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Ann [662]
3 years ago
14

What may be expected when K < 1.0? Choose the THREE correct statements. The concentration of one or more of the reactants is

small. The concentration of one or more of the products is small. The reaction will not proceed very far to the right. The reaction will generally form more reactants than products.
Chemistry
1 answer:
mestny [16]3 years ago
4 0

Answer:

The concentration of one or more of the products is small.

The reaction will not proceed very far to the right.

The reaction will generally form more reactants than products  

Explanation:

We often write

K =[Products]/[Reactants]

Thus, if K is small

  • We have fewer products than reactants
  • We have more reactants than products
  • The position of equilibrium lies to the left

A. is wrong. Usually, if K < 1, the concentration of reactants is greater than that of the products.

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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:

<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
The compound potassium nitrate is a strong electrolyte. write the reaction when solid potassium nitrate is put into water:
Alenkinab [10]
<span>There is no chemical reaction between potassium nitrate and water. Potassium nitrate dissolves in water, which is a physical change.</span>
7 0
3 years ago
11.0 L of hydrogen and 5.52 L of oxygen are exploded together in a reaction tube. What volume of water vapor was formed, at STP?
Marysya12 [62]

Answer:

11.0 L

Explanation:

The equation for this reaction is given as;

2H2  +  O2  -->  2H2O

2 mol of H2 reacts with 1 mol of O2 to form 2 mol of H2O

At STP;

1 mol = 22.4 L

This means;

44.8 L of H2 reacts with 22.4 L of O2 to form 44.8 L of H2O

In this reaction, the limiting reactant is H2 as O2 is in excess.

The relationship between H2 and H2O;

44.8 L = 44.8 L

11.0 L would produce x

Solving for x;

x = 11 * 44.8 / 44.8

x = 11.0 L

4 0
3 years ago
Determine the empirical formula of an oxide of silicon from the following: mass of crucible= 18.20g, mass of crucible +silicon =
lyudmila [28]
The required empirical formula of the silicon oxide is SiO2.
4 0
3 years ago
What were the purposes of this document? Check all that apply. to extend the rights of sovereignty to all states in the Union to
Dmitrij [34]

Answer:

Sorry that I'm late but the correct answers are to dissolve the political bonds between Louisiana and the united states and to  declare that Louisiana was a free and independent state, or D and E

Explanation:

7 0
3 years ago
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