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... a Students describe their reasoning that connects the evidence, along with the assumption that theories and laws that describe their natural world operate today as they did in the past and will continue to do so in the future, to construct an explanation for how the patterns of outermost electrons a ...
... a Students describe their reasoning that connects the evidence, along with the assumption that theories and laws that describe their natural world operate today as they did in the past and will continue to do so in the future, to construct an explanation for how the patterns of outermost electrons a ...
1 Intro / Review : Chemical Kinetics
... Also, the orientations of the reactant molecules during the collision must allow for the rearrangement of reactant bonds to form product bonds. Essential knowledge 4.B.3: A successful collision can be viewed as following a reaction path with an associated energy profile. Enduring understanding 4.D: ...
... Also, the orientations of the reactant molecules during the collision must allow for the rearrangement of reactant bonds to form product bonds. Essential knowledge 4.B.3: A successful collision can be viewed as following a reaction path with an associated energy profile. Enduring understanding 4.D: ...
m5zn_1ed95c16cede0b1
... 2- Calculate the mass of all elements in 8 g of Na3PO4. (Na = 23, P= 31, 0 = 16) → Mwt = (23 x 3) + 31 + (16x4) = 164 g/mol Mass of sodium = (23x3/164) x 8 = 3.36585 g. Mass of phosphorus = = (31/164) x 8 = 1.5122 g Mass of oxygen = = (16x4/164) x 8 = 3.1219 g Note: the sum of all masses should be e ...
... 2- Calculate the mass of all elements in 8 g of Na3PO4. (Na = 23, P= 31, 0 = 16) → Mwt = (23 x 3) + 31 + (16x4) = 164 g/mol Mass of sodium = (23x3/164) x 8 = 3.36585 g. Mass of phosphorus = = (31/164) x 8 = 1.5122 g Mass of oxygen = = (16x4/164) x 8 = 3.1219 g Note: the sum of all masses should be e ...
Chapter 11 Chemical Calculations
... Divide the g/mol fo reach element by the atomic mass 36.00 g/mol C /12 g/atom = 3 atoms/mole = C3 6 g/molH / 1g/atom H = 6 atoms/mol = H6 Molecular formula = C3H6 Key calculation Given % composition and molar mass be able to determine the molecular formula. This can be done by two different procedur ...
... Divide the g/mol fo reach element by the atomic mass 36.00 g/mol C /12 g/atom = 3 atoms/mole = C3 6 g/molH / 1g/atom H = 6 atoms/mol = H6 Molecular formula = C3H6 Key calculation Given % composition and molar mass be able to determine the molecular formula. This can be done by two different procedur ...
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... element or compound) they weigh exactly the same number of grams as the relative atomic mass of the element or compound 1 mole of atoms of molecules of any substance will have a mass in ...
... element or compound) they weigh exactly the same number of grams as the relative atomic mass of the element or compound 1 mole of atoms of molecules of any substance will have a mass in ...
Unit Two Objectives
... It VERY important to remember that Standard Atmospheric Pressure is 101.3 kPa, or 1 atmosphere, or 760 mm Hg (torr). The boiling point for liquids AT THIS PRESSURE is the NORMAL BOILING POINT. 3. Interpret a phase diagram of a substance at any given temperature and pressure. The TRIPLE POINT is wher ...
... It VERY important to remember that Standard Atmospheric Pressure is 101.3 kPa, or 1 atmosphere, or 760 mm Hg (torr). The boiling point for liquids AT THIS PRESSURE is the NORMAL BOILING POINT. 3. Interpret a phase diagram of a substance at any given temperature and pressure. The TRIPLE POINT is wher ...
Week 6 Review 2014-15
... All zeros to the right of numbers aren’t significant if there’s no decimal EX: 1200 is 2s.f., 1000 is 1. (trailing zeros) All zeros to the left of numbers aren’t significant if there’s no number in front EX: 0.0045 is 2s.f. (Leading zeros) All zeros after a number and a decimal are significant EX: 1 ...
... All zeros to the right of numbers aren’t significant if there’s no decimal EX: 1200 is 2s.f., 1000 is 1. (trailing zeros) All zeros to the left of numbers aren’t significant if there’s no number in front EX: 0.0045 is 2s.f. (Leading zeros) All zeros after a number and a decimal are significant EX: 1 ...
1.2 The Mole Concept
... • Molar Mass: the mass of one mole of a substance (units: g mol-1) • Note: because this is the mass of 1 mole this means that this is the mass of 6.02 x 1023 atoms of a element or 6.02 x 1023 molecules/formula units of a compound! ...
... • Molar Mass: the mass of one mole of a substance (units: g mol-1) • Note: because this is the mass of 1 mole this means that this is the mass of 6.02 x 1023 atoms of a element or 6.02 x 1023 molecules/formula units of a compound! ...
Chemical equilibrium and the kinetic theory of gases
... specific reaction conditions). The ideal gas equation enables a simple prediction of these volumes to be made. In different manufacturing locations, different units may be used for pressure, volume (and even temperature; for instance, °F is still in common use in the USA), so before the ideal gas eq ...
... specific reaction conditions). The ideal gas equation enables a simple prediction of these volumes to be made. In different manufacturing locations, different units may be used for pressure, volume (and even temperature; for instance, °F is still in common use in the USA), so before the ideal gas eq ...
Ch6.Thermochem - Mr. Fischer.com
... A state function (aka function of state) is a property whose value depends only on the state of the system, not how it achieved that state The state of the system is specified by the pressure, temperature, and composition of the system. P, V, & T are state functions. E is a state function. q a ...
... A state function (aka function of state) is a property whose value depends only on the state of the system, not how it achieved that state The state of the system is specified by the pressure, temperature, and composition of the system. P, V, & T are state functions. E is a state function. q a ...
Chapter 07 Notes - Mr. Julien`s Homepage
... Conservation of Matter and Stoichiometry 3. The conservation of atoms in chemical reactions leads to the principle of conservation of matter and the ability to calculate the mass of products and reactants. As a basis for understanding this concept: b. Students know the quantity one mole is set by de ...
... Conservation of Matter and Stoichiometry 3. The conservation of atoms in chemical reactions leads to the principle of conservation of matter and the ability to calculate the mass of products and reactants. As a basis for understanding this concept: b. Students know the quantity one mole is set by de ...
Chem EOC Review Cumulative Free Response
... a) A 7.0 liter balloon at room temperature (22oC) contains hydrogen gas. If the balloon is carried outside to where the temperature is –3.0oC, what volume will the balloon occupy? b) A 5.0 liter tank of oxygen gas is at a pressure of 3 atm. What volume of oxygen will be available if the oxygen is us ...
... a) A 7.0 liter balloon at room temperature (22oC) contains hydrogen gas. If the balloon is carried outside to where the temperature is –3.0oC, what volume will the balloon occupy? b) A 5.0 liter tank of oxygen gas is at a pressure of 3 atm. What volume of oxygen will be available if the oxygen is us ...
LESSON 23: Exploding Bags
... the structure or composition of the materials change. Chemical reactions occur around us all the time. When a chemical change is complete, the resulting substance(s) is/are different from the original substance(s). The substance or substances that start a chemical reaction are called reactants. The ...
... the structure or composition of the materials change. Chemical reactions occur around us all the time. When a chemical change is complete, the resulting substance(s) is/are different from the original substance(s). The substance or substances that start a chemical reaction are called reactants. The ...
Summer Assignment Packet
... a. __C4H6(g) + __O2(g) → __CO2(g) + __H2O(l) b. __NH3(g) + __O2(g) → __NO2(g) + __H2O(l) c. __PCl3(l) + __H2O(l) → __H3PO3(aq) + __HCl(aq) d. __Ca3P2(s) + __H2O(l) → __Ca(OH)2(aq) + __PH3(g) e. __C4H8(OH)2(l) + __O2(g) → __CO2(g) + __H2O(l) f. __NH3(g) + __NO(g) → __N2(g) + __H2O(l) g. __KClO3(s) → ...
... a. __C4H6(g) + __O2(g) → __CO2(g) + __H2O(l) b. __NH3(g) + __O2(g) → __NO2(g) + __H2O(l) c. __PCl3(l) + __H2O(l) → __H3PO3(aq) + __HCl(aq) d. __Ca3P2(s) + __H2O(l) → __Ca(OH)2(aq) + __PH3(g) e. __C4H8(OH)2(l) + __O2(g) → __CO2(g) + __H2O(l) f. __NH3(g) + __NO(g) → __N2(g) + __H2O(l) g. __KClO3(s) → ...
Thermochemistry
... increases to 28.1°C. Assuming that the calorimeter absorbs only a negligible quantity of heat, that the specific heat capacity of the solution 4.18 J/°C x g, and that the density of the final solution is 1.0 g/mL, calculate the enthalpy change per mole of BaSO4. (∆H=specific heat x mass of solution ...
... increases to 28.1°C. Assuming that the calorimeter absorbs only a negligible quantity of heat, that the specific heat capacity of the solution 4.18 J/°C x g, and that the density of the final solution is 1.0 g/mL, calculate the enthalpy change per mole of BaSO4. (∆H=specific heat x mass of solution ...
Electrochemistry
... Now, cancel the species that appear on both sides of the equation: Finally, since we know that the reaction takes place in basic solution, we must add 2 OH – ions to both sides of the equation to neutralize the 2 H+ ions on the right giving 2 additional H2O. The final net ionic equation, balanced fo ...
... Now, cancel the species that appear on both sides of the equation: Finally, since we know that the reaction takes place in basic solution, we must add 2 OH – ions to both sides of the equation to neutralize the 2 H+ ions on the right giving 2 additional H2O. The final net ionic equation, balanced fo ...
Chapter 2
... form molecules and compounds Molecule—two or more atoms bonded together (e.g., H2 or C6H12O6) Compound—two or more different kinds of atoms bonded together (e.g., C6H12O6) ...
... form molecules and compounds Molecule—two or more atoms bonded together (e.g., H2 or C6H12O6) Compound—two or more different kinds of atoms bonded together (e.g., C6H12O6) ...
CHEM 481. Assignment 0. Review of General Chemistry. Answers
... 5. Name and give symbols for (a) three elements that are metals, (b) four elements that are nonmetals, (c) and two elements that are metalloids. In each case locate the element in the periodic table by giving the group and period in which the element is found. Metalloids are: B, Si, Ge, As, Sb, Te; ...
... 5. Name and give symbols for (a) three elements that are metals, (b) four elements that are nonmetals, (c) and two elements that are metalloids. In each case locate the element in the periodic table by giving the group and period in which the element is found. Metalloids are: B, Si, Ge, As, Sb, Te; ...
Section 4.8: Acid-Base Reactions
... reagent (the titrant), that is required to react completely with an exact amount of another reagent, is measured precisely. The method presumes that the titration reaction goes to completion. In most titrations the concentration of titrant is known, in which case the exact amount of analyte, the sub ...
... reagent (the titrant), that is required to react completely with an exact amount of another reagent, is measured precisely. The method presumes that the titration reaction goes to completion. In most titrations the concentration of titrant is known, in which case the exact amount of analyte, the sub ...
Stoichiometry
Stoichiometry /ˌstɔɪkiˈɒmɨtri/ is the calculation of relative quantities of reactants and products in chemical reactions.Stoichiometry is founded on the law of conservation of mass where the total mass of the reactants equals the total mass of the products leading to the insight that the relations among quantities of reactants and products typically form a ratio of positive integers. This means that if the amounts of the separate reactants are known, then the amount of the product can be calculated. Conversely, if one reactant has a known quantity and the quantity of product can be empirically determined, then the amount of the other reactants can also be calculated.As seen in the image to the right, where the balanced equation is:CH4 + 2 O2 → CO2 + 2 H2O.Here, one molecule of methane reacts with two molecules of oxygen gas to yield one molecule of carbon dioxide and two molecules of water. Stoichiometry measures these quantitative relationships, and is used to determine the amount of products/reactants that are produced/needed in a given reaction. Describing the quantitative relationships among substances as they participate in chemical reactions is known as reaction stoichiometry. In the example above, reaction stoichiometry measures the relationship between the methane and oxygen as they react to form carbon dioxide and water.Because of the well known relationship of moles to atomic weights, the ratios that are arrived at by stoichiometry can be used to determine quantities by weight in a reaction described by a balanced equation. This is called composition stoichiometry.Gas stoichiometry deals with reactions involving gases, where the gases are at a known temperature, pressure, and volume and can be assumed to be ideal gases. For gases, the volume ratio is ideally the same by the ideal gas law, but the mass ratio of a single reaction has to be calculated from the molecular masses of the reactants and products. In practice, due to the existence of isotopes, molar masses are used instead when calculating the mass ratio.