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3 years ago

Which of the following is not an example of a polymer? (2 points)

1 answer:

3 0

Answer is "sucrose".

Polymer is a large molecule which is made from repeating units. The smallest repeating unit is called as monomer. Polystyrene, nylon and PVC are examples for polymers. But sucrose is a disaccharide which is made from glucose and fructose. Hence, sucrose is not an example of polymer.

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Chemists use the periodic table to organize the elements. What are some ways in which this organization is useful? Select all th

max2010maxim [7]

Answer: Another useful feature of the periodic table is that most tables provide all the information you need to balance chemical reactions at a glance. The table tells each element's atomic number and usually its atomic weight. The typical charge of an element is indicated by its group.

Explanation:

7 0

2 years ago

Read 2 more answers

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:

Part A: nicotine

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 $C_5H_7N_1=C_5H_7N$

Part B: caffeine

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

What is the wavelength of the matter wave associated with an electron (me= 9.1 x 10-31 kg) moving with a speed of 2.5 x 107 m/s?

Novay_Z [31]

Lambda = h\ Mv
lambda = 6.624 x 10^-34 / 9.1 x 10^-31 x 2.5 x 10^7
lambda = 2.9 x 10^-11 is your wavelength

3 0

3 years ago

Is silver tarnishing a physical change or a chemical change?

Shalnov [3]

Answer:

chemical

Explanation:

Some bronze objects tarnish to a dark brown color. Chemical properties can be identified by the changes they produce. The change of one substance into another substance is called a A piece of wood burning, an iron fence rusting, and a silver spoon tarnishing are all examples of chemical changes.

6 0

3 years ago

Select the correct answer

amm1812

Answer:

C The experiment shows that the red substance experienced a chemical change.

Explanation:

Apparently, adding heat caused the red substance to decompose into a gas and a metallic liquid. If it were simply a phase change, the original red substance could be expected to return when the temperature cooled. Because the substance apparently decomposed, it is clearly not an element. At no point in the experiment is there any evidence of a plasma being formed.

The observed decomposition is a chemical change.

8 0

3 years ago