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hodyreva [135]
3 years ago
15

You are a researcher for a golf club manufacturer. You are given two identical looking cubes of a metal alloy. You are informed

that they are made of the exact same material, but one is a crystalline, while the other is amorphous. It is your job to determine which one is amorphous because this one is more stress-resistant and is useful in reinforcing golf clubs. Which of the following is the best way to determine which is which?
A. Determine the density of each cube. The more dense one is the amorphous solid.

B. Determine the density of each cube. The less dense cube is the amorphous solid.

C. Melt both cubes and look for a broader range of melting temperatures. The one that melts over a broader range of temperatures is the amorphous solid.

D. Melt both cubes and measure the range of melting temperatures. The one that melts over a narrower range of temperatures is the amorphous solid.
Chemistry
1 answer:
Rashid [163]3 years ago
7 0

Answer:

C.Melt both cubes and look for a broader range of melting temperatures. The one that melts over a broader range of temperatures is the amorphous solid.

Explanation:

Amorphous solids is one that do not have a fixed melting points but melt over a wide range of temperature due to the irregular shape hence its name. Contrariwise crystalline solids, have a fixed and sharp melting point.

This comes in handy to solve the riddle. We can characterise the pair with the melting point property.

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Answer:

. CuO

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Since oxygen also has -2 the valency will cancel out the +2 on Cu leaving CuO

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68.3 grams of sodium hydroxide reacts with 78.3 grams of magnesium nitrate. ____ grams of magnesium hydroxide will form from thi
Vera_Pavlovna [14]

Answer:

30.8 grams of magnesium hydroxide will form from this reaction, and magnesium nitrate is the limiting reagent.

Explanation:

The reaction that takes place is:

  • 2NaOH + Mg(NO₃)₂ → 2NaNO₃ + Mg(OH)₂

Now we <u>convert the given masses of reactants to moles</u>, using their respective <em>molar masses</em>:

  • 68.3 g NaOH ÷ 40 g/mol = 1.71 mol NaOH
  • 78.3 g Mg(NO₃)₂ ÷ 148.3 g/mol = 0.528 mol Mg(NO₃)₂

0.528 moles of Mg(NO₃)₂ would react completely with (0.528 * 2) 1.056 moles of NaOH. There are more than enough NaOH moles, so NaOH is the reagent in excess and <em>Mg(NO₃)₂ is the limiting reagent.</em>

Now we <u>calculate how many Mg(OH)₂ are produced</u>, using the <em>moles of the limiting reagent</em>:

  • 0.528 mol Mg(NO₃)₂ * \frac{1molMg(OH)_2}{1molMg(NO_3)_2} = 0.528 mol Mg(OH)₂

Finally we convert Mg(OH)₂ moles to grams:

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