Well, clearly the calculated value for the number of hydrating water molecules would increase above its true level, because the total weight loss would be greater than expected. This is of course undesirable, but may usually be avoided by careful application of the experimental procedures. The signs to look for include
<span>(a) loss of water of hydration usually occurs at a considerably lower temperature than decomposition of the salt, because the water molecules are not strongly bonded in the hydrated complex. Dehydration typically occurs in a broad range of temperatures, typically from 50°C to around 200°C, whereas decomposition of the dehydrated salt generally takes place at temperatures over 200°C and in some case over 1000°C. So dehydration should be performed with care - avoid over-heating the sample in order to ensure that all the water has been driven off. </span>
<span>(b) dehydration often results in a change of appearance of the sample, particularly the colour and particle size of crystalline hydrates. However, decomposition may be accompanied by an additional change at higher temperatures, which gives a warning of its occurrence. </span>
<span>(c) if it is suspected that decomposition is occurring, or that dehydration is not complete, exploratory runs of varying duration at a given temperature may be carried out. There are two criteria to judge the effectiveness of the procedure </span>
<span>(i) the weight of the sample decreases to a constant stable value: this is a sign that dehydration is complete and decomposition - which is usually a much slower process - is not occurring. </span>
<span>(ii) the calculated number of molecules of water lost should take an integer value. If it differs by more than, say, 0.1 from an integer than it is probable that one of these two undesirable effects is present. Some hydrates lose water in steps through intermediate compounds with a lower level of hydration. These may provide plateaus where the weight loss is stable but dehydration is not complete. These will, in general, not provide an integer value for the number of water molecules present (because the calculation is based on the assumption that the residual sample is completely dehydrated salt).</span>
A density of a substance is constant. It is an extensive property, meaning it does not depend on the amount of substance because it is a ratio of mass to volume. No matter how much of each there is, they would always have a fixed ratio called density. For lead, the density is
Density = mass/volume
Density = 23.94 g/ 2.10 cm³
Density = 11.4 g/cm³
Answer:
The answer to your question is T2 = 610.9 °K
Explanation:
Data
Volume 1 = V1 = 30 cm³
Temperature 1 = T1 = 420 °K
Pressure 1 = P1 = 110 kPa
Volume 2 = V2 = 40 cm³
Temperature 2 = T2 = ?
Pressure 2 = P2 = 120 kPa
Process
To solve this problem use the combined gas law.
P1V1/T1 = P2V2/T2
-Solve for T2
T2 = P2V2T1 / P1V1
-Substitution
T2 = (120 x 40 x 420) / (110 x 30)
-Simplification
T2 = 2016000 / 3300
-Result
T2 = 610.9 °K
