JEET SCIENCE ACADEMY CHOWK AZAM (LAYYAH) TIME 2.5 Hr CHEMISTRY 1st YEAR CH # 1‚2‚9‚10 Marks: 85 G.Super . 1 Name…………………………………………...... Objective ROLL NO. ………………………………. Q.NO.1. Encircle the correct answer? (1×17=17) 1. Atoms of which one of the follelement have independent existance: (A) Flourine (B) Krypton (C ) Oxygen (D) Nitrogen 2.18g glucose is dissolved
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This is important for the students to appreciate the knowledge of chemistry that is still new for themselves. Personally‚ I think that this chapter is an interesting chapter as it revealed the way of scientist produces the material around me. It also gives me new knowledge of the uses of chemical substances that I usually found in the laboratories. I hope‚ by learning this chapter‚ I will be more interested in learning chemistry as it will help me in the future. All the equations from this chapter
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atoms can rotate and vibrate with respect to each other. These vibrations and rotations also have discrete energy levels‚ which can be considered as being packed on top of each electronic level. UV/Vis spectroscopy is routinely used in analytical chemistry for thequantitative determination of different analytes‚ such as transition metal ions‚ highly conjugated organic compounds‚ and biological macromolecules. Spectroscopic analysis is commonly carried out in solutions but solids and gases may also
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Candidate name…………………………………………………………………….. MINISTRY OF EDUCATION MASALA SECONDARY SCHOOL BOARD MOCK EXAMINATION 2014 SCIENCE 5124/3 PAPER 3 (CHEMISTRY) 1hour 15minutes Additional materials: Monday 30th July 2014. Answer paper INSTRUCTIONS TO CANDIDATES Write your name‚ centre number and candidate number at the top of this page all separate Attempt all questions in SECTION A and only TWO questions in SECTION B SECTION A (45 marks)
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CHEMISTRY PROJECT WORK SESSION -2013-2014 ROLL NO- CLASS-XII ‘A’ SUBMITTED BY SUBMITTED TO ROHIT LAKHERA Mrs. ANITA BISHT CERTIFICATE This is hereby to certify that‚ the original and genuine investigation work has been carried out to investigate about the subject matter and related data collection
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In the 18th and 19th centuries scientists wrestled with identifying and describing the nature of the “stuff” that produced change. One concept that became popular for a while was that of “caloric” (what we now call heat). “Caloric was originally conceived of as a quantity that would flow from a hotter object to a cooler one that would warm up as a result. It answered the need for a way for the cause of warming to get from here to there. Not only did caloric serve as a cause for warming‚ it
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gas that occupies 1.00 dm3 at a pressure of 2.00 bar. If the gas is compressed isothermally at constant external pressure‚ Pext‚ so that the final volume is 0.500 dm3‚ what is the smallest value Pext can have? Calculate the work involved using this value of Pext. Solution For compression to occur‚ the value of Pext must be at least as large as the final pressure of the gas. P1 = 2.00 bar‚ V1 = 1.00 dm3‚ V2 = 0.500 dm3‚ P2 = ? = 4.00 bar This is the smallest value of Pext can be to compress
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Combustion of Acetylene (ethyne) Rx: Back Why I don’t do acetlyene explosions in our lab.... Combustion of alkanes A quick review.... • Combustion reactions of alkanes consist only of C and H (hydrocarbon) • require O2(g) as a reactant • produce CO2(g) and H2O(g) and a large amount of energy!! Oxygen can be the limiting reagent which can lead to Incomplete combustion. For complete combustion of a hydrocarbon‚ oxygen must be in excess. If there isn’t sufficient oxygen‚ incomplete combustion occurs
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Chemistry Ozone Depletion Ozone depletion describes two distinct but related phenomena observed since the late 1970s: a steady decline of about 4% per decade in the total volume of ozone in Earth’s stratosphere (the ozone layer)‚ and a much larger springtime decrease in stratospheric ozone over Earth’s polar regions. The latter phenomenon is referred to as theozone hole. In addition to these well-known stratospheric phenomena‚ there are also springtime polartropospheric ozone depletion events. The
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3.9 = 9.0 *153.14 + 75.099 = 228.239 -> 228.24 FINAL ROUNDED ANSWER *150.15 - 87.991 = 62.159 -> 62.16 FINAL ROUNDED ANSWER 3. Multiplication and Division *The answer is rounded to the same number of significant figures as the number with the least number of significant figures. Examples *2.8 (2 sig figs) x 1.05 (3 sig figs) = 2.94 -> 2.9 FINAL ROUNDED ANSWER. *1.75 (3 sig figs) x 35.01(4 sig figs) = 61.2675 -> 61.3 FINAL ROUNDED ANSWER. Units *S.I. Units used for
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