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guajiro [1.7K]
2 years ago
15

Discuss the microstructural transformation of eutectic reaction , hypo eutectic and hyper eutectic. Include temperature, reactio

ns, phase development and phase diagrams
Chemistry
1 answer:
Ket [755]2 years ago
7 0

The phase development of the eutectic reaction includes the equilibrium point of Liquid, crystals of A, and crystals of B

<h3>What is the Eutectic Reaction?</h3>

This refers to the three-phase reaction that occurs when a liquid forms after cooling.

Hence, we can see that the hypo eutectic reaction occurs when there is a homogenous mixture of substances that undergoes a reaction where it melts and solidifies.

The hyper eutectic reaction has to do with the metallic alloy that contains a composition that has exceeded its eutectic point.

Read more about eutectic reactions here:

brainly.com/question/23160517

#SPJ1

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The answer is A, between 0 and 7.

In a pH scale from 0 to 14, we can groups these numbers into acidic, neutral, and alkaline. 7 is the neutral pH value, therefore, 0-7 is always acidic, and 7-14 is alkaline.
The smaller the number is, the more acidic the solution will be. This applies same in alkalis, the larger the pH value is, the more alkaline the solution is.

We can measure the pH of solution with many methods, the easiest way include using a pH paper, more advanced and accurate methods includes using a pH meter.
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Please help, need answer
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Dylan changed a formless pile of sand into a sand castle. What other natural materials can be changed into art
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3 years ago
Aluminum and oxygen react according to the following equation: 4Al + 3O2 -&gt; 2Al2O3 In a certain experiment, 4.6g Al was react
stiv31 [10]

Answer:

Percent yield: 78.2%

Explanation:

Based on the reaction:

4Al + 3O₂ → 2Al₂O₃

<em>4 moles of Al produce 2 moles of Al₂O₃</em>

<em />

To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:

(Actual yield (6.8g) / Theoretical yield) × 100

Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:

4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.

As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:

0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = <em>0.0852 moles of Al₂O₃</em>,

In grams (Molar mass Al₂O₃ = 101.96g/mol):

0.0852 moles of Al₂O₃ × (101.96g / mol) =

<h3>8.7g of Al₂O₃ can be produced (Theoretical yield)</h3>

Thus, Percent yield is:

(6.8g / 8.7g) × 100 =

<h3>78.2% </h3>
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