safety regulations

nyfuruev94
timer Asked: May 2nd, 2017

Question Description

these questions' answrrs will be from the book Legal Liabities


legal liabilities in safety loss prevention book



this is the link in googole books


https://books.google.com/books?id=OgHAScZT9MMC&pri...

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To find enthalpy changes of the following reactions. 2A(g) 2B(g) ∆ = 120Kj B(g) A(g) ∆ =-60Kj A(g) C(s) ∆ =-170Kj 17. amount of carbon in 40.0kg of anthracite. 1 mole of the sample =3083g/mol Grams in the sampe=40*1000=40000g %carbon content is 92.1%, so the amount in the sample is 0.921*40= 36.84kg 16. The molar mass of methanol is 32grams 32grams produces 764Kj So, 35.81grams will produce 855kj. 35.81 grams. 15.The formation of 51.0g of O2 results in the absorption of 438kJ of heat. 14. when 1.60mol of H2 reacts, 387kj is released. 13. all statements are true except, when energy is transferred as heat from the surrounding to the system, ∆H is negative. 12. specific heat= number of joules used/(∆temperature*grams) 305J/(∆temperature change*37.6g)=0.128J ∆temperature*37.6=2382.8125 ∆temperature change=63.3727 So final temperature is 20.0+63.3727=83.37. 11.heat energy = (specific heat capacity)*(∆temperature change*grams) =(1.75)*(168*2430)=714420J 10.heat energy = (specific heat capacity)*(∆temperature change*grams) 263J=(0.903)*(22.5*grams of sample 22.5*grams =180.509 Grams=8.0226g 9. mass=24.5g,temperature change =15.7,heat energy=90.4J. Specific heat capacity=90.4/(15.7*24.5)=0.235 Hence, silver is the answer. 8. grams=21.59g, heat energy used=1435J,final temperature=181.5 C. Temperature change=154.4 Specifc heat capacity=1435/(154.4*21.59)=0.43047J/(g.C) 7. The amount of heat required to increase the temperature of 1g of a substance by 1 0C. 3. Endothermic exothermic Making popcorn in a microwave a burning match Boiling water The reaction inside a chemical pack Burning rocket fuel. 6. To convert the following energy units. 1kj=0.239kilocalorie 13kj=3107.075 calories. 123kcal=514.632kj 5. a) 938kj=938000j b)2530kcal=10585.52kj c)6.27*10^6j=1498.565kcal 2. Classify the following phase changes as endothermic or exothermic. Endothermic exothermic Solid to liquid liquid to solid Liquid to gas gas to liquid Solid to gas gas to solid 1.Classify each point with correct information about kinetic energy and potential energy. A=max PE,min KE B=max KE, min PE C=max PE, min KE. Which of the following represents kinetic energy? Wind, riding a bicycle, an apple falling from a tree. 14. If 506mol of octane combusts, what volume of carbon dioxide is produced at 38.C and 0.995 atm. Forever 2 moles of octane you have, you get 16 moles of CO2. So divide 506 moles of octane by two, and then multiply by 16 to find the number of moles of carbon dioxide. Number of moles of CO2=4048 Use the equation PV = nRT, or Pressure x Volume = moles x R x Temperature (in Kelvin) .995 atm x V = number of CO2 moles x .08206 x 311 V=(4048*0.08206*311)/0.995=103826.766cm3=0.1038L 12. first we need to compute moles of each gas Moles of Argon=3.04/39.95=0.0761 Moles of Krypton=1.12/85=0.0132 Pressure acted by Argon, P(Ar)=nRT/V=[(0.0761)(0.08206)(298)]/(2.50)=0.744 atm Pressure acted by Krypton, P(Kr)=nRt/V=[(0.0132)(0.08206)(298)]/(2.50)=0.129 atm Total pressure acted=0.744+0.129=0.873 atm. 13. Pressure acted by a third gas=[(0.200)(0.08206)(304)]/(9.30)=0.536 atm P(total)=P1 + P2 +P3= 0.244+0.540+0.536=1.32 atm. 9. PV=nRT, T=PV/nR.[(0.039610)(0.08206)(T)]/(887.1*10^-3)=0.96 atm. T= 0.96/(3.66407*10^-3)=262K T=262-253= 90 C. 8. Directly proportional P and n inversely proportional P and V V and n P and T T and V P and V 5. The volume of a fixed amount of gas is directly proportional to its absolute temperature. 4. Pressure is inversely proportional to volume, so P1V1/T1=P2V2/T2 P2=(PIV1T2)/(T1V2)=(1.14*2.30*304)/(2.90*10^2*1.03)=2.668 atm. 7. Volume is directly proportional to the number of moles. V and n (directly proportional). Initial number of moles n1=(32.4*331.8*10^-3)/4.265L= 2.52 moles. An ideal gas in a sealed container has an initial volume of 2.35L. At constant pressure, it is cooled to 24.00 degrees, where its final volume is 1.75L. what is the initial temperature? V1/T1=V2/T2, T1=(V1T2)/V2=(2.35*297)/1.75=398.83K= 125.80C 1. To label the curve, starting from the origin, solid only, solid and liquid, liquid only, liquid and gas, gas only. In that order. 4. volume of a gas is directly proportional, so PIV1=P2V2, P2=(P1V1)/V2=(5.31*2.84)/5.22=2.89 atm. 3. If the pressure in the manometer was equal to atmospheric, h would be 0.0 The gas pressure is 773-55=718mm. 5. q=mc*temperature change=(2.55*1000*168*1.75)=749700J. 2. 1520mmHg,1.5 atm, 494mmHg, 0.25 atm.(from highest to lowest). 3. To start, you will need the equation q=∆Hvap(mass/molecular mass). You can also convert to moles first. So using the values you provided: 3.02g*(1/18.01)*(40.66)=6.82Kj 4. molar mass of ethanol=46grams Molar heat of vaporization=38.6kj/mol Therefore (241.8/38.6)moles require 241.8kj. 6.2642moles, requires 241.8kj The number of gram heated is(6.2642*46)=288.155g. 7. which molecules can hydrogen bond? H2 and HF 8. Cl2 exhibits only London dispersion because it's nonpolar, all the rest are polar 6. only NF3 and CIF have polar interactions. 9. H2O hydrogen bonding ,CH3Cl and CO Dipole dipole and dispersion,H2 Dispersion only 11. CH3OH,CH3CI, CH4 (from the highest boiling point to the lowest boiling point) 10. Hydrogen bonding, London Dispersion Forces, and Dipole-Dipole interactions 2. density-determines if a substance floats or sinks, relates a mass to its volume, is a physical property. 12. CH4, CH3CI, CH3OH.(from highest vapor to lowest) 11. P(total)=P1+P2+P3, Partial pressure of Ne=11.3-(4.15+1.36)=5.79 atm.
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