____C3H8 + ____O2 ____CO2 + ____H2O Suppose 1.5 moles of propane was consumed in the reaction‚ answer the next three problems. How many grams is 1.5 moles C3H8 (propane)? How many grams CO2 is formed? How many molecules of H2O are formed? If 6.75 moles of O2 and 3.25 moles C3H8 were combined‚ how many grams H2O will be produced? Which is the limiting reagent? If 100 grams each of O2 and C3H8 were combined‚ how many moles CO2 will be produced? Which is the excess reagent? How
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Initial Buret Reading (mL) | Final Buret Reading (mL) | Volume of MnO4- solution used (mL) | 1 | 10.00 | 9.70 | 20.60 | 10.90 | 2 | 10.00 | 20.60 | 31.50 | 10.90 | 3 | 10.00 | 31.60 | 42.50 | 10.90 | Table 2: Mole ratio calculations Titration # | Moles of Oxalate Pipetted | Moles of Permanganate Used |
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increased temperature causes an increase in kinetic energy. The higher kinetic energy causes more motion in molecules which break intermolecular bonds and escape from solution. Q.5What is Henry’s law explain with an example. AnsHenry’s law states mole fraction χ of the gas Q.2What is interstitial solid solutions. AnsIn certain alloys the atoms of smaller size occupy the voids or interstices between atoms of solid. such are called interstitial solids example of such solid are Tungsten carbide
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KMnO4 dispensed 28.5ml 26.3ml Moles of KMnO4 5.7E-4 5.26E-4 Moles of C2O42- 1.425E-3 1.315E-3 Table 2: (Second Titration) C2O42- Analysis Sample 1 Sample 2 Molarity of KMnO4 0.02m 0.02m Weight of Sample 0.237g 0.225g Final Buret Reading 4.2ml 4.5ml Initial Buret Reading 0ml 0ml Volume of KMnO4 dispensed 4.2ml 4.5ml Moles of KMnO4 8.4E-5 9.0E-5 Moles of C2O42- 2.1E-4 2.25E-4 Calculations: 1.) 2 KMnO4 + 5 K2C2O4 + 8 H2SO4 = 2 MnSO4 + 10 CO2 + 8 H2O + 6 K2SO4 Find moles of C2O42- Trial 1: KMnO4 =
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complex and bind more tightly to EBT than Ca2+ b. What is the color change at endpoint? The color change at endpoint is sky blue/light blue. 5) A 50.0 ml water sample requires 16.33 ml of 0.0109 M Na2H2Y to reach EBT endpoint. a. Calculate the moles of hardening ions in the water sample. 1 mol of Na2H2Y = 1 mol of CaCO3 If mol of Na2H2Y = 0.0109 M x 0.01633 L = 1.78 x 10-4 mol then there are 1.78 x 10-4 mol of CaCO3 b. Express the hardness concentration in mg CaCO3 / L sample. (1.78
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Making up Hydrogen Peroxide Volume required 250 cm3 and concentration required 0.1 moldm-3 Given concentration of H2O2 = 1.7 moldm¬-3 Number of moles (n¬¬¬¬¬¬¬¬¬¬¬) = Concentration (moldm-3) x Volume (dm-3) = 0.1 x 0.25 = 0.025 mol Volume (dm-3) = Number of Moles (n) X 1000 Concentration (moldm-3) = (0.025/1.7) x 1000 = 14.7 cm3 Distilled water required: 250 cm3 – 14.7
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smallest particle‚ mole. Measuring the mass was a primary difficulty at that time since one mole of a substance was unable to weigh without using developed technology. Even though‚ it was clear that everything was made out of a small unit‚ there was no evidence that could determine what it was and how to measure it. So‚ as most scientific ideas had to be created with indirect evidence‚ it was difficult for theories to develop further. Therefore‚ the discovery of the mass of one mole‚ Avogadro’s number
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Topic 1.2 AMOUNT OF SUBSTANCE The mole Reacting masses and atom economy Solutions and titrations The ideal gas equation Empirical and molecular formulae Ionic equations Mill Hill County High School THE MOLE Since atoms are so small‚ any sensible laboratory quantity of substance must contain a huge number of atoms: 1 litre of water contains 3.3 x 1025 molecules. 1 gram of magnesium contains 2.5 x 1022 atoms. 100 cm3 of oxygen contains 2.5 x
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Rutherford Gold-Foil Experiment Shot alpha particles at gold foil; the particles either went straight through‚ or bounced back at large angles Atoms‚ Molecules‚ Formula Units‚ Mole‚ Mole Conversions‚ Percent Composition Identifying atoms‚ molecules‚ and formula units Atoms Molecules Formula units Understanding one mole
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calcium in water (b) Adding silver nitrate solution to sodium chloride solution A36.2 (p.36-7) (a) Rate of formation of NO2(g) == 0.24 mol dm–3 s–1 (b) From the equation‚ mole ratio of NO2 to N2 = 2 : 1. Rate of consumption of N2(g) =× Rate of formation of NO2(g) =× 0.24 mol dm–3 s–1 = 0.12 mol dm–3 s–1 (c) From the equation‚ mole ratio of NO2 to O2 = 2 : 2 = 1 : 1. Rate of consumption of O2(g) = Rate of formation of NO2(g) = 0.24 mol dm–3 s–1 A36.3 (p.36-12) (a) Instantaneous rate of reaction
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