(a) simplete ratio of atoms of each element in a compound (1) 1 (b) (1) × 12 (1) (or definition in terms of moles) 2 (c) CHO = 29 = 6 (1) C6 H6 O6 (1) 2 (d) CH4O+ O2 CO2+2H2O (1) 1 [6] 3. (a) (i) (1) = 1.64(1) allow 1.63 to 1.64 PV = nRT (1) V = (1) if no × 3 CE allow use of p = 100 if answer in dm3 (1) allow 0.162 to 0.166 allow conseq on moles CH2NO2 5 (ii) V = V1 × (1) allow conseq on vol of gas products in (i) = 0.410 (m3) (1) allow 0.4 to
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temporary accounting employment agency known as Tempters. Today you have been asked to work at Moondollars‚ a small coffee supplies store that operates in the city and is owned by Adrian Bartos. Your task here is to complete the accounting cycle for Moondollars for the month of June 20XX. To assist you in this task‚ Adrian tells you to read the company’s accounting policies and procedures. Note that you will be required to follow these policies and procedures when completing the accounts for Moondollars
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followed: Mg + 2HCL H2 + Mg2+(aq) + 2Cl-(aq) In this experiment there is a one to one relationship between the number of moles of hydrogen gas evolved and the moles of magnesium metal consumed in the reaction. Therefore in the finding of the experiment moles of H2 evolved is equal to the moles of Mg consumed‚ and atomic weight of Mg is equal to the weight of Mg consumed per moles of H2 evolved. Procedure 1st. Obtain a 600ml beaker‚ add 300ml of water 2nd. add 30ml of HCl (2M) to the beaker
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penalise for missing units 3 b) Moles of solution A 8.5 x 0.25 ½ = 0.002125 ½ moles 1000 c)i) Moles of HCl in 25cm3 of solution D: HCl + NaOH NaCl + H2O 1 : 1 ½ Moles of HCl in 25cm3 = 0.002125 ½ ( Mole ratio) ii) Moles of HCl on 100cm3 of solution D 25 0.002125 100x 0.002125 ½ = 0.0085 moles ½ 25 iii) Moles of HCl in 100cm3 Solution B: 0.5 moles 1000cm3 0.5 x 100cm3 = 0.05moles
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bandwagon. I once experienced bandwagon and it had a ripple effect on my life of negative consequences. One gloomy Friday night my friends and I were at a football game. They suggested since it was somewhat cold that we leave early and go to eat in Adrian then come back. "We will be back before the game even ends" they said. I knew that my dad wanted me to stay at the football game‚ but I was the only one who could drive and my friends were all depending on me. What should I have
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Objective: To determine relative molecular mass of a soluble base‚ X2 CO3 by carrying out an acid-base titration with the following reaction ‚ knowing the amount of hydrochloric acid used and the amount of substance Z used. Hypothesis: The X in substance Z is a group 1 element because substance Z is a soluble metal carbonate and would most likely be sodium or potassium because these elements are commonly used. Materials 100 cm3 beaker‚ 250 cm3 beaker‚ 250 cm3 volumetric flask with stopper
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and the average change in mass of burner for each of the fuels. The averages were based on five trials. PropanolTheoretical Heat of Reaction: Bonds Broken Bonds Formed Type of Bond Bond Energy (KJ/mole) Number of Bonds Total (KJ/mole) Type of Bond Bond Energy (KJ/mole) Number of Bonds Total (KJ/mole) C-H 412 7 2884 C=O 802 6 4812 C-C 348 2 696 H-O 463 8 3704 O=O 496 5 2480 C-O 366 1 366 O-H 463 1 463 Total 6889 Total 8516 Table 5: Shows the theoretical heat change in enthalpy for Propanol From
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weight of Ferrocene 0.225 g Moles of Ferrocene 1.21 * 10 ^ -3 mol Initial volume of acetic anhydride 1.00 mL Moles of acetic anhydride 0.0110 mol Initial volume of phosphoric acid 0.300 mL Initial moles of phosphoric acid 5.15*10^-3 mol Limiting reagent Ferrocene Moles of limiting reagent 1.21*10^-3 mol Final weight of product 0.288g Theoretical yield 29.49 Moles of product 3.73*10^-4 mol % yield 97% melting point (°C) of product 128°C -131°C Calculations: Moles of Ferrocene 0.225g * mol
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50V 4. Convert the following masses into moles. A. 135 grams of Aluminum. Molar Mass of Aluminum = 26.98g/mol # moles Al = 135g x 1 mole = 5.0 moles 26.98g B. 1.0 grams of Copper. Copper Molar Mass = 63.55g/mol # moles Cu = 1.0g x 1 mole = 0.016 moles 63.55g 5. Convert the following moles into masses. A. 0.160 moles of Magnesium. Mg Molar Mass = 24.31g/mol
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= 65 + 32 + (16 x 4) =161 * Mass of water in 1 mole of ZnSO4 = Mr ZnSO4 x Mass of water Mass of anhydrous salt * Mas of water in 1 mole of ZnSO4 = 161 x 5.79 = 129.29 7.21 * Moles of water in 1 mole of ZnSO4 = Mass of water in one mole of ZnSO4
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