Answer:Osmotic pressure is the minimum amount of pressure a solution must exert in order to prevent from crossing a barrier by osmosis. Solute molecules have difficulty crossing semipermeable membranes, so the more solutes that are in a solution, the higher the osmotic pressure will be. Between 30% sucrose and 60% sucrose, 60% sucrose will have a greater osmotic pressure than 30% because it has a higher percentage of solutes. However, since sucrose has a higher potential to cross semipermeable membranes and is more absorbable than magnesium sulfate, magnesium sulfate would have a higher osmotic pressure than 60% sucrose even though 60% sucrose has higher molecules.
Explanation:
Properties of a compound is completely different from their elements.
Water is composed by hydrogen and oxygen.
For example, the boiling point of oxygen is - 183 °C and hydrogen is - 253 °C, meanwhile, water has a boiling point of 100°C
Another example is when you put a burning wooden splint into oxygen, it burns more brightly. Put it in hydrogen, you may hear a "pop" sound, or even explode when large amount of hydrogen. But if u put a burning splint in water, it goes off.
This is the thing called the mass number of the thing
Answer:
They are most likely solid
Explanation:
solid is a physical property
Answer:
a) ![\Delta S](https://tex.z-dn.net/?f=%5CDelta%20S%3C0)
b) entropy of the sistem equal to a), entropy of the universe grater than a).
Explanation:
a) The change of entropy for a reversible process:
![\delta S=\frac{\delta Q}{T}](https://tex.z-dn.net/?f=%5Cdelta%20S%3D%5Cfrac%7B%5Cdelta%20Q%7D%7BT%7D)
![\Delta S=\frac{Q}{T}](https://tex.z-dn.net/?f=%5CDelta%20S%3D%5Cfrac%7BQ%7D%7BT%7D)
The energy balance:
![\delta U=[tex]\delta Q- \delta W](https://tex.z-dn.net/?f=%5Cdelta%20U%3D%5Btex%5D%5Cdelta%20Q-%20%5Cdelta%20W)
If the process is isothermical the U doesn't change:
![0=[tex]\delta Q- \delta W](https://tex.z-dn.net/?f=0%3D%5Btex%5D%5Cdelta%20Q-%20%5Cdelta%20W)
![\delta Q= \delta W](https://tex.z-dn.net/?f=%5Cdelta%20Q%3D%20%5Cdelta%20W)
![Q= W](https://tex.z-dn.net/?f=Q%3D%20W)
The work:
![W=\int_{V1}^{V2}P*dV](https://tex.z-dn.net/?f=W%3D%5Cint_%7BV1%7D%5E%7BV2%7DP%2AdV)
If it is an ideal gas:
![P=\frac{n*R*T}{V}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7Bn%2AR%2AT%7D%7BV%7D)
![W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV](https://tex.z-dn.net/?f=W%3D%5Cint_%7BV1%7D%5E%7BV2%7D%5Cfrac%7Bn%2AR%2AT%7D%7BV%7D%2AdV)
Solving:
![W=n*R*T*ln(V2/V1)](https://tex.z-dn.net/?f=W%3Dn%2AR%2AT%2Aln%28V2%2FV1%29)
Replacing:
![\Delta S=\frac{n*R*T*ln(V2/V1)}{T}](https://tex.z-dn.net/?f=%5CDelta%20S%3D%5Cfrac%7Bn%2AR%2AT%2Aln%28V2%2FV1%29%7D%7BT%7D)
![\Delta S=n*R*ln(V2/V1)}](https://tex.z-dn.net/?f=%5CDelta%20S%3Dn%2AR%2Aln%28V2%2FV1%29%7D)
Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:
![\Delta S](https://tex.z-dn.net/?f=%5CDelta%20S%3C0)
b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.