The smallest part of a compound that still acts like that compound is

How many moles are equivalent to 55.5g of nano3

Construct the hypothetical phase diagram for metals a and b between room temperature (20c) and 700c, given the following information: * the melting temperature of metal a is 480c. • the maximum solubility of b in a is 4 wt% b, which occurs at 420c. • the solubility of b in a at room temperature is 0 wt% b. • one eutectic occurs at 420c and 18 wt% b–82 wt% a. • a second eutectic occurs at 475c and 42 wt% b–58 wt% a. • the intermetallic compound ab exists at a composition of 30 wt% b–70 wt% a, and melts congruently at 525c.• the melting temperature of metal b is 600c. • the maximum solubility of a in b is 13 wt% a, which occurs at 475c. • the solubility of a in b at room temperature is 3 wt% a.

Problem #3 (ch. 1, problem 15)the ideal gas law provides one way to estimate the pressure exerted by a gas on a container. the law isí‘ťí‘ť=푛푛푛푛푛푛푉푉mo re accurate estimates can be made with the van der waals equationí‘ťí‘ť=í‘›í‘›í‘›í‘›í‘›í‘›í‘ ‰í‘‰â’푛푛푟푟â’푞푞푛푛2í‘ ‰í‘‰2where the term nb is a correction for the volume of the molecules and the term an2/v2is a correction for molecular attractions. the values of a and b depend on the type of gas. the gas constant is r, the absolutetemperature is t, the gas volume is v, and the number of moles of gas molecules is indicated by n. if n = 1 mol of an ideal gas were confined to a volume of v = 22.41 l at a temperature of 0â°c (273.2k), it would exert a pressure of 1 atm. in these units, r = 0.0826.for chlorine gas (cl2), a = 6.49 and b = 0.0562. compare the pressure estimates given by the ideal gas law and the van der waals equation for 1 mol of cl2 in 22.41 l at 273.2 k. what is the main cause of the difference in the two pressure estimates, the molecular volume or the molecular attractions?