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Jobisdone [24]
1 year ago
9

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

Chemistry
1 answer:
AlekseyPX1 year ago
7 0

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).

To learn more about Homogenization from the given link:

brainly.com/question/18271118

#SPJ4

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3241004551 [841]
With your mind. Boom. Lol I'm sorry I don't know I just need points
4 0
3 years ago
Suppose you need to prepare 136.9 mL of a 0.315 M aqueous solution of NaCl.
Varvara68 [4.7K]

Answer:

2.52 g NaCl

Explanation:

(Step 1)

To find the mass, you first need to find the moles NaCl. This value can be found using the molarity ratio:

Molarity = moles / volume (L)

After you convert mL to L, you can plug the given values into the equation and simplify to find moles.

136.9 mL / 1,000 = 0.1369 L

Molarity = moles / volume

0.315 M = moles / 0.1369 L

0.0431 = moles

(Step 2)

Now, you can use the molar mass to convert moles to grams.

Molar Mass (NaCl): 22.990 g/mol + 35.453 g/mol

Molar Mass (NaCl): 58.443 g/mol

0.0431 moles NaCl           58.443 g
------------------------------  x  -------------------  =  2.52 g NaCl
                                            1 mole

4 0
2 years ago
The standard free energy of formation, ΔG∘f, of a substance is the free energy change for the formation of one mole of the subst
OLEGan [10]

Answer:

B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG∘f=−451.0 kJ/mol

D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol

Explanation:

The spontaneity of a reaction  is given by the value of the standard Gibbs free energy of the reaction (ΔG°rxn). The more negative is the ΔG°rxn, the more spontaneous is a reaction.

The ΔG°rxn can be calculated using the following expression:

ΔG°rxn = ∑np × ΔG°f(products) − ∑nr × ΔG°f(reactants)

By definition, the standard Gibbs free energy of formation of simple substances in their most stable state is zero. That is why, in the reaction of formation of a compound ΔG°rxn = ΔG°f(product).

<em>Based on the standard free energies of formation, which of the following reactions represent a feasible way to synthesize the product? </em>

<em>     A. N₂(g) + H₂(g) → N₂H₄(g); ΔG°f=159.3 kJ/mol. </em>

<em>     </em>Not feasible. ΔG°rxn = ΔG°f(product) > 0.

    <em>B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG°f=−451.0 kJ/mol</em>

    Feasible. ΔG°rxn = ΔG°f(product) < 0.

    <em>C. 2 C(s) + 2 H₂(g) → C₂H₄(g); ΔG°f=68.20 kJ/mol</em>

    Not feasible. ΔG°rxn = ΔG°f(product) > 0.

    <em>D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol</em>

    Feasible. ΔG°rxn = ΔG°f(product) < 0.

3 0
3 years ago
The element Oxygen, O, is an example of which of the following?
Anon25 [30]
It is an example of a molecule 
5 0
4 years ago
Read 2 more answers
Which two elements have similar physical and chemical properties?
guapka [62]

Explanation:

Elements in the same group have same number of valence electrons. And we know, the elements which have same number of valence electrons, have similar physical and chemical properties. Hence, the elements in the same group have similar physical and chemical properties.

5 0
3 years ago
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