(a)
Interpretation:
The coloured boxes that represent four non-metal elements are to be identified.
Concept introduction:
The periodic table is an arrangement of elements according to their properties,
The characteristic properties of non-metals are as follows:
1. Non-metals, unlike metals, can be solid, liquid or gas.
2. Non-metal oxides are acidic in nature.
3. Non-metals are poor conductors of heat and electricity.
4. Non-metals have a tendency to gain electrons to form anions.
5. Non-metals are non-malleable.
6. Non-metals are not ductile.
7. Non-metals do not exhibit sonority.
(b)
Interpretation:
The coloured boxes that represent two metal elements are to be identified.
Concept introduction:
The periodic table is an arrangement of elements according to their properties, atomic number, and electronic configurations.
The characteristic properties of metals are as follows:
1. Metals are hard and shiny in appearance. Except for mercury, all metals are solid.
2. Metallic oxides are basic in nature.
3. Metals are good conductors of heat and electricity
4. Metals have a tendency to lose electrons to form cations.
5. Metals are malleable. They can be beaten into thin sheets
6. Metals are ductile. They can be drawn into wires.
7. Metals exhibit sonority.
(c)
Interpretation:
The coloured boxes that represent three elements that are gaseous at room temperature are to be identified.
Concept introduction:
A periodic table is an arrangement of elements based on their atomic number, properties and electronic configuration. The table is arranged into groups and periods. The elements which are metallic in nature, occupy the large lower-left portion of the table. The non-metals occupy the small upper-right portion of the table. Metalloids like along the staircase line. Elements which appear in the same group have similar behaviour.
(d)
Interpretation:
The coloured boxes that represent three elements that are solid at room temperature are to be identified.
Concept introduction:
A periodic table is an arrangement of elements based on their atomic number, properties and electronic configuration. The table is arranged into groups and periods. The elements which are metallic in nature, occupy the large lower-left portion of the table. The non-metals occupy the small upper-right portion of the table. Metalloids like along the staircase line. Elements which appear in the same group have similar behaviour.
(e)
Interpretation:
A pair of elements that will form a covalent compound is to be determined.
Concept introduction:
Covalent compounds are formed by the interaction of two or more non-metal elements. In covalent compounds, the covalent bonds are formed by the sharing of electrons between the atoms instead of their transfer from one atom to another.
(f)
Interpretation:
Another pair of elements that will likely form covalent compounds is to be determined.
Concept introduction:
Covalent compounds are formed by the interaction of two or more non-metal elements. In covalent compounds, the covalent bonds are formed by the sharing of electrons between the atoms instead of their transfer from one atom to another.
(g)
Interpretation:
The coloured boxes that represent a pair of elements that will likely form an ionic compound with the formula
Concept introduction:
Ionic compounds are formed by the interaction of metal elements with non-metal elements. In an ionic bond formation, there is a transfer of electrons between atoms. The metal elements have a tendency to lose electrons in order to gain stability, whereas non-metals acquire stability by gaining the electrons. The ions thus formed attract each other due to strong electrostatic force between them to form ionic compounds.
(h)
Interpretation:
The coloured boxes that represent another pair of elements that will likely form an ionic compound with the formula
Concept introduction:
Ionic compounds are formed by the interaction of metal elements with non-metal elements. In an ionic bond formation, there is a transfer of electrons between atoms. The metal elements have a tendency to lose electrons in order to gain stability, whereas non-metals acquire stability by gaining the electrons. The ions thus formed attract each other due to strong electrostatic force between them to form ionic compounds.
(i)
Interpretation:
The coloured boxes that represent a pair of elements that will likely form an ionic compound with the formula
Concept introduction:
Ionic compounds are formed by the interaction of metal elements with non-metal elements. In an ionic bond formation, there is a transfer of electrons between atoms. The metal elements have a tendency to lose electrons in order to gain stability, whereas non-metals acquire stability by gaining the electrons. The ions thus formed attract each other due to strong electrostatic force between them to form ionic compounds.
(j)
Interpretation:
The coloured boxes that represent a pair of elements that will likely form an ionic compound with the formula
Concept introduction:
Ionic compounds are formed by the interaction of metal elements with non-metal elements. In an ionic bond formation, there is a transfer of electrons between atoms. The metal elements have a tendency to lose electrons in order to gain stability, whereas non-metals acquire stability by gaining the electrons. The ions thus formed attract each other due to strong electrostatic force between them to form ionic compounds.
(k)
Interpretation:
The coloured box that represents an element that forms no compound is to be determined.
Concept introduction:
A periodic table is an arrangement of elements based on their atomic number, properties and electronic configuration. The table is arranged into groups and periods. The elements which are metallic in nature, occupy the large lower-left portion of the table. The non-metals occupy the small upper-right portion of the table. Metalloids like along the staircase line. Elements which appear in the same group have similar behaviour.
(l)
Interpretation:
The coloured boxes that represent a pair of elements whose compounds exhibit the law of multiple proportions are to be determined.
Concept introduction:
Law of multiple proportions states that, if two elements can combine to form more than one compound, the masses of one element that combines with a fixed mass of the other element are in the ratio of small whole numbers.
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Chapter 2 Solutions
CHEMISTRY: MOLECULAR...V2SSM <CUSTOM>
- Part C IN H N. Br₂ (2 equiv.) AlBr3 Draw the molecule on the canvas by choosing buttons from the Tools (for bonds and + e (×) H± 12D T EXP. L CONT. דarrow_forward9. OA. Rank the expected boiling points of the compounds shown below from highest to lowest. Place your answer appropriately in the box. Only the answer in the box will be graded. (3) points) OH OH بر بد بدید 2 3arrow_forwardThere is an instrument in Johnson 334 that measures total-reflectance x-ray fluorescence (TXRF) to do elemental analysis (i.e., determine what elements are present in a sample). A researcher is preparing a to measure calcium content in a series of well water samples by TXRF with an internal standard of vanadium (atomic symbol: V). She has prepared a series of standard solutions to ensure a linear instrument response over the expected Ca concentration range of 40-80 ppm. The concentrations of Ca and V (ppm) and the instrument response (peak area, arbitrary units) are shown below. Also included is a sample spectrum. Equation 1 describes the response factor, K, relating the analyte signal (SA) and the standard signal (SIS) to their respective concentrations (CA and CIS). Ca, ppm V, ppm SCa, arb. units SV, arb. units 20.0 10.0 14375.11 14261.02 40.0 10.0 36182.15 17997.10 60.0 10.0 39275.74 12988.01 80.0 10.0 57530.75 14268.54 100.0…arrow_forward
- A mixture of 0.568 M H₂O, 0.438 M Cl₂O, and 0.710 M HClO are enclosed in a vessel at 25 °C. H₂O(g) + C₁₂O(g) = 2 HOCl(g) K = 0.0900 at 25°C с Calculate the equilibrium concentrations of each gas at 25 °C. [H₂O]= [C₁₂O]= [HOCI]= M Σ Marrow_forwardWhat units (if any) does the response factor (K) have? Does the response factor (K) depend upon how the concentration is expressed (e.g. molarity, ppm, ppb, etc.)?arrow_forwardProvide the structure, circle or draw, of the monomeric unit found in the biological polymeric materials given below. HO OH amylose OH OH 행 3 HO cellulose OH OH OH Ho HOarrow_forward
- OA. For the structure shown, rank the bond lengths (labeled a, b and c) from shortest to longest. Place your answer in the box. Only the answer in the box will be graded. (2 points) H -CH3 THe b Нarrow_forwardDon't used hand raitingarrow_forwardQuizzes - Gen Organic & Biological Che... ☆ myd21.lcc.edu + O G screenshot on mac - Google Search savings hulu youtube google disney+ HBO zlib Homework Hel...s | bartleby cell bio book Yuzu Reader: Chemistry G periodic table - Google Search b Home | bartleby 0:33:26 remaining CHEM 120 Chapter 5_Quiz 3 Page 1: 1 > 2 > 3 > 6 ¦ 5 > 4 > 7 ¦ 1 1 10 8 ¦ 9 a ¦ -- Quiz Information silicon-27 A doctor gives a patient 0.01 mC i of beta radiation. How many beta particles would the patient receive in I minute? (1 Ci = 3.7 x 10 10 d/s) Question 5 (1 point) Saved Listen 2.22 x 107 222 x 108 3.7 x 108 2.22 x 108 none of the above Question 6 (1 point) Listen The recommended dosage of 1-131 for a test is 4.2 μCi per kg of body mass. How many millicuries should be given to a 55 kg patient? (1 mCi = 1000 μСi)? 230 mCiarrow_forward
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