JAMB UTME - Chemistry - 2019

A secondary alkanol can be oxidized to give an

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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.

If acidified Potassium Dichromate(VI) (K${}_2$Cr${}_2$O${}_7$) acts as oxidizing agent, color changes from

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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.

For the general equation of the nature

XP + yQ ⇌ mR + nS, the expression for the equilibrium constant is

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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.

Which quantum divides shells into orbitals?

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The quantum that divides shells into orbitals is the "Azimuthal" quantum number, also known as the "angular momentum" quantum number. The azimuthal quantum number determines the shape of an electron's orbital, which is a region in space where there is a high probability of finding an electron. It describes the angular momentum of an electron in an atom and the number of subshells within a given shell. Each subshell is associated with a specific shape, and can hold a certain number of electrons. The azimuthal quantum number is represented by the letter "l" and can have integer values ranging from 0 to (n-1), where "n" is the principal quantum number. Each value of "l" corresponds to a different subshell shape: - l = 0 corresponds to an "s" subshell, which is spherical in shape. - l = 1 corresponds to a "p" subshell, which has a dumbbell shape with two lobes. - l = 2 corresponds to a "d" subshell, which has a more complex shape with four lobes and a doughnut-like ring. - l = 3 corresponds to an "f" subshell, which has an even more complex shape with eight lobes. The number of orbitals within a subshell is equal to 2l+1. For example, a "p" subshell (l = 1) has three orbitals (2l+1 = 3), which are labeled as "px", "py", and "pz". In summary, the azimuthal quantum number determines the shape of the electron's orbital and the number of subshells within a given shell, and it is represented by the letter "l".

Which of the following pairs cannot be represented with a chemical formula?

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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.

Which of the following reactions is an oxidation process?

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An element Z contains 80% of ${}_8^{16}$Z and 20% of ${}_8^{18}$Z. Its relative atomic mass is

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R.A.M of Z = $16\left(\frac{80}{100}\right)+18\left(\frac{20}{100}\right)$
= $12.8+3.6$
= 16.4

When ammonia and hydrogen ion bond together to form ammonium ion, the bond formed is called

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When ammonia and hydrogen ion go into bonding, they form ammonium ion by combining with a dative/coordinate covalent bond.

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. If the molar mass of the compound is 180. Find the molecular formula.

[H = 1, C = 12, O = 16]

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The molecular formula of a compound is determined by the number of atoms of each element present in the molecule. To find the molecular formula, we need to determine the number of atoms of each element in the compound. First, we convert the percent composition to grams. For example, 40.0% carbon means 40.0 g of carbon per 100 g of compound. Then we divide the number of grams of each element by the molar mass of each element. For example, 40.0 g of carbon divided by the molar mass of carbon (12 g/mol) gives us 3.33 mol of carbon. Next, we convert the number of moles of each element to the number of atoms by multiplying the number of moles by Avogadro's number (6.022 x 10^23 atoms/mol). Finally, we balance the numbers of atoms of each element by dividing them by the smallest number of atoms of all the elements and rounding to the nearest whole number. In this case, the smallest number of atoms is 2, which is the number of hydrogen atoms. So, we divide the number of atoms of carbon and oxygen by 2 to balance the numbers of atoms of all the elements. Therefore, the molecular formula of the compound is C6H12O6.

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

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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.

By what amount must the temperature of 200cm${}^3$ of Nitrogen at 27°C be increased to double the pressure if the final volume is 150cm${}^3$ (Assume ideality)

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Using the ideal gas law and equation:
$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
$\frac{P_1\times 200{\text{cm}}^3}{300\text{K}}=\frac{2P\times 150{\text{cm}}^3}{T_2}$
Cross multiply:
$T_2=\frac{300\times 150\times 2P}{200\times P}$
$=450\text{K}$ or ${177}^{\circ }\text{C}$
Don't forget to convert to ${}^{\circ }\text{C}$

Which of the following is the best starting material for the preparation of oxygen? Heating of trioxonitrate (v) with

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The IUPAC name of the compound CF${}_3$CHBrCl is

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The velocity, V of a gas is related to its mass, M by (k = proportionality constant)

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Recall:
V = $\sqrt{\frac{3RT}{M}}$
$\therefore V\propto \frac{1}{\sqrt{M}}$
$V=\frac{k}{\sqrt{M}}$
V = $\frac{k}{M^{\frac{1}{2}}}$

Consider the reaction

A${}_{\left(s\right)}$ + 2B${}_{\left(g\right)}$ → 2C${}_{\left(aq\right)}$ + D${}_{\left(g\right)}$

What will be the effect of a decrease in pressure on the reaction?

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Given: The equation below

A${}_{\left(s\right)}$ + 2B${}_{\left(g\right)}$ → 2C${}_{\left(aq\right)}$ + D${}_{\left(g\right)}$

Since we have a higher number of moles of gaseous species on the LHS, i.e 2, a decrease in pressure will favor the forward reaction.

Which of the following is a physical change?

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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.

The IUPAC name for CH${}_3$CH${}_2$COOCH${}_2$CH${}_3$ is

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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.

SO${}_3$ is not directly dissolved in water in the industrial preparation of H${}_2$SO${}_4$ by the contact process because

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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

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Ionization energy and electron affinity increase across a period, and decrease down a group.

Elements X and Y have electronic configurations 1s${}^2$2s${}^2$2p${}^4$ and 1s${}^2$2s${}^2$2p${}^6$3s${}^2$3p${}^1$ respectively. When they combine, the formula of the compound formed is

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When chlorine water is exposed to bright sunlight, the following products are formed

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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.

The oxidation state(s) of nitrogen in ammonium nitrite is/are

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Ammonium nitrite = NH${}_4$NO${}_2$
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.

Which of the following conditions will most enhance the spontaneity of a reaction?

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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.

At 27°C, 58.5g of sodium chloride is present in 250cm${}^3$ of a solution. The solubility of sodium chloride at this temperature is?

(molar mass of sodium chloride = 111.0gmol${}^{-1}$)

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Given the Mass of the salt = 58.5g
Volume = 250 cm${}^3$ = 0.25 dm${}^3$
Mass concentration = $\frac{Mass}{Volume}$
= $\frac{58.5}{0.25}$ = 234 gdm${}^{-3}$
Solubility (in moldm${}^{-3}$ = $\frac{234}{111}$
= 2.11 moldm${}^{-3}$
$\approxeq$ 2.0 moldm${}^{-3}$

When the end alkyl groups of ethyl ethanoate are interchanged, the compound formed is

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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.

Which of the following gases contains the least number of atoms at s.t.p?

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At standard temperature and pressure (s.t.p), all gases have the same number of atoms or molecules. What changes between them is the volume they occupy, and this is dependent on their molecular mass and the number of moles. Comparing the number of moles between the gases listed above, 7 moles of argon will contain the most number of atoms, followed by 4 moles of chlorine, then 3 moles of ozone, and finally 1 mole of butane would contain the least number of atoms. In summary, the number of atoms in a gas sample depends on the number of moles, but at s.t.p, the volume occupied by each gas depends on its molecular mass and the number of moles.

Consider the equation below:

Cr${}_2$O${}_7^{2-}$ + 6Fe${}^{2+}$ + 14H${}^{+}$ $\to$ 2Cr${}^{3+}$ + 6Fe${}^{3+}$ + 7H${}_2$O.

The oxidation number of chromium changes from

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Cr${}_2$O${}_7^{2-}$ + 6Fe${}^{2+}$ + 14H${}^{+}$ $\to$ 2Cr${}^{3+}$ + 6Fe${}^{3+}$ + 7H${}_2$O
The oxidation of Cr in Cr${}_2$O${}_7^{2-}$ :
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

Na${}_2$CO${}_3$ + 2HCl $\to$ 2NaCl + H${}_2$O + CO${}_2$

The indicator most suitable for this reaction should have a pH equal to

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Methyl orange is the best indicator for the reaction with range 3.1 - 4.4.

The following are isoelectronic ions except

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Two or more ions are said to be isoelectronic if they have the same electronic structure and the same number of valence electrons.
Na${}^{+}$ = 10 electrons = 2, 8
Mg${}^{2+}$ = 10 electrons = 2,8
O${}^{2-}$ = 10 electrons = 2,8
Si${}^{2+}$ = 12 electrons = 2,8,2
$⟹$ Si${}^{2+}$ is not isoelectronic with the rest.

What mass of magnesium would be obtained by passing a current of 2 amperes for 2 hours, through molten magnesium chloride?
[1 faraday = 96500C, Mg = 24]

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Current (I) = 2A; Time (t) = 2 hours = 7200 secs
Q = It
= 2 x 7200 = 14400C
1 F = 96500C
x = 14400C
x = $\frac{14400}{96500}$
= 0.15F
Mg${}^{2+}$ + 2e${}^{-}$ $\to$ Mg
2F $\to$ 24g
0.15F $\to$ x
2x = 24 x 0.15
x = $\frac{24\times 0.15}{2}$
= 1.8g

Which of the following describes the chemical property of acids?

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Which two gases can be used for the demonstration of the fountain experiment?

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Two gases that can be used in the study of fountain experiment is ammonia gas and hydrogen chloride gas. The experiment introduces concepts like solubility and the gas laws at the entry level.

In the reaction:

M + N → P

ΔH = +Q kJWhich of the following would increase the concentration of the product?

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Increasing the temperature would increase the concentration of the product, P. The reaction rate, or the speed at which the reaction occurs, is influenced by temperature. An increase in temperature raises the kinetic energy of the reacting molecules, making it easier for them to collide and react. This leads to a higher rate of reaction and a higher concentration of the product, P. Adding a suitable catalyst can also increase the reaction rate, but it does not directly affect the concentration of the product. Increasing the concentration of P does not affect the reaction itself, but is a result of the reaction having taken place. Decreasing the temperature would slow down the reaction rate and reduce the concentration of the product.

200cm${}^3$ of 0.50mol/dm${}^3$ solution of calcium hydrogen trioxocarbonate (IV) is heated. The maximum weight of solid precipitated is

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To solve this problem, we need to use the concept of stoichiometry and the solubility product constant (Ksp) of calcium hydrogen trioxocarbonate (IV). First, we need to write the balanced equation for the reaction that occurs when the solution of calcium hydrogen trioxocarbonate (IV) is heated: Ca(HCO3)2(s) → CaCO3(s) + H2O(g) + CO2(g) From the balanced equation, we can see that 1 mole of calcium hydrogen trioxocarbonate (IV) produces 1 mole of calcium carbonate. Therefore, we need to determine the number of moles of calcium hydrogen trioxocarbonate (IV) in the solution: Number of moles = concentration x volume Number of moles = 0.50 mol/dm³ x 0.2 dm³ Number of moles = 0.1 mol Since 1 mole of calcium hydrogen trioxocarbonate (IV) produces 1 mole of calcium carbonate, the number of moles of calcium carbonate produced will also be 0.1 mol. Next, we need to use the solubility product constant (Ksp) of calcium carbonate to determine the maximum amount of solid that can be precipitated: Ksp = [Ca²⁺][CO3²⁻] Ksp = 3.3 x 10⁻⁹ (at 25°C) At the maximum amount of solid precipitated, all the calcium carbonate formed will have precipitated, and the concentration of calcium ions and carbonate ions will be equal. Therefore, we can assume that the concentration of calcium ions and carbonate ions is both x. Substituting into the Ksp expression: Ksp = x² 3.3 x 10⁻⁹ = x² x = 5.74 x 10⁻⁵ mol/dm³ The mass of calcium carbonate precipitated can now be calculated: Mass = number of moles x molar mass Mass = 0.1 mol x 100.1 g/mol Mass = 10.01 g Therefore, the maximum weight of solid precipitated is approximately 10 g. Note that this calculation assumes that all the calcium carbonate precipitated as a solid, which may not always be the case in a real-world experiment. Additionally, this calculation does not take into account any losses due to filtration or other experimental errors.

Hydrocarbons which will react with Tollen's reagent conform to the general formula

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Elements in the periodic table are arranged in the order of their

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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

Which of the following statements does not show Rutherford's account of Nuclear Theory? An atom contains a region

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Rutherford's account of Nuclear theory does not include the fact that atoms contain a massive region and cause deflection of from projectiles.

If the cost of electricity required to discharge 10g of an ion X${}^{3+}$ is N20.00, how much would it cost to discharge 6g of ion Y${}^{2+}$?

[1 faraday = 96,500C, atomic masses are X = 27, Y = 24]

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X${}^{3+}$ + 3e${}^{-}$ $\to$ X
3F = 27g
xF = 10g
$\frac{x}{3}=\frac{10}{27}⟹x=\frac{10}{9}F$
$\frac{10}{9}$F $\equiv$ N20.00
1F is equivalent to x
$\frac{1}{\frac{10}{9}}=\frac{x}{20}$
$\frac{9}{10}=\frac{x}{20}⟹x=N18.00$
1F is equivalent to N18.00.
Y${}^{2+}$ + 2e${}^{-}$ $\to$ Y
2F = 24g
xF = 6g
x = $\frac{6\times 2}{24}=\frac{1}{2}F$
1F = N18.00
$\frac{1}{2}$F = $\frac{1}{2}\times N18.00$
= N9.00

Which of the following sets of operation will completely separate a mixture of sodium chloride, sand and iodine?

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The set of operations that will completely separate a mixture of sodium chloride, sand, and iodine is: - filtration, to separate the sand and iodine from the sodium chloride - evaporation to dryness, to concentrate the sodium chloride solution and remove any remaining water - sublimation, to separate the iodine as a solid from the remaining sodium chloride By using these operations, you can separate each component of the mixture into separate, pure forms. The order of the operations is important because each step must be done in a way that effectively separates the components and does not interfere with subsequent steps.