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3 years ago

A compound is 53.31 % C , 11.18 % H , and 35.51 % O by mass. What is its empirical formula? Insert subscripts as needed.

Chemistry

1 answer:

Amanda [17]3 years ago

5 0

Answer:

The empirical formula is C2H5O

Explanation:

Step 1: Data given

Suppose the mass of the compound = 100.0 grams

The compound contains:

53.31 % C = 53.31 grams C

11.18 % H = 11.18 grams H

35.51 % O = 35.51 grams O

Step 2: Calculate moles

moles = mass / molar mass

Moles C = 53.31 grams / 12.01 g/mol

Moles C = 4.439 moles

Moles H = 11.18 grams / 1.01 g/mol

Moles H = 11.07 moles

Moles O = 35.51 grams / 16.0 g/mol

Moles O = 2.22 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

C: 4.439 / 2.22 = 2

H: 11.07 / 2.22 = 5

O: 2.22/2.22 = 1

C2H5O

The empirical formula is C2H5O

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What is height? Give an example.

Sav [38]

Height is defined as the distance from the bottom to the top of something or the highest point or the greatest degree. An example of height is 5'8". An example of height is the top of Mount Everest.

5 0

3 years ago

A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at

Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0

3 years ago

1 C8H10(l) +21/2O2(g) → 8CO2(g) + 5H2O(g), Hcomb= ? Hf for C8H10(l) = +49.0kJ/mol C8H10(l) Use the balanced combustion reaction

nikklg [1K]

Answer:

$H_{comb}=-4406kJ/mol$

Explanation:

Hello,

In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

$H_{comb}=8*\Delta _fH_{CO_2}+5\Delta _fH_{H_2O}-\Delta _fH_{C_8H_{10}}$

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

$H_{comb}=8*(-393.5kJ/mol)+5(-241.8kJ/mol)-49.0kJ/mol\\\\H_{comb}=-4406kJ/mol$

Best regards.

4 0

4 years ago

In each pair, choose the larger of the indicated quantities or state that the samples are equal:

masha68 [24]

0.4 mol of $O_3$ molecules has larger mass than 0.4 mol of O atoms

<h3>What are molecules?</h3>

The smallest component of a substance that possesses both its chemical and physical characteristics. One or more atoms make up molecules.

They may have the same atoms (for example, an oxygen molecule has two oxygen atoms) or different atoms if they have more than one (a water molecule has two hydrogen atoms and one oxygen atom). Biological molecules like DNA and proteins can include thousands of atoms.

the lowest recognised unit into which a pure material can be divided while retaining its chemical makeup and attributes is made up of two or more atoms.

Learn more about Molecules

brainly.com/question/1078183

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5 0

1 year ago

What is the weight of a 90kg man standing on the moon, where gmoon = 1.64 m/s2?

Doss [256]

Answer:

The weigth of a 90kg man standing on the moon is 147.6 N (option C)

Explanation:

Weight is called the action exerted by the force of gravity on the body.

The mass (amount of matter that a body contains) of an object will always be the same, regardless of where it is located. Instead, the weight of the object will vary according to the force of gravity acting on it.

The formula that allows you to calculate the weight of any body is:

W = m*g

where:

W = weight measured in N.
m = mass measured in kg.
g = acceleration of gravity measured in m/s². The acceleration of gravity g is the same for all objects that fall due to gravitational attraction, whatever their size or composition. For example, as an approximate value on Earth, g = 9.8 m/s².

In this case, the mass m has a value of 90 kg and the gravity g has a value of 1.64 m/s², which is the value of the acceleration of gravity of the moon. Then:

W=90 kg* 1.64 m/s²

W= 147.6 N

Finally, the weigth of a 90kg man standing on the moon is 147.6 N (option C)

3 0

4 years ago