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Shweta Thakur posted an Question

March 02, 2021 • 19:59 pm 30 points

Please solve the third question -568 ki=4568 j es of the solution = 200 + 400= 600g atm o 4

2es -4-18 J Kg QmxCxar 2284 mxC 700 x 4-18 r= (), heat produced (ai), heat Producca S68 kJ = 4568 J 4-568 kI=4568 J es of the solution = 200 + 400= 600g Atm O 4508 4568 XC600 x4 ROBLEMS FOR PRACTICE 250 g sample or octane (CgHi8) is burned in excess of oxygen in a Do meter is893 k3 ctane (CgHig) is burned in excess of o 1-250 g sample of octa cess of oxygen in a bomb calorimeter of the calorimeter is8-9. Jehe Calorimeter rises trom 294-05 K to 300-78 K. If heat capacity of the cao eof octane 93 kJK) fi ehe calorimeter nises irOm 294-05 K to 300-78 K. If heat capacity of lpy combustion of the sample of octane. )1s dissolved in 125 g of water in a coffee-cup calormeter, value of q for the calorimeter. ransferred to the calorimeter. Also calculte the enthalpy combusu ! to the calorimeter. Also calculte the ent 20-0g of ammonium nitrate (NHNO) is dissolved in 125 g of water g of water in a coffee-cup calorimeter, thn ormperature falls rom 296-5 K to 286.4 K. Find the value of q for the calonie lls from 296-5 K to 286-4 K. Find the val Treat heat capacity Treat heat capaclyy or water as the heat capacity of the calorimeter and its cOnte / of water as the heat capac: acity of the calorimeter and its contents). 0-16 g OI Inetnane was subjected to combustion at 27°C in a bomb calorimeter system. 1ne tenp the calorimeter s methane at (1) constant volume, and (i) constant pressure. The thermal 17.7 kJ K (R = 8-314 kJ Kmol-H of methane was subjected to combustion at 27°C in a bomb c calorimeter system. The temperature system (including water) was found i d to rise by(0-5°C)Calculate the heat of combustion o ungwater) was found to rise by(0-5°C)Calculate the heat of combustion o al capacity of the calorimeter system is 2.-5-28 kJ -5-28 k 1. Heat transferred = 60-1 kJ, A-H = 5481 1 kJ mol =-885 kl mol, q,=--890 kJ mol HINTS FOR DIFFICULT PROBLE re = 300-78-294-05 K = 6-73 K calorimeter= Heat capacity of the calorimeter x Rise in temp. 8-93 kJ K) (6-73 K) = 60-1 k + 18= 114 g mol kJ mol l14kJ mol- = 5481-1 kJ mol calorimeter, the bheat gaine a/ mass of th 600x 4-18 heat transferredt When 20-0 temperature fall1 3. 0-16g o ANSWERS l14kJ mol= 5481-1 4-184 J = 5282

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