www.igcse.at.ua ORGANIC CHEMISTRY OIL and its many useful PRODUCTS The origin of oil Crude oil is formed from organic material of the remains of plant and animal organisms that lived millions of years ago. These remains form sediments eg at the bottom of seas‚ and become buried under layers of sedimentary rock. They decay‚ without air (oxygen)‚ under the action of heat and pressure to form crude oil over millions of years. It is a fossil fuel because it is formed from
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Temperature‚ T (°C) RESULT Preassure‚ P (bar) | | | | | | Measure slope‚ dT/dP | Calculated slope‚Tvg/hfg | Gauge | Absolute | Increase (°C) | Decrease(°C) | AverageTavg (°C) | AverageTavg (K) | | | 0.1 | 1.1 | 104.0 | 106.6 | 105.30 | 378.45 | - | 0.260 | 0.2 | 1.2 | 107.3 | 109.3 | 108.30 | 381.45 | 0.3 | 0.243 | 0.3 | 1.3 | 109.8 | 111.8 | 110.80 | 383.95 | 0.25 | 0.227 | 0.4 | 1.4 | 112.0 | 113.9 | 112.95 | 386.10 | 0.22 | 0.213 | 0.5 | 1.5 | 114.3 | 115.9 | 115.10
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Directions: The following questions involve more practice with radioactive decay half-life. Complete the problems to the best of your ability. This assignment is due by next WEDNESDAY‚ November 16th. 1. If 100.0 g of carbon-14 decays until only 25.0 g of carbon is left after 11 460 y‚ what is the half-life of carbon-14? a. Calculate how many half-lives have passed during the decay of the 100.0 g sample. 100 grams/2 50 grams/2 25grams…2 half lives have passed b. Solve for the half-life
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Chemistry 12 Thought lab Part 1 (Procedure) 1) 1s22s22p23s23p24s23d24p25s24d25p26s24f25d26p2 2) When ml can only equal 1‚ each energy level can have only 1 orientation so according to the exclusion principle only 2 electrons with opposite spins can be in each orbital. So in order to get to element 30 you would need to go all the way to the 6p orbitals. In other words‚ every two electrons would necessitate going to a new orbital. 1 | 1s1 | 11 | 1s2 2s2 2p2 3s2 3p2 4s1 | 21 | 1s2 2s2 2p2 3s2 3p2
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9.2 a) 120 b) If the blue balloon expands‚ the angle between red and green balloons decreases. c) Nonbonding (lone) electron pairs exert greater repulsive forces than bonding pairs‚ resulting in compression of adjacent bond angles. 9.12 a) Both molecules would be symmetrical because all four surrounding atoms are the same. In a symmetrical tetrahedron‚ the four bond angles are equal to each other‚ with values of 109.5°. The H-C-H angles in CH4 and the O-Cl-O angles in ClO4- will have
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Apeejay School Kolkata (WEST BENGAL) DEPARTMENT OF CHEMISTRY This is to satisfy that Ritika Kataruka bearing Roll no _______ Studyng in class XII (Science) of this institution has successfully completed the Chemistry Project. “To Study the quantity of casein present in different samples of Milk” Based on the syllabus of C.B.S.E. council for the session 2013-2014 AISSCE under my guidance and supervision and has given satisfactory account
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مدرستنا الثانوية اإلنجليزية‚الشارقة OUR OWN ENGLISH HIGH SCHOOL‚ SHARJAH A GEMS SCHOOL CHEMISTRY WORKSHEET TOPIC: METALS AND NON-METALS GRADE:8 Note: Revise the symbols and valency of metals and non-metals. Q1.What would you observe when you put a) Some zinc pieces into blue copper sulphate solution? b) Some copper pieces into green ferrous sulphate solution? c) An iron nail into blue copper sulphate solution? Q2.Complete and balance the following equations: a) Zn + O2 b) Pb + O2 c) Cu
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REVIEW a. I‚. • • C • • c. • C 3. The effective nuclear charge is a. equal to the suns of the charges of the protons in the nucleus b. equal to the suns of the charges of the protons in the nucleus minus the sum of the electrons in the Outer shell c. lest than tire sum of the charges of the protons in the nucleus due to shielding by the electrons in the outer shell d. lest than the sum of the charges of the protons in tire nucleus due to nisielding by the
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Title: Zinc and Copper II Sulfate Lab Purpose: To determine which mole ratio of Zinc and Copper II Sulfate produces the greatest temperature change in degrees celsius. Background: This experiment will be looking for color change‚ temperature change‚ and precipitation change. Some background knowledge I know is how to balance equations. I also know side effect of a chemical change‚ in this experiment there was a color and temperature change. Other background knowledge is using and applying the
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Compleximetric Determination of Water Hardness Caindec‚ Patricia Ysabel B. Water hardness is a measure of the amount of calcium and magnesium carbonate dissolved as Ca 2+ and Mg 2+ in water. There are no health hazards associated with water hardness‚ however‚ it causes scaling‚ as well as forming of soap suds. Compleximetric titration is one of the best ways of measuring total water hardness using a standard ethylenediaminetetraacetic acid (EDTA) solution. EDTA solution is used as it has the
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