It could be either sublimation, freezing or condensation. If i had more information i could tell you which
Hey there!:
Given the mass of PbCl(OH) :
0.135 Kg = 0.135 Kg*(1000g / 1Kg) = 135 g
Molecular mass of PbCl(OH) = 207+35.5+16+1 = 259.5 g / mol
Atomic mass of Pb = 207 g/mol
Hence mass of Pb in 135 g PbCl(OH) :
(207 g Pb / 259.5 g PbClOH) * 135g PbClOH =
0.79768 * 135 => 107.68 g of Pb
For Pb2Cl2CO3 :
Given the mass of Pb2Cl2CO3 :
0.135 Kg = 0.135 Kgx(1000g / 1Kg) = 135 g
Molecular mass of Pb2Cl2CO3 = 2*207+2*35.5+12+3*16 = 545 g / mol
Mass of Pb present in 1 mol (=545 g / mol) of Pb2Cl2CO3 = 2*207 = 414 g
Hence mass of Pb in 135 g Pb2Cl2CO3:
(414 g Pb / 545 g PbClOH) * 135g PbClOH =
0.75963 * 135 => 102.55 g of Pb2Cl2CO3
Hope that helps!
Compare the density of the object in question to the density of water. If its density is less than water, it will float. For example, oak floats because its density is 0.7 g/cm³. If the density of an object is greater than water, it will sink.
The heat that creates this temperature change coming from change in the internal energy of the system as per as first law of thermodynamics.
<h3>What is Boyle's law ?</h3>
A law stating that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.
As we know, Boyle's law only works when the gas is kept at a constant temperature
Here,
When volume of gases decreased, it means work done has occurred on the system, so the work done is used for raising internal energy of the gas and the other is released as the thermal energy.
So,
According to 1st law of thermodynamics,
we know Q = ΔU + W i.e, change in internal energy and work done. So this is a reason. Changing temperature occurs.
Learn more about Internal enrgy here ;
brainly.com/question/11278589
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Saturated solution is a solution in which no more solute can be dissolved in the solvent. When saturated solution cools, the solution began precipitate from the solution, because under lower temperature, usually, less amount solute can be dissolved in the solvent.
