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3 years ago

In terms of the atmosphere, which of the following best describes the oxygen that is present?

1 answer:

6 0

The atmosphere is considered homogeneous. It isn’t exactly on the smallest scales but that doesn’t matter. Homogenous means the composition will be the same in any sample taken from the substance. And clearly, the atmosphere is mostly gas. So the last answer is right

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Which of the following is the simplest ketone used as an organic solvent?

Lyrx [107]

Answer: Option (d) is the correct answer.

Explanation:

A ketone is an organic functional group that contains a carbon and oxygen atom bonded together through a double bond, that is, C=O.

For example, acetone $CH_{3}-CO-CH_{3}$ is a ketone.

Whereas a hydrocodone is a drug which is used to relieve from moderate to sever pain.It is mostly combined with other drugs and resulting in a chemical formula $C_{18}H_{21}NO_{3}.C_{4}H_{6}O_{6}$ .

A camphor is a volatile white color substance with chemical formula $C_{10}H_{16}O$ . It has aromatic smell and its taste is bitter.

A menthone is also an organic compound with chemical formula $C_{10}H_{18}O$ .

Thus, we can conclude that out of the given options the simplest ketone used as an organic solvent is acetone.

3 0

3 years ago

What are two different kinds of rock that are easily weathered by carbonic acid?

horsena [70]

Limestone and marble are the two rocks that are easily weathered by carbonic acid. Two kinds of weathering mostly affect the rocks. and they are physical weathering and chemical weathering. The above mentioned two rocks are highly affected by carbonic acid. The appeareance of these kind of rocks change their structure due to erosion very regularly. Caronation is the type of chemical weathering that affects the rocks limestone and marble. The carbonic acide is formed by the reaction of carbon dioxide in the air and water in the rivers. This carbonic acid results in weathering.

8 0

4 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

photosynthesis in plants require chloroplasts and light energy. identify two materials plants also use in this process

Nataliya [291]

I would assume carbon and water if not Im sorry.

5 0

3 years ago

Match the characteristics and elements to their group.

Luba_88 [7]

Answer:

Kr is a Noble Gas. Na is an alkali metal. F is halogen.

Group 17 is halogens. Inert is Noble Gases. Odourless and colourless is Noble Gases. Alkali metals do not occur freely in nature. Alkali metals are malleable

Explanation:

8 0

3 years ago