200 Things to Know to Pass the Chemistry Exam 1. Protons are positively charged (+) with a mass of 1 amu. Example: Which has the greatest nuclear charge? Cl-35 Ar-40 K-39 Ca-40 2. Neutrons have no charge and a mass of 1 amu. 3. Electrons are small and are negatively charged (-) with a mass of almost 0 amu.. 4. Protons & neutrons are in an atom’s nucleus (nucleons). Which has the greatest number of nucleons? Sn-119 Sb-122 Te-128 I-127 5. Electrons are found in “clouds”
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Chemistry Reference Tables Name Avogadro’s number Value 6.022 × 1023 particles/mole 0.0821 L atm mole K L mmHg 62.4 mole K 8.314 L kPa mole K 1.00 atm = 101.3 kPa = 760. mmHg = 760. torr 0°C or 273K 22.4 L Symbol Value Gas constant (R) Standard pressure Standard temperature Volume of 1 mole of any gas at STP Thermodynamic Constants Heat of fusion of water Heat of vaporization of water H f (water) Hv (water) 334 J/g 2‚260 J/g J for ice‚ 2.02 J for steam‚ g°C g°C J for liquid 4.18
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Biology Key Questions – Lesson #1 1. How many electrons does Mg2+ have? Answer: Mg2+ has 10 electrons. A Magnesium atom has 12 electrons‚ 2 of which are valence electrons. The atom will lose the 2 outermost electrons in order to stabilize its electron shell. The nucleus maintains its 12 protons and becomes a Mg2+ ion. 2. Given that the half-life of the radioisotope carbon-14 is 5730 years‚ how useful do you think this isotope would be for dating bones that are over a million years
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Importance of water use in the steel industry Water’s ever-growing importance in the industrial manufacturing arena has been demonstrated by an increasing concern regarding the sufficiency of both its quantity and quality for use in industrial applications. In general‚ water is used in steelmaking as a: coolant for equipment‚ furnaces‚ and intermediate steel shapes; cleansing agent to remove scale from steel products; source of steam; medium for lubricating oils and cleaning solutions;
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Investigating Chemical Equilibrium Date: 30th April 2013 Due Date: 15th May 2013 Prepared For: M. Seraji Prepared by: Andrea Odunze Abstract Many reactions proceed to a state of equilibrium. A chemical reaction at equilibrium‚ where the rates of the forward reaction and reverse reaction are equal‚ looks like this: A + B AB There are three factors‚ according to Le Chatelier’s principle‚ that affect the equilibrium position and equilibrium constant. These are the concentrations of products
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INTERNATIONAL PUBLIC SCHOOL‚ BHOPAL HOLIDAY – HOMEWORK (2013-2014) CLASS-X ENGLISH Section- A: BBC: Reading Comprehension: Ex.1‚ 2 & 3. Section- B: BBC: Writing Skills: E-mail 1 & 2; Letter writing 1 & 2; Speech 1 & 2; Article 1& 2. Section- C: BBC: Grammar: Preposition‚ Voice‚ Speech‚ Tenses. Section- D: 1. Value based: Which poem do you appreciate the most – ‘The frog and the nightingale’ or ‘Mirror’? Why? Write the literary devices used in that poem. Write about the poet. (150 words) 2. Extrapolatory:
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Last Name ____________________First Name _______________________ Ignore: 5‚ 7‚ 19‚ 30 1. | How many grams of CaCl2 (molar mass = 111.0 g/mol) are needed to prepare 4.44 L of 0.500 M CaCl2 solution? | | A) 369 g B) 271 g C) 258 g D) 296 g E) 246 g | 2. | An aqueous solution of ammonium sulfate is allowed to react with an aqueous solution of lead(II) nitrate.The complete ionic equation contains which of the following species (when balanced in standard form)? | | A)
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0.88g H2O × 0.68% = 0.005984g H2O 0.88g ±0.005984g H2O CuSO4: 63.55 + 32.07 + 16.00(4) = 159.62g1 mol (n)(Hydrated CuSO4): 2.12g CuSO4 ×1 mol CuSO4(s)159.62g CuSO4(s) = 0.01328 mol Hydrated CuSO4 H2O: 16.00 + 1.01(2) = 18.02g1 mol (n)( H2O): 0.88g H2O ×1 mol CuSO4(s)18.02g CuSO4(s) = 0.04883 mol H2O 0.01328 mol CuSO4 0.01328 mol = 1 CuSO4 0.04883 mol CuSO4 0.01328 mol = 4 H2O Presenting Processed
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carbon dioxide Element | Symbol | Valency | Hydrogen | H | 1 | Oxygen | O | -2 | Carbon | C | 4 | Calcium | Ca | 2 | Iron | Fe | 2 | Aluminium | Al | 3 | Sodium | Na | 1 | Potassium | K | 1 | Magnesium | Mg | 2 | Zinc | Zn | 2 | Copper | Cu | 2 | Silver | Ag | 1 | Lead | Pb | 2 | Chloride | Cl | -1 | Sulphate | SO4 | -2 | Hydroxide | OH | -1 | Carbonate | CO3 | -2 | Nitrate | NO3 | -2 | Phosphate | PO4 | -3 | Hydrogen Carbonate | HCO3 | -1 | Acids and Bases
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Question # 1: You are titrating 50 ml. of nitrous acid with sodium hydroxide (0.15 M). Equivalence is reached at 22.0 ml of NaOH delivered. Calculate the pH of the solution in the flask: a- before the beginning of the titration - after the delivery of 5 ml. of titrant b- at half-equivalence c-at equivalence d-after delivery of 23 ml. of titrant Part A) Reaction is: HNO2 + OH- NO2- + H2O Ka for Nitrous acid = 7.2E-4 Kw = Ka*Lb Kb = 1.0E-14/7.2E-4 Kb = 1.38E-11 Moles of base:
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