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    Question Paper

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    CHEMISTRY (Theory) Time allowed : 3 hours General Instructions: (i) All questions are compulsory. Maximum Marks : 70 (ii) Marks for each question are indicated against it. (iii) Question numbers 1 to 8 are very short-answer questions and carry 1 mark each. (iv) Question numbers 9 to 18 are short-answer questions and carry 2 marks each. (v) Question numbers 19 to 27 are also short-answer questions and carry 3 marks each. (vi) Question numbers 28 to 30 are long-answer questions and carry 5 marks

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    13.80 mL x = 0.01380 L Moles of potassium acid phthalate = = = 0.000735 Moles of NaOH solution = Moles of (HKC8H4O4) x = 0.000735 mol x = 0.000735 mol Molarity (NaOH) = = = 0.053 M TRIAL II Volume of NaOH solution = Final reading of buret - Initial reading of buret =27.20 mL– 13.80 mL = 13.40 mL Converting the Volume (mL) to Volume (L) Volume (liters) = Volume (mL) x = 13.40 mL x = 0.01340 L Moles of potassium acid phthalate

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    Practice Final

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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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    3.09 Honors Chem Online

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    water = (Mass of hydrate) – (mass of dehydrate) Mass of water = (answer to #1) – (answer to #3) Mass of water = (5.000g) – (4.500g) Mass of water = 0.500g 5. Convert the mass of water to moles of water. To do this‚ we need the molar mass (from 2.04 and 3.09-molar mass is the mass‚ in grams‚ of one mole of your

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    Chemistry

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    9-12 Unit 7 II: Unit Title: Mole Concept III. Unit Length: 7 days (on a 90 min. per day block schedule) IV. Major Learning Outcomes: Students should be able to: Mole Concept • Calculate formula mass. • Convert representative particles to moles and moles to representative particles. (Representative particles are atoms‚ molecules‚ formula units‚ and ions.) • Convert mass of atoms‚ molecules‚ and compounds to moles and moles of atoms‚ molecules‚ and compounds

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    (Oh) 2.4.2 Mol Of Ca3p2

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    appropriate response. 1. What is a mole ratio? 2. What piece of information is needed to convert grams of a compound to moles of the same compound? 3. Phosphine (PH3) can be prepared by the hydrolysis of calcium phosphide‚ Ca3P2: Ca3P2 + 6 H2O 3 Ca (OH)2 + 2 PH3 a) One mole of Ca3P2 produces 2 mol of PH3. b) One gram of Ca3P2 produces 2 g of PH3. c) 3 moles of Ca(OH)2 are produced for each 2 mol of PH3 produced. d) The mole ratio between phosphine and calcium

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    CHEM 165

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    corrodes and rust is a byproduct. (16 pts.) 5. Determine the mass of iron consumed in the reaction. Show your work. (16 pts.) 7.75-5.5g=2.25g 6. Calculate the number of moles of iron consumed. Show your work. (16 pts.) 2.25G X 1MOL FE/55.85g= 0.040286 = 0.040286=0.0403 moles Fe Formula: Mass Fe used X 1 mol Fe = Moles of iron used Molar mass of Fe 7. Determine the mass of product formed. Show your work. (16 pts.) 3.52g-.086g=2.66g Formula: (Mass of

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    of lithium plus the watch glass *37.41 g of Watch glass *122 cm3 of hydrogen gas collected Calculations from method 1 Moles of hydrogen produced: Assuming that 1 mole of gas occupies 24000 cm3 at pressure and room temperature! I will now calculate the number of moles of the hydrogen gas produced. I will do that by dividing the volume of gas produced by 24000cm3 : Moles (n) = Volume (cm") 24000 cm" So: n = 122 cm" = 0.005083 mol of H2 24000 cm" Equation: 2 Li(s) + 2 H2O (L)2LiOH(Ag) +

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    11 g Questions 1. Calculate the value of “n”. Show your work. Include an example of ALL calculations involving solving for “n”. 3.02 g of hydrated copper sulfate is heated to drive off water. The dry sample has a mass of 2.11 g. What is the mole ratio between copper (II) sulfate‚ CuSO4‚ and water‚ H2O? Mass of Hydrate 3.02 g Mass of Dry Sample - 2.11g Mass of Water 0.91 g MM CuSO4= 63.55 + 32.07 + 4(16.00)= 159.62 g/mol 2.11 g CuSO4 (1 mol CuSO4

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    General Chemistry I

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    Test 2 Name___________________________________ MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Find the combination that will give a gas as a product A. AgNO3(aq) + NaI(aq) B. HCl(aq) + Na2CO3 (aq) C. MgCl2(aq) + K2CO 3 (aq) D. CaSO4 (aq) + (NH4)3 PO4(aq) 1) A) A B) C C) B D) D 2) 2) The balanced molecular equation for complete neutralization of H2 SO4 by KOH in aqueous solution is __________. A) 2H+ (aq) + 2OH- (aq)

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