Effect of Two-Step Homogenization on the Evolution of Al3Zr Dispersoids in Al-0.3Mg-0.4Si-0.2Zr Alloy

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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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A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

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Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$