ooo Verizon i 12:54 PM 3. A5.00 g mass of a metal was heated to 100.0 Cand then
ID: 517638 • Letter: O
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ooo Verizon i 12:54 PM 3. A5.00 g mass of a metal was heated to 100.0 Cand then plunged into 100.0gof water at 24.0 c. The temperature of the resulting mixture became 28.0"c. cs for water- 4.184 JgC a) How many joules did the water absorb? b How many joules did the metal lose? c) What is the specific heat of the metal? d) What is the heat capacity of the 5.00gsample? 4. Calculate the number of joules of heat released when 72.5 grams of water at 95.0 C cools to 28.0 C, specific heat capacity of water 4.184 Mg C. 5. What is the specific heat of aluminum if the temperature ofa 284gsample of aluminum is increased by 8 C when 207 Jof heat is added? 6. A 13.5 g sample of gold is heated, then placed in a calorimeter containing 60.0 g of water. Temperature of water increases from 19.00 Cto 20.00 C. The specific heat of gold is 0.130 Jg C. What was the initial temperature of the gold metal sample? 7. A5.00 g sample ofacetylene undergoes complete combustion in an oxygen bomb calorimeter. The calorimeter has a beat capacity of6.49 kJ/C. The maximum temperature increase is recorded as 950 C. Find the molar heat of combustion for acetylene, C,H,. (Ans: -3.21 x 0 k/mol) 8, Solve the followi ess's law a. Calculate the AH for the following reaction: b. Calculate the AH for the following reaction: 2C1, SnCl NO NO, Sn You are given these two equations: You are given these three equations: Sn Cl: SnCI, AH 325 kJ O, 20 AH +495 kJ Snci 186 kJ 20, 30, AH -427 NO o, NO, O, AHI 99 kJ 9. Use a standard enthalpies of formation table to determine the change in enthalpy for each of these a) NaOHOs) HCl(g) 10. What is the change in enthalpy when 9.75 g of aluminum reacts with excess ammonium nitrate (NHNO, according to the equation: 3NHLNO 3N2+6H20 Al 20, AH 2030 99%Explanation / Answer
3.
From energy conservation:
Heat absorbed by water = heat lost by metal
a.
heat absorbed by water = Mw*Cp*dT
Mw = 100 gm
Cp = 4.184 J/g-C
dT = 28 - 24 = 4 C
heat absorbed by water = 100*4.184*4 = 1673.6 J
b.
heat lost by metal = 1673.6 J
c.
Heat lost by metal = m*Cm*dTm
m = 5 gm
Cm = ?
dTm = 100 - 28 = 72 C
Cm = Heat lost/(m*dTm) = 1673.6/(5*72) = 4.65 J/g-C
d.
Heat capacity = Specific heat of 1 kg metal = 4650 J/C
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