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HighFour Chemistry Category C: Grades 9 – 10 Round 1 Thursday, September 17, 2015 The use of calculator is required. tin, Sn Elements in the periodic table are grouped according to their physical and chemical properties of elements. There are periodic variations in the elements’ physical and chemical behaviors. The periodic trend for atomic radius is clear: it increases from top to bottom, and decreases from left to right. Increasing atomic radius Increasing atomic radius Answer #1: Explanation: In the excerpt, the element with the largest atomic radius is tin (Sn). Answer #2: Solution: 6.43 atm The problem is an application of the ideal gas law, . Use the universal gas constant, R = (provided in Table 2 of the Chemistry Data Booklet) ( )( )( ) The pressure of the gas in the container is 6.43 atm. Answer #3: Explanation: phosphorus, P This problem requires a little bit more of analysis. Neutral atoms should have the same number of protons and electrons. Species W, X, and Z all have unequal numbers of protons and electrons. Specie Y is therefore the neutral atom. To determine the element, we know that the no. of protons is equivalent to the atomic number because the number of protons remains unchanged during reactions. Referring to Table 24 (Periodic Table) of the Chemistry Data Booklet, the element with an atomic number of 15 is phosphorus, P. HighFour Chemistry Category C: Grades 9 – 10 Round 1 Thursday, September 17, 2015 The use of calculator is required. Answer #4: Solution: 4825.765 grams Au This is a direct application of the formula Au (from the Periodic Table) is 196.97 g/mol. ( Answer #5: Solution: . The atomic weight of )( ) 521.7 nm This problem is a direct application of the formula . The velocity or speed of light is provided in Table 2 of the Chemistry Data Booklet: . ( ) The green light has a wavelength of 521.7 nm. Answer #6: Solution: KCN To determine the empirical formula of this compound, convert the mass of the elements using the atomic weights: K – 39.10; C – 12.01; N – 14.01. After calculating the no. of moles for each element, we find the mole ratios by dividing with the least number of moles. Element Mass, g Atomic Wt., g/mol Mole, mol Potassium, K 60.1 39.10 1.54 Carbon, C 18.4 12.01 1.53 Nitrogen, N 21.5 14.01 1.53 Mole Ratio The ratio of atoms of K:C:N is therefore 1:1:1. Therefore, the empirical formula of the compound is KCN. HighFour Chemistry Category C: Grades 9 – 10 Round 1 Thursday, September 17, 2015 The use of calculator is required. Answer #7: Explanation: titanium (IV) chloride Ionic compounds are made up of cations and anions. In this case, metal cations take their names from the elements (titanium). Since titanium is a transition metal, it occurs as Ti+2 and Ti+4. In the Stock system, Roman numerals are used to designate different cations. The nonmetallic anions is named by taking the first part of the element name (chlorine) and adding “-ide”. The compound is titanium (IV) chloride. Answer #8: Explanation: triple bond A triple bond is formed when 2 atoms share three pairs of electrons between them. The Lewis structure for acetylene is: H:C C:H 8e- 8eor H−C≡C−H Answer #9: Solution: 58.33 mL The problem is an application of dilution of solutions; the formula is (VC)before = (VC)after ( )( ) ( )( ) To prepare 350 mL of 1.25 M H2SO4 solution from the stock solution, about 58.33 mL of 7.50 M H2SO4 solution. Answer #10: Solution: 5090.91 light year The Avogadro’s number is a constant that can be found in Table 2 of the Chemistry Data Booklet: . If the thickness of a single -3 piece of paper is 8.00 x 10 cm, and the number of pages in the book is 6.02x1023; the thickness of the book is: ( ( ) )( ) HighFour Chemistry Category C: Grades 9 – 10 Round 1 Thursday, September 17, 2015 The use of calculator is required. Answer #11: Explanation: 7 valence eValence electrons are the outermost electrons in an atom; and these electrons are the ones involved in chemical bonding. The valence electron of an atom can be determined thru its electron configuration. Bromine (Br) has an atomic number of 35, its electric configuration is: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5. The outermost shell, n = 4, has 7 electrons. Therefore, the number of valence electrons of bromine (Br) is 7. Answer #12: Solution: 92.51% 73Li The algebraic expression for determining the % abundance of 73Li is: (ave. atomic mass) = (%abundance x atomic mass, ) + (%abundance x atomic mass, ) ( ) The fraction of abundance of 73Li is 0.9251. Therefore, the percentage abundance of 73Li (0.9251 x 100%) is 92.51%. Answer #13: Solution: 27.2 grams NaNO3 Use the molarity formula to determine the no. of grams of NaNO3 crystals. The molecular weight of NaNO3 (Na = 22.99; N = 14.01; O = 16.00) is 85 g/mol. ( )( )( ( ) ) To prepare a 400-mL 0.800 M NaNO3 solution, 27.2 grams of the crystals should be dissolved in 400 mL of distilled water. HighFour Chemistry Category C: Grades 9 – 10 Round 1 Thursday, September 17, 2015 The use of calculator is required. Answer #14: Solution: 22.01 J/mol·K With the formula provided, the entropy of this change phase is: This means that when 1 mole of ice melts at 0°C, there is an increase in entropy of 22.0 J/K. Answer #15: Solution: 24.45 L/mol STP stands for standard temperature (25°C) and pressure (1 atm), as provided in Table 2 of the Chemistry Data Booklet. To solve for the molar volume, use the formula . )( ( ( Answer #16: Solution: ) ) 8 mols HNO3 Based on the partially-balanced chemical equation, to balance out the 3 moles of Cu atoms on the left side of the reaction, there should be 3 moles of Cu ( in Cu(NO3)2 ) on the right side of the reaction. Hence, 3Cu + ?HNO3 ⟶ 3Cu(NO3)2 + ?NO + 4H2O To balance the 8 H atoms on the right side, 8 moles of HNO3 is required. To finalize the balancing, 2 moles of NO is required to balance out the O and N atoms. The balanced chemical reaction is therefore: 3Cu + 8HNO3 ⟶ 3Cu(NO3)2 + 2NO + 4H2O It requires 8 moles of nitric acid (HNO3) to balance out the reaction. HighFour Chemistry Category C: Grades 9 – 10 Round 1 Thursday, September 17, 2015 The use of calculator is required. Answer #17: Explanation: I2 (iodine gas) Reviewing the data provided, it is clear that iodine gas (I2) has the largest molecular weight, and consequently, has the greatest strength of van der Waals’ forces. It would need a higher energy to break apart these forces. This translates into a higher boiling point. In conclusion, iodine gas (I2) has the highest boiling point. This conclusion can be supported by a table where their actual boiling points are provided. Answer #18: Solution: Compound MW, g/mol Boiling Point, °C CH4 16.05 -164 C2H6 30.08 -89 F2 39.0 -188 C3H8 44.11 -42 C4H10 58.14 -0.5 Cl2 70.90 -34 Br2 159.8 59 I2 253.8 185 40% carbon atoms To determine the % composition of the carbon atom, it would be best to tabulate all data: Atomic # of Mass Element % Composition Weight Atoms Equivalent Carbon, C 12.01 6 72.06 40 Hydrogen, H 1.01 12 12.12 6.72 Oxygen, O 16.00 6 96.00 53.28 180.18 100 HighFour Chemistry Category C: Grades 9 – 10 Round 1 Thursday, September 17, 2015 The use of calculator is required. Answer #19: Explanation: zinc (Zn) [Ar] is called the argon core and represents 1s22s22p63s23p6. For an electron configuration of [Ar]4s23d10, the total electrons in the element is 30. The number of electrons in a ground-state element is also equal to the atomic number of the element. Therefore, the element is zinc (Zn). Answer #20: Solution: 257.96 g/mol The molecular weight of H5P3O10 is (H=1.01; P=30.97; O=16.00) 257.96 g/mol.