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Yuliya22 [10]
3 years ago
9

I need help on this one pliz

Chemistry
1 answer:
Artyom0805 [142]3 years ago
8 0
The answer would be 16.04.
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In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous methane and gaseous water in a 0.778 L
sergejj [24]

Answer:

the water concentration at equilibrium is

⇒ [ H2O(g) ] = 0.0510 mol/L

Explanation:

  • CH4(g) + H2O(g) ↔ CO(g) + 3H2(g)

∴ Kc = ( [ CO(g) ] * [ H2 ]³ ) / ( [ CH4(g) ] * [ H2O(g) ] ) = 0,30

  • equilibrium:

⇒ [ CO(g) ] = 0.206 mol / 0.778 L = 0.2648 mol/L

⇒ [ H2(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L

⇒ [ CH4(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L

replacing in Kc:

⇒ ((0.2648) * (0.2404)³) / ([ H2O(g) ] * 0.2404 ) = 0.30

⇒ 0.0721 [ H2O(g) ] = 3.679 E-3

⇒ [ H2O(g) ] = 0.0510 mol/L

7 0
3 years ago
What is the mole fraction of O2O2 in a mixture of 15.1 gg of O2O2, 8.19 gg of N2N2, and 2.46 gg of H2H2
SVETLANKA909090 [29]

Answer:

Mole fraction O₂=  0.43

Explanation:

Mole fraction is the moles of gas/ total moles.

Let's determine the moles of each:

Moles O₂ → 15.1 g / 16 g/mol = 0.94

Moles N₂  →  8.19 g / 14 g/mol = 0.013

Moles H₂  → 2.46 / 2 g/mol = 1.23

Total moles = 2.183

Mole fraction O₂= 0.94 / 2.183 → 0.43

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How many chlorine atoms will correctly complete the data table ?
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Answer:

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Explanation:

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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:

<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
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