What Is The Formula For Zinc Iodide – 1.1 The Mole Concept and Avogadro’s Constant 1.2 Formulas 1.3 Chemical Equations 1.4 Gas Mass-to-Volume Ratios in Chemical Reactions 1.5 Solutions (See IB Program Guide for details)
Examples of concentration of a solution: a) Describe how you could prepare cm3 of a 2.50 mol dm-3 sodium hydroxide solution with the starting ingredients listed below. Support your explanation with an appropriate calculation. i) NaOH(s) ii) 6.0 mol dm-3 NaOH(aq) Add 10.0 g of NaOH to enough water to make 1 mL of solution.
What Is The Formula For Zinc Iodide
Examples of concentration of a solution: a) Describe how you could prepare cm3 of a 2.50 mol dm-3 sodium hydroxide solution with the starting ingredients listed below. Support your explanation with an appropriate calculation. i) NaOH(s) ii) 6.0 mol dm-3 NaOH(aq) Measure 41.7 cm3 of the stock solution and dilute to cm3.
Zinc Iodide Chemical Compound Zinc Iodine Stock Illustration 769733377
5 Examples of solution concentration: b) Calculate the amount of hydrochloric acid contained in 2.00 dm3 of 1.50 mol dm-3 HCl (water).
6 Examples of solution concentrations: c) What mass of potassium permanganate KMnO4(s) is contained in cm3 of 0.35 mol dm-3 KMnO4(aq)?
7 Examples of solution concentration: d) What will be the concentration of the solution that is formed when 200 cm3 of 3.00 mol dm-3 HCl (aqueous) is mixed with 300 cm3 of 1.50 mol dm-3 HCl (aqueous)?
8 Example 1 (IB 2005): When a small amount of a gas with a strong smell, such as ammonia, is released into the air, it can be detected several meters away in a short time. Use kinetic molecular theory to explain why this happens. Gases consist of randomly moving molecules. When a new gas (such as ammonia) enters a mixture of gases (such as air), its molecules will move rapidly away from the point of release and will be visible at a distance for a short time.
Solved] This Questions Are Part Of My Experiment Lab #5 That I In Trouble…
9 Example 1 (IB 2005): When a small amount of a gas with a strong smell, such as ammonia, is released into the air, it can be detected several meters away for a short time. b) State and explain how the time required to detect a gas changes with increasing temperature. Since the kinetic energy and therefore the speed of the gas molecules will increase as the temperature increases, the time required to detect the gas is reduced.
10 Example 2 (IB 2005): mass percentage of hydrocarbon C = 85.6% and H = 14.4%. a) Calculate the empirical formula of a hydrocarbon. CH H Mass per 100 g 85.6 g 14.4 g Moles per 100 g 85.6 ÷ 12.01 = 7.13 mol 14.4 ÷ 1.01 = 14.3 mol The simplest ratio 7.13 ÷ 7, 13 = 1172.3 Empirical formula = 1172.3 ÷
Example 2 (IB 2005): hydrocarbon mass percentage C = 85.6% and H = 14.4%. b) A hydrocarbon sample weighing 1 g at a temperature of 273 K and a pressure of 1.01 x 105 Pa (1.00 atm) has a volume of dm3. Calculate the molar mass of the hydrocarbon. PV = nRT, therefore n = PV ÷ RT n=(1.01×105 Pa)(0.399×10-3 m3) ÷ (8.31 J mol-1 K-1)(273 K) n= mol Molar mass M = Mass ( g) ÷ Quantity (mol) = 1 g ÷ mol = 56.3 g mol-1
12 Example 2 (IB 2005): mass percentage of hydrocarbon C = 85.6% and H = 14.4%. b) A hydrocarbon sample weighing 1 g at a temperature of 273 K and a pressure of 1.01 x 105 Pa (1.00 atm) has a volume of dm3. ii) Derive the molecular formula of a hydrocarbon. The molar mass of the empirical formula is equal to (2 x 1.01) = g mol-1. The molecular weight, 56.3 g mol-1, is about four times 14.03, so the molar formula must be (CH2)4 or C4H8.
Binary Cpd Wkst
13 Pr.3 (IB 2003): Sodium reacts with water as follows: 2Na(s) + H2O 2NaOH(aq) + H2(g) 1.15 g of sodium reacts completely with water. The resulting solution is diluted to 250 cm3. Calculate the concentration in mol dm-3 of the formed sodium hydroxide solution. As usual, we need to convert to moles: 1.15 g Na is 1.15 g ÷ g mol-1 = mol Na From the balanced equation, we see that a mole of Na produces a mole of NaOH. Now we know the volume of the solution (250 cm3, dm3) and the amount of solute (mol NaOH) Concentration = amt. of dissolved substances ÷ vol. solution = mol ÷ dm3 = mol dm-3
Example 4 (IB 2004): 100 cm3 of ethane, C2H4, is burned in 400 cm3 of oxygen, producing carbon dioxide and some liquid water. Part of the oxygen remains unreacted. a) Write the equation for the complete combustion of ethane. C2H4(g) O2(g) CO2(g) H2O(l) 3 2 2
Example 4 (IB 2004): 100 cm3 of ethene, C2H4, is burned in 400 cm3 of oxygen to form carbon dioxide and liquid water. Part of the oxygen remains unreacted. b) Calculate the volume of carbon dioxide formed and the volume of oxygen remaining. C2H4(g) O2(g) 2CO2(g) H2O(l) Initially: 100 cm cm cm no gas Finally: cm cm cm3 (100 cm3 of ethene will react with 300 cm3 of oxygen to form 200 cm3 of carbon dioxide, if the volumes gas, measured under the same conditions, according to Avogadro’s hypothesis)
16 Ave. 5 (IB 2004): a) Write the equation for the formation of zinc iodide from zinc and iodine.
Write The Balanced Chemical Equation For The Following And Identify The Type Of Reaction In Each Case:(a) Potassium Bromide(aq) + Barium Iodide(aq) → Potassium Iodide(aq) + Barium Bromide(s)(b) Zinc Carbonate(s) → Zinc
To calculate the formulas, you need to remember that iodine is diatomic (I2), and the charges of zinc and iodide ions are +2 and -1: Zn(s) + I2(s) ZnI2(s)
Pr. 5 (IB 2004): b) g of zinc reacts with g of iodine to form zinc iodide. Calculate the amount (in moles) of zinc and iodine and thus determine which reagent is in excess. 100.0 g Zn ÷ g mol-1 = mol Zn g I2 ÷ g mol-1 = mol I2 According to the balanced chemical equation: Zn(s) + I2(s) ZnI2(s) mol Zn will form mol ZnI mol I2 will form a mole of ZnI2. Iodine is the limiting agent because the reaction stops when iodine reacts to form moles of ZnI2. Zinc is an agent in excess.
Pr. 6 (IB 2004): g of hydrated crystals of sodium carbonate, Na2CO3xH2O, was dissolved in water and added to dm cm3 of this solution, cm3 of hydrochloric acid with a concentration of mol dm-3 was neutralized. a) Write the equation for the reaction between sodium carbonate and hydrochloric acid. Acid + carbonate salt + water + CO2 Hydrochloric acid forms chlorides Na2CO3(aq) + HCl(aq) NaCl(aq) + H2O(l) + CO2(g) We will learn about this and more later 2 2
20 Ave. 6 (IB 2004): g of hydrated crystals of sodium carbonate, Na2CO3xH2O, was dissolved in water and added to dm cm3 of this solution, cm3 of hydrochloric acid with a concentration of mol dm-3 was neutralized. b) Calculate the molar concentration of sodium carbonate solution neutralized with hydrochloric acid. Moles of HCl = 0.1 mol dm-3 x 48.8×10-3dm3 = 4.88×10-3mol = 4.880 mmol 1mol Na2CO3 reacts with 2 moles of HCl, so… 2.440 mmol Na2CO3 reacts with mmolHCl = Concentration2x10-3mol Na2.4x03mol -3dm3 = mol d m -3
Solved Lab 5 Worksheet: Synthesis And Decomposition Of Zinc
21 Ave. 6 (IB 2004): g of hydrated crystals of sodium carbonate, Na2CO3xH2O, was dissolved in water and added to dm cm3 of this solution, cm3 of hydrochloric acid with a concentration of mol dm-3 was neutralized. c) Determine the mass of sodium carbonate neutralized with hydrochloric acid, and therefore the mass of sodium carbonate in dm3 of the solution. Start with the answer to part b… 2.44×10-3 mol Na2CO3 x g mol-1 = g This is the mass of 25 cm3, so the mass of 1000 cm3 is g x 1000 ÷ 25 = g
Pr. 6 (IB 2004): g of hydrated crystals of sodium carbonate, Na2CO3xH2O, was dissolved in water and added to dm cm3 of this solution, cm3 of hydrochloric acid with a concentration of mol dm-3 was neutralized. d) Calculate the mass of water in the hydrated crystals and thus find the value of x. Mass of H2O = g – g = g Moles of H2O= g ÷ g mol-1= mol Moles of Na2CO3=10.34 g ÷ g mol-1= mol x= mol ÷ mol= 10 Na2CO310H2O
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