Mid Term & Revision Notes PDF

# Relations - Boylis law P.V=PU₂ - Combine (n= Const ) - Agas, a certain gas - Given or Req. densing - Charle's law 뿏= - Avog's law =쁨 - Ideal Gras law. - PV=ORT - Units - 1atm = 760mmHg= 760 torr = 101.3 kB - 1 lit = 10³ m³ = loc²= loml - Gay's law = 11 = 103243 - Given or Req. mass...

# Relations - Boylis law P.V=PU₂ - Combine (n= Const ) - Agas, a certain gas - Given or Req. densing - Charle's law 뿏= - Avog's law =쁨 - Ideal Gras law. - PV=ORT - Units - 1atm = 760mmHg= 760 torr = 101.3 kB - 1 lit = 10³ m³ = loc²= loml - Gay's law = 11 = 103243 - Given or Req. mass - Conditions # Laws - Dalton's law - P=ΣΡ - Garhamis law [ Rate of diffusion] - Taten = HR = R - Rate ↑ عکس timed - Chemical Reaction. - N₂ + 3H2 → 2NH3 # Tricks - Real gases (a, b, VDW const.) - PTUT - A gas collected over water ېوقال - PT = Pg + Pv - Po₂=*=ar High P - at STP - Compressibity factor. # Relations: - PV_1 = PV_2 - n_1.T_1 n_2.T_2 # Q: - Gas in ballon occupies 2.51 at 300k. At What temp. K will the ballon expand to 7.5L? - Pui Povz / AT - 2.5 = 7.5 →T2=900k - 300 K - 2.2 # Q: - Two Flasks A and B of equal Capacity Contain equal masses of th and CHy gas. respectively FlaskA at temp. 300k While Flask B at book. The ratio of Pa to Pois - PAVA PEXE - 300 Мени 600 - - 2PA P - 300 $600 - PA - 4:1 - B # Q: - Flask x filled with 20g Chy gas at 1000 and another identical flask Y is filled with 40 g O₂ at the same temp. What is the Correct statment:. - P in two is identical - Pox > Py - Px < Py - Px = ½ Py - 20 - Pxle ружу - ←nx - Px=Py - 2 # Q: - First of two gases has/ 14 times molar vol. A as much as the second at same T. then Pi/P₂ = ? - 4:11:4 - - 4 P - - P₂ - 2 - 1:4 # Ideal gas law: - PV=nRT - density: d= PM/RT # Q: - What is the mass in gm of 151 ammonial(NH at 27℃ and 900 mmHg? - PV=nRT (27+273) - 900 em - 760 - M =?? - 151 - M = 14+3 - m = g # Q: - IP 6.3 mg of Bxtly is contained in Plask 385 ml at 258 and 11 mmHg. What is the hydride formula? (Nut B= 10.8) - al BH - b) BH₃ ↑ B₂ Ho di By Ho - PV=nRT→25+273 - - 760 alm - 1 - → - m = - 6.3×103 - 385 - X163 - M= 27.6 g/gmole - Вань - d) a & b # Q: - An unknown diatomic gas has a density of 3.184g/l at s TP. What is the identity of the gas? (Na:23. (1:355) - a) Naci - b) O2 - c) O2 - d)JNz - d= PM - RT - 3.164 = 1*M - Nescaped - → - 0082 243 - M= 40.8 - - 2 - - Cl₂ CIO₂ # Q: - Density of O2 at 27 C and later -81L - - ← - 1x 32 - = 1-3g/L - 0.0827 (27+273) # Q: - Density of a gas at STP = 1.429 g/L then gas identity (S:32) - al) O2 - b) S - c) CHy - d) a&b - 1 - d=PM - RT 273 - 1.429 - 0082 - O2 - - Sxxx # Q: - A 2.24L Cylinder Containing O2 at STP. is found to have leakage. When the leakage was stopped, the pressure dropped to thong - The number of moles escaped gas -- - 570 mmitg - = - - 27 - % - ( - mmity - Nescaped n,-72=? - 3 - → - Di - ni = - Pui =^,RT, - 1atm x 2.24 = - 1*0.082 - 02 - x273 - 0.10006 mole. The - 760 - 570 - → - ^2= - 0.075045 - 0-10006 - 2 - Nescaped = 0.025015 mole. # Dalton's Law - PT=2Pi - Pi=Yi PT - UT=Vi - If the Partial Pressure of ages at STP equals 604 mmHg, then the mole % of this gas equals - Pi - Yi= - 604 - 760 - % = Yi x 100 - - = - Pressure of Oz in air at STP = 2atm - 0.2 - 으므 - 20 Lit. Mixture Contains 40% by mole O₂ and 30% by mole H₂, at T= 25℃,P=750 mm Hg. - then Vol. of Oz in mixture equal - 20 - L - If 21 of Oz gas at 760 mmHg mixed with 3L of N2 gas at 140mmHg then Pressure af AL mix. equal --- atm, at Const. T - PT* VT = Voz Poz + UNZPN 2 - PT x 4 = 2 * 760 + 3 x 740 - 760 - 760 - PT=1.23 - atm # Chemical reaction: - If 28 g of Al is Corroded in HCl to release hydrogen gas - 2 Als+ 6HCl→ 2AlCl3 + 3H2 - (S) - (aq) - (5) - (g) - Vol. of Hydrogen at 258 and 800milly = -- - L - Sol - (A1:27) - 2A1 3H2 - 28 - n= - X ?? 1.555 mole - 2.7 - 240882 - PV=nRT - 29. - - 800 ?? 1.555 0.082 - V=36.11L - - 2 - - 760 - - - - - - Hydrogen Cyanide is prepared by - CHu + NH3 + 3/20₂ → HCN + 3H20) - (9) - (9) - Vol. of HCN=---L, Which can be obtained from 201 CHy, 201 Nilz, and 2010월 at Const. T.P - Soli - - 20 - 201 - 201 - 20 - X - ?? 20x1 = 13.3 - 3/2 - - - L - L.R - 20/120/15 - HEN = 13.3L # Real gases: - (P+anz) (V-nb) =nRT - V2 - Z= PV/ART d = PM - Compressibility factor of Carbon at high P is greater than one (X) # Q: - The P of 2 g mole O₂ gas of Vol. 1L at o℃ is --- atm (a=1.362, b=0.215) - sol. - (P+ (1362) (2)²) (1-2x0215) = 2*0.082 - - 273 - L↳ Shift Solve - P= # Gharham's law - - - - - - - - - ①:. rate of diffusion of O₂ is ④ that of H₂ - - - - - 22 - - 2 - Ног - 32 - - - - 1 - - 13 - - 16 - ↑ If Gax x has density 4 times that of gas y - ther - - - - rx - - dy - - dx - - √4x - - - # R.H Ratio between Int Air Cond. - Ρω τo Ρως - Evaporation for - R.H= - PT= - - - - - - Condensation كحلي باك - R.L=100%, - خلي باك + 100 % - Ρω Ρω - ③ may be Condensat - R.H₁= - Priv - PTェニ - V2 = ?? - Assume - T= - Tュニ - m - T₁= - Μω = -- - Cord - Pw, Vi - تنا ايه بس بيقول may be - no Cond. - PW2 VL - Tューレ - = - → - Pus= - Pws - Pus= - T - If Plu>F - اعمل اي 2 - steps - steps الخطوت - DR.HI= PWI - Pws, - Pr₁=Po,+B→ - evi - = - Pa - 12 - V2 - لازم - K - P - P - 2] PT₁ = Pa,+ Po, d - Madd - = πωπω, - Pws, - (PW) = Pws - Mwadd - = Owaday - X18 - 3] Pavi - Paz V2 - V2 - K - PT₂ = Pa₂+ PW₂→82 - 3] PVI = P22 - Ti - V2 - T₂ - If Ruz > Ρω, - - - - 5] R.Hz = Puz - Pus - Ι (πω - DrDina - R.H - Ρω, νι - RT - J - خلی باك لازم تحول الوحدات - 4] πω, = Ρωινι - لازم تحول - RTI - writs - If Ρω₂ < Ρω - Πω₂ = Ρω2U2 - يبقى مفيس Contest - RT - ④ - - 5 - nwi = PW, VI/RT - 'wcond = (πω,-nw2)x18 - nw2 = PW2V2/RTE # Thermo Chemistry - Summary - - - # Thermo Chemistry - * Theoritical Questions:- - exothermic Theat evolved /-ve - Heat of Pormation may be Tendothermic / heat absorbed/que - Formation of Imole Compound from its element - ↳ Reaction of Co and Os to produce Co₂ XX (not Hp) - Reaction of Carbon and Oz to produce Coz ur (HP) - Heat of formation of any element = zero - L↳ Hfo, zero - exothermic - Heat of Combustion always - heat released - -ve value - - Combustions Complete burning of Imole Hydro Carbon to produce Co2 and H20 - Înot co - H combustion of Kagdrogen = - 68.4 kcal/mole - It combustion of Carbon = - -34.2 kcal/gm - = - -94-4 kcal/mole - Hf - CO₂ - = - -4-9 kcal/gm - HCamb. per mole H₂< Hamb per mole Carbon - Hcomb per gm He > Нсеть per gm Carbon - → Heat of reaction measured by Calorimeter - → Heat of reaction Enthalpy change - → Air is prefered in indusrty (more economical) - → Combustion efficiency (2) is inversly prop. with amount of air used - - →17= head of Combustion - → - - - - @movent of fud + aèr 1 - All Carbon Convert to CO₂ - → Specific head Capacity of cor at STP = 1.0035 J/gm² - → - - - - of Copper = - 0.325 J/gmc - of Walter = - 4.18 J/gmE - → Heat of Combustion of 1 gm H2 = 2.5-3 heat of Combustion of 19m N.G - → He of 1 gm H2 = 4 Hc of 1gm Carbon. # Heat of Combusion Callcubet - ancelle - Por 1 Paraffinic Rule - [ Rule 157] - Ie of each CH₂ = 157 kcal حفظ - C - - - - - - - - - - CH2] - @Calculate Calorific value of Octane (C8H18) if you know Calorific value of Hexane (CH4) = 18kcal/gm - Sol. - 드의 CV) CHIN - - - - - VC - 6시2+14 - - c) Gaia = c + - - - 2x [CH2] = 1118kcal+ 2x157 - = - - 1432 kcal - - C.V) COHIO - - 1432 - 8X12+18 - - - - - - - - - - - - - - - - - # Walter rule :- - Cancelled - for - Credit - [CHO] - For Compounds (Sugers) - Oc = (no. & c - noof 02) +94.4+ (10.4 H2)x684 - - - 2 - = - - - - - - - QiCalculate Civ for GH128 - 9=[6- - - 1]94.4+(를)×68.4 - - - C.V= - - - 2 - = - 693.6 kcal - 693.6 - C.V= 6x12+12+6x16. - - = - kcal - 3.85 - gm # حسب المركب Heat of Combustion حاسب - لا خلى بالك - الوحدات ) - M - 1] Paraffinic Rule [0157 kcal] - - - - ويطلب الثاني بدلالة الأموال ويكون الفرق بينهم عدد في - I CHEI - - - - - - - - - - q - C - + [cy] - C2H6 - -157 kcal - 2] Welter's Rule [ C12 H22O11] - - - - - - - - - - - - - - - - - - - - - - q = [ no & c - no 40 ] *-944 kcal - - + - - C - - 2 - [no.fㅂ]* - 68.4 kcal - - - 2 - = [12-#]*-94.4+ [2]*-68.4 - 3 Hessis law - HE - де=9= 2nite - Products - نيكون معطى معادلة - - Σπί HPR # Inits Thermo - ΔΗ Canth. or ac of H2 = 68:4 kcal - Camb - mole - amount of heat produced from burning Az 19m Hz - Calorific value = c. - =- - 34.2 kcal - M - - Lo tue vcelue. - gm - DH Comb. - or gre of Carbon = €94.4 kcal - mole - C.V= r - = 7.86 kcal - the value - M - gm - Steall - 1kcal 4.18 KJ - ラ - 1 Cal= 4.18 J - Specific heat Capacity of water= 4.18 - = - - gm. K - 1 Cal - - gm- - → - Molar heat capacity of watter - = - 18 Col - gmolek - → CP = Cu +R - ⇒ CP-Cu = R - large per mode - فل بالك - Heat comb - Co - Small per gm - CV - + Small - The device used to determine the heat associated with a chemical reaction is --- - Calorimeter - Specific heat Capacity of:. - Air at STP equal --- - 4.0035 J/gn & - Copper equal --- - 0.385 J/gm² - Water "Liquid) equal --- - 4.1813 J/gm & - Green House gases like --- and-- - CO2 H20 Cни - DHC of H₂ per gram = - 2.5-3 - DHC of N'G - per - gram - DHC of H₂ per gram = - 4) - DHC of Carbon - per - gram - Cement - + - Green Concept - Questions + Answer # Complete the followmy.. - Transesterification reaction for biodiesel productine - -Triglycerides + Alcohol Callalist brodiesel & Glyceroll - Hydrogen needs extensive Safety measures due die its Leakage explosions and associaded hazand - The main components of OPC are GS..., S., -GA., and SYAF - Hydrogen Could be produced by methian pyrolysis, Coal gasification : While green hydrogen mainly produced by electrolysis of water method. OR Steam reforming of bio-based - Steam reforming of N.G is keaction of N.G with Stean tas and the gas produced is Synthesis gas # Iron slag Cement has low cost - Di. Calcium Silicate is responsibike for Settling and arly stisnigthin ordinary Portland Cement. - Green Concept is any-approach to decrease GHG and the following are examples of green alternatives - biodiesel green H21 and green Concrete - Removal of humidity in pre-treatment of biodiesd morder to avoid Saponification - Separation process in biodiesel industry is done to remove byproducts such as glyceral Soap. excess-alcohol + and watter traces...... - Rapid hardening Cement differs from portland Cement as its_very fime = 99.5% pass mesh 30c # The percentage of Sulpther trouxide should not exceed 3.5/.. - The excess presence of the Volatile materials in Cement Cause -Crsasking. (Cracks Prmation - In green Concrete. Super plasticizes is added to Save water (water reducing admixtures) - Pyrolysis is a process that involves heating of biomass in absence of oxygen ... - In trans- esterification. sthand ... is the most used alcohol because of its low Cost., However, greater Conversions into biodiesel Can be reached using Hethangl... - .. is used as fluxing agent in OPC industry... - OPC must has very fine size, Where - 98% % of particules must pass screen, of -200.- mesh. # Pur (T) or (F - Ethanol is the most common used calcatalin Tranester Pication process (T) - Gypsum is added to Cliniker to retarded Cohesion of Cement (T) - Yellow H₂ produced by electrolysis with nuclear energy (F) - Rapid hardening Cement is fine Which 98% Should pass Hirough mesh 200 (F) - The decreasc in hydraulic factor means slow Setting and low early Strength (T) - Blue hydrogen is produced from NG steam reforming and released to air as syn gas (FI - B40 is biodiesel Contains 40% of biodiesel fuel (T. - Buildings made from OPC needs less energy than building made from green Concrete (F) - Gypsum added to rotary furnace in Cement industry (F) # Imp - Different Colors:. - Color method/Source released to - Brown H₂ Coal atmosphere - Grey H₂ Steam reformung air (as raw Synthesis) gas - Black H₂ Coal gasification emissions - Blue H₂ N.G steam reforming OR Partial oxidation of refrey reside Captured by C and storage. - Green H₂ electrolysis of water (Renewable) OR electricity Steam reforming of bio-based - Turquoise H₂ methane pyrolysis - Pink/Purple H₂ electrolysis with nuclear energy - Yellow H₂ electrolysis with Solar power - White H₂ naturally occurs underground H₂ difficult to exploit # Manufacturing Process - Block flow diagram - Raw Material - S - Crushing - Grinding - Reaction in kiln - Clinker Clinker cooling - Clinker grinding - Cement final product - ↑ Atack # Triglycerides, Alcohol - Pre-treatment - Reaction "catalyst" - Separation - By - products - Water - Washing - impuritie - Drying - Storge and Distribution - W2T - FAMES - () - Simplified Block Flowsheet for Production of Biodiesel - (C) - Tale-3002 # Mid-terms Exam - # Academic Year: - Spring 2024 - Term Exam: - Second Term - Department: - Preparatory - Date: - 26/3/2024 - Course Code: - CHEG001 - Course Title: - Chemistry - Time: - 1 hour - Full Mark: - 20 - # General Instructions: - The exam Is composed of 3 sections - All 3 sections are to be answered in the space provided in the ANSWER SHEET except for the figures of section 3 which are To be answered in the Place provided in the exam sheet. - The 3 sections are marked out of 75 marks. This will be multiplied by a factor and the full mark will be out of 20 marks. # Section 1: (Multiple Choice) - Q1: a b c d - Q2: a b c d - Q3: a b c d - Q4: a b c d - Q5: a b c d - Q6: a b c d - Q7: a b c d - Q8: a b c d - Q9: a b c d - Q10: a b c d # ANSWER SHEET Section 2: V or X - Q 51: - Q 52: - Q 53: - Q 54: - Q 55: - Q 56: - Q 57: - Q 58: - Q 59: - Q 60: # SECTION 3: (Fill in the blanks) - Q 101: - liter - Q 102: - - 9 - Q 103: - mol - Q104: - kg - Q 105: - Q 106: - Q 107: a). - - b) - g - c) - - d) - g - e) .... - liter - mmHg - Q 108: - Kcal/g - Q 109: - g - Q 110: a) - - b). # Q: - What is the pressure, in Pa, inside a 1 m³ cylinder containing 10 kg of H2(0) at 25 °C? - a) [1x10'x8.31x25]/[1x10³] - b) [5x10x8.31x298]/[1] - c) [1x8.31x25]/[1x10³] - d) [5x10x8.31x298]/[1] - The combustion reaction of pentyl alcohol, CsH1OH and oxygen can be represented by the unbalanced equation below. - CSHOH+.... O₂→..... CO₂ + ..... H₂O - What is the coefficient for O₂? - a) 6.5 - b) 6 - c) 7.5 - d) 8 - The number of moles of air necessary for complete combustion of ethyl vinyl ketone, CH,CHO Is equal to, gmol: - a) 6.5 - b) 7 - c) 32.5 - d) 35 # SECTION 2: True of False - Read the following statements carefully. Put T in front of the correct statements and X in front of the wrong ones. Transfer your answers to the Answer Sheet on page 1 (10 marks) - 51- If the temperature in Kelvin of 2 dm³ of an Ideal gas is doubled and its pressure increased by a factor of four, the final volume will remain 2 dm³. - 52- Equal masses of gases occupy the same volume at constant température and pressure. - 53- The mass of 1.12 Ilters of gas Y at STP is found to be 6.23 g. The molar mass of gas Y is 125. - 54- Oxygen will have a density of 1.1 g/m³ at 25, °C and 0.85 atm. - 55- If equal masses of hydrogen and carbon dioxide will be placed together in the same container, their partial pressure will be equal. - 56- In van der Waals equation, a and b are constant that depends on gas type only. - 57- The pressure a gas would exert under Ideal conditions is always greater than the observed pressure of a real gas. - 58- Carbon releases more energy per gram than hydrogen when burned in air. - 59- The cause of the greenhouse effect is carbon monoxide gas. - 60- Heat of combustion of CO2 equals heat of formation of C. # V₁=................. liter - Pw.......... mmHg - Pd.............mmHg - V₂= 150 liter - Pe..... - %RH₂ = - ....... mmHg - % - 112. Explain the effect of pressure on the compressibility factor. - (3 marks) - 113. Draw a Block diagram for the production of Biodiesel. - (3.marks) # Mid-term 2022-2023 (main stream) - Sec. 1 - ] PTVT = VN2PN₂ + Vo₂ Poz - PT - = - (250 x103 * 720/160) + (280x103 ¥ 650/760) - 2KHCO3 → K₂O₃ + CO2 + H20 - - n = - 353 - = - 3.53 - 39+1+12+3x16 - مش - معانا - KHCO3 - 2 - CO2 H20 - 1 +1 - 3.53 - →3.53 mole - PV=nRT - 880 * V = 3°S3 * 0082 * (520+273) - 760 - 3] - - ⇒ - 2 - Mi - - M - 4 - a - Same - molecules - P - 5 - 6 - b - XX - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 550x M - - - - - - 400 cm²³² * 44 = - - M=32 - - - - - - - - - - - - - - - - - CH3CHO + O2→2002 + 2H2O - - 2 - b - 2.5 - - - - - q - ONC₂H₃OH = CHOOH - + 2 [-157] ०÷८१७) - = - 174 + 2*-157 = -488 kmole. - CV = - 488 - =⊕8 kcal/goto - 3×12+8+16 - - - - - - - - - - - 07-9 - heat of Combustion - - x-157 - +7-