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8 How many grams of nitrogen are in 250 g of (NH4)2SO4 (A) 530 g (B) 130 g (C) 0190 g (D) 265 g (E) 140 g 9 Nitrogen forms a number of oxides Which of the following oxides is 64% nitrogen (A) N2O5 (B) N2O4 (C) N2O3 (D) N2O2 (E) N2O 10 Sodium sulfate forms a number of hydrates A sample of a hydrate is heated until all the water is removed What is the formula of the original hydrate if it loses 43% of its mass when heated (A) Na2SO4 H2O (B) Na2SO4 2H2O (C) Na2SO4 6H2O (D) Na2SO4 8H2O (E) Na2SO4 10H2O 11 3Cu(s) + 8HNO3(aq) 3Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l) Copper metal reacts with nitric acid according to the above equation A 030-mol sample of copper metal and 100 mL of 12 M nitric acid are mixed in a flask How many moles of NO gas will form (A) 0060 mol (B) 0030 mol (C) 0010 mol (D) 020 mol (E) 010 mol 12 Gold(III) oxide, Au2O3, can be decomposed to gold metal, Au, plus oxygen gas, O2 How many moles of oxygen gas will form when 221 g of solid gold(III) oxide is decomposed The formula mass of gold(III) oxide is 442 (A) 0250 mol (B) 0500 mol (C) 150 mol (D) 100 mol (E) 0750 mol

i k RT For a binary system comprised of species 1 and 2, this becomes: Qi = exp Q1 = exp BII(P - P p t ) RT

(146)

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Activity coefficients [yi in Eq (141)] are evaluated from models for G~ as discussed in Sec 122 Thus for data at constant T we presume the availability of a correlation giving:

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e The one formula is double the formula of the other Thus, the smaller molecule dimerizes to produce the larger molecule You get 1 point if you combined two of the smaller molecules Total your points There are 9 points possible

- = g(xl, x 2 , , x ~ - ~ ) (const T)

The nature of dewpoint and bubblepoint calculations is evident from the examples of Secs 104 and 105 for Raoult's law and modified Raoult's law All such calculations made by the gammdphi formulation require iteration because of its complex functionality:

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At the moderate pressures where the gammdphi approach to VLE is appropriate, activity coefficients are assumed independent of pressure The need for iteration is evident, for example, in a BUBL P calculation of { y i ]and P, which requires values of Qi that are functions of P and {yi}Simple iterative procedures, described in the following paragraphs, apply to BUBL P, DEW P, BUBL T, and DEW T calculations Equation (141) may be solved for either yi or xi:

xiyi = 1 and xixi = 1, these equations may be summed to give:

The mole is the amount of substance that contains the same number of particles as exactly 12 g of carbon-12 Avogadro s number is the number of particles per mole, 6022 1023 particles A mole is also the formula (atomic, molecular) mass expressed in grams If you have any one of the three moles, grams, or particles you can calculate the others The empirical formula indicates which elements are present and the lowest whole-number ratio The molecular formula tells which elements are present and the actual number of each Be able to calculate the empirical formula from percent composition data or quantities from chemical analysis Stoichiometry is the calculation of the amount of one substance in a chemical equation by using another one Always use the balanced chemical equation in reaction stoichiometry problems Be able to convert from moles of one substance to moles of another, using the stoichiometric ratio derived from the balanced chemical equation In working problems that involve a quantity other than moles, sooner or later it will be necessary to convert to moles The limiting reactant is the reactant that is used up first Be able to calculate the limiting reactant by the use of the mol/coefficient ratio Percent yield is the actual yield (how much was actually formed in the reaction) divided by the theoretical yield (the maximum possible amount of product formed) times 100% A solution is a homogeneous mixture composed of a solute (species present in smaller amount) and a solvent (species present in larger amount) Molarity is the number of moles of solute per liter of solution Be able to work reaction stoichiometry problems using molarity Always use the balanced chemical equation in reaction stoichiometry problems

Solved for P , they become:

LOMBARDI RULE #7

(1411)

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