Wanted quantity

In a problem, identify given and wanted quantities that are related by a Per expression. Set up and solve the problem by dimensional analysis. 

A conversion factor is used to change a quantity of either unit to an equivalent amount of the other unit. The conversion follows a unit path (Path) from the given quantity (Given) to the wanted quantity (Wanted). A unit path may have any number of steps, but you must know the conversion factor for each step in the path. 

We set up this intermediate answer —intermediate because it is not the wanted quantity—to show that it does have meaning. If you had calculated it, you would have learned that 672 hours is the same as 28 days. Intermediate quantities are always real quantities with real units if the calculation setup follows the unit path. 

Step 2 Identify and write down the units of the wanted quantity. 

Begin by identifying the Given and Wanted quantities for the °F-to-°C conversion. Then write the equation (Equation 3.7), solved for the wanted quantity. 

If the given and wanted quantities are related by one or more Pea expressions, solve the problem by If the given and wanted quantities are related by an algebraic equation, solve the problem by... 

Write the Per/Path. Solve the equation for the wanted quantity. 

Write the calculation setup, including units. Start with the given quantity. Multiply by one or more conversion factors until the wanted quantity is reached. Substitute all given values. Include units. 

Solve by algebra because the Given and Wanted quantities are related by an equation in which the Wanted quantity is the only unknown. 

Wanted quantity (Wanted) Section 3.4 Cubic centimeter (cm ) Kilogram (kg), gram (g) Liter (L)... 

Plan the solution. Recall that Plan, printed as you see it here, means to complete the first three steps in the problem-solving procedure (1) Write down what is Given. (2) Write down what is Wanted. (3) Decide how to solve the problem. If the given and wanted quantities are related by a Per expression, use dimensional analysis write the Per/Path. If the given and wanted quantities are related by an algebraic equation, use algebra by solving the equation for the Wanted quantity. Plan your solution now. 

In this case, Equation 4.4 gives you a Per relationship between the Given and Wanted quantities, 1 atm = 14.69 psi. Therefore, the problem is solved by dimensional analysis. 

Changing from units to mass or vice versa is a two-step dimensional analysis conversion Change the given quantity to moles, then change moles to the wanted quantity. 

Several introductory comments may help you learn how to solve stoichiometry problems. The problem-solving strategy from Section 3.9 is used repeatedly. You will soon see that a series of Per relationships link the Given and Wanted quantities in all problems in this chapter. Therefore, the problems are solved by dimensional analysis. 

Now Plan the problem. Identify the Given and Wanted quantities and the Per/Path. 

The conversion factor comes from the Per relationship between the given and wanted quantities, using their coefficients in the equation. 

The given quantity is moles of heptane. In other words, Step 1 of the stoichiometry path is completed. The wanted quantity of Oj is to be expressed in grams. Therefore, you need its molar mass for the mol g conversion. 

The starting steps will help you Plan your strategy for solving this problem. Our Plan will be for a single dimensional-analysis setup from the given quantity to the wanted quantity. 

Equation 14.3 is the key to this problem. Plan the problem. Solve the equation for the wanted quantity, substitute the given information, and find the answer. Watch your units. 

R. As with all problems to be solved algebraically, first solve the equation for the wanted quantity. Then substitute the quantities that are given, including units, and calculate the answer. 

First notice that you are given amount (n), pressure (P), and temperature (T). You are asked to find volume (V). Thus the mathematical connection between the Given and Wanted is the ideal gas equation, solved for the wanted quantity. The given measurements are clearly identified, but another given, R, calls for a decision. Which value of R should you use The choice is based on the units of pressure. 

This time pressure is given in torr, so you use 62.4 L torr/mol K for R. Molar mass is the Wanted quantity, and it is the only unknown in the PV = mRT/MM form of the ideal gas equation (Equation 14.8). Complete the example. 

In the ideal gas equation method, there are two procedures (1) If the given quantity is a gas, the ideal gas equation is solved for n to change the given volume to moles. The problem is completed by the second and third steps in the stoichiometry path. (2) If the wanted quantity is a gas, the moles of wanted quantity are calculated by the first and second steps in the stoichiometry path. The ideal gas equation is then solved for V to convert the moles of gas to liters. The section describing the ideal gas equation method is identified by a green bar in the inside margin, as next to this paragraph. 

Step 2 Use the molar volume to calculate the wanted quantity by all three steps of the stoichiometry path. 

The stoichiometry path may be summarized as given quantity mol given mol wanted wanted quantity. In a gas stoichiometry problem, the first or third step in the path is a conversion between moles and liters of gas at a given temperature and pressure. If you are given volume, you must convert to moles if you find moles of wanted substance, you must convert to volume. These conversions are made with the ideal gas equation, PV = nRT. You have already made conversions like these. For example, in Example 14.3, you calculated the volume occupied by 0.393 mol N2 at 24°C and 0.971 atm. You used the ideal gas equation solved for V. 

In a gas stoichiometry problem, either the given quantity or the wanted quantity is a gas at specified temperature and pressure. The problem is usually solved in two steps, the order of which depends on whether the gas volume is the wanted quantity or the given quantity. 

This volume ratio is useful for stoichiometry problems when both the given and wanted quantities are gases measured at the same temperature and pressure. For example, consider the reaction 3 H2(g) + N2(g) 2 NH3(g). Let s calculate the volume of ammonia that will be produced by the reaction of 5 liters of N2, with both gases measured at STP. The equation coefficients, interpreted for gas volumes, tell us... 

Your ability to solve the gas problems in this chapter depends largely on your algebra skills. Most students find it easiest to determine what is wanted and then solve the ideal gas equation for that variable. If the wanted quantity is density or molar volume, solve the equation for the combination of variables that represents the desired property. Then substitute the known variables, including units, and calculate the answer. Units are important If they don t come out right, you know there is an error in the algebra. 

This is a straightforward equation-type problem in which the equation is solved for the wanted quantity, known values are substituted, and the answer is calculated. In... 

Equation 15.7 is already solved for the wanted quantity, so the problem may be solved by direct substitution. For AT, substitute Tf - Tj. As always, include units in your calculation setup. 

Solve Equation 16.11 for the wanted quantity, substitute, and calculate the... 

This time the first stoichiometry step begins with a volume of solution of known molarity rather than the mass of a solid. Notice also that both Given and Wanted quantities are solution volumes in milliliters. Does that suggest anything to you Remember Example 16.18, which we solved using volume in mL and molarity as mmol/mL. It s easier that way. Complete the example. 

The second group includes reactions (1) and (2) (Table 65), involving lanthanum fluorides (Hildenbrand and Lau, 1995). Since there are two such reactions and, correspondingly, two sought enthalpies of atomization (for LaF and Lap2) and since the AatH°(BaF, 0) value is known, the wanted quantities can be determined from the enthalpies of reactions (1) and (2), ArH°(l, 0) and ArH (2, 0), respectively. We can then calculate the enthalpy of atomization of Lap2 by the equation... 

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