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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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N =- triple bond -= C – O

<span>So there is only single bond between C and O therefore the hybridization of C is sp.</span>

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2 years ago
Calculate the empirical formula for each stimulant based on its elemental mass percent composition. a. nicotine (found in tobacc
Ira Lisetskai [31]

This an incomplete question, here is a complete question.

Calculate the empirical formula for each stimulant based on its elemental mass percent composition.

a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%

b. caffeine (found in coffee beans): C 49.48%, H 5.19 %, N 28.85% and O 16.48%

Answer:

(a) The empirical formula for the given compound is C_5H_7N

(b) The empirical formula for the given compound is C_4H_5N_2O

Explanation:

<u>Part A: nicotine </u>

We are given:

Percentage of C = 74.03 %

Percentage of H = 8.70 %

Percentage of N = 17.27 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 74.03 g

Mass of H = 8.70 g

Mass of N = 17.27 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{74.03g}{12g/mole}=6.17moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{8.70g}{1g/mole}=8.70moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{17.27g}{14g/mole}=1.23moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.23 moles.

For Carbon = \frac{6.17}{1.23}=5.01\approx 5

For Hydrogen  = \frac{8.70}{1.23}=7.07\approx 7

For Nitrogen = \frac{1.23}{1.23}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

The empirical formula for the given compound is C_5H_7N_1=C_5H_7N

<u>Part B: caffeine</u>

We are given:

Percentage of C = 49.48 %

Percentage of H = 5.19 %

Percentage of N = 28.85 %

Percentage of O = 16.48 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{49.48g}{12g/mole}=4.12moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.19g}{1g/mole}=5.19moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{28.85g}{14g/mole}=2.06moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{16.48g}{16g/mole}=1.03moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.03 moles.

For Carbon = \frac{4.12}{1.03}=4

For Hydrogen  = \frac{5.19}{1.03}=5.03\approx 5

For Nitrogen = \frac{2.06}{1.03}=2

For Nitrogen = \frac{1.03}{1.03}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The empirical formula for the given compound is C_4H_5N_2O_1=C_4H_5N_2O

6 0
3 years ago
Which part of a chemical formula indicates the number of copies of molecules or compounds? A. plus signs B. subscripts C. coeffi
nadya68 [22]
I think it is C. coefficients but I don't know what subscripts are
3 0
3 years ago
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