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2 years ago

If there are 3.10 moles of O, how many moles of each of the compounds are present?

Chemistry

1 answer:

Burka [1]2 years ago

3 0

Explanation:

The question pretty much requires us to find the amount of moles of each compounds based on the number of moles of O given.

H2SO4

1 mol of H2SO4 contains 4 mol of O

x mol of H2SO4 would contain 3.10 mol of O

x = 3.10 * 1 / 4 = 0.775 mol of H2SO4

C2H4O2

1 mol of C2H4O2 contains 2 mol of O

x mol of C2H4O2 would contain 3.10 mol of O

x = 3.10 * 1 / 2 = 1.55 mol of C2H4O2

NaOH

1 mol of NaOH contains 1 mol of O

x mol of NaOH would contain 3.10 mol of O

x = 3.10 * 1 / 1 = 3.10 mol of NaOH

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3 years ago

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Answer:

D. Electrical energy is changed into mechanical energy

Explanation:

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3 years ago

Which of the following is not a simple machine

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C: screw I believe so

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3 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

Olympic cyclists fill their tires with helium to make them lighter. Assume that the volume of the tire is 860 mL , that it is fi

nata0808 [166]

Answer:

a) the mass of air = 8.24 grams

b) the mass of helium = 1.14 grams

c) the mass difference = 7.10 grams

Explanation:

Step 1: Data given

Volume of the tire = 860 mL

Total pressure = 120 psi

Temperature = 26°C

molar mass of air = 28.8 g/mol

Step 2: Convert psi to atm

(

120 psi) (1 atm / 14.7 psi) = 8.163

Step 4: Calculate moles

PV = nRT

⇒ with P = the pressure = 8.163 atm

⇒ with V = the volume = 860 mL = 0.860 L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ with R = the universal gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 26 °C = 299 Kelvin

n = (8163*0.860)/(0.08206*299)

n = 0.2861 moles of gas

Step 5: Calculate the mass of air in an air-filled tire.

Mass = moles * molar mass

Mass = (0.2861 moles of gas) (28.8 g/mol)

Mass = 8.24 grams

Step 6: Calculate the mass of helium in a helium- filled tire.

mass of helium = 0.2861 moles of gas * 4 g/mol)

mass of helium = 1.14 grams

Step 7: What is the mass difference between the two?

Δmass= 8.24 grams - 1.14 grams

Δmass= 7.10 grams

8 0

3 years ago