Conversion factors water volume

You can use density as a conversion factor between volume and mass. For example, given that water has a density of 1.00 g/cm, what is the mass of 350 cm of water ... 

U.S. regulations define this standard as foUows proof spirit shaU be held to be that alcohoHc Hquor which contains one-half its volume of alcohol of a specific gravity of 0.7939 at 15.6°C ie, the figure for proof is always twice the percent alcohol content by volume. For example, 100° proof means 50% alcohol by volume. In the United Kingdom as weU as Canada, proof spirit is such that at 10.6°C alcohol weighs exactiy twelve-thirteenths of the weight of an equal bulk of distiUed water. A proof of 87.7° indicates an alcohol concentration of 50%. A conversion factor of 1.142 can be used to change British proof to U.S. proof. 

Stability. Lower explosive limit in air, 7.3% by volume sensitive to adiabatic compression (Angus Chemical Co., 1998) forms an explosive sodium salt which bursts into flame on contact with water (Budavari, 1998) Octanol/waterpartition coefficient (P) log P, -0.35 (Hansch et al., 1995) Conversion factor -, mg/m = 2.50 x ppm... 

The conversion factors needed to express the concentration in ppm hemin (1 ppm - 1 pg/g) and the dilution factor must also be taken into consideration. The dilution factor is the total extraction fluid volume (ml) divided by the sample weight (g). The total extraction fluid volume includes the water content of the sample plus the amount of aqueous acetone solution. Most cooked meats have -70% water. Thus, for a 2 g sample containing 1.4 ml water, the total extraction volume - 1.4 ml + 9 ml acetone solution, and the dilution factor = 10.4 ml/2 g sample = 5.2. Pearson and Tauber (1984) rounded the dilution factor to 5, but for best results the decimal should be retained, Also, if the sample water content is significantly different from 70%, the dilution factor should be recalculated as appropriate. 

A 4.028 m solution of ethylene glycol in water contains 4.028 mol of ethylene glycol per kilogram of water. To find the solution s molarity, we need to find the number of moles of solute per volume (liter) of solution. The volume, in turn, can be found from the mass of the solution by using density as a conversion factor. 

You are probably well aware that the pressures at great depths under the ocean are extremely large, and the deeper you go, the greater the pressure that is exerted. Let s start small and look at a beaker filled with water. The beaker has a diameter of 10 cm, and the height of the water is 20 cm. How much pressure does the water exert on the bottom of the beaker To answer this question we simply need to recall that P = FI A. The force involved is the weight of the water contained in the beaker. The weight of the water can be calculated from the mass of the water, and the mass of the water can be calculated from the volume, using density as a conversion factor. 

Because virtually all stoichiometric calculations involve moles (abbreviated mol) of material, molarity is probably the most common concentration unit in chemistry. If we dissolved 1.0 mol of glucose in enough water to give a total volume of 1.0 L, we would obtain a 1.0 molar solution of glucose. Molarity is abbreviated with a capital M. Notice that, because molarity has units of moles per liter, molar concentrations are conversion factors between moles of material and liters of solution. 

The first formulation is useful as a conversion factor between grams of solute and milliters of solution. The second is more useful to calculate the concentration of a solution. For example what is the percent (w/v) of a solution prepared by dissolving 25 g of glucose in enough water to give a total volume of 500 mL ... 

Various units are used for expressing pressures (see Chapter 1, Footnote 8). A pressure of one standard atmosphere, or 0.1013 MPa, can support a column of mercury 760 mm high or a column of water 10.35 m high. As indicated in Chapter 1, the SI unit for pressure is the pascal (Pa), which is 1 N m-2 an SI quantity of convenient size for hydrostatic pressures in plants is often the MPa (1 MPa = 10 bar = 9.87 atm). (An extensive list of conversion factors for pressure units is given in Appendix II, which also includes values for related quantities such as RT.) Pressure is force per unit area and so is dimensionally the same as energy per unit volume (e.g., 1 Pa = 1 N m-2 = 1J m-3). Vw has the units of m3 mol-1, so VWP and hence /aw can be expressed in J mol-1. 

Being asked to convert from volume into mass is the tip-off that we can use the density of water as a conversion factor in solving this problem. We can find water s density on a table of densities, such as Table 8.2. 

The conversion factor on the right allows us to convert from mass in grams to volume in milliliters. First, however, we need to convert the given unit, kilograms, to grams. Then, after using the density of water to convert grams to milliliters, we need to convert milliliters to liters. 

We assume that the volume of liquid water is zero compared to that of steam. How do we calculate the volume of the steam What is the conversion factor between L-atm and J ... 

One of the most fundamental measurements that one has to make is the concentration of solid material (whether clay, flint, stone, or feldspar) in a suspension or slip. There are many ways in which this concentration can be expressed, but the method universally adopted in industry is that of slip density, expressed as the weight of slip per unit volume. In metric units this quantity will be in grams per ml., or in British units ounces per pint. The conversion factor of 20 stems from the fact that 1 pint of water x g per ml. = 20jc oz per pint. 

The basic units are those into which all values are converted when a calculation is conducted (kg for mass-bound substances and kJ for forms of energy). To facilitate the comparability with the inventory data, entry units were defined using a conversion formula and established as display units. In the inventory (Table 3.2) most of the substances were expressed in kg except for water and gas-oil, for which volume display units (m water and m gasoil, respectively) were defined using the density as conversion factor. Similarly, kWh was selected as display unit for electricity, wind power, and natural gas. 

To find the mass percent, substitute into the equation for mass percent. You need the mass of ethanol and the mass of water. Obtain the mass of water from the volume of water by using the density as a conversion factor. 

P-11 Metric Practice Guide for the Compressed Gas Industry. Guide sets forth guidelines concerning units that are to be used to express (1) the volume and/or mass of gas in compressed gas containers, (2) pressure, and (3) container water capacities. It includes volume, pressure, mass, and other conversion factors pertinent to the compressed gas industry (6 pages). 

The concentration of CaCOj (55 ppm) means 55.0 g CaCOj/IO g water it is a conversion factor between g water and g CaCOj. We first convert the volume of water (1.50 L) into mass of water in grams ... 

A water tank leaks water at the rate of 3.9 mL/h. If the tank is not repaired, what volume of water in liters will it leakin ayear Showyour setup for solving this. Hint Use one conversion factor to convert hours to days and another to convert days to years, and assume that one year has 365 days. 

When 2 moles of Na react with water at 25°C and 1 atm, the volume of formed is 24.5 L. Calculate the work done in joules when 0.34 g of Na reacts with water under the same conditions. (The conversion factor is 1 L atm = 101.3 J.)... 

Body temperamre, ambient pressure,. saturated with water vapor (BTPS) occurs when the person tested exhales a volume of gas at body temperature (37 °C). When collected in the spirometer, this volume rapidly cools to approach the lower ambient temperature and contracts. TTiis reduced volume must be multiplied by the appropriate BTPS conversion factor to correct it to what it should be at normal body temperature (Table 11). 

Estimate the pressure necessary to melt water at — 10.0 C if the molar volume of liquid water is 18.01 mL and the molar volume of ice is 19.64 mL. AS for the process is -1-22.04 J/K and you can assume that these values remain relatively constant with temperature. You will need the conversion factor 1 L-bar = 100 J. 

To convert volume fractions into molar concentrations, a choice is needed for the size of a lattice site h. Here we have used b = 0.5 nm, which is a compromise between the size of water molecules and surfactant segments. For a given molecule this conversion depends on the molar volume (v), which we will express in units b, and using the quantity N as the number of segments in a molecule (Table 5.1). For example, to convert the volume fraction for the ions one needs to multiply by roughly 10. For the surfactant the conversion factor is dose to unity. 

Use the density of water in a conversion factor to obtain the volume of water (and hence, the volume of the glass vessel). 

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