JAMB UTME - Chemistry - 2019

A solution X, on mixing with AgNO${}_3$ solution gives a white precipitate soluble in aqueous NH${}_3$, a solution Y, when also added to X, also gives a white precipitate which is soluble when heated solutions X and Y respectively contain

Detalles de la respuesta

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

Detalles de la respuesta

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

Detalles de la respuesta

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.

2-methylprop-1-ene is an isomer of

Detalles de la respuesta

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.

The emission of two successive beta particles from the nucleus ${}_{15}^{32}P$ will produce

Detalles de la respuesta

Explanation:

Which of the following alkaline metals react more quickly spontaneously with water?

Detalles de la respuesta

The alkaline earth metals (Be, Mg, Ca, Sr, Ba, and Ra) are the second most reactive metals in the periodic table, and, like the Group 1 metals, have increasing reactivity in the higher periods. Beryllium (Be) is the only alkaline earth metal that does not react with water or steam, even if metal is heated to red heat. Additionally, beryllium has a resistant outer oxide layer that lowers its reactivity at lower temperatures.

Magnesium shows insignificant reaction with water, but burns vigorously with steam or water vapor to produce white magnesium oxide and hydrogen gas:

Mg(s) + 2 H2O(l) ⟶ Mg(OH)2(s) + H2(g)

A metal reacting with cold water will produce metal hydroxide. However, if a metal reacts with steam, like magnesium, metal oxide is produced as a result of metal hydroxides splitting upon heating.

The hydroxides of calcium, strontium and barium are only slightly water-soluble but produce sufficient hydroxide ions to make the environment basic, giving a general equation of:

M(s) + 2 H2O(l) ⟶ M(OH)2(aq) + H2(g)

• If metals react with cold water, hydroxides are produced.
• Metals that react with steam form oxides.
• Hydrogen is always produced when a metal reacts with cold water or steam.

Which of the following describes the chemical property of acids?

Detalles de la respuesta

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)

Detalles de la respuesta

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 represents the kind of bonding present in ammonium chloride?

Detalles de la respuesta

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.

The following are isoelectronic ions except

Detalles de la respuesta

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.

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

Detalles de la respuesta

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.

Which quantum divides shells into orbitals?

Detalles de la respuesta

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 factors will speed up the rate of evolution of carbon (iv) oxide in the reaction below?

2HCl + CaCO${}_3$ → CaCl${}_2$ + H${}_2$O + CO${}_2$

Detalles de la respuesta

The following factors increase a reaction rate

- Increase in concentration of reactants

- Increase in temperature

- Addition of catalyst

- Increase in the surface area of reactant(s)

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

Detalles de la respuesta

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

Which of the following statements about catalyst is false?

Detalles de la respuesta

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.

Which of the following reactions is an oxidation process?

Detalles de la respuesta

Which of the following does not support the fact that air is a mixture?

Detalles de la respuesta

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.

Which of the following could not be alkane?

Detalles de la respuesta

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.

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

Detalles de la respuesta

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

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

Detalles de la respuesta

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.

In the reaction between sodium hydroxide and tetraoxosulphate (VI) solutions, what volume of 0.5 molar sodium hydroxide would exactly neutralize 10cm${}^3$ of 1.25 molar tetraoxosulphate (vi) acid?

Detalles de la respuesta

Equation of reaction : 2NaOH + H${}_2$SO${}_4$ → Na${}_2$SO${}_4$ + 2H${}_2$O

Concentration of a base, CB = 0.5M

Volume of acid, V${}_A$ = 10cm${}^3$

Concentration of an acid, C${}_A$ = 1.25M

Volume of base, V${}_B$ = ?

Recall:
$\frac{C_A{V}_A}{C_B{V}_B}=\frac{n_A}{n_B}$ ... (1)
N.B: From the equation,
$\frac{n_A}{n_B}=\frac{1}{2}$

From (1)
$\frac{1.25\times 10}{0.5\times V_B}=\frac{1}{2}$
$\frac{12.5}{0.5V_B}=\frac{1}{2}$
25 = 0.5V${}_B$

VB = 50.0 cm${}^3$

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

Detalles de la respuesta

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.

Which process(es) is/are involved in the turning of starch iodide paper blue-black by chlorine gas?

Detalles de la respuesta

The process involved in the turning of starch iodide paper blue-black by chlorine gas is option number 3: chlorine oxidizes the iodide ion to produce iodine which attacks the starch to give the blue-black color. When chlorine gas comes in contact with iodide ions on the starch iodide paper, it oxidizes the iodide ion to form iodine. The iodine that is produced in this reaction is then able to react with the starch present on the paper to form a blue-black complex. This blue-black complex is formed due to the arrangement of the starch molecules and the iodine atoms in a way that causes them to absorb light at a specific wavelength, giving the blue-black color. Therefore, the blue-black color that is observed on the starch iodide paper is due to the reaction between iodine and starch, which is made possible by the oxidation of iodide ions by chlorine gas.

The shapes of water, ammonia, carbon (iv) oxide and methane are respectively

Detalles de la respuesta

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

Detalles de la respuesta

The IUPAC nomenclature of the compound

H${}_3$C - CH(CH${}_3$) - CH(CH${}_3$) - CH${}_2$ - CH${}_3$

Detalles de la respuesta

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

Detalles de la respuesta

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

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

Detalles de la respuesta

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.

Hydrogen diffused through a porous plug

Detalles de la respuesta

Hydrogen gas (H2) diffuses faster than oxygen gas (O2) through a porous plug. This is because the rate of diffusion of a gas through a porous plug is inversely proportional to the square root of its molar mass. Since the molar mass of hydrogen (2 g/mol) is much smaller than that of oxygen (32 g/mol), the rate of diffusion of hydrogen through a porous plug is much faster than that of oxygen. To be more specific, the ratio of the diffusion rates of two gases through a porous plug is given by the equation: Rate of diffusion of gas A / Rate of diffusion of gas B = √(Molar mass of gas B / Molar mass of gas A) Using the molar masses of hydrogen and oxygen, we get: Rate of diffusion of hydrogen / Rate of diffusion of oxygen = √(32 g/mol / 2 g/mol) = √16 = 4 Therefore, hydrogen diffuses through a porous plug four times as fast as oxygen. Thus, the correct answer is: four times as fast as oxygen.

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

Detalles de la respuesta

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 pollutants will lead to the depletion of ozone layer?

Detalles de la respuesta

The pollutant that leads to the depletion of the ozone layer is chlorofluorocarbon (CFCs). CFCs are man-made chemicals that were widely used in the past as refrigerants, solvents, and propellants. When CFCs are released into the atmosphere, they rise into the stratosphere, where they come into contact with ozone molecules. The chlorine atoms in CFCs react with ozone, breaking apart the ozone molecules and causing a reduction in the overall amount of ozone in the stratosphere. This process continues until all of the ozone-depleting chlorine atoms have been depleted. The resulting decrease in ozone in the stratosphere leads to an increase in the amount of harmful ultraviolet radiation that reaches the Earth's surface, which can have negative impacts on human health and the environment.

Which important nitrogen-containing compound is produced in Haber's process?

Detalles de la respuesta

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.

A certain hydrocarbon on complete combustion at s.t.p produced 89.6dm${}^3$ of CO${}_2$ and 54g of water. The hydrocarbon should be

Detalles de la respuesta

In the question above an Hydrocarbon combust to give CO2 and H20

Let Hydrocarbon be
CxHy + x+Y/4O2= xCO2 + Y/2H2O

Mass of C0=44g and H2O=18g
at STP vol= 22.4
Therefore, 1mole of CO2 contains 44g and 22.4dm³ at STP

1mole = 22.4dm³
xmole = 89.6dm³
Cross multiplying x=89.6/22.4 =4mole of CO2 produce

1mole of H2O = 18g
Xmole = 56g
Cross multiplying
X = 56/18 = 3mole of H20
Then....
CxHy + X + y/4O2 = 4CO2+ 3H2O

Balancing
C4H6 + 6O2 = 4CO2 + 3H2O

In the reaction:

M + N → P

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

Detalles de la respuesta

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.

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

Detalles de la respuesta

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 will give a precipitate with an aqueous solution of copper (I) chloride?

Detalles de la respuesta

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

Detalles de la respuesta

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.

X is a substance which liberates CO${}_2$ on treatment with concentrated H${}_2$SO${}_4$. A warm solution of X can decolorize acidified KMnO${}_4$. X is

Detalles de la respuesta

It should be noted that for X to liberate CO${}_2$, X must be a carbonate or an oxalate. Since X decolorizes KMnO${}_4$, X must be an oxalate. 
Therefore, X is H${}_2$C${}_2$O${}_4$.

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]

Detalles de la respuesta

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

How many electrons will be found in the nucleus of an atom with mass number 23 and 17 neutrons?

Detalles de la respuesta

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.