the liquid state‚ they can occur in all three main phases‚ gas‚ liquid‚ and solid. For example‚ the water in the air is a liquid solute dissolved in a gas solvent; solid solutions include brass‚ which is made from combining zinc with copper 3. Describe two examples of colloids (see Table 132). A colloid consists of small particles 1nm to 1000nm in size that are suspended in a solid‚ liquid or gas. Examples include common products such as gel‚ a solid network extending throughout a liquid‚ or
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[Problems 1‚ 3‚ 4‚ and 5 due: Wednesday‚ September 19th‚ 2012] Problem # 1: Solubility of formaldehyde in a solvent is measured by a spectrophotometer operating at 570 nm. The data collected are optical density versus concentration expressed in grams per liter. Concentration‚ c 100 300 500 600 700 Optical density‚ d 0.086 0.269 0.445 0.538 0.626 Based on Beer’s law‚ obtain the best estimates of the coefficients for a linear relation c = bo + b1 d and plot the equation obtained together with the data. NOTE:
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Solvent: A liquid substance capable of dissolving other substances c. Mixture: a substance consisting of two or more substance mixed together d. Solution: A homogeneous mixture of two or more substances e. Emulsion: a colloid in which both phases are liquids f. Suspension: a mixture in which fine particles are suspended in a fluid where they are supported g. Colloidal dispersion: a colloid that has continuous liquid phase in which a solid is suspended in a liquid. 5. How
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Evaporation Several factors determine how fast a sample of liquid will evaporate. The volume of the sample is a key factor. A drop of water takes less time to evaporate than a liter of water. The amount of energy supplied to the sample is another factor. In this lab‚ you will investigate how the type of liquid and temperature affect the rate of evaporation. Problem: How do intermolecular forces affect the evaporation rates of liquids? Pre-Lab: Read the ENTIRE LAB first then answer the following
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also two different containers to place the creatures in. Lastly‚ the person performing the experiment should choose two different liquids to place the creatures in. The first step in the actual experiment is to open the two packages containing the polymer creatures and then placing each of them aside for later. Next‚ fill the the two containers with the two different liquids of choice‚ which in this case is water and vinegar. Finally‚ after filling the containers‚ place one of the two polymer creatures
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generates through a novel microfluidic phenomenon. Reverse electrowetting is a process by which liquid motion is converted into electrical energy. It can generate a power up to 103Wm−2‚ by this technology. Reverse electrowetting (REWOD) concept At first‚ The principle of electrowetting is defined in here. The shape of a liquid droplet on a solid layer is characterized by the contact angle θ between the liquid and the solid layer (Figure 1). Electrowetting is the phenomenon whereby an electric field
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Longitudinal and transverse waves Sound is transmitted through gases‚ plasma‚ and liquids as longitudinal waves‚ also called compression waves. Through solids‚ however‚ it can be transmitted as both longitudinal waves and transverse waves. Longitudinal sound waves are waves of alternating pressure deviations from the equilibrium pressure‚ causing local regions of compression and rarefaction‚ while transverse waves are waves of alternating shear stress at right angle to the direction of propagation
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The simulation and analysis of liquid droplets with respect to heat capacity and droplet size Abstract Define Using Molecular dynamic simulation a cluster of particles was modelled and their behaviour analysed with respect to temperature change. As proof of a specific state i.e. solid‚ the Mean squared displacement was calculated and illustrated with respect to time. This occurred for varying temperatures. From the internal energy of the system per time step‚ the heat capacity was determined
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Part I: 1. Calculate the energy change (q) of the surroundings (water) using the enthalpy equation qwater = m × c × ΔT. We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL. qwater = m × c × ΔT m = mass of water = density x volume = 1 x 26 = 26 grams ΔT = T(mix) - T(water) = 38.9 - 25.3 = 13.6 q(water) = 26 x 13.6 x 4.18 q(water) = 1478 Joules SPECIFIC HEAT: qmetal = -205 J = 15.363 g X c X (27.2 - 100.3 C) c
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Well Engineering & Construction 24 Kilometers Hussain Rabia Index Well Engineering & TOC Previous Next Table of Contents Chapter 1 : Pore Pressure Chapter 2 : Formation Integrity Tests Chapter 3 : Kick Tolerance Chapter 4 : Casing Functions & Types Chapter 5 : Casing Design Principles Chapter 6 : Cementing Chapter 7 : Drilling Fluids Chapter 8 : Practical Rig Hydraulics Chapter 9 : Drill Bits Chapter 10 : Drillstring Design Chapter 11 : Directional Drilling Chapter 12 : Hole Problems
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