Chemistry: A Molecular Approach (Tro) Chapter 9 Chemical Bonding I: Lewis Theory 1) Which of the following represent the Lewis structure for N? A) [pic] B) [pic] C) [pic] D) [pic] E) [pic] Answer: A 2) Which of the following represent the Lewis structure for Cl? A) [pic] B) [pic] C) [pic] D) [pic] E) [pic] Answer: B 3) Which of the following represent
Free Chemical bond Covalent bond Ionic bond
the reaction) Examples: (1) Calibration of the calorimeter given that: ΔHrxn = -55.8 kJ/mol and nLR = 0.005 mol qrxn = ΔHrxn x nlimiting reagent qrxn = -55.8 [kJ/mol] x 0.005 [mol] = -279 J qcal = -(219 J) = 279 J (2) Determining the qrxn of a given chemical reaction: NH3 (aq) + H+ (aq) NH4+ (aq) And given that: ΔT = 3.5 °C and Ccal=111.6 J/°C qrxn = - Ccal ΔT + mcsolid ΔT qrxn = -( 111.6 [kJ/°C] x 3.5 [°C]) = -390.6 J qcal = -(-390.6 J) = 390.6 Determining the Ccal. Ccal = qcal
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Lab Report 3: Combustion of Magnesium and Specific Heat of a Metal C4C Jeffrey Silvin Fall 2017 Major Kittle Chem 100 T1 Introduction The purpose of this lab was to determine an experimental value for the heat of formation of MgO with Hess’s Law and then use the result to find percent error. To do so‚ approximately 50 mL of 1.0 M HCl was added to a calorimeter. Initial temperature was measured and then 0.25 g of Mg was added. After the reaction is completed the maximum temperature was recorded.
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53J/g*C Tinitial = (27.1+23.8)/2 = 25.45 qrxn = -(100g)(4.18)(38.43-24.45)+(-12.53x12.98) =-5400J/.1mol(1J/1000kJ) = -54.0 kJ/mol Tinitial = (26.0-24.5)/2 qrxn =-(100g)(4.18)(26.26-25.25)+(-12.53x1.01)= -40.9J/.1mol(1J/1000kJ) = -4.09kJ/mol Tinitial = (23.9+22.9)/2 =23.4 qrxn =-(100g)(4.18)(29.48-23.4)+(-12.53x 6.08)= -2620J/.1mol(1/1000kJ)= -26.2 kJ/mol Verify Hess’s Law 1. HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l) H+ + Cl- + Na+ + OH- → Na+ +Cl- + H2O (l)
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you know why this site sucks? necausealskdjflaksdjfalskjdfalsdkjfsd asdfaskldjf asdlkfjasldkf my mom sdfjsadlkfjl lakjdsflkasjdflkasdjflaskjdflaksjdf dlkfjasdlkfj ldkjfalskdjf lskjflaskdjf lfkjsldkfjl lk kj lk lj lkj lkj lkj l;kj lkj l;kj lkjhgasdf asdf we fa b t wsr r tg g dfg e t teh arefg adf g wer g es hts rt hs rt hs rht srt h lkfdjlask lkfjsa;dlkf d;lfkjasdfl dkfl jdflkjasdlkfj asdlkfj sldkf Citation Page "Drive-Thru 101: History." Dr. Drive-Thru: In Order
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Calculate ΔGf° for COCl2 at 25°C. (ΔGf° for CO(g) = –137.3 kJ/mol at 25°C) PRIVATE "<INPUT TYPE=\"radio\" NAME=\"null_12_0\" VALUE=\"3\" DISABLED>" MACROBUTTON HTMLDirect A. –66.7 kJ/mol PRIVATE "<INPUT TYPE=\"radio\" NAME=\"null_12_0\" VALUE=\"1\" DISABLED>" MACROBUTTON HTMLDirect B. 188 kJ/mol PRIVATE "<INPUT TYPE=\"radio\" NAME=\"null_12_0\" VALUE=\"4\" DISABLED>" MACROBUTTON HTMLDirect C. 274.6 kJ/mol PRIVATE "<INPUT TYPE=\"radio\" NAME=\"null_12_0\"
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xam 2 [pic] Summer 2011 Chemistry 1411 EXAM # 2 Activity Series of Metals in Aqueous Solution [pic] CHEM 1411 Exam # 2A Name:________________________________ (Chapters 4‚5‚6‚ and 7) Score: [pic] Part I- ( 3 points each) - Please write your correct answer next to each question number‚ DO NOT CIRCLE. ____ 1. Compare the rates of effusion of CO2 & SO2‚ rCO2/rSO2 A. CO2 will effuse 1.45 times
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bar T2 = ? V2 = ? mdot = ? V2 = sqrt(V1*V1+2*(h1-h2)) mdot = rho*A2*V2 = A2*V2/v2 T2 382.60 m/s 0.22 kg/s 111.81 degC Superheated steam at nozzle entry T1 150 deg C P1 1.5 bar s1 7.420 kJ/kg K h1 2773 kJ/kg K V1 10 m/s P2 A2 1 bar 0.001 m2 At Nozzle exit P2 = 1 bar Rev & Adiabatic s1 = s2 = 7.420 kJ/kg K h2 (Superheated because s2 > sg2 (7.359) s 7.360 7.614 7.420 s 7.360 7.614 7.420 s 7.360 7.614 7.420 h 2676 2777 2699.86 v 1.696 1.937 1.753 T 100 150 111.81 v2 T2 Example
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produce by combining 0.2036g Mg and 5.00mL H₂SO₄‚ ice was melted and readings were taken by using an ice calorimeter. Readings were taken before during and after the reaction were completed. The data taken shows a value of -405 KJ/mol while the theoretical value was -483.7 KJ/mol. Calculating the experimental value taken with the actual theoretical value it shows a 16% error in the experiment. A possible explanation for this error could be due to heat escaping the open test tube during the Mg reacting
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The Advanced Placement Examination in Chemistry Part II - Free Response Questions & Answers 1970 to 2007 Thermodynamics Teachers may reproduce this publication‚ in whole or in part‚ in limited print quantities for non-commercial‚ face-to-face teaching purposes. This permission does not apply to any third-party copyrights contained within this publication. Advanced Placement Examination in Chemistry. Questions copyright© 1970-2007 by the College Entrance
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