Answer: <u><em>True</em></u>
Explanation:
<u><em>Q. 10g of white powder reacts with 10g of clear liquid. The reaction bubbles and changes color producing a black liquid that has a mass of 13g. What can be ...</em></u>
Physical Properties<span>: </span>Physical properties<span> can be observed or measured without changing the composition of matter. </span>Physical properties<span> are used to observe and describe matter. so physical changes are the change in temperature of the land and the evaporation of water and change humidity of the air. chemical change is the ripening of the orange</span>
Is true. Nitrogen gas behaves more like an ideal gas as the
temperature increases. Under normal conditions such as normal pressure and temperature
conditions , most real gases behave qualitatively as an ideal gas. Many
gases such as air , nitrogen , oxygen ,hydrogen , noble gases , and some heavy
gases such as carbon dioxide can be treated as ideal gases within a reasonable tolerance. Generally,
the removal of ideal gas conditions tends to be lower at higher temperatures and lower density (that is at lower pressure ), since the work made by the intermolecular
forces is less important compared to the kinetic energy<span> of the particles, and the size of the molecules is less important
compared to the empty space between them. </span><span>The ideal gas model
tends to fail at lower temperatures or at high pressures, when intermolecular
forces and intermolecular size are important.</span>
Explanation:
P1V1 = nRT1
P2V2 = nRT2
Divide one by the other:
P1V1/P2V2 = nRT1/nRT2
From which:
P1V1/P2V2 = T1/T2
(Or P1V1 = P2V2 under isothermal conditions)
Inverting and isolating T2 (final temp)
(P2V2/P1V1)T1 = T2 (Temp in K).
Now P1/P2 = 1
V1/V2 = 1/2
T1 = 273 K, the initial temp.
Therefore, inserting these values into above:
2 x 273 K = T2 = 546 K, or 273 C.
Thus, increasing the temperature to 273 C from 0C doubles its volume, assuming ideal gas behaviour. This result could have been inferred from the fact that the the volume vs temperature line above the boiling temperature of the gas would theoretically have passed through the origin (0 K) which means that a doubling of temperature at any temperature above the bp of the gas, doubles the volume.
From the ideal gas equation:
V = nRT/P or at constant pressure:
V = kT where the constant k = nR/P. Therefore, theoretically, at 0 K the volume is zero. Of course, in practice that would not happen since a very small percentage of the volume would be taken up by the solidified gas.
