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4 years ago

Using the periodic table, how would you find elements with chemical properties similar to helium?

1 answer:

3 0

The state of the helium in its natural form is gaseous and is a chemical element of colorless aspect and belongs to the group of noble gases. The atomic number of helium is 2. The chemical symbol of helium is He. For the following we focus on those elements and relate it with similar chemical properties. Then we find that; Neon, Hydrogen, Boron and Carbon are related to helium, either by proximity in their atomic number or period or by their group.

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True or false? An organism may play more than one role in a food web

almond37 [142]

FALSE

Organisms may play only one role in an ecosystem. ... Each of the organisms in an ecosystem fills the energy role of producer, consumer, or decomposer.

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3 0

3 years ago

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Which factors help you identify the solute and the solvent in a solution? Select all that apply. In a solution, the solvent is p

KatRina [158]

The answers would be:

In a solution, the solvent is present in a greater amount.

In a solutions, the solute dissolves in a solvent.

In general, these are the best answers. The solute is what is being dissolved and the solvent is what dissolves. A solvent comes in greater amounts in a solution and it is the dissolving agent.

For example, sugar and water.

To make a sugar water solution, you will need to dissolve sugar in water. Sugar is the solute in this case because it is what is being dissolved. The water is the solvent, because it dissolves the sugar.

If you had more sugar than water, then you cannot make a solution.

4 0

3 years ago

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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

Consider the reaction CaCO3(s) --> CaO(s) + CO2(g)

tankabanditka [31]

Answer:

Kc = [CO2], that is to say it is equal to the concentration of CO2

Explanation:

It is a heterogeneous equilibrium since the substances that participate in the reaction are in different phases

In the heterogeneous limestone decomposition reaction:

CaCO3(s) --> CaO(s) + CO2(g)

The equilibrium constants are:

Kc = [CO2(g)]; Kp = PCO2(g); Kc = Kp (R T)^ −(1−0) = Kp (R T)^ −1

The equilibrium situation is not affected by the amount of solid or liquid, as long as these substances are present.

The equilibrium constant is independent of the amounts of solids and liquids in equilibrium.

4 0

3 years ago

Filtration definition

dezoksy [38]

Answer:

the action or process of filtering something.

8 0

3 years ago

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