Answer:
The temperature should be higher than 437.9 Kelvin (or 164.75 °C) to be spontaneous
Explanation:
<u>Step 1:</u> Data given
ΔH∘=20.1 kJ/mol
ΔS is 45.9 J/K
<u>Step 2:</u> When is the reaction spontaneous
Consider temperature and pressure = constant.
The conditions for spontaneous reactions are:
ΔH <0
ΔS > 0
ΔG <0 The reaction is spontaneous at all temperatures
ΔH <0
ΔS <0
ΔG <0 The reaction is spontaneous at low temperatures ( ΔH - T*ΔS <0)
ΔH >0
ΔS >0
ΔG <0 The reaction is spontaneous at high temperatures ( ΔH - T*ΔS <0)
<u>Step 3:</u> Calculate the temperature
ΔG <0 = ΔH - T*ΔS
T*ΔS > ΔH
T > ΔH/ΔS
In this situation:
T > (20100 J)/(45.9 J/K)
T > 437.9 K
T > 164.75 °C
The temperature should be higher than 437.9 Kelvin (or 164.75 °C) to be spontaneous
Answer : The final temperature of the metal block is, 
Explanation :
As we know that,
![m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
.................(1)
where,
q = heat absorbed or released
= mass of aluminum = 55 g
= mass of water = 0.48 g
= final temperature = ?
= temperature of aluminum =
= temperature of water =
= specific heat of aluminum = 
= specific heat of water= 
Now put all the given values in equation (1), we get
![55g\times 0.900J/g^oC\times (T_{final}-25)^oC=-[0.48g\times 4.184J/g^oC\times (T_{final}-25)^oC]](https://tex.z-dn.net/?f=55g%5Ctimes%200.900J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-25%29%5EoC%3D-%5B0.48g%5Ctimes%204.184J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-25%29%5EoC%5D)
Thus, the final temperature of the metal block is,
Answer:
The monosaccharides, amino acids, bile salts, vitamins, and other nutrients are absorbed by the cells of the intestinal lining
Explanation:
Explanation:
The correct answer is C.
Through runoff on Earth's surface. Hope this helped!
Answer:
Explanation:
A 12.48 g sample of an unknown metal, heated to 99.0 °C was then plunged into 50.0 mL of 25.0 °C water. The temperature of the water rose to 28.1 Go to calculating final temperature when mixing two samples of water ... Problem #1: A 610. g piece of copper tubing is heated to 95.3 °C and placed in an ... The two rings are heated to 65.4 °C and dropped into 12.4 mL of water at 22.3 °C. ... Problem #4: A 5.00 g sample of aluminum (specific heat capacity = 0.89 J g¯1
