Question

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CIO), is mixed with 0.300 L 0.190 M Nal? Assume What mass of precipitate will form i 1.50 L of highly concentrated P the reaction goes to completion. mass of precipitate: Write the balanced neutralization reaction that occurs between H, SO, and KOH in aqueous solution. Phases are optional. neutralization reaction: Suppose 0.250 L of 0.450 MH,SO, is mixed with 0.200 L of 0.230 M KOH. What concentration of sulfuric acid remains after neutralization? concentration: MH,SO A 50.86 g sample of a substance is initially at 29.1 °C. After absorbing 2455 J of heat, the temperature of the substance is 134.6 "C. What is the specific heat (c) of the substance? The balanced combustion reaction for C, H, is 2C,H,O) + 150,(8) 1200,() + 6H2O(l) + 6542 If 5.900g CH is burned and the heat produced from the burning is added to 5691 g of water at 21 'C, what is the final temperature of the water? final temperature: 41.78199 Given the reactions, X(s) + 0,(8) XO(8) XCO,(s) - XO(s) + CO, (g) AH = -761.7 kJ AH = +405.3 kJ what is AH for this reaction? X(s) + 0,(9) + CO,®) — XCO, (S) AH = -989.4 Consider the reaction. 2 Fe, 0, 4Fe +30, AHin = +824.2 kJ The formation of 20.0 g of O2 results in the absorption of 172 kJ of heat. the absorption of 5490 kJ of heat. the absorption of 515 kJ of heat. the release of 172 kJ of heat. the release of 5490 kJ of heat. the release of 515 kJ of heat.

Answer 1

* The balanced equation for the reaction I Pb(ClO3)₂(aq) + 2Nai (aq) → pbla(s) +2 Naelo, (aq) alow Number of moles of Nal - 0.300* 0.19 = 0.057 mole 50 Number 2 of moles of Pola (Preciptate) → 00057 => 0.0285 of pble = No of moles x molar mass Now Mass = 000285 X461 = 13013 gm

* Here substract reaction to reactione. Now we will get Rinal reaction - DH - DH, - DHz - - - 761.7 - 40503 = 1167 kJ

for 246 * Balanced combustion reaction ?CHE (1) +1502 (9) ► 1220249] + 6H2001)+6542 kJ => CHE = 6x12 + 6 x 1 . . = 78 Now the equation indicates that Rx78 = 1569 benzene will Produce 6542 kJ By using proportions we can calculate X 6542 KJ = 5.99 . 1569 X-2470421 KJ => 7474 & Heat = mass xDT * 40185/gº DT-247421 J . (5691gx 40185/6) » 10040° final temparature = 21+ 10.40 = 31.40°C

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