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Ajụjụ 1 Ripọtì
Which of the following reactions is an oxidation process?
Ajụjụ 2 Ripọtì
Which of the following metals is the most essential in the regulation of blood volume, blood pressure and osmotic equilibrium?
Akọwa Nkọwa
The metal that is most essential in the regulation of blood volume, blood pressure, and osmotic equilibrium is sodium. Sodium is a key electrolyte that helps maintain the balance of fluids in the body, including blood volume and blood pressure. Sodium ions are positively charged and are attracted to negatively charged ions, such as chloride (Cl-) and bicarbonate (HCO3-), which together help regulate the pH of the blood. Sodium is also essential for maintaining osmotic equilibrium, which refers to the balance of solutes between cells and the extracellular fluid. Osmotic equilibrium is critical for proper cellular function and is regulated by the movement of water and electrolytes, including sodium, in and out of cells. While the other metals listed (zinc, manganese, and iron) are important for various functions in the body, such as enzyme activity and oxygen transport, they are not directly involved in regulating blood volume, blood pressure, and osmotic equilibrium in the same way that sodium is. Therefore, the answer is not options 1, 2, or 4, and the correct answer is: sodium.
Ajụjụ 3 Ripọtì
Sulphur exists in six forms in the solid state. This property is known as
Akọwa Nkọwa
The property of sulfur existing in six different forms in the solid-state is known as allotropy. Allotropy is a phenomenon where an element can exist in multiple forms, called allotropes, that have different physical and chemical properties but are composed of the same atoms. These different forms arise due to differences in the arrangement of atoms or molecules within the substance. In the case of sulfur, it can exist in multiple solid-state allotropes, including rhombic, monoclinic, and plastic sulfur, among others. Each of these allotropes has a different crystal structure, melting point, and other physical and chemical properties, even though they are all composed of sulfur atoms. Allotropy is a common phenomenon observed in many elements, including carbon, oxygen, and phosphorus, among others.
Ajụjụ 4 Ripọtì
Which of the following represents the kind of bonding present in ammonium chloride?
Akọwa Nkọwa
Ammonium chloride contains both ionic and covalent bonds. In ammonium chloride, the ammonium ion (NH4+) is positively charged and the chloride ion (Cl-) is negatively charged. These ions are held together by ionic bonds, which are formed between positively and negatively charged ions. However, the bond between the hydrogen atom in the ammonium ion and the nitrogen atom in the ammonium ion is also a covalent bond. This type of covalent bond is known as a dative covalent bond, or a coordinate covalent bond, because the electron pair being shared is supplied by one atom only (the nitrogen atom in this case). So, the kind of bonding present in ammonium chloride is both ionic and dative covalent. In simple terms, ammonium chloride contains both ionic bonds between its positive and negative ions, and a dative covalent bond between the hydrogen atom and nitrogen atom within the ammonium ion.
Ajụjụ 5 Ripọtì
A solution X, on mixing with AgNO3 solution gives a white precipitate soluble in aqueous NH3 , a solution Y, when also added to X, also gives a white precipitate which is soluble when heated solutions X and Y respectively contain
Ajụjụ 6 Ripọtì
How many electrons will be found in the nucleus of an atom with mass number 23 and 17 neutrons?
Akọwa Nkọwa
Electrons are not found in the nucleus of an atom. The nucleus of an atom only contains protons and neutrons, while electrons are located outside the nucleus in the electron cloud. The mass number of an atom is equal to the sum of the number of protons and the number of neutrons in the nucleus. Therefore, if an atom has a mass number of 23 and 17 neutrons, then the number of protons in the nucleus can be calculated as: Protons = Mass number - Neutrons Protons = 23 - 17 Protons = 6 This means that the nucleus of the atom contains 6 protons. The number of electrons in a neutral atom is equal to the number of protons, so the atom also contains 6 electrons in the electron cloud surrounding the nucleus. In summary, the answer is that there are 6 protons and 6 electrons in the atom.
Ajụjụ 7 Ripọtì
The IUPAC name of the compound CF3 CHBrCl is
Ajụjụ 8 Ripọtì
For the general equation of the nature
XP + yQ ⇌ mR + nS, the expression for the equilibrium constant is
Akọwa Nkọwa
The expression for the equilibrium constant for the general equation XP + yQ ⇌ mR + nS is: Kc = [R]m[S]n / [P]x[Q]y where Kc is the equilibrium constant, [R] and [S] are the concentrations of the products, and [P] and [Q] are the concentrations of the reactants, all raised to the stoichiometric coefficients (m, n, x, y) in the balanced equation. This equation is known as the equilibrium constant expression and it represents the ratio of the concentrations of the products and reactants at equilibrium for a particular chemical reaction. The equilibrium constant is a measure of how far a reaction proceeds towards completion, with a larger value indicating a greater extent of reaction. The equilibrium constant expression is derived from the law of mass action, which states that the rate of a chemical reaction is proportional to the product of the concentrations of the reactants raised to their stoichiometric coefficients. At equilibrium, the rates of the forward and reverse reactions are equal, and the equilibrium constant expression represents the ratio of the rate constants for these two reactions. Therefore, the correct expression for the equilibrium constant for the general equation XP + yQ ⇌ mR + nS is Kc = [R]m[S]n / [P]x[Q]y.
Ajụjụ 9 Ripọtì
The IUPAC name for CH3 CH2 COOCH2 CH3 is
Akọwa Nkọwa
The IUPAC name for the given molecule is ethyl propanoate. To arrive at the IUPAC name, we first identify the longest continuous chain of carbon atoms, which in this case is a 4-carbon chain (propane). We then identify and name the substituent groups attached to this chain, which are a methyl group (CH3) attached to the second carbon atom and an ethoxy group (OC2H5) attached to the third carbon atom. The ethoxy group is named as an ethyl group, and the entire molecule is named as ethyl propanoate, following the standard IUPAC naming conventions for esters.
Ajụjụ 10 Ripọtì
Which of the following pairs cannot be represented with a chemical formula?
Akọwa Nkọwa
The pair that cannot be represented with a chemical formula is air and bronze. Air is a mixture of several gases, primarily nitrogen (N₂) and oxygen (O₂), with small amounts of other gases such as argon (Ar), carbon dioxide (CO₂), and neon (Ne). Since air is a mixture and not a pure substance, it cannot be represented by a chemical formula. Bronze, on the other hand, is an alloy composed mainly of copper (Cu) and tin (Sn) with small amounts of other metals. The composition of bronze can vary depending on the specific alloy, but it can be represented by a chemical formula such as CuSn. Sodium chloride (NaCl) is a compound composed of sodium (Na) and chlorine (Cl) in a fixed ratio of 1:1, and it can be represented by a chemical formula. Similarly, copper (Cu) and sodium chloride (NaCl) can each be represented by a chemical formula. Cu is an element, so its chemical formula is simply its symbol, while NaCl is a compound with a fixed ratio of sodium and chlorine atoms. Caustic soda (sodium hydroxide, NaOH) and washing soda (sodium carbonate, Na₂CO₃) are both compounds that can be represented by chemical formulas. NaOH consists of one sodium atom, one oxygen atom, and one hydrogen atom, while Na₂CO₃ consists of two sodium atoms, one carbon atom, and three oxygen atoms.
Ajụjụ 11 Ripọtì
Which important nitrogen-containing compound is produced in Haber's process?
Akọwa Nkọwa
The important nitrogen-containing compound that is produced in Haber's process is NH3, which is also known as ammonia. Haber's process is a chemical process used to produce ammonia by reacting nitrogen gas (N2) and hydrogen gas (H2) under high pressure and temperature in the presence of an iron catalyst. The reaction between nitrogen and hydrogen produces ammonia as the main product, along with some nitrogen and hydrogen gases that do not react. NH3 is an important compound that is widely used in industry for the production of fertilizers, plastics, and other chemical products. It is also used as a cleaning agent, a refrigerant, and a fuel for engines. In addition, NH3 is an essential compound for life, as it is a key component of amino acids, which are the building blocks of proteins.
Ajụjụ 12 Ripọtì
Hydrogen bond is a sort of
Akọwa Nkọwa
Hydrogen bond is a covalent intermolecular bond that exists between hydrogen and highly electronegative elements like nitrogen, oxygen and fluorine.
Ajụjụ 13 Ripọtì
A radioactive nucleus has a half-life of 20 years, starting with 100,000 particles, how many particles will be left exactly at the end of 40 years
Akọwa Nkọwa
The half-life of a radioactive nucleus is the time it takes for half of its particles to decay. This means that after 20 years, 100,000 particles will become 50,000 particles. After 40 years, we can find the number of particles remaining by counting the number of half-lives that have passed. Since 40 years is double the half-life of 20 years, this means that two half-lives have passed, so the number of particles will be halved twice. Starting with 100,000 particles: - After 1 half-life (20 years), there will be 50,000 particles remaining. - After 2 half-lives (40 years), there will be 25,000 particles remaining. So, exactly at the end of 40 years, there will be 25,000 particles remaining.
Ajụjụ 14 Ripọtì
Which of the following does not support the fact that air is a mixture?
Akọwa Nkọwa
The option that does not support the fact that air is a mixture is "the constituents of air are in a fixed proportion by mass". Air is a mixture of different gases, primarily nitrogen (78%) and oxygen (21%), with small amounts of other gases such as carbon dioxide, argon, and neon. The proportion of each gas in air is not fixed and can vary depending on the location and other factors. For example, the amount of carbon dioxide in air can increase in areas with high levels of pollution, while the proportion of oxygen can decrease at high altitudes. Therefore, the composition of air is not in a fixed proportion by mass. On the other hand, the fact that air cannot be represented with a chemical formula and its constituents can be separated by physical means support the fact that air is a mixture. A chemical formula represents a pure substance, and since air is a mixture of gases, it cannot be represented by a single formula. Air can be separated into its individual components through physical means such as distillation or filtration, which is a characteristic of mixtures.
Ajụjụ 15 Ripọtì
When the end alkyl groups of ethyl ethanoate are interchanged, the compound formed is
Akọwa Nkọwa
The compound formed when the end alkyl groups of ethyl ethanoate are interchanged is ethyl propanoate. This is because ethyl ethanoate consists of two parts: the "ethyl" group and the "ethanoate" group. The ethyl group is a two-carbon chain, and the ethanoate group is a combination of a one-carbon chain and a carbonyl group (C=O) that is also attached to an oxygen atom. When the end alkyl groups are interchanged, the "ethyl" group is moved from the second carbon to the first carbon of the ethanoate group, and the "propanoate" group is formed. The "propanoate" group consists of a three-carbon chain and the carbonyl group. Therefore, the resulting compound is ethyl propanoate, which has a chemical formula of CH3CH2COOCH2CH3. This compound is commonly used as a flavoring agent and has a fruity odor reminiscent of pears.
Ajụjụ 16 Ripọtì
Which of the following properties increases from left to right along the period but decreases down the group in the Periodic Table?
I. Atomic Number ii. Ionization energy iii. Metallic character iv. Electron affinity
Akọwa Nkọwa
Ionization energy and electron affinity increase across a period, and decrease down a group.
Ajụjụ 17 Ripọtì
By what amount must the temperature of 200cm3 of Nitrogen at 27°C be increased to double the pressure if the final volume is 150cm3 (Assume ideality)
Akọwa Nkọwa
Using the ideal gas law and equation:
P1V1T1=P2V2T2
P1×200cm3300K=2P×150cm3T2
Cross multiply:
T2=300×150×2P200×P
=450K
or 177∘C
Don't forget to convert to ∘C
Ajụjụ 18 Ripọtì
Which of the following statements does not show Rutherford's account of Nuclear Theory? An atom contains a region
Akọwa Nkọwa
Rutherford's account of Nuclear theory does not include the fact that atoms contain a massive region and cause deflection of from projectiles.
Ajụjụ 19 Ripọtì
Which of the following is a physical change?
Akọwa Nkọwa
A physical change refers to a change in a substance that does not result in a change in its chemical composition. Out of the options provided, freezing ice cream is a physical change. This is because when ice cream is frozen, it changes from a liquid state to a solid state without any chemical reaction occurring. Exposing white phosphorus to air is a chemical change, as it reacts with oxygen in the air to form a new substance, phosphorus oxide. Burning kerosene is also a chemical change, as it undergoes combustion to form new substances, such as carbon dioxide and water vapor. Dissolving calcium in water is a physical change, as it simply involves the physical mixing of two substances without any chemical reaction occurring. Therefore, the only option that is a physical change is freezing ice cream.
Ajụjụ 20 Ripọtì
At 27°C, 58.5g of sodium chloride is present in 250cm3 of a solution. The solubility of sodium chloride at this temperature is?
(molar mass of sodium chloride = 111.0gmol−1 )
Akọwa Nkọwa
Given the Mass of the salt = 58.5g
Volume = 250 cm3
= 0.25 dm3
Mass concentration = MassVolume
= 58.50.25
= 234 gdm−3
Solubility (in moldm−3
= 234111
= 2.11 moldm−3
≊
2.0 moldm−3
Ajụjụ 21 Ripọtì
Consider the equation below:
Cr2 O2−7 + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2 O.
The oxidation number of chromium changes from
Akọwa Nkọwa
Cr2
O2−7
+ 6Fe2+
+ 14H+
→
2Cr3+
+ 6Fe3+
+ 7H2
O
The oxidation of Cr in Cr2
O2−7
:
Let the oxidation of Cr = x;
2x + (-2 x 7) = -2 ⟹
2x - 14 = -2
2x = 12 ; x = +6
Hence, the change in oxidation of Cr = +6 to +3
Ajụjụ 22 Ripọtì
The combustion of carbon(ii)oxide in oxygen can be represented by equation.
2CO + O2 ? 2CO2
Calculate the volume of the resulting mixture at the end of the reaction if 50cm3 of carbon(ii)oxide was exploded in 100cm3 of oxygen
Akọwa Nkọwa
Ajụjụ 23 Ripọtì
When ammonia and hydrogen ion bond together to form ammonium ion, the bond formed is called
Akọwa Nkọwa
When ammonia and hydrogen ion go into bonding, they form ammonium ion by combining with a dative/coordinate covalent bond.
Ajụjụ 24 Ripọtì
Which of the following could not be alkane?
Akọwa Nkọwa
An alkane is a type of hydrocarbon with only single bonds between the carbon atoms. It follows the general formula CnH2n+2, where "n" is the number of carbon atoms in the molecule. To determine whether a molecule is an alkane or not, we can calculate its molecular formula and check if it fits the general formula of alkane. Out of the given options, the third one (C7H14) cannot be an alkane. To see why, let's use the general formula of alkane, which is CnH2n+2. For C7H14 to be an alkane, it should have 2n+2 = 2(7) + 2 = 16 hydrogen atoms. However, C7H14 has only 14 hydrogen atoms, which means it does not follow the general formula of alkane. Therefore, C7H14 cannot be an alkane. The other options are as follows: - C4H10: This is butane, which is an alkane with four carbon atoms. - C5H12: This is pentane, which is an alkane with five carbon atoms. - C8H18: This is octane, which is an alkane with eight carbon atoms. In summary, the molecule C7H14 cannot be an alkane because it does not follow the general formula of alkane, while the other options are all examples of alkanes.
Ajụjụ 25 Ripọtì
The shapes of water, ammonia, carbon (iv) oxide and methane are respectively
Ajụjụ 26 Ripọtì
What technique is suitable for separating a binary solution of potassium chloride and potassium trioxochlorate (V)?
Akọwa Nkọwa
Fractional crystallization is the most suitable technique for separating a binary solution of potassium chloride and potassium trioxochlorate (V). This is because fractional crystallization is a process that separates a mixture of substances based on their solubility in a solvent at a particular temperature. In this case, potassium chloride and potassium trioxochlorate (V) have different solubilities in a solvent such as water at different temperatures. By carefully controlling the temperature, the solubility of each compound can be selectively increased or decreased, allowing them to be separated by crystallization. The less soluble compound will form crystals first and can be separated from the more soluble compound, which remains in the solution. Therefore, fractional crystallization can be used to separate potassium chloride and potassium trioxochlorate (V) in a binary solution.
Ajụjụ 27 Ripọtì
2-methylprop-1-ene is an isomer of
Akọwa Nkọwa
2-methylprop-1-ene is an isomer of 3-methyl but-1-ene and 2-methyl but-1-ene. An isomer is a molecule that has the same molecular formula as another molecule, but a different arrangement of atoms. In this case, 2-methylprop-1-ene has the molecular formula C4H8, and so do 3-methyl but-1-ene and 2-methyl but-1-ene. The difference between these three molecules is in the arrangement of the carbon and hydrogen atoms. 2-methylprop-1-ene has a branched structure with a double bond between the first and second carbon atoms. 3-methyl but-1-ene is also a branched molecule, but the double bond is between the second and third carbon atoms. Similarly, 2-methyl but-1-ene has a double bond between the first and second carbon atoms, but it has a different branching pattern. On the other hand, pent-2-ene has five carbon atoms, so it has a different molecular formula than 2-methylprop-1-ene. Therefore, 2-methylprop-1-ene is an isomer of 3-methyl but-1-ene and 2-methyl but-1-ene, but not of pent-2-ene, because it has the same molecular formula and a different arrangement of atoms compared to the other two isomers.
Ajụjụ 28 Ripọtì
A secondary alkanol can be oxidized to give an
Akọwa Nkọwa
A secondary alkanol is an alcohol with two carbon atoms attached to the carbon bearing the hydroxyl group (-OH). Secondary alkanols can be oxidized by a strong oxidizing agent, such as potassium dichromate (K2Cr2O7), to give an alkanone. During the oxidation process, the oxygen atom from the oxidizing agent replaces the hydroxyl group of the secondary alkanol to form a carbonyl group (C=O) in the alkanone. Since alkanones contain a carbonyl group, they are also known as ketones. Therefore, the answer to the question is alkanone, as secondary alkanols can be oxidized to form ketones.
Ajụjụ 29 Ripọtì
Elements in the periodic table are arranged in the order of their
Akọwa Nkọwa
Elements in the periodic table are arranged in the order of their atomic numbers. The atomic number of an element is the number of protons in the nucleus of an atom of that element. The elements are arranged in order of increasing atomic number from left to right and from top to bottom in the periodic table. The elements in each row, also known as a period, have the same number of electron shells, while the elements in each column, also known as a group or family, have the same number of valence electrons. This arrangement makes it possible to predict the chemical and physical properties of an element based on its position in the periodic table. Therefore, the correct answer is: - atomic numbers
Ajụjụ 30 Ripọtì
The heat of formation of ethene, C2 H4 is 50 kJmol−1 , and that of ethane, C2 H6 is -82kJmol−1 . Calculate the heat evolved in the process:
C2 H4 + H2 → C2 H6
Akọwa Nkọwa
The heat evolved in a chemical reaction can be calculated by subtracting the heat of formation of the reactants from the heat of formation of the products. In this case, the reactants are ethene (C2H4) and hydrogen (H2), and the product is ethane (C2H6). The heat of formation of ethene is 50 kJ/mol and that of hydrogen is 0 kJ/mol (because hydrogen is a reference element). The heat of formation of ethane is -82 kJ/mol. So, the heat evolved in the reaction is given by: Heat evolved = (Heat of formation of products) - (Heat of formation of reactants) = (-82 kJ/mol) - (50 kJ/mol + 0 kJ/mol) = -82 kJ/mol - 50 kJ/mol = -132 kJ/mol. Therefore, the heat evolved in the process is -132 kJ.
Ajụjụ 31 Ripọtì
In the reaction between sodium hydroxide and tetraoxosulphate (VI) solutions, what volume of 0.5 molar sodium hydroxide would exactly neutralize 10cm3 of 1.25 molar tetraoxosulphate (vi) acid?
Akọwa Nkọwa
Equation of reaction : 2NaOH + H2 SO4 → Na2 SO4 + 2H2 O
Concentration of a base, CB = 0.5M
Volume of acid, VA = 10cm3
Concentration of an acid, CA = 1.25M
Volume of base, VB = ?
Recall:
CAVACBVB=nAnB
... (1)
N.B: From the equation,
nAnB=12
From (1)
1.25×100.5×VB=12
12.50.5VB=12
25 = 0.5VB
VB = 50.0 cm3
Ajụjụ 32 Ripọtì
Which of the following conditions will most enhance the spontaneity of a reaction?
Akọwa Nkọwa
The condition that will most enhance the spontaneity of a reaction is when ΔH is negative (i.e., the reaction releases heat) and ΔS is positive (i.e., the reaction increases the disorder or randomness of the system). This is because a negative ΔH indicates that the reaction releases energy, which is favorable for a spontaneous reaction, while a positive ΔS indicates that the system becomes more disordered, which is also favorable for spontaneous reactions. Among the given options, the first condition of a negative and greater ΔH than ΔS is the best option for enhancing the spontaneity of a reaction. The other options have either a positive ΔH or a zero ΔS, which is not favorable for spontaneous reactions.
Ajụjụ 33 Ripọtì
Which of the following statements about catalyst is false?
Akọwa Nkọwa
The false statement about catalysts is: "catalysts do not alter the mechanism of the reaction and never appear in the rate law." Catalysts are substances that speed up chemical reactions without being consumed in the process. They achieve this by reducing the activation energy needed for the reaction to occur. Enzymes are a type of biological catalysts. In a chemical reaction, a catalyst is not consumed and does not appear in the overall balanced equation. However, catalysts can alter the mechanism of a reaction by providing an alternative pathway with a lower activation energy. This alternative pathway can have a different rate-determining step, which means that the presence of the catalyst can change the rate law of the reaction. Therefore, the statement that catalysts do not alter the mechanism of the reaction and never appear in the rate law is false.
Ajụjụ 34 Ripọtì
Which of the following will give a precipitate with an aqueous solution of copper (I) chloride?
Akọwa Nkọwa
Ajụjụ 35 Ripọtì
The oxidation state(s) of nitrogen in ammonium nitrite is/are
Akọwa Nkọwa
Ammonium nitrite = NH4
NO2
NH+4
: Let the oxidation number of Nitrogen = x
x + 4 = 1 ⟹
x = 1 - 4
x = -3
NO−2
: x - 4 = -1
x = -1 + 4 ⟹
x = +3.
The oxidation numbers for Nitrogen in Ammonium Nitrite = -3, +3.
Ajụjụ 36 Ripọtì
X is a substance which liberates CO2 on treatment with concentrated H2 SO4 . A warm solution of X can decolorize acidified KMnO4 . X is
Akọwa Nkọwa
It should be noted that for X to liberate CO2
, X must be a carbonate or an oxalate. Since X decolorizes KMnO4
, X must be an oxalate.
Therefore, X is H2
C2
O4
.
Ajụjụ 37 Ripọtì
If the cost of electricity required to discharge 10g of an ion X3+ is N20.00, how much would it cost to discharge 6g of ion Y2+ ?
[1 faraday = 96,500C, atomic masses are X = 27, Y = 24]
Akọwa Nkọwa
X3+
+ 3e−
→
X
3F = 27g
xF = 10g
x3=1027⟹x=109F
109
F ≡
N20.00
1F is equivalent to x
1109=x20
910=x20⟹x=N18.00
1F is equivalent to N18.00.
Y2+
+ 2e−
→
Y
2F = 24g
xF = 6g
x = 6×224=12F
1F = N18.00
12
F = 12×N18.00
= N9.00
Ajụjụ 38 Ripọtì
A cell shorthand notation can be written as A / A+ // B2+ /B. The double slash in the notation represents the
Akọwa Nkọwa
The double slash in the cell shorthand notation represents the salt bridge. A salt bridge is a component of an electrochemical cell that connects the two half-cells and allows the flow of ions between them. It consists of an inert electrolyte solution (usually a salt) that is placed between the two half-cells. The purpose of the salt bridge is to maintain electrical neutrality in each half-cell by allowing the flow of ions to balance the charge buildup in the half-cells. In the cell shorthand notation, the double slash "//" represents the salt bridge that connects the two half-cells of the electrochemical cell. The first half-cell is represented on the left-hand side of the slash and the second half-cell is represented on the right-hand side of the slash. The anode (where oxidation occurs) is represented on the left side, and the cathode (where reduction occurs) is represented on the right side. Therefore, the correct answer is option number 3: salt bridge.
Ajụjụ 39 Ripọtì
The molecular shape and bond angle of water are respectively
Akọwa Nkọwa
The shape of water molecule = Bent/ V- shaped
The bond angle of water = 104.5°/ 105°
Ajụjụ 40 Ripọtì
If acidified Potassium Dichromate(VI) (K2 Cr2 O7 ) acts as oxidizing agent, color changes from
Akọwa Nkọwa
Potassium Dichromate (VI), when it is acidified, acts as an oxidizing agent. When this happens, the color changes from orange to green. This is because the orange color of the potassium dichromate is due to the presence of Cr(VI) ions, which are oxidized to Cr(III) ions. The green color that is produced is due to the formation of chromium(III) ions. In this reaction, the dichromate ions are being oxidized, which means that they are losing electrons, and the chromium ions are being reduced, which means that they are gaining electrons. The transfer of electrons causes the color change from orange to green.
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