Ch u1l18 unit_test_answers
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This document contains a unit test with multiple chemistry questions on stoichiometry. It includes 3 questions (labeled Q25, Q26, Q27) that involve calculating molar ratios, empirical and molecular formulas, and masses of reactants and products from balanced chemical equations. The test also contains questions on calculating molarity (Q29), moles and volumes of gases (Q30) from given values like mass and temperature. Overall, the test assesses students' abilities to set up and solve various stoichiometry calculations that are fundamental to chemistry.
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![UNIT TEST
Unit 1 Stoichiometry – Teacher Marked
Questions
Q25
C H O
Mol 40.00/12.01 = 3.331 Mol 6.73 / 1.01 = 6.663 Mol 53.27 / 16.00 = 3.330
(1 mark)
RatioC: H: O = 1:2:1
(1 mark)
Empirical formula=CH2O
(1 mark)
Mr CH2O = 30
(1 mark)
Molecularformula= 90 / 30 = 3 = (CH2O)3 = C3H6O3
(1 mark)
Q26
Mol KCl = 15.62 / [39.10 + 35.45] = 0.2095 mol
(1 mark)
Mol H2SO4 = (3/4) × 0.2095 = 0.1571 mol
(1 mark)
Mass H2SO4 = 0.1571 × [(1.01 × 2) + 32.07 + (16.00 × 4)] = 15.41g
(1 mark)
RatioKCl : CrO2Cl2 = 4:2 = 2:1
(1 mark)
Mol CrO2Cl2 = 0.2095 / 2 = 0.1048 mol
(1 mark)
Mass CrO2Cl2 =0.1048 × [52.00 + (16.00 × 2) + (35.45 × 2)] = 16.23g
(1 mark)](https://image.slidesharecdn.com/chu1l18unittestanswers-210815155539/85/Ch-u1l18-unit_test_answers-1-320.jpg)
![Q27
Mol KCl = 0.8114 / (39.10 + 35.45) = 0.01088 mol
(1 mark)
Mol K2Cr2O7 =0.2471 / [(39.10 × 2) + (52.00 × 2) + (16.00 × 7)] = 0.0008399 mol
(1 mark)
RatioKCl : K2Cr2O7 = 4:1
0.01088 mol KCl wouldneed 0.00272 mol K2Cr2O7 so K2Cr2O7 = limitingreagent(getsusedup)
(1 mark)
Mol CrO2Cl2 =0.0008399 × 2 = 0.0016798 mol
Mass CrO2Cl2 = 0.0016798 × [52.00 + (16.00 × 2) + (35.45 × 2)] = 0.2602g
(1 mark)
Mol K2SO4 = 0.0008399 × 3 = 0.0025197 mol
Mass K2SO4 = 0.0025197 × [(39.10 × 2) + 32.07 + (16.00 × 4)] = 0.4391g
(1 mark)
MolesH2O = 0.0008399 × 3 = 0.0025197 mol
Mass H2O = 0.0025197 × [(1.01 × 2) + 16.00] = 0.0454g
(1 mark)
Q29
Mol NaOH = 0.8442 × (12.85 / 1000) = 0.01085 mol
(1 mark)
RatioNaOH : CHCl3 = 7:2
Mol CHCl3 = 0.01085 × (2/7) = 0.0031 mol
(1 mark)
Concentration=0.0031 / (18.45/1000) = 0.1680 mol dm-3
(1 mark)
Q30
Mol Cl2 = 1068 / (35.45 × 2) = 15.06 mol
(1 mark)](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Okay, here are the steps:
1) Convert the mass percentages to grams of each element in 100 g of the compound:
K: 24.75% = 24.75 g
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O: 40.51% = 40.51 g
2) Calculate the moles of each element:
K: 24.75 g / 39.10 g/mol (molar mass of K) = 0.634 mol
Mn: 34.77 g / 54.94 g/mol (molar mass of Mn) = 0.634 mol
O: 40.51 g / 16.00 g/mol (molar mass of O)
PS F4 Set 1
The document provides a marking scheme for a chemistry exam with questions on the structure of atoms and chemical equations. It includes:
1) 19 multiple choice and short answer questions on topics like chemical formulas, moles, molarity, and balancing chemical equations.
2) The answers and explanations for each question are provided along with the number of marks for each.
3) The marking scheme is broken into three papers covering the structure of atoms, chemical formulas and equations, and diffusion.
The marking scheme provides a comprehensive review of key concepts tested on the chemistry exam and the level of understanding and work required to earn marks for each question.
Mole concept
The document discusses moles, molar mass, molar concentration, and stoichiometry. It defines a mole as 6.02 x 1023 particles, whether atoms, molecules, ions, or electrons. Molar mass is the mass of one mole of a substance in grams. Molar concentration, or molarity, expresses the moles of solute per liter of solution. Stoichiometry uses mole ratios from balanced chemical equations to calculate amounts of reactants and products.
Chapter_3_Mass_Relationships.ppt
1. The document discusses mass relationships in chemical reactions including atomic mass, molar mass, molecular mass, formula mass, and percent composition.
2. It defines the mole as the amount of a substance containing 6.022x1023 elementary entities, such as atoms or molecules. Molar mass is the mass in grams of one mole of a substance.
3. Examples are provided for calculating the number of atoms, moles of substances, and masses involved in chemical reactions using molar mass and mole ratios from balanced chemical equations.
Ib- Topic 1 Stoichiometric - Chemistry HL
- A physical change involves no new substances being formed and includes processes like melting, freezing, evaporation and sublimation. Heat is either absorbed or released during physical changes.
- A chemical change results in new chemical substances being formed through atoms rearranging. The products of chemical changes often have very different properties than the reactants. Examples include combustion, corrosion and electrolysis.
- Matter can exist in three physical states - solid, liquid and gas. Solids have a fixed shape and volume, liquids have a fixed volume but not shape, and gases have neither a fixed shape nor volume.
Ch3 stoichiometry
This document discusses stoichiometry, which is the quantitative relationships between reactants and products in chemical reactions. Some key points covered include:
- Stoichiometric amounts refer to the exact molar amounts of reactants and products in a balanced chemical equation.
- Atomic masses are given in atomic mass units (amu) which is based on carbon-12 having a mass of exactly 12 amu.
- Average atomic masses take into account the natural abundances of isotopes.
- The mole is the unit used to express amounts of substances in chemistry and 1 mole contains 6.022x1023 elementary entities.
- Molar mass is the mass in grams of 1 mole of a substance and is equal to the sum of the
05a the mole
The document discusses the mole concept and related terms. It defines the mole as the unit used to count particles in chemistry and equals 6.022 x 1023 particles. It explains that one mole of an element equals its atomic mass in grams. Formula mass is the sum of atomic masses of elements in a formula unit, while molar mass refers to grams per mole of a pure substance. It provides examples of calculating moles, mass, and grams in chemical equations and reactions.
#17 Key
1. The document contains 7 chemistry problems involving calculation of equilibrium constants (Keq and Kc) for various chemical reactions.
2. Problem 3 calculates Keq as 1.84x10-2 for the decomposition of HI gas into H2 and I2 gases at 425C, given the partial pressures of reactants and products at equilibrium.
3. Problem 7 involves a multiple step chemical equilibrium problem, calculating the initial and equilibrium partial pressures and concentrations of gases in a reaction between CO2, H2 and H2O, and determining the equilibrium constant Kp.
Trabajo encargado analisis de alimentos
The document provides information on the molecular structure, formula, and properties of various classes of organic compounds including monosaccharides, disaccharides, amino acids, vitamins, alkaloids, steroids, and more. Each compound lists its molecular formula, weight, composition, and other physical properties.
PERCENTAGE COMPOSITION, EMPIRICAL AND MOLECULAR FORMULAS.ppt
Here are the steps to solve these sample problems:
1. Empirical Formula: 0.44 g H = 0.044 mol H = 2 mol H
6.92 g O = 0.433 mol O = 1 mol O
Empirical Formula = H2O
Molecular Formula = H2O (same as empirical)
2. Empirical Formula: 36.48% Na = 36.48 g Na = 1.5 mol Na
25.41% S = 25.41 g S = 1 mol S
38.11% O = 38.11 g O = 2.375 mol O
Empirical Formula = NaSO2
Molar mass of NaSO2 = 105 g/mol
HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.10.04 = 4.pdf
HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.1*0.04 = 4*10-3(mol) When added
10ml NaOH n(NaOH) used = 0.1 * 0.01 = 1*10-3(mol) n(HClO3) left = 3*10-3(mol) [HClO3] =
3*10-3/ 0.05 = 0.06 (M) [ClO3-] = 1*10-3/ 0.05 = 0.02(M) pH=pKa + log[ClO3-]/[HClO3] =
7.42 - 0.477 = 6.943 When added 20ml NaOH n(NaOH) used = 0.1 * 0.02 = 2*10-3(mol)
n(HClO3) left = 2*10-3(mol) [HClO3] = 2*10-3/ 0.06 = 0.033 (M) [ClO3-] = 2*10-3/ 0.06 =
0.033(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0 = 7.42 When added 30ml NaOH
n(NaOH) used = 0.1 * 0.03 = 3*10-3(mol) n(HClO3) left = 1*10-3(mol) [HClO3] = 1*10-3/
0.07 = 0.0143 (M) [ClO3-] = 3*10-3/ 0.07 = 0.0429(M) pH=pKa + log[ClO3-]/[HClO3] =
7.42 + 0.477 = 7.897 When added 40ml NaOH n(NaOH) used = 0.1 * 0.04 = 4*10-3(mol)
n(HClO3) left = 0(mol) [ClO3-] = 4*10-3/ 0.08 = 0.05(M) Kb = Kw/Ka = 2.63 *10-7 [OH-] =
Kb[ClO3-] =2.63*10-7*0.05 =1.14*10-4pOH = 3.94 pH = 10.06
Solution
HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.1*0.04 = 4*10-3(mol) When added
10ml NaOH n(NaOH) used = 0.1 * 0.01 = 1*10-3(mol) n(HClO3) left = 3*10-3(mol) [HClO3] =
3*10-3/ 0.05 = 0.06 (M) [ClO3-] = 1*10-3/ 0.05 = 0.02(M) pH=pKa + log[ClO3-]/[HClO3] =
7.42 - 0.477 = 6.943 When added 20ml NaOH n(NaOH) used = 0.1 * 0.02 = 2*10-3(mol)
n(HClO3) left = 2*10-3(mol) [HClO3] = 2*10-3/ 0.06 = 0.033 (M) [ClO3-] = 2*10-3/ 0.06 =
0.033(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0 = 7.42 When added 30ml NaOH
n(NaOH) used = 0.1 * 0.03 = 3*10-3(mol) n(HClO3) left = 1*10-3(mol) [HClO3] = 1*10-3/
0.07 = 0.0143 (M) [ClO3-] = 3*10-3/ 0.07 = 0.0429(M) pH=pKa + log[ClO3-]/[HClO3] =
7.42 + 0.477 = 7.897 When added 40ml NaOH n(NaOH) used = 0.1 * 0.04 = 4*10-3(mol)
n(HClO3) left = 0(mol) [ClO3-] = 4*10-3/ 0.08 = 0.05(M) Kb = Kw/Ka = 2.63 *10-7 [OH-] =
Kb[ClO3-] =2.63*10-7*0.05 =1.14*10-4pOH = 3.94 pH = 10.06.
Propiedades fq compuestos varios
This document provides physical property data for various organic and inorganic substances in two tables:
Table A1 lists molecular weights, melting points, boiling points, densities, and heat capacities for compounds like acetaldehyde, acetic acid, benzene, carbon dioxide, ethanol, methane, and water.
Table A2 gives heats of formation and combustion for several compounds, including acetic acid, acetaldehyde, acetone, acetylene, and ammonia.
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Ch u1l18 unit_test_answers
- 1. UNIT TEST Unit 1 Stoichiometry – Teacher Marked Questions Q25 C H O Mol 40.00/12.01 = 3.331 Mol 6.73 / 1.01 = 6.663 Mol 53.27 / 16.00 = 3.330 (1 mark) RatioC: H: O = 1:2:1 (1 mark) Empirical formula=CH2O (1 mark) Mr CH2O = 30 (1 mark) Molecularformula= 90 / 30 = 3 = (CH2O)3 = C3H6O3 (1 mark) Q26 Mol KCl = 15.62 / [39.10 + 35.45] = 0.2095 mol (1 mark) Mol H2SO4 = (3/4) × 0.2095 = 0.1571 mol (1 mark) Mass H2SO4 = 0.1571 × [(1.01 × 2) + 32.07 + (16.00 × 4)] = 15.41g (1 mark) RatioKCl : CrO2Cl2 = 4:2 = 2:1 (1 mark) Mol CrO2Cl2 = 0.2095 / 2 = 0.1048 mol (1 mark) Mass CrO2Cl2 =0.1048 × [52.00 + (16.00 × 2) + (35.45 × 2)] = 16.23g (1 mark)
- 2. Q27 Mol KCl = 0.8114 / (39.10 + 35.45) = 0.01088 mol (1 mark) Mol K2Cr2O7 =0.2471 / [(39.10 × 2) + (52.00 × 2) + (16.00 × 7)] = 0.0008399 mol (1 mark) RatioKCl : K2Cr2O7 = 4:1 0.01088 mol KCl wouldneed 0.00272 mol K2Cr2O7 so K2Cr2O7 = limitingreagent(getsusedup) (1 mark) Mol CrO2Cl2 =0.0008399 × 2 = 0.0016798 mol Mass CrO2Cl2 = 0.0016798 × [52.00 + (16.00 × 2) + (35.45 × 2)] = 0.2602g (1 mark) Mol K2SO4 = 0.0008399 × 3 = 0.0025197 mol Mass K2SO4 = 0.0025197 × [(39.10 × 2) + 32.07 + (16.00 × 4)] = 0.4391g (1 mark) MolesH2O = 0.0008399 × 3 = 0.0025197 mol Mass H2O = 0.0025197 × [(1.01 × 2) + 16.00] = 0.0454g (1 mark) Q29 Mol NaOH = 0.8442 × (12.85 / 1000) = 0.01085 mol (1 mark) RatioNaOH : CHCl3 = 7:2 Mol CHCl3 = 0.01085 × (2/7) = 0.0031 mol (1 mark) Concentration=0.0031 / (18.45/1000) = 0.1680 mol dm-3 (1 mark) Q30 Mol Cl2 = 1068 / (35.45 × 2) = 15.06 mol (1 mark)
- 3. Volume Cl2 =nRT / P = 15.06 × 8.31 × [273+25] / 89589 = 0.4163 m3 (1 mark) RatioCl2:O2 = 10:5 = 2:1 Volume O2 = 0.4163 / 2 = 0.2081 m3 (1 mark) [Total: 50 marks]