Effect of Two-Step Homogenization on the Evolution of Al3Zr Dispersoids in Al-0.3Mg-0.4Si-0.2Zr Alloy Al3Zr nano-particles can be introduced in Al-Mg-Si 6xxx alloys to improve their elevated temperature behavior and recrystallization resistance. The effect of two-step homogenization treatments on
the precipitation of Al3Zr dispersoids in Al-0.3Mg-0.4Si-0.2Zr alloy was investigated and compared to
<h3>What is
Homogenization?</h3>
Any of a number of methods, including homogenization and homogenisation, are used to uniformly combine two liquids that are insoluble in one another. To do this, one of the liquids is changed into a state in which very minute particles are evenly dispersed across the other liquid. The process of homogenizing milk, in which the milk fat globules are equally distributed throughout the remaining milk and reduced in size, is a classic example. In order to create an emulsion, two immiscible liquids (i.e., liquids that are not soluble in all amounts one in another) must be homogenized (from "homogeneous"; Greek, homos, same + genos, kind)[2] (Mixture of two or more liquids that are generally immiscible).
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I believe the correct answer from the choices listed above is option D. Catalysts lower the activation energy of a chemical reaction. It <span>is a substance which speeds up a reaction, but is chemically unchanged at the end of the reaction. It provides another pathway for the reaction to occur.</span>
Answer: The volume of gas is 3020 ml
Explanation:
According to ideal gas equation:
P = pressure of gas = 821.4 torr = 1.08 atm (760 torr = 1atm)
V = Volume of gas in L = ?
n = number of moles = 
R = gas constant =
T =temperature =
Thus volume of gas is 3020 ml
Answer:
0.576M and 0.655m
Explanation:
<em>...Dissolves 15.0g of styrene (C₈H₈) in 250.mL of a solvent with a density of 0.88g/mL...</em>
<em />
Molarity is defined as moles of solute (Styrene in this case) per liter of solution whereas molality is the moles of solute per kg of solvent. Thus, we need to find the moles of styrene, the volume in liters of the solution and the mass in kg of the solvent as follows:
<em>Moles styrene:</em>
Molar mass C₈H₈:
8C = 12.01g/mol*8 = 96.08g/mol
8H = 1.005g/mol* 8 = 8.04g/mol
96.08g/mol + 8.04g/mol = 104.12g/mol
Moles of 15.0g of styrene are:
15.0g * (1mol / 104.12g) = 0.144 moles of styrene
<em>Liters solution:</em>
250mL * (1L / 1000mL) = 0.250L
<em>kg solvent:</em>
250mL * (0.88g/mL) * (1kg / 1000g) = 0.220kg
Molarity is:
0.144 moles / 0.250L =
<h3>0.576M</h3>
Molality is:
0.144 moles / 0.220kg =
<h3>0.655m</h3>
