Q27:
Heat lost by metal is gained by the water.
m1s1(∆T)1 = m2s2(∆T)2
150.0g×4.18J/g.℃ × (23.3- 20.0)℃ = 150.0g×s2×(80.0-23.3)℃
s2 = 4.18×3.3/56.7 = 0.2433J/g.℃
Answer: 0.24J/(g.℃)
Q28:
Sn + SnO2 + 92kJ 2SnO
That means +92kJ is the heat required to produce 2 mole of SnO. The reaction is endothermic in nature i.e. ∆H > 0.
(1/2)Sn + (1/2)SnO2 SnO. ∆H = 92kJ/2 = + 46.kJ
The enthalpy change for production of 1.00mol of SnO = 46kJ
Answer: 46kJ
solve both pls -> XCIO DE D Question 27 A 150.0-g sample of metal at 80.0°C...
A 150.0-g sample of metal at 80.0°C is added to 150.0 g of H20 at 20.0°C. The temperature rises to 23.3°C. Assuming that the calorimeter is a perfect insulator, what is the specific heat of the metal? (Specific heat of H2O is 4.18 J/g. °C.) A) -0.48 J/g. °C B) 0.24 J/g °C C) 0.48 J/g °C D) 0.72 J/g °C E) 0.96 J/g. "C
1. (10 pts) A 150.0 g sample of a metal at 75.0 °C is added to 150.0 g H20 at 15.0 °C (s.h. = 4.184 J/g °C). The temperature of the water rises to 18.3 °C. Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water.
. A 150.0 g sample of a Metal was heated to 95.0°C. When the hot metal was placed into 100.0 g of water in a calorimeter, the temperature of the water increased from 20.0°C to 35.0°C. The specific heat of water is 4.184 J/g °C. a) What is the specific heat of the metal? Kb) What would the final temperature be if the mass of water was 150.0 q?
A piece of metal of mass 35.0 g at 100.0°C was placed in 150.0 g of water at 20.0 °C. After stirring, the final temperature of the water and the metal is 23.8°C. What is the specific heat capacity of the metal? (specific heat capacity for H2O = 4.184 J/g °C) O-0.89 J 8°C 19.6 J/g °C 1.96J/g °C O 0.89 J/g °C
1. Prepare for It! Prelab Question A A 235.0 g sample of metal is heated to 100.0°C and poured into a calorimeter containing 50.0 g of water at 20.5°C. The equilibrium temperature of the water and metal is 30.5°C. Using the specific heat of water, 4.18 J/g C, determine the specific heat of the metal from equation 3. Use equation 5 to find the approximate molar mass of the metal. Show your work! Use back if needed. 2. Prepare for...
QuesLIUI A 170.0-g sample of metal at 83.00'C is added to 170.0 g of H 200) at 15.00°C in an insulated container, The temperature rises to 18.16°C. Neglecting the heat capacity of the container, what is the specific heat capacity of the metal? The specific heat capacity of H 2011) is 4.18 J/("C). a 20.5 JgC) 5.4.18 °C) C 85.6J/(g°C) 0.0.204 J/g °C) -0.204 J/(g°C) Question 2 of 21
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