Answer:
Explanation:
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In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:
Thus, we insert mass, specific heat and temperatures to obtain:
In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:
Now, we plug in to obtain:
NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.
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Answer :
- Boiling point of the sugar solution will be higher than that of water's boling point.
- Freezing point of the sugar solution will be lower than that of water's freezing point.
Explanation:
- Boiling point of a liquid is defined as temperature at which vapor pressure of liquid becomes equal to the atmospheric pressure.
Boiling point of solution is always higher than that of the pure solvent
Vapor pressure increases with increase in temperature which means sugar solution will be heated more to make vapor pressure equal to atmospheric pressure.
- Freezing point is defined as temperature at which solid and liquid phase are at equilibrium or temperature at which vapor pressure of liquid becomes equal to the vapor pressure in its solid phase.
Freezing point of solution is always lower than that of the pure solvent.
Lower the temperature, lower will be the vapor pressure which sugar solution solution will get freeze at lower temperature than that of the water.
The balanced equation for the decomposition of solid lead iv oxide is as follows: 2PbO2 = 2PbO + O2.
Lead IV oxide decompose to give lead ll oxide and oxygen. Lead iv oxide is thermally unstable and it usually decomposes into oxygen and lead ll oxide when heated. Lead ll oxide is more stable than lead lV oxide.
