Find the volume of 2.40 mol of gas whose temperature is 50.0 °C and whose pressure is 202 kPa. 6. How many moles of gas are contained in a 50.0 L cylinder at a pressure of 10100 kPa and a temperature of 35.0 °C? 7. Determine the number of grams of carbon dioxide in a 450.6 mL tank at 1.80 atm and minus 50.5 °C.
(b) Calculate the volume of 0.2M UO3- needed to react with 20.00 cm3 of 0.1M Cr2O72-. 3. 24.40 g of hydrated iron(II) sulphate, FeSO4.xH2O was dissolved and made up to 1.0 dm3 of aqueous solution, acidified with sulphuric acid. 25.00 cm3 of the solution was titrated with 20.00 cm3 of 0.022M potassium manganate(VII) solution for complete oxidation. a) Write the equation for the reaction.
How many grams of CO2 would be liberated by complete thermal decomposition of the MgCO3 in the sample? b) How much would the residue weigh? 3. Describe the location of the hottest portion of a burner flame. The tip of the inner blue flame is the hottest portion.
The average bond enthalpies for O—O and O==O are 146 and 496 kJ mol−1 respectively. What is the enthalpy change, in kJ, for the reaction below? H—O—O—H(g) ® H—O—H(g) + ½O==O(g) A. – 102 B. + 102 C. + 350 D. + 394 (1) 7.
| A) | CH3CH2O– | B) | CH3CH2O+ | C) | CH3CH2OH2+ | D) | CH3CH2OH3+ | 5. | Which one of the following mechanistically depicts the protonation of methanol by hydrogen bromide? | A) | A | B) | B | C) | C | D) | D | 6. | Give the molecular formula of the compound shown below: | A) | C8H16O | B) | C9H18O | C) | C10H18O | D) | C10H20O | 7. | The most stable resonance contributor of this would be: | A) | A | B) | B | C) | C | D) | D | 8.
Calculate the surface area of the circle formed (πd2/4): Surface area = .785 cm2 2. Calculate the number of molecules on the top layer. We must convert the surface area in centimeters squared to nanometers squared and then multiple that by the surface area of a sodium stearate molecule. Convert the surface area of the circle formed (#1) to molecules per layer using the matrix below: Answer = 4.76*10^14 molecules/top layer 3. Calculate the concentration of grams of sodium stearate per milliliter of diluted solution.
Name: Anh Lan Do Course: CHE-131-92L Date: Feb 24th, 2010 General Chemistry 1 Experiment 4 Percent Composition of a Compound Professor Robert M. Cady Questions: 1. According to the law of definite composition, elements form compounds in the same ratios by mass each time the compound is produced. Using the theoretical percent by mass of magnesium in MgO, calculate the mass of oxygen that will combine with 12.25g of Mg and the mass of MgO that will be formed. 2. Which compound has the greatest percentage by mass of Magnesium?
The molar mass of a compound or atom is the mass of 1 mole of anything; this is relative to the atomic mass from the periodic table. The percentage yield from an experiment would be the actual yield divided by the theoretical yield multiplied by 100. We need to know how to balance chemical equations due to law of conservation of mass. Being able to turn moles into mass and mass into moles through the equation “m=nM” rearranged to find out other components. We need to be able to find the limiting reagent and be able to go through the process of gravimetric stoichiometry.
The angle of the drop height was at 90 degrees and the length of the string was at 300 mm. Size of Mass ( grams) | Trial 1 (seconds) | Trial 2 (seconds) | Trial 3 (seconds) | Average (seconds) | Period (seconds) | 10 grams | 25.00 seconds | 24.90 seconds | 25.21 seconds | 25.04 seconds | 1.252 seconds | 100 grams | 25.86 seconds | 25.93 seconds | 25.81 seconds | 25.87 seconds | 1.2935 seconds | 500 grams | 26.53 seconds | 26.30 seconds | 26.44 seconds | 26.42 seconds | 1.321 seconds | 1,000 grams | 25.82 seconds | 26.46 seconds | 25.91 seconds | 26.06 seconds | 1.303 seconds | 4.) Angle or Drop Height: Hypothesis: I predict that the larger the angle of drop, the longer period of the pendulum. The independent variable is the drop height. The mass of the pendulum was 100 grams and the string measured 300 mm.
The suffix -ic is given to the compound that contains the cation with the highest charge. To create the formula, you switch the charges, and that tells you how many of each element you will need. For example: Type 2 Ion Stock Name Latin Name Co2+ Cobalt (II) Ion Cobaltous Ion Co3+ Cobalt (III) Ion Cobaltic Ion Type 2 Compounds Stock System Latin System Charge of Metal Original Elements Co2O3 Cobalt (III) Oxide Cobaltic