giment Calorimetry Lab Sec. Name heated to 99.9\"C in Desk No. An 18.50-s sample
ID: 1010479 • Letter: G
Question
giment Calorimetry Lab Sec. Name heated to 99.9"C in Desk No. An 18.50-s sample of aluminum is heated to 99.9°C in a hor water minum sampie is quiekly transferred t rium temperature of the aluminun sample bot water bath 1. An 18.50-s sample of aluminum i is quickdy transferred to 50.0 ml of water at 23.8C contained thermal equilibrum is reached. The tn a calorimeter. The thermal equib pecide heat of aluminum? of the aluminum sample plus water mixture is 294°C what is the s reaction peformed in a glass calorimeter be greaser or recorded temperaalorimeter? Explain. Assume glass to be a better conductor of heat than 2. Will the recorded temperature change for an exotbermic less than that in a styrofoam "coffee cup" calorimeter? Explein. Assume peutralization reaction is exothermic. For the measurement of the enthalpy of neutralization, AH, for (beaker or styrofoam cup). How will this unaccounted for heat 3. An acid-base peutralization the reaction, heat is inevitably lost to the calorimeter loss affect the reported value for the enthalpy of neutralization for the reaction? Explain. 4. Three student chemists measured 50.0 mL of 1.00M NaOH in separate styrofoam "coffee cuP" calarimeters (Part B) Andrew added 50.0 mL of 1.10 M HCI to his solution of NaOH: Lyndsay added 45.5 mL of 1.10 M HCI (equal moles) to her NaOH solution. Dale added 50.0 mL of 1.00 M HCl to his NaOH solution. Each stadent recorded the tempera- ture change and calculated the enthalpy of neutralization. Two of the chemists will report, within experimental error, the same temperature change for the HCI/NaOH reaction Identify the two students and explain. Experiment 21 251Explanation / Answer
1.
Specific heat capacity of water, C = 4.186 J/goC
1 ---> Al
2 --> water
since density of water = 1g/mL , mass of water = 50 g
use:
m1*c1*delta T1 = m2*C2*delta T2
18.5 * C1*(99.9 - 29.4) = 50*4.186*(29.4-23.8)
C1 = 0.899 J/goC
Answer: 0.899 J/goC
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