Unit 1 Test - Atomic Structure and Properties

To spectrophotometrically determine the mass percent of cobalt in an ore containing cobalt and some inert materials, solutions with known [Co2+] were prepared and the absorbance of each of the solutions was measured at the wavelength of optimum absorbance. The data were used to create the calibration plot, shown below. [couldn't upload the graph but it is, 2018 ap chem frq form b] A 0.630g sample of the ore is completely dissolved in concentrated HNO3. The mixture is diluted with water to a final volume of 50.00mL. Assume that all the cobalt in the ore sample is converted to Co2+. b) calculate the number of moles of Co2+ in the 50.00mL solution.

0.0004mol - 0.00045mol 0.008mol/L * 0.05L = 0.0004mol 0.009mol/L * 0.05L = 0.00045mol

To spectrophotometrically determine the mass percent of cobalt in an ore containing cobalt and some inert materials, solutions with known [Co2+] were prepared and the absorbance of each of the solutions was measured at the wavelength of optimum absorbance. The data were used to create the calibration plot, shown below. [couldn't upload the graph but it is, 2018 ap chem frq form b] A 0.630g sample of the ore is completely dissolved in concentrated HNO3. The mixture is diluted with water to a final volume of 50.00mL. Assume that all the cobalt in the ore sample is converted to Co2+. a) what is the [Co2+] in the solution if the absorbance of a sample of the solution is 0.47?

0.008M -- 0.009M using the graph find an estimation of around 0.008M -- 0.009M

A new element with atomic number 116 was discovered in 200. In 2023 it was named livermorium, Lv. Although Lv is radioactive and short-lived, its chemical properties and reactivity should follow periodic trends. d) Shown below is a hypothetical mass spectrum for a sample of Lv containing 10 atoms. Using the information in the graph, determine the average atomic mass of Lv in the sample to four significant figures. [graph on other side]

292.5 amu To solve multiply the number of atoms by it's atomic mass and average out. 291.2 * 2 = 582.4 292.2 * 3 = 876.6 293.2 * 5 = 1466 ---------------- = 29250 amu 29250/10 = 292.5 amu

A new element with atomic number 116 was discovered in 200. In 2023 it was named livermorium, Lv. Although Lv is radioactive and short-lived, its chemical properties and reactivity should follow periodic trends. a) Write the electron configuration for the valance electrons of Lv in the ground state.

7s² 7p⁴ Write out full electron configuration of Lv: [Rn] 7s^2 7p^4 5f^14 6d^10 Looking at the periodic table, we can see that Lv should have 6 valance electrons. Since the end of the configuration has too many. We move down to 7s² 7p⁴. This also aligns with Lv's position in the 7th period of the periodic table.

A new element with atomic number 116 was discovered in 200. In 2023 it was named livermorium, Lv. Although Lv is radioactive and short-lived, its chemical properties and reactivity should follow periodic trends. c) The first ionization energy of polonium, Po, is 812 kJ/mol. Is the first ionization energy of Lv expected to be greater than, less than or equal to that of Po? Justify your answer in terms of Coulomb's law.

Less than, because the atomic radius of Lv is larger, so there are less coulombic attractions.

A new element with atomic number 116 was discovered in 200. In 2023 it was named livermorium, Lv. Although Lv is radioactive and short-lived, its chemical properties and reactivity should follow periodic trends. b) According to periodic properties what would be the most likely formula for the product obtained when Lv reacts with H2?

LvH2 Elements in Group 16 typically form compounds with hydrogen in a 1:2 ratio. The general formula for the compound formed between an element in Group 16 and hydrogen is EH2, where E represents the element.

To spectrophotometrically determine the mass percent of cobalt in an ore containing cobalt and some inert materials, solutions with known [Co2+] were prepared and the absorbance of each of the solutions was measured at the wavelength of optimum absorbance. The data were used to create the calibration plot, shown below. [couldn't upload the graph but it is, 2018 ap chem frq form b] A 0.630g sample of the ore is completely dissolved in concentrated HNO3. The mixture is diluted with water to a final volume of 50.00mL. Assume that all the cobalt in the ore sample is converted to Co2+. c) Calculate the mass percent of Co in the 0.630g sample of the ore.

Mass percent of Co = (mol C * molar mass of C)/mass of sample [ (0.0004 * 58.9)/0.630 ] * 100 = 3.74% [ (0.00045 * 58.9)/0.630 ] * 100 = 4.21% (correct me if I am wrong, I am unsure if this is correct)

How many carbon atoms are contained in 2.8g of C2H4? a) 1.2 * 10²³ b) 3.0 * 10²³ c) 6.0 * 10²³ d) 1.2 * 10²⁴ e) 6.0 * 10²⁴

a) 1.2 * 10²³ Find the molar mass of C2H4: 2(12.01) + 4(1.01) = 28.06 g/mol convert to moles: 2.8g * 1mol/28.06g = 0.0998 mol C2H4 Using the mole ratio (The ratio of carbon atoms to the entire C2H4 molecule is 2:1) convert to moles of carbon: 0.0998 * 2/1 = 0.1996 Convert to atoms using Avogadro's number: 0.1996 * (6.0 * 10²³) ≈ 1.2 * 10²³

Which of the following represents the ground state electron configuration for the Mn3+ ion? (atomic number Mn = 25) a) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴ b) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s² c) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s² d) 1s² 2s² 2p⁶ 3s² 3p⁶ e) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s¹

a) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴ The Mn3+ ion has a charge of +3, indicating the loss of three electrons from the neutral Mn atom's configuration. Starting with the ground state electron configuration of a neutral manganese atom: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s² subtracting 3 electrons you get: a) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴

Of the elements listed, which has the lowest first-ionization energy? a) Cs b) Ag c) Pb d) Br e) Se

a) Cs - Ionization energy tends to increase from left to right across a period. - Ionization energy tends to decrease from top to bottom within a group. Considering these trends, Cs is located at the bottom of the periodic table, making it likely to have the lowest first-ionization energy among the elements listed.

Of the elements listed, which has the largest atomic radius? a) Cs b) Ag c) Pb d) Br e) Se

a) Cs Use Use the periodic table and the period trends. Atomic radius decreases as you move from left to right across a period and increases as you move down a group.

Of the following electron configurations of neutral atoms, which represents an atom in an excited state? a) 1s² 2s² 2p⁵ b) 1s² 2s² 2p⁵ 3s² c) 1s² 2s² 2p⁶ 3s¹ d) 1s² 2s² 2p⁶ 3s² 3p² e) 1s² 2s² 2p⁶ 3s² 3p⁵

b) 1s² 2s² 2p⁵ 3s² An atom is in an excited state when one or more electrons are elevated to a higher energy level than the ground state. In its ground state, an atom organizes its electrons in the lowest energy arrangement possible, which means filling the lower energy levels before the higher ones. Upon careful examination of the options, it is evident that choice B features electrons occupying higher energy levels, even without completely filling the preceding levels. This characteristic signifies that B is the correct answer.

Based on the periodic trends and the data in the table below, which of the following are the most probably values of the atomic radius and the first ionization energy for potassium, respectively? ELEMENT | ATOMIC RADIUS | FIRST IONIZATION ENERGY ------------------------------------------------------------------------ Ca | 194pm | 590 kJ/mol K | ---- | ---- a) 242 pm, 633 kJ/mol b) 242 pm, 419 kJ/mol c) 120 pm, 633 kJ/mol d) 120 pm, 419 kJ/mol

b) 242 pm, 419 kJ/mol Use periodic table and periodic trends. - Ionization energy tends to increase from left to right across a period. - Ionization energy tends to decrease from top to bottom within a group. - Atomic radius decreases as you move from left to right across a period -Atomic radius increases as you move down a group.

Each student in a class placed a 2.00 g sample of a mixture of Cu and Al in a beaker and placed the beaker in a fume hood. The students slowly poured 15.0 mL of 15.8 M HNO3 into their beaker. The reaction between the copper in the mixture and the HNO3 is represented by the equation below. The students observed that a brown gas was released from the beakers and that the solutions turned blue, indicating the formation of CU2+. The solutions were then diluted with distilled water to known volumes. Cu(s) + 4HNO3 (aq) --> Cu(NO3)2(aq) + 2NO3(g) +2H2O(l) The students determined that the reaction produced 0.010mol of CU(NO3)2. Based on the measurement, what was the percent of Cu by mass in the original 2.00g sample of the mixture? a) 16% b)32% c)64% d)96%

b) 32% From the equation of reaction, 1 mole of Cu(NO₃) is produced from 1 mole of copper. Therefore, 0.010 moles of Cu(NO₃) will be produced from 0.010 mole of copper. Molar mass of copper = 64 g/mol mass of copper = number of moles * molar mass mass of copper = 0.01 mol * 64 g/mol = 0.64 g Percentage by mass of copper in the 2.00 g sample = (0.64/2.00) * 100% Percentage mass of copper in the sample = 32%

Consider atoms of the following elements. Assume that the atoms are in the ground state which atom contains only one electron in the highest occupies energy level? a) S b) Ga c) Pb d) Br

b) Ga Find the electron configuration of each element: S: [Ne] 3s² 3p⁴ Ga: [Ar] 3d¹⁰ 4s² 4p¹ Pb: [Xe] 6s² 4f¹⁴ 5d¹⁰ 6p² Br: [Ar] 3d¹⁰ 4s² 4p⁵ Ga has only one electron in its highest energy level, so the answer is: b) Ga

According to the information in the table below, a 1.00 g sample of which of the following contains the greatest mass of oxygen? a) Na_2O b) MgO c) K_2O d) CaO

b) MgO find molar masses of each compound: ​Na2O = 62.98 g/mol MgO = 40.31 g/mol ​K2O = 94.20 g/mol CaO = 56.08 g/mol divide each by the molar mass of oxygen (16.00) and multiply by 1 gram to convert to grams: ​Na2O = 16/62.98 = 0.25g MgO = 16/40.31 = 0.40g ​K2O = 16/94.20 = 0.17g CaO = 16/56.08 = 0.29g Which ever has the highest grams of oxygen is the answer: b) MgO

A sample containing atoms of C and F was analyzed using x-ray photoelectron spectroscopy. The portion of the spectrum showing the 1s peaks for atoms of the two elements is shown below. Which of the following correctly identifies the 1s peak for the F atoms and provides an appropriate explanation? [graph on other side] a) Peak X, because F has a small first ionization energy than C has b) Peak X, because F has a greater nuclear charge than C has c) Peak Y, because F is more electronegative than C is d) Peak Y, because F has a smaller atomic radius than C has

b) Peak X, because F has a greater nuclear charge than C has Looking at the options we can rule out option A and C because A is incorrect about the period trend and C is irrelevant. It is B because the binding energy of electrons is influenced by the strength of the nuclear charge holding the electrons. Since fluorine (F) has a greater nuclear charge than carbon (C), the electrons in F will experience a stronger attraction to the nucleus, resulting in a higher binding energy and a peak at a higher energy level (Peak X).

The mass spectrum of an average sample of a pure element is shown in the figure below. Which of the following is the identity of the element? [graph is shown on other side (except ignore the bar on the 96, that should'nt be there)] a) Y b) Zr c) Nb d) Th

b) Zr To solve take the percent abundance and multiply by the atomic mass and add it together. 90 * 0.5 = 45 91 * 0.1 = 9.1 92 * 0.2 = 18.4 94 * 0.2 = 18.8 ---------------- ≈ 91.3 amu Look at periodic table and find the element that is the closest to what you calculated, and that is your answer! Zr = 92.224 ≈ 91.3 b) Zr

A compound contains 1.1o mol of K, 0.55 mol of Te, and 1.65 mol of O. What is the simplest formula of this compound? a) KTeO b) KTe2O c) K2TeO3 d) K2TeO6 e) K4TeO6

c) K2TeO3 Take the smallest amount of moles, and divide by every element: K: 1.10/0.55 = 2 Te: 0.55/0.55 = 1 O: 1.65/0.55 = 3 c) K2TeO3

For element X represented below, which of the following is the most likely explanation for the large difference between the second and third ionization energies? X(g) --> X+(g) + e- IE_1 = 740kJ/mol X+(g) --> X2+(g) + e- IE_2 = 1450 kJ/mol X2+(g) --> X3+(g) +e- a) the effective nuclear charge decreases with successive ionizations b) The shielding of out electrons increases with successive ionizations c) The electrons removed during the third ionization is, on average, much closer to the nucleus than the first two electrons removed were. d) the ionic radius increases with successive ionizations.

c) The electrons removed during the third ionization is, on average, much closer to the nucleus than the first two electrons removed were.

Based on the ionization energies of element X given in the table below, which of the following is the most likely empirical formula of an oxide of element X? IONIZATION LEVELS | IONIZATION ENERGY(kJ/mol) first | 577 second | 1816 third | 2745 fourth | 11577 fifth | 14482 a) XO2 b) X2O c) X2O3 d) X2O5

c) X2O3 When an element forms an oxide, it tends to lose electrons. The significant increase in ionization energy from IE3 to IE4 suggests the removal of electrons from a different energy level, indicating the element's transition from the third to the fourth energy level. Considering this pattern, the empirical formula of the oxide is likely to involve the loss of three electrons.

Of the elements listed, which has the highest electronegativity? a) Cs b) Ag c) Pb d) Br e) Se

d) Br use periodic table and the periodic trends

The photoelectron spectra represented below shows the energy required to remove a 1s electron from a nitrogen atom and from an oxygen atom. Which of the following statements best accounts for the peak in the upper spectrum (Nitrogen N) being to the right of the peak in the lower spectrum (Oxygen O). [Binding energy (eV) decreases to the right on the graphs. The two graphs show that Nitrogen has a lower binding energy than Oxygen.] a) Nitrogen atoms have a half filled p subshell b) There are more electron-electron repulsions in oxygen atoms than in nitrogen atoms. c) Electrons in the p subshell of oxygen provide more shielding than electrons in the p subshell of nitrogen atoms. d) Nitrogen atoms have a smaller nuclear charge than oxygen atoms.

d) Nitrogen atoms have a smaller nuclear charge than oxygen atoms. Use the periodic table and the period trends. Nuclear charge increases left to right across a period and decreases down a group.

Based on the information represented in the table below, and periodic trends, which of the following is the best hypothesis regarding the oxide(s) forbed by Rb? ELEMENT | KNOWN OXIDES ------------------------------------------- H | H2O, H2O2 Li | Li2O, Li2O2 Na | Na2O, Na2O2, NaO2 K | K2O, K2O2, KO2 a) Rb will form only Rb2O b) Rb will form only Rb2O2 c) Rb will form only Rb2O and Rb2O2 d) Rb will form Rb2O, Rb2O2, and RbO2

d) Rb will form Rb2O, Rb2O2, and RbO2 As we move down the group from H to K, there is a progression in the formation of oxides, peroxides, and superoxides. Based on this trend, we can infer that Rb will continue this pattern. Therefore, the hypothesis is: d) Rb will form Rb2O, Rb2O2, and RbO2

The mass spectrum of element X is presented in the diagram above. Based on the spectrum, which of the following can be concluded about element X? [graph on the other side] a) X is a transition metal and each peak represents an oxidation state of that metal b) X contains five electron sub-levels c) The atomic mass of X is 90 d) The atomic mass of X is between 90 and 92

d) The atomic mass of X is between 90 and 92 By examining the choices, we can immediately eliminate the first two options as irrelevant. When comparing options C and D, the graph clearly indicates that the atomic mass of element X is situated in the range of 90 to 92. Therefore, option D is the accurate conclusion.