Answer:
The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)
Explanation:
Step 1: Data given
Initial temperature = 10.0 °C
Final temperature = 25.0 °C
Energy required = 30000 J
Mass of the object = 40.0 grams
Step 2: Calculate the specific heat capacity of the object
Q = m* c * ΔT
⇒With Q = the heat required = 30000 J
⇒with m = the mass of the object = 40.0 grams
⇒with c = the specific heat capacity of the object = TO BE DETERMINED
⇒with ΔT = The change in temperature = T2 - T2 = 25.0 °C - 10.0°C = 15.0 °C
30000 J = 40.0 g * c * 15.0 °C
c = 30000 J / (40.0 g * 15.0 °C)
c = 50 J/g°C
The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)
Answer:
At the top of Group 11 above silver and gold.
Period 4
Explanation:
When a water vapor condenses, heat is being released from the process. This heat is called latent heat of vaporization since the phase change happens without any change in the temperature. This value is constant per mole of a substance as a function of pressure and temperature. For this problem, we are given the heat of vaporization at a certain T and P. We use this value to calculate the total heat released from the process. We calculate as follows:
Total heat released: 32.4 g ( 1 mol / 18.02 g ) (40.67 kJ / mol) = 73.12 kJ
Therefore, 73.12 kJ of heat is released from the condensation of 32.4 g of water vapor.
Answer:
A simple example of decomposition reaction is hydrolysis of water where a water molecule is broken down into hydrogen and oxygen gas.
