charge on an ion formed when an atom gains two electrons? -2 The Mole 1) Find the molar mass of zinc 65g/mol 2)Find the molar mass of sulfuric acid H2 SO4 98g/mol 3) How many moles of sodium chloride are present in 117g of Na2 CI 2 moles 4) I have 54g of water (H2O) and 84 g of iron (Fe) do I have more moles of water or of iron I have more miles of water (54 g = 3 moles H20 ‚ 84g = 1.5 moles Fe) Arrangement of Electrons 1) Draw diagrams to show the electron
Premium Atom Ion Chemical element
Chem Final Exam Combination Combination reactions‚ also known as synthesis reactions‚ involve the combination of two or more substances into a single new substance: A + B ---> AB Many metal oxides‚ for example‚ absorb carbon dioxide (CO2) from the air to form carbonates: CaO(s) + CO2(g) ---> CaCO3(s) Decomposition Decomposition reactions represent the reverse of the combination process in that one substance breaks down into two or more substances: AB ---> A + B Water‚ or H2O‚ for example
Premium Chemical reaction Energy Temperature
(iii) How much excess reagent remains after the reaction is complete? Answers: (i) Solution for determining the limiting reagent Determine the moles of Al2S3 and H2O aluminum sulfide: 15.00 g ÷ 150.158 g/mol = 0.099895 mol water: 10.00 g ÷ 18.015 g/mol = 0.555093 mol Divide each mole amount by equation coefficient aluminum sulfide: 0.099895 mol ÷ 1 mol = 0.099895 water: 0.555093 mol ÷ 6 mol = 0.0925155 which is less than 0.099895 (for aluminum sulfide) Therefore
Premium Sodium Reagent Stoichiometry
Chapter 3: Stoichiometry 3: Stoichiometry 5: Thermochemistry 8: Covalent Bonding and Molecular Structure 15: Chemical Equilibrium 16: Acids and Bases 3.2 Stoichiometry and Compound Formulas 3.1 The Mole and Molar Mass 3.2 Stoichiometry and Compound Formulas 3.3 Stoichiometry and Chemical Reactions 3.4 Stoichiometry and Limiting Reactants 3.5 Chemical Analysis Chapter Summary Chapter Summary Assignment Reference Tools Periodic Table Molarity Calculator Molar Mass Calculator Unit
Premium Chemistry Chemical reaction Oxygen
Heats of Reaction – Hess’s Law Overview Hess’s Law states that the energy change for a reaction depends on the enthalpy of the reactants and products and is independent of the pathway of the reaction. In this experiment you will use calorimetry to measure the heats of reaction for three reactions: Reaction 1 Reaction Equation NaOH(s) → Na+(aq) + OH-(aq) Solid NaOH is dissolved in water 2 Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → Na+(aq) + Cl-(aq) + H2O(l) Solutions of NaOH and HCl are mixed 3 NaOH(s)
Premium Chemical reaction Energy Sodium hydroxide
molecules Mass number & atomic number Atomic structure (neutrons‚ protons & electrons) Mole concept & conversion Avogadro’s concept Empirical & molecular formulae Isotopes FHSC1114 Physical Chemistry Centre for Foundation Studies‚ UTAR Objectives To define relative atomic masses of atoms & molecules To define & determine mass no. & atomic no. To determine no. of neutrons‚ protons & electrons To understand mole concept & Avogadro’s concept To determine the empirical & molecular formulae Atomic
Premium Atom Atomic number Isotope
performed. The heat given off by these two reactions was used to determine the stoichiometric ratio and the limiting reactants in each experiment. Introduction: Coefficients in a balanced equations show how many moles of each reactant is needed to react with each other and how many moles of each product that will be formed. Stoichiometry allows us to calculate the amount of reactants needed and also the amount of product. The major basis of stoichiometry is formed by the law of definite proportions
Premium Chemical reaction Sodium hydroxide
of NaOH used Average volume of NaOH Calculations 1. Moles of NaOH = 0.500 mol/L X 0.010525 L = 5.26 x 10-3 mol 2. NaOH + HCl --> H2O + NaCl NaOH: HCl 1:1 ratio Moles of HCl = Moles of NaOH = 5.26 X 10-3 mol 3. Molarity of HCl = 5.26 x 10-3 mol / 0.010 L = 0.526 M 4. Moles of NaOH = 0.500 mol/L x 0.016275 L = 8.14 X 10-3 mol 5. CH3COOH + NaOH --> H2O + CH3COONa CH3COOH: NaOH 1:1 ratio Moles of CH3COOH = Moles of NaOH = 8.14 X 10-3 mol 6. Molarity of CH3COOH = 8.14 x 10-3
Premium Chlorine Acetic acid Sodium chloride
3)3 Moles for Mn(acac)3: 0.00904mol MW of MnCl2.4H2O: 143.856 gmol Mass ofMnCl2.4H2O: 1.3 g Moles of acetyl acetone: 0.0499 mol Manganese (II) chloride tetra hydrate and acetyl acetone reacts in a 1:1 stoichiometric ratios and therefore Manganese (II) chloride tetra hydrate is clearly a limiting reagent for the reaction above. The maximum amount of Manganese (II) chloride tetra hydrate that can be formed is 0.00904 mol. Mass of Mnacac3 (g) = number of moles × Molecular
Premium Stoichiometry Thin layer chromatography Chemical reaction
reaction. Let us represent reactant molecules by A and product molecules by B. Then we can re-write the above equation in the following from: A B (1) This is a very simple reaction where A molecules are converted into B molecules or one mole of A disappears for each mole of B forms. As the time progresses‚ the number A molecules decrease and B molecules increase that is shown in the following figure. 100 80 No. of molecules B molecules 60 40 20 0 0 2 4 Time 6 8 10 A molecules When we say “rate”‚
Premium Chemical reaction Reaction rate Chemistry