -
Step 3 β Remember the state at STP.
Fluorine is a gas at STP, so the particles should be spread far apart. -
Step 4 β Choose the diagram.
The correct diagram must show:
β’ pairs of identical atoms
β’ spread out like a gas
That matches choice (1).
For particle diagrams:
- Gas β particles far apart
- Diatomic element β atoms in pairs
- Same element β same type/color of atoms
Question No. 39 | Most Soluble Compound in Water
Step-by-Step Solution
-
Step 1 β Use Table F: Solubility Guidelines.
This question is testing whether each ionic compound is soluble or insoluble in water. -
Step 2 β Recall one of the most important Table F rules.
All nitrates (NO3β) are soluble. -
Step 3 β Check the nitrate compound.
AgNO3 contains the nitrate ion, so it is soluble in water. -
Step 4 β Eliminate the others.
Chromates, carbonates, and phosphates are generally insoluble unless paired with Group 1 ions or ammonium.
So:
β’ Ag2CrO4 β not very soluble
β’ CuCO3 β insoluble
β’ Cu3(PO4)2 β insoluble -
Step 5 β Choose the most soluble compound.
The correct answer is AgNO3.
Memorize these from Table F:
- All nitrates are soluble
- Carbonates are usually insoluble
- Phosphates are usually insoluble
- Chromates are usually insoluble except with Group 1 or NH4+
Question No. 40 | Lowest Freezing Point
Step-by-Step Solution
-
Step 1 β Recall the freezing point depression rule.
The more dissolved particles in a solution, the lower the freezing point. -
Step 2 β Compare the concentration.
A 2.0 M solution has more dissolved particles than a 1.0 M solution of the same type. -
Step 3 β Compare NaCl and glucose.
NaCl is an electrolyte, so it dissociates into 2 ions: Na+ and Clβ.
C6H12O6 is a nonelectrolyte, so it stays as whole molecules. -
Step 4 β Determine which solution has the most particles.
2.0 M NaCl gives the greatest total number of dissolved particles. -
Step 5 β Connect this to freezing point.
Greatest number of particles β greatest freezing point depression β lowest freezing point.
For colligative properties:
- More particles = lower freezing point
- Electrolytes make more particles than nonelectrolytes
- Higher molarity = more particles
Question No. 41 | Heat Needed to Vaporize Water
Step-by-Step Solution
-
Step 1 β Identify the phase change.
The sample is being vaporized, which means liquid water changes to gas. -
Step 2 β Use the heat of vaporization of water.
At 373 K and 1 atm, waterβs heat of vaporization is:
2260 J/g -
Step 3 β Use the formula.
q = mHv -
Step 4 β Substitute the values.
q = (12.5 g)(2260 J/g)
q = 28,250 J -
Step 5 β Match the closest answer choice.
28,250 J β 28,300 J
For phase-change energy:
- Melting/freezing: q = mHf
- Boiling/condensing: q = mHv
- No temperature change during the phase change
Question No. 42 | Fastest Reaction Rate
Step-by-Step Solution
-
Step 1 β Recall the factors that increase reaction rate.
A reaction goes faster when:
β’ temperature is higher
β’ surface area is greater -
Step 2 β Compare the temperatures.
303 K is higher than 273 K, so reactions at 303 K are faster. -
Step 3 β Compare powdered iron and strip iron.
Powdered iron has much more surface area than a strip, so it reacts faster with the acid. -
Step 4 β Combine both best conditions.
The fastest rate happens with:
powdered iron at 303 K
Reaction rate increases with:
- higher temperature
- greater surface area
- higher concentration
- presence of a catalyst
Question No. 43 | System in a Sealed Flask of Water
Step-by-Step Solution
-
Step 1 β Notice the flask is sealed.
Because the flask is sealed, water vapor cannot escape. -
Step 2 β Think about what happens to liquid water in a sealed container.
Some liquid water evaporates into vapor, and some vapor condenses back into liquid. -
Step 3 β Recognize the type of equilibrium.
When evaporation and condensation happen at the same rate, the system reaches physical equilibrium. -
Step 4 β Eliminate the other choices.
It is not chemical equilibrium because no new substances are forming.
It is not βcondensation onlyβ or βevaporation onlyβ because both processes happen.
If a liquid is in a sealed container:
- evaporation happens
- condensation happens
- eventually the two rates become equal
- that is physical equilibrium
Question No. 44 | Time When Equilibrium Is Reached
Step-by-Step Solution
-
Step 1 β Recall how equilibrium looks on a graph.
A system is at equilibrium when the concentrations stop changing and become constant. -
Step 2 β Look at the graph carefully.
The N2O4 concentration decreases and then becomes flat.
The NO2 concentration increases and then becomes flat. -
Step 3 β Identify a time on the flat part of the graph.
The graph is clearly flat by about 70 s. -
Step 4 β Choose a time when equilibrium is definitely established.
The best answer is 70 s.
On a concentration-versus-time graph:
- equilibrium does not mean equal concentrations
- equilibrium means constant concentrations
- look for the flat, horizontal part of the graph
Question No. 45 | Temperature Change and Equilibrium Shift
Step-by-Step Solution
-
Step 1 β Look at where the energy is written.
In this equation, energy is on the reactant side. -
Step 2 β Identify the forward reaction type.
If energy is a reactant, the forward reaction is endothermic. -
Step 3 β Apply Le ChΓ’telierβs principle.
Increasing temperature adds heat to the system.
The system shifts in the direction that uses up the added heat. -
Step 4 β Determine the shift.
Since the forward reaction uses heat, the equilibrium shifts to the right. -
Step 5 β Predict what happens to HI.
HI is a product, so if the system shifts right, the concentration of HI increases.
For equilibrium and heat:
- If heat is on the left, forward reaction is endothermic
- If temperature increases, equilibrium shifts to use the added heat
- Shift right β product concentration increases
Question No. 46 | Naming an Organic Compound
Step-by-Step Solution
-
Step 1 β Find the longest continuous carbon chain.
When we trace the longest unbroken path of carbon atoms in the structure, the longest chain contains 5 carbons.
So the parent name is pentane, not hexane. -
Step 2 β Identify the branches.
The structure has two methyl groups attached to the main chain. -
Step 3 β Number the chain correctly.
We number from the end that gives the lowest possible set of numbers for the branches.
This gives branch positions 2 and 3, not 3 and 4. -
Step 4 β Write the name.
Two methyl groups on a pentane chain at carbons 2 and 3 gives:
2,3-dimethylpentane -
Step 5 β Eliminate the other choices.
Choices (1) and (2) use hexane as the parent chain, but the longest chain here is only 5 carbons. β
Choice (3) gives branch numbers 3 and 4, but naming rules require the lowest set of locants. β
For organic naming:
- Always choose the longest continuous carbon chain
- Then identify the branches
- Number the chain to give the branches the lowest possible numbers
- If there are 2 identical branches, use the prefix di-
Question No. 47 | Oxidation State of Chlorine
Step-by-Step Solution
-
Step 1 β Assign the known oxidation states.
Potassium, K, is a Group 1 metal, so:
K = +1
Oxygen is usually:
O = β2 -
Step 2 β Use the formula KClO3.
There are 3 oxygens, so total oxygen charge is:
3(β2) = β6 -
Step 3 β Let chlorine be x.
Since the compound is neutral, the total charge must equal 0:
(+1) + x + (β6) = 0 -
Step 4 β Solve for x.
1 + x β 6 = 0
x β 5 = 0
x = +5
For oxidation-number questions:
- Group 1 metals are always +1
- Oxygen is usually β2
- The total oxidation numbers in a neutral compound must add up to 0
Question No. 48 | Acid-Base Theory and the Reverse Reaction
Step-by-Step Solution
-
Step 1 β Look at the reverse reaction.
The reverse reaction is:
NH4+ + OHβ β NH3 + H2O -
Step 2 β Recall the BrΓΈnsted-Lowry acid definition.
A BrΓΈnsted-Lowry acid is a substance that donates an H+ ion. -
Step 3 β Identify which reactant donates H+.
In the reverse reaction, NH4+ gives an H+ to OHβ, forming H2O. -
Step 4 β Choose the acid in the reverse reaction.
The H+ donor is NH4+.
For BrΓΈnsted-Lowry acids and bases:
- Acid = H+ donor
- Base = H+ acceptor
- In a reversible reaction, the roles can switch in the reverse direction
Question No. 49 | Change in Hydronium Ion Concentration
Step-by-Step Solution
-
Step 1 β Recall the pH scale rule.
Every increase of 1 pH unit means the hydronium ion concentration becomes 10 times smaller. -
Step 2 β Find the pH change.
The pH changes from 4.0 to 7.0, which is an increase of:
3 pH units -
Step 3 β Apply the factor of 10 for each unit.
A 3-unit increase means the hydronium concentration changes by:
10 Γ 10 Γ 10 = 1000 -
Step 4 β Decide whether it increases or decreases.
Since the pH goes up, the solution becomes less acidic, so the hydronium concentration must decrease.
On the pH scale:
- Each pH change of 1 = factor of 10
- Higher pH = lower [H3O+]
- Lower pH = higher [H3O+]
Question No. 50 | Identifying a Radioactive Emission
Step-by-Step Solution
-
Step 1 β Observe the path of the emission.
In the diagram, the radiation travels straight through the electric field without bending upward or downward. -
Step 2 β Use the electric field idea.
Charged particles are deflected by electric fields:
β’ positive particles bend toward the negative plate
β’ negative particles bend toward the positive plate -
Step 3 β Identify what is not deflected.
If the path does not bend at all, the emission must be neutral. -
Step 4 β Match the neutral radiation.
Gamma radiation has no charge, so it is not deflected in an electric field. -
Step 5 β Eliminate the other choices.
Alpha particles are positive. β
Beta particles are negative. β
Positrons are positive. β
Only gamma radiation is neutral. β
Quick comparison of radioactive emissions:
- Alpha β positive, deflects one way
- Beta β negative, deflects the opposite way
- Gamma β neutral, goes straight
- Positron β positive, deflects like alpha
Question No. 51 | Percent Composition by Mass
Step-by-Step Solution
-
Step 1 β Write the percent composition formula.
% by mass = (mass of part / total mass) Γ 100 -
Step 2 β Identify the mass of Cu and the total mass.
In CuSO4, the copper contributes about 63.5 g (or about 64 g), and the total gram-formula mass is 160. g/mol. -
Step 3 β Substitute into the formula.
% Cu = (63.5 / 160.) Γ 100 -
Step 4 β Calculate.
% Cu = 39.7%
This is usually rounded to about 40.0%.
For percent composition:
- Use part Γ· whole Γ 100
- The βpartβ is the mass of the element you are asked about
- The βwholeβ is the total gram-formula mass of the compound
Question No. 52 | Why K-39 and K-40 Are Isotopes
Step-by-Step Solution
-
Step 1 β Recall what isotopes are.
Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons. -
Step 2 β Use potassiumβs atomic number.
Potassium always has atomic number 19, so both K-39 and K-40 have 19 protons. -
Step 3 β Compare the neutrons.
K-39 has 39 β 19 = 20 neutrons.
K-40 has 40 β 19 = 21 neutrons.
Isotopes always:
- belong to the same element
- have the same atomic number
- have different mass numbers because the neutrons differ
Question No. 53 | Electron Energy in Different Shells
Step-by-Step Solution
-
Step 1 β Recall the energy-level pattern.
Electrons in shells farther from the nucleus have higher energy. -
Step 2 β Compare shell 1 and shell 4.
The first shell is much closer to the nucleus than the fourth shell. -
Step 3 β State the comparison.
Therefore, an electron in the first shell has less energy than an electron in the fourth shell.
As you move to shells farther from the nucleus:
- electron energy increases
- electrons are held less tightly
Question No. 54 | Numerical Setup for Atomic Mass of Potassium
Step-by-Step Solution
-
Step 1 β Use a weighted average.
Atomic mass is found by multiplying each isotopeβs mass by its decimal abundance and then adding. -
Step 2 β Convert the percentages to decimals.
93.26% β 0.9326
0.01% β 0.0001
6.73% β 0.0673 -
Step 3 β Write the setup.
(38.964)(0.9326) + (39.964)(0.0001) + (40.962)(0.0673)
Weighted average formula:
- mass Γ decimal abundance
- then add all isotope contributions
- do not use the percent directly unless you divide by 100 at the end
Question No. 55 | Why the Equation Is Balanced
Step-by-Step Solution
-
Step 1 β Count nitrogen atoms on the reactant side.
N2 has 2 nitrogen atoms. -
Step 2 β Count hydrogen atoms on the reactant side.
3H2 has 3 Γ 2 = 6 hydrogen atoms. -
Step 3 β Count atoms on the product side.
2NH3 has:
β’ 2 nitrogen atoms
β’ 2 Γ 3 = 6 hydrogen atoms -
Step 4 β Compare both sides.
Both sides have 2 nitrogen atoms and 6 hydrogen atoms.
A balanced equation must have:
- the same number of each kind of atom on both sides
- coefficients changed, not subscripts
Question No. 56 | Evidence the Reaction Is Exothermic
Step-by-Step Solution
-
Step 1 β Look at where the energy appears.
The 91.8 kJ is written on the product side of the equation. -
Step 2 β Interpret that placement.
When energy is shown as a product, the reaction is releasing energy. -
Step 3 β Connect this to the term exothermic.
A reaction that releases heat is exothermic.
Heat in equations:
- energy on the right β exothermic
- energy on the left β endothermic
Question No. 57 | Moles of Hydrogen Required
Step-by-Step Solution
-
Step 1 β Use the mole ratio from the balanced equation.
3 mol H2 : 2 mol NH3 -
Step 2 β Set up the proportion.
x / 10.0 = 3 / 2 -
Step 3 β Solve for x.
x = 10.0 Γ (3/2)
x = 15.0 mol H2
For mole-mole conversions:
- always start with the balanced equation
- use the coefficients as the mole ratio
Question No. 58 | Significant Figures in 300. K
Step-by-Step Solution
-
Step 1 β Look carefully at the number.
The temperature is written as 300. with a decimal point. -
Step 2 β Apply the sig-fig rule.
When a whole number ends with a decimal point, the trailing zeros are significant. -
Step 3 β Count the significant figures.
In 300., all three digits count.
So there are 3 significant figures.
Sig fig shortcut:
- 300 usually has 1 sig fig
- 300. has 3 sig figs
- 300.0 has 4 sig figs
Question No. 59 | Force of Collisions at 300 K vs 450 K
Step-by-Step Solution
-
Step 1 β Recall what temperature means for particles.
Higher temperature means particles have more kinetic energy. -
Step 2 β Connect kinetic energy to collisions.
Faster particles collide more forcefully. -
Step 3 β Compare 300 K and 450 K.
Since 450 K is higher, collisions at 450 K are more forceful.
Therefore, at 300 K the collisions are less forceful.
Temperature and particles:
- higher temperature β faster particles
- faster particles β more forceful collisions
Question No. 60 | Conditions for More Ideal Gas Behavior
Step-by-Step Solution
-
Step 1 β Recall when real gases act most ideally.
Gases behave more like ideal gases at:
β’ high temperature
β’ low pressure -
Step 2 β Explain why.
High temperature helps particles overcome attractions.
Low pressure spreads particles farther apart, so their own volume matters less. -
Step 3 β State the required changes.
The temperature should increase and the pressure should decrease.
Real gases are most ideal at:
- high temperature
- low pressure
Question No. 61 | Combined Gas Law Volume Calculation
Step-by-Step Solution
-
Step 1 β Use the combined gas law.
P1V1/T1 = P2V2/T2 -
Step 2 β Solve for V2.
V2 = (P1V1T2) / (T1P2) -
Step 3 β Substitute the values.
V2 = (100. Γ 320. Γ 450.) / (300. Γ 60.0) -
Step 4 β Calculate.
V2 = 800. mL
For combined gas law problems:
- temperature must be in Kelvin
- match pressure units on both sides
- solve algebra first, then substitute carefully
Question No. 62 | Class of Compound for Reactant 1
Step-by-Step Solution
-
Step 1 β Identify reactant 1 from its structure.
The structure shows a carbon attached to:
β’ a double-bonded oxygen
β’ an OH group -
Step 2 β Recognize the functional group.
A compound containing the βCOOH group belongs to the class called a carboxylic acid. -
Step 3 β Use the Regents Table R wording.
On Table R, this class is named organic acid.
Common Table R clue:
- βCOOH β organic acid
- βOH β alcohol
- βCOOβ β ester
Question No. 63 | Element That Makes Reactant 2 Organic
Step-by-Step Solution
-
Step 1 β Recall the basic definition of an organic compound.
In Regents Chemistry, organic compounds are compounds that contain carbon. -
Step 2 β Look at reactant 2.
Reactant 2 contains the symbol C, which stands for carbon. -
Step 3 β State the element.
The element that makes reactant 2 an organic compound is carbon.
Organic chemistry almost always centers around carbon-containing compounds.
Question No. 64 | Electrons Shared Between Carbon Atoms
Step-by-Step Solution
-
Step 1 β Identify the bond between the two carbon atoms.
In ethanol, the two carbon atoms are connected by a single covalent bond. -
Step 2 β Recall what a single covalent bond means.
A single covalent bond represents 1 shared pair of electrons. -
Step 3 β Convert shared pair to number of electrons.
One shared pair means 2 electrons are shared.
Bond-electron quick rule:
- single bond = 2 electrons
- double bond = 4 electrons
- triple bond = 6 electrons
Question No. 65 | Why Propanoic Acid and Ethyl Methanoate Are Isomers
Step-by-Step Solution
-
Step 1 β Recall the definition of isomers.
Isomers are compounds that have the same molecular formula but different structural formulas. -
Step 2 β Find the molecular formula of propanoic acid.
CH3CH2COOH = C3H6O2 -
Step 3 β Find the molecular formula of ethyl methanoate.
HCOOCH2CH3 = C3H6O2 -
Step 4 β Compare their structures.
Even though both have the same molecular formula, their atoms are connected differently:
β’ propanoic acid is an organic acid
β’ ethyl methanoate is an ester
Always separate these two ideas:
- Molecular formula = how many of each type of atom
- Structural formula = how the atoms are connected
Topics Covered in the Chemistry Regents Exam
The Regents Chemistry exam assesses a wide range of topics typically covered in high school chemistry courses. Students preparing for the January 2026 exam should review the following major concepts:
- Atomic structure and periodic trends
- Chemical bonding and molecular geometry
- Stoichiometry and mole calculations
- Chemical reactions and reaction types
- Acids, bases, and pH calculations
- Energy changes and thermochemistry
- Equilibrium and reaction rates
- Electrochemistry and redox reactions
How to Use This Exam Paper for Effective Practice
To maximize your preparation for the Regents Chemistry exam, follow this recommended study approach:
1. Attempt the Questions First
Try solving each question independently before checking any solutions. This helps build exam confidence and improves problem-solving skills.
2. Analyze Step-by-Step Solutions
Review detailed explanations to understand the reasoning behind each answer. Focus on chemical principles rather than memorizing answers.
3. Identify Weak Areas
Track topics where mistakes occur frequently and review those concepts in your chemistry notes or textbooks.
4. Practice Similar Problems
Use additional Regents practice exams to strengthen problem-solving speed and accuracy.