Chemistry Questions Answers – Flashcards
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One drop of water weighs 0.040 g. Calculate how many molecules are there in one drop, taking the molar mass of water as exactly 18 g þmol^-1. A 1.3 × 10^21 B 2.4 × 10^22 C 3.3 × 10^22 D 3.9 × 10^22
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A. 0.04g x 18 = 0.0022 moles 0.0022 x 6.02x10^23 = 1.337x10^21
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Calculate to 3 sig figs the mass of 1 ? 4 mole of Li2O.
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7.468g 6.94+6.94+15.99 = 29.87 g/mol 29.87/4 = 7.468
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Calculate to 3 sig figs the mass of 2.60 × 10^22 molecules of dinitrogen monoxide.
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1.90g 6.02 / .260 = 23.153 (14+14+16) / 23.153 = 1.90g
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Calculate how many moles are in 4000g of Nickel Sulfate
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25.848 moles 4000g / (58.6934 +32.06 + (16x4)) = 25.848
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3.01 × 10^25 molecules of a gas has a mass of 6.40 kg. Determine its molar mass.
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128 g/mol 6.4x1000 = 6400g 6.02/301 = 0.02 6400 x 0.02 = 128 g/mol
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Of the following, the only empirical formula is A C12H22O11 B c6h12o6 C C6H6 D C2H4
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A.
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Determine the percentage by mass of silver in silver sulfide (Ag2S). A 33.3% B 66.7% C 77.1% D 87.1%
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D. 107.86 x 2 + 32.06 = 247.78 g/mol 107.86 x 2 = 215.72 215.72/247.78 = 0.87106 = 87.1%
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2.40 g of element Z combines exactly with 1.60 g of oxygen to form a compound with the formula ZO2. Determine the relative atomic mass of Z. A 24.0 B 32.0 C 48.0 D 64.0
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C. 1.6 g / 16 = 0.1 0.1 / 2 = 0.05 0.05 x 2.4 = 48g/mol
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A certain compound has a molar mass of about 56 g mol-1. All the following are possible empirical formulas for this compound except: A CH2 B CH2O C c3h4o D CH2N
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B. CH2O is the only one with a molar mass that isn't a factor of 56.
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When magnesium is added to aqueous silver nitrate, the following reaction takes place Mg(s) + 2 AgNO3 (aq) -> 2 Ag(s) + Mg(NO3)2 (aq) What mass of silver is formed when 2.43 g of magnesium is added to an excess of aqueous silver nitrate. A 107.9 g B 21.6 g C 10.8 g D 5.4 g
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B. 2.43 / 24.305 g/mol = 0.09998 0.09998 x 2 = 0.199958 0.199958 x 107.868 = 21.569 g
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Freon-12 (used as coolant in refrigerators), is formed as follows: 3CCl4(l) +2SbF3(s) -> 3CCl2F2(g) +2SbCl3(s) 150 g tetrachloromethane is combined with 100 g antimony(III) fluoride to give Freon-12 (CCl2F2). (a) Identify the limiting and excess reagents. (b) Calculate how many grams of Freon-12 can be formed. (c) How much of the excess reagent is left over?
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a) Limiting reagent is SbF3, CCl4 is in excess b) 101.5 g c) 20.9 g a) CCl4: 150g/156gmol = 0.987 0.987 / 3 = 0.329 SbF3: 100g / 178.7 gmol = 0.2795 0.2797 / 2 = 0.2795 SbF3 is limiting b) 100g x 1/178.7gmol x 3/2mol x 120.7gmol = 101.5g c) 100g x 1/178.7gmol x 3/2mol x 153.8 = 129.09g 150g-129.09g = 20.9 g
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What volume would 3.20 g of sulfur dioxide occupy at s.t.p.? A 0.56 dm3 B 1.12 dm3 C 2.24 dm3 D 4.48 dm3
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B. 3.2g / 64.06gmol = 0.04995 0.04995 x 22.4 = 1.118dm = 1.12dm
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A mixture of 20 cm3 hydrogen and 40 cm3 oxygen is exploded in a strong container. After cooling to the original temperature and pressure (at which water is a liquid) what gas, if any, will remain in the container?
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30cm unreacted. In water, hydrogen and oxygen has a 2:1 ratio. Thus, 20cm of H reacts with 10cm of O, leaving 30cm behind.
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To three significant figures, calculate how many methane molecules are there in 4.48 dm3 of the gas at standard temperature and pressure?
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1.20x10^23 4.48/22.4 = 0.2 0.2 x 6.02x10^23 = 1.20x10^23
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If 3.00 dm3 of an unknown gas at standard temperature and pressure has a mass of 6.27 g, Determine the molar mass of the gas.
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46.816 gmol 3 / 22.4 = 0.13392 6.27g / 0.13392mol = 46.816 gmol
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A sealed flask contains 250 cm3 of gas at 35C and atmospheric pressure. The flask is then heated to 350C. The pressure of the gas will increase by a factor of about A 2 B 10 C 250 D 585
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A.
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Sodium phosphate has the formula Na3PO4. Determine the concentration of sodium ions in a 0.6 mol dm-3 solution of sodium phosphate? A 0.2 mol dm-3 B 0.3 mol dm-3 C 0.6 mol dm-3 D 1.8 mol dm-3
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D. There are 3 Na ions in ever sodium phosphate so.. 0.6 x 3 = 1.8
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What volume of a 0.5 mol dm-3 solution of sodium hydroxide can be prepared from 2 g of the solid? A 0.05 litres B 0.1 litres C 0.4 litres D 0.5 litres
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B. 2g / 40gmol = 0.05 0.05 mol / 0.5M = 0.1 liters
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500 cm3 of 0.500 mol dm-3 NaCl is added to 500 cm3 of 1.00 mol dm-3 Na2CO3 solution. Calculate the final concentration of Na+ ions in solution.
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1.25 mol dm-3 500/1000 = 0.5dm 0.5 x 0.5M = 0.25 0.5 x 1M = 0.5 0.5 x 2= 1 1+0.25 = 1.25 mol dm-3
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How would you prepare 1.2 dm3 of a 0.40 mol dm-3 solution of hydrochloric acid starting from a 2.0 mol dm-3 solution?
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Measure out 240 cm3 of 2.0 mol dm-3 hydrochloric acid (0.48 moles) and make this up to 1.2 dm3 of solution. 1.2 x 0.4 = 2 x ? 0.48 = 2? ? = 0.24dm = 240cm
