Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water
Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.
From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water
At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
= (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)-
= 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
= (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
= 268.356(T - 27.8)
Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C
Answer: 28.4 °C
Answegood
Explanation:this answer is two it has letters
Answer:
Zymogen ActivationZymogens are activated by snipping the bonds between two or more amino acids, rather like cutting a balloon string so that it floats away. When the bonds are cut, the enzyme changes its conformation, its 3-D structure, so that the active site is free and able to become active.
Explanation:
Answer:
Sp3 hybridization
Explanation:
The NH3 molecule, which consists of one lone pairs and three bond pair of electron on its valance shell due to lone pair bond pair repulsion makes bond angle of 107.5°resulting distorted tetrahedral geometry.
Hybridization =no. of bond pair +lone pair=3+1=4=sp3 hybridization
