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Chemistry 103 Module 1 Exam Questions and Answers Latest 2025/2026 COMPLETE FREQUENTLY MOST TESTED QUESTIONS WITH COMPLETE SOLUTION /GET IT 100% ACCURATE!(Graded A+) $10.49
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Chemistry 103 Module 1 Exam Questions and Answers Latest 2025/2026 COMPLETE FREQUENTLY MOST TESTED QUESTIONS WITH COMPLETE SOLUTION /GET IT 100% ACCURATE!(Graded A+)

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Chemistry 103 Module 1 Exam Questions and Answers Latest 2025/2026 COMPLETE FREQUENTLY MOST TESTED QUESTIONS WITH COMPLETE SOLUTION /GET IT 100% ACCURATE!(Graded A+) Chemistry 103 Module 1 Exam Questions and Answers Latest 2025/2026 COMPLETE FREQUENTLY MOST TESTED QUESTIONS WITH COMPLETE SOLUTION...

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  • January 30, 2025
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Chemistry 103 Module 2 Exam g g g g




Questions and Verified Answers / A+ Grade g g g g g g




1. 2.1: MOLECULAR WEIGHT: A compound is made up of two or more elements
g g g g g g g g g g g g




combined in a definite ratio that is represented by a molecular formula.Each of these elem
g g g g g g g g g g g g g g g




ents has a certain atomic weight, which can be found in the periodic table.The sum of the
g g g g g g g g g g g g g g g g g




atomic weights of the atoms in the molecular formula is called the formula weight or mole
g g g g g g g g g g g g g g g




cular weight or formula mass.
g g g g




2. Ca3(PO4)2: calcium phosphate g g g




Molecular Weight = g g




3 Ca= 3 x 40.08=120.24
g g g g




2 P=2 x 30.97=61.94
g g g




8 O=8 x 16.00=128.00
g g g




Total 310.18
g




3. 2.2: MOLES: Chemical compounds react with one another in amounts that are base
g g g g g g g g g g g g




d on their molecular weights; this chemically reactive amount of compound is called a
g g g g g g g g g g g g g g




mole.

4. moles: = grams / molecular weight
g g g g g




1g/g1
4

,5. Calculate the number of moles in 10.0 grams of each of the following compoun
g g g g g g g g g g g g g




ds:: Ca3(PO4)2 10.0 g ÷ 310.18 = 0.0322 mol (3 sig fig because of 10.0 g)
g g g g g g g g g g g g g g g




C3H5O2Cl 10.0 g ÷ 108.52 = 0.0921 mol
g g g g g g g




Al2(SO4)3 10.0 g ÷ 342.17 = 0.0292 mol
g g g g g g g




Ca3(PO4)2 0.0500 mol x 310.18 = 15.5 g g g g g g g g




C3H5O2Cl 0.0500 mol x 108.52 = 5.43 g
g g g g g g g




Al2(SO4)3 0.0500 mol x 342.17 = 17.1 g
g g g g g g g




6. 2.3: PERCENT COMPOSITION: he molecular formula represents the definite ratio of
g g g g g g g g g g




elements in a compound. The weight of each element present in the compound represe
g g g g g g g g g g g g g g




nts a certain percentage of the total weight of the compound.The percentage of each ele
g g g g g g g g g g g g g g g




ment present in a compound is called the % composition of the compound.
g g g g g g g g g g g g




7. The percentage of an element present in a compound can be calculated as s
g g g g g g g g g g g g g




hown below:: % of an element = weight of element / molecular weight of compoun
g g g g g g g g g g g g g g




d x 100
g g




2g/g1
4

,8. Ca3(PO4)2: 3 Ca = 3 x 40.08= 120.24
g g g g g g g g




2 P = 2 x 30.97= 61.94
g g g g g g




8 O = 8 x 16.00 = 128.00
g g g g g g g




3g/g1
4

, 310.18



% Ca = (120.24 ÷ 310.18) x 100 = 38.76%
g g g g g g g g g




%P = (61.94 ÷ 310.18) x 100 = 19.97%
g g g g g g g g




%O = (128.00 ÷ 310.18) x 100 = 41.27%
g g g g g g g g




9. 2.4: EMPIRICAL FORMULA: If the formula of a compound is known, the % composit
g g g g g g g g g g g g g




ion of the compound can be determined. This process can be done in reverse: The formu
g g g g g g g g g g g g g g g




la of the compound can be determined if the % of each element present in the compoun
g g g g g g g g g g g g g g g g




d is known. The formula calculated from % composition
g g g g g g g g




is known as the empirical formula (or the simplest formula). The actual molecular formu
g g g g g g g g g g g g g




la is some multiple of this simplest formula, which is determined by knowing the molecu
g g g g g g g g g g g g g g




lar weight. g




10. To determine the empirical formula:: (1) Divide each element % by its exact ato
g g g g g g g g g g g g g




mic weight to give a set of numbers.
g g g g g g g




(2) Divide the smallest of this set of decimal numbers into each of the numbers (includin
g g g g g g g g g g g g g g g




g itself) to yield a second set of numbers.
g g g g g g g g




(3a) Round off each of the second set of decimal numbers to a whole number. OR
g g g g g g g g g g g g g g g




4g/g1
4

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