Answer:
They affect the weathering of molecules and atoms.
Explanation:
Answer:
1. Ice at 0 degrees C.
2. N₂ at STP.
3. N₂ at STP.
4. Water vapor at 150 degrees C and 1 atm.
Explanation:
First, we need to remember that entropy (S) is a <em>measure of how spread out or dispersed the energy of a system is among the different possible ways that system can contain energy</em>. The greater the dispersal, the greater is the entropy.
When the temperature is increased, the energies associated with all types of molecular motion increase. Consequently, the entropy of a system always increases with increasing temperature.
With this in mind, we consider the pairs:
1. Since the ice at 0ºC has a greater temperature than the ice at -40 ºC, the first has the higher entropy.
2. The N₂ at STP (that is, 1 atm and 25 ºC) has higher entropy than N₂ at 0ºC and 10 atm because it has a higher temperature and less pressure, which allows a greater dispersal of energy by the molecules of the gas.
3. The N₂ at STP has a higher entropy since it has a higher temperature than N₂ at 0ºC, even though it the first has a lower volume (24,4 L vs. 50 L).
4. The water vapor at 150 ºC and 1 atm have a higher temperature and a lower pressure. This means that its molecules will have an increased molecular motion than the molecules of water vapor at a lower temperature and higher pressure. Therefore, the first has the highest entropy.
Weathering would be chemical because it's more natural than physical
Answer:
The answer is 0.01 M
Explanation:
The problem is solved by applying the expression for ionic product of water as follows:
Kw = [H₃O⁺] [OH⁻]
Where Kw is the ionic product of water and it is 1.10⁻¹⁴ M at⁴ 25ºC. As [OH⁻]= 1.10⁻¹² M, [H₃O⁺] will be:
[H₃O⁺]= Kw/ [OH⁻]= 1.10⁻¹⁴M/1.10⁻¹²M= 0.01 M
