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    Water Hardness

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    Solution: 0.7533g of Na2EDTA mixed with 500ml DI water Table A: Standardization of EDTA Trial | Start Volume | End Volume | Amount of EDTA | 1 | 3.90ml | 29.85ml | 25.95ml | 2 | 5.35ml | 28.35ml | 23.00ml | 3 | 0.45ml | 23.50ml | 23.05ml | Moles

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    you can buy bananas by the pound. The third and last way to quantify or measure matter is to determine its volume. For example‚ people by gasoline by the gallon. Chemists use a unit that is a certain number of particles. This unit is known as the mole. A mole is the quantity of anything that has the same number of particles found in 12.000 grams of carbon-12. That number of particles is Avogadro’s Number which is about 6.02 x 1023 ‚ representative particles. That is an SI unit for measuring the

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    analysis of alum lab

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    point and mole ratio of the water to the anhydrous. Every substance has unique characteristics which help to determine the identity of it. The actual values of the properties can be compared to the experimental values. In real life‚ this can be used to determine the unknown medicines. Hypothesis: The hypothesis is that the alum can be verified by finding the properties; the actual melting point of alum is 92.5(http://cookeatshare.com/popular/melting-point-of-alum) and the actual mole ratio of

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    Properties of Gases

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    Chapter 6 – Properties of gases lecture notes 1. Gas phase Gases have neither definite shape or volume 1) volume changes with pressure 2) volume changes with temperature 3) gases are miscible 4) gases are generally MUCH less dense than liquids 2. Atmospheric pressure 1 atm = 760 torr (mm of Hg) F= ma F = force m = mass a = acceleration P = F/A - ma/A P = pressure

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    find: i. Mass ii. Mole iii. Divide each mole by lowest mole value iv. Whole number ratio of each element 2. Examples: v. 75% C‚ 25% H Basis: 100g a. Mass: C = 75g‚ H = 25g b. Recall from Table T: mole = massatomic mass Mole C = 75g12g/mole = 6.25 moles Mole H = 251 g/mole = 25 moles c. Mole C = 6.25 moles6.25 moles = 1 Mole H = 25 moles6.25 moles = 4 d. CH4 vi. 13% Mg‚ 87%

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    Chapter 3 Stoichiometry

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    Stoichiometry calculation • Limiting reactant • Theoretical yield‚ actual yield and percentage yield 1 2 Mole Concept No. of Moles = Molarity (M) • Molarity (molar concentration) is the number of moles of a solute that is contained in 1 liter of solution Mass (g) molar mass (g/mol) No. of Moles = Molarity (mol/L)  volume (L) Molarity (M) = Amount of solute (Mol) Volume of solution (L) • 1 mole contains 1 Avogadro’s number (6.022 x 1023) 3 Example: Saline Water Concentration Typical seawaters contain

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    Molar Mass

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    What is molar mass?Molar mass is the weight of one mole (or 6.02 x 1023 molecules) of any chemical compounds. Molar masses of common chemical compounds that you might find in the chemistry laboratory can range between 18 grams/mole for compounds like water to hundreds of grams per mole for more complex chemical compounds.The lightest possible chemical that one can have under normal conditions is hydrogen gas‚ or H2. There is no limit to how heavy a chemical compound can be - it is not uncommon for macromolecules (large

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    and aromatics‚ timber and slaves” if they sponsor his travels. In these journal entries columbus’s writes to the King and Queen every detail about the voyage. He begins in October when he he arrives and meets as what he descirbes are the indians as naked and painted. In November he begins capturing the indians and their possessions. By December columbus had what he orignially promised to the King and Queen of Spain. But is it really a capture if the Indians are willing? In the absract made by tghe

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    0.6‚ 0.7‚ 0.8 and 0.9 to form several complexes. Using the UV-Vis Spectrometer‚ the absorbances are taken at wavelengths 530‚ 545‚ 578‚ 622 and 640 nm due to the emergence of several complex compositions. The values of corrected absorbances and the mole fraction of en (X) are plotted to obtain a curve. From these curves‚ an intersection from the 2 trend lines are determined‚ denoting the values of x‚ needed for the determination of n. The process led the group with similar values of n which is 1‚

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    Unit 9 Final Lab Report

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    1. If atomic mass of Mg atom is 24 g‚ find mass of 1 Mg atom. 2. Find mass of 1 molecule C2H6. (C=12‚ H=1) 3. Find mole of 6‚9 g Na. (Na=23) 4. Find mass of 0‚2 mol P4 . (P=31) 5. Find mole of 4‚48 liters O2 under normal conditions. 6. Find mass of Fe in the compound including 4‚8x1023 O atoms ;Fe3O4 . 7. 1. Find relation between number of molecules of given matters; 8. I. C2H2 that includes 2mol H atom 9. II. CH4 that includes N atoms (N is Avogadro number) 10. III. C3H4 that includes 1‚5 N C atoms

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