QUESTION 3 How would 250 ml of 0.15 M KNO3 solution be prepared?​

The combustion of 4.7 kg of pure octane ([tex]C_8H_{18[/tex]) produces approximately 15 kg of carbon dioxide ([tex]CO_2[/tex]).

1. Start by writing the balanced equation for the combustion of octane ([tex]C_8H_{18[/tex]):

  [tex]C_8H_{18[/tex] + 12.5O2 → [tex]8CO_2[/tex] + [tex]9H_2O[/tex]

  This equation shows that for every 1 mole of octane burned, 8 moles of carbon dioxide and 9 moles of water are produced.

2. Determine the molar mass of octane ([tex]C_8H_{18[/tex]):

  The molar mass of carbon (C) is approximately 12.01 g/mol.

  The molar mass of hydrogen (H) is approximately 1.008 g/mol.

  Calculating the molar mass of octane: (8 * 12.01 g/mol) + (18 * 1.008 g/mol) ≈ 114.23 g/mol.

3. Calculate the number of moles of octane in 4.7 kg:

  Number of moles = mass (in grams) / molar mass

  Moles of octane = (4.7 kg * 1000 g/kg) / 114.23 g/mol ≈ 41.11 mol

4. Determine the number of moles of carbon dioxide produced:

  From the balanced equation, we know that for every mole of octane burned, 8 moles of carbon dioxide are produced.

  Moles of carbon dioxide = 41.11 mol octane * 8 mol CO2 / 1 mol octane ≈ 328.88 mol

5. Calculate the mass of carbon dioxide produced:

  Mass = moles * molar mass

  Mass of carbon dioxide = 328.88 mol * (12.01 g/mol + 2 * 16.00 g/mol) ≈ 7,883.51 g ≈ 7.88 kg

6. Express the answer using two significant figures:

  The mass of carbon dioxide produced is approximately 7.88 kg when 4.7 kg of octane is burned.

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Answer:

true

Explanation:

atomic nuclei consist of electrically positive proton and electrically neutral neutrons. These are held together by the strongest known fundamental force, called the strong force.

The nucleus of every atom contains protons. This statement is true.

Protons are positively charged subatomic particles, which are one of the fundamental components of an atom, along with neutrons and electrons. Protons play a crucial role in determining the identity of an element. They determine the atomic number of an element.

The atomic number is used to arrange elements in the periodic table and is used as a basis for defining the number of electrons in an atom of that element. The arrangement and combination of protons, along with neutrons, determine the atom's mass and stability.

In summary, protons are an essential component of the nucleus in all atoms, making the statement true.

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According to the engineer's design, 14.67 moles of N2 would be produced in the reaction.

To determine the amount of N2 that would be produced according to the engineer's design, we need to understand the stoichiometry of the reaction between dinitrogen tetroxide (N2O4) and hydrazine (N2H4).

The balanced chemical equation for the reaction is:

N2H4 + N2O4 → N2 + 2H2O

From the balanced equation, we can see that one mole of N2H4 reacts with one mole of N2O4 to produce one mole of N2. Therefore, the mole ratio between N2H4 and N2 is 1:1.

Given that the engineer designed the rocket to hold 1.35 kg of N2O4, we need to convert this mass to moles using the molar mass of N2O4. The molar mass of N2O4 is approximately 92.01 g/mol.

Moles of N2O4 = Mass of N2O4 / Molar mass of N2O4

= 1.35 kg / 92.01 g/mol

= 14.67 mol

Since the mole ratio between N2H4 and N2 is 1:1, the number of moles of N2 produced would be the same as the number of moles of N2O4, which is 14.67 mol.

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Answer:

Ba + Cr + O₄

Answer:

A Negative Charge

Explanation:

Positive Charges Repel

Positive and Negative Charges Attract.

Negative Charges Repel.

Answer:

Tcs foods are foods that pose a greater risk of causing foodborne illness if not prepared.

Non Tcs foods on the other hand, are foods that are less likely to support the growth of bacteria and have a lower risk of causing foodborne illness.

Answer:

A liquid of unknown composition: Unknown liquid

A liquid of known composition: Known liquid

A solid-liquid mixture of unknown composition: Unknown solid-liquid mixture

A solid of known composition: Known solid

The energy of combustion for one mole of butane to be approximately 2888.81 kJ/mol.

To calculate the energy of combustion for one mole of butane (C4H10), we need to use the information provided and apply the principle of calorimetry.

First, we need to convert the mass of the butane sample from grams to moles. The molar mass of butane (C4H10) can be calculated as follows:

C: 12.01 g/mol

H: 1.01 g/mol

Molar mass of C4H10 = (12.01 * 4) + (1.01 * 10) = 58.12 g/mol

Next, we calculate the moles of butane in the sample:

moles of butane = mass of butane sample / molar mass of butane

moles of butane = 0.367 g / 58.12 g/mol ≈ 0.00631 mol

Now, we can calculate the heat released by the combustion of the butane sample using the equation:

q = C * ΔT

where q is the heat released, C is the heat capacity of the calorimeter, and ΔT is the change in temperature.

Given that the heat capacity of the bomb calorimeter is 2.36 kJ/°C and the change in temperature is 7.73 °C, we can substitute these values into the equation:

q = (2.36 kJ/°C) * 7.73 °C = 18.2078 kJ

Since the heat released by the combustion of the butane sample is equal to the heat absorbed by the calorimeter, we can equate this value to the energy of combustion for one mole of butane.

Energy of combustion for one mole of butane = q / moles of butane

Energy of combustion for one mole of butane = 18.2078 kJ / 0.00631 mol ≈ 2888.81 kJ/mol

Therefore, the energy of combustion for one mole of butane is approximately 2888.81 kJ/mol.

In conclusion, by applying the principles of calorimetry and using the given data, we have calculated the energy of combustion for one mole of butane to be approximately 2888.81 kJ/mol.

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The true statements about catalysts are the statement 1,2 and 3.

1. Catalysts increase the rate of reaction: Catalysts facilitate chemical reactions by providing an alternative reaction pathway with lower activation energy. They enhance the rate of the reaction without being consumed in the process.

2. Catalysts behave as reactants in the reaction mixture: Catalysts participate in the reaction by interacting with the reactants. They form temporary bonds with the reactant molecules, leading to the formation of an intermediate complex that ultimately results in the desired products.

3. Catalysts decrease the activation energy of a reaction: Catalysts lower the energy barrier required for a reaction to occur by providing an alternative pathway with a lower activation energy. This enables the reactants to overcome the energy barrier more easily, thus increasing the reaction rate.

4. Catalysts show no physical change at the end of the reaction: Catalysts are not consumed or permanently altered in the reaction. They remain chemically unchanged and are available to participate in subsequent reaction cycles.

The statement "Catalysts are required in large concentrations in a reaction" is false. Catalysts work effectively even in small concentrations, as their role is to facilitate the reaction rather than being directly involved in the stoichiometry of the reaction.

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The bullet has a kinetic energy of approximately 344.45 joules (J), while the ocean liner has a kinetic energy of approximately 676,200,000 joules (J). As we can see, the ocean liner has significantly more kinetic energy than the bullet due to its larger mass and velocity.

To calculate the kinetic energy of an object, we use the formula:

Kinetic Energy (Ek) = 0.5 * mass * velocity^2

Let's calculate the kinetic energy for both the bullet and the ocean liner:

For the bullet:

Mass (m) = 0.0020 kg

Velocity (v) = 415 m/s

Ek-bullet = 0.5 * 0.0020 kg * (415 m/s)^2

Ek-bullet = 0.5 * 0.0020 kg * 172225 m^2/s^2

Ek-bullet = 344.45 J

For the ocean liner:

Mass (m) = 6.9 * 10^7 kg

Velocity (v) = 14 m/s

Ek-ocean liner = 0.5 * (6.9 * 10^7 kg) * (14 m/s)^2

Ek-ocean liner = 0.5 * (6.9 * 10^7 kg) * 196 m^2/s^2

Ek-ocean liner = 676200000 J

Therefore, the bullet has a kinetic energy of approximately 344.45 joules (J), while the ocean liner has a kinetic energy of approximately 676,200,000 joules (J). As we can see, the ocean liner has significantly more kinetic energy than the bullet due to its larger mass and velocity.

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Based on the given bright-line spectra and the observed wavelengths in the mixture's spectrum, the elements G and J are the ones present in the mixture.

From the given bright-line spectra and the spectrum of the mixture, we can determine the elements present in the mixture by comparing the specific wavelengths observed. Examining the bright-line spectra, we can identify that G has a distinct wavelength at 650 nm, J at 600 nm, L at 550 nm, and M at 500 nm.

Looking at the spectrum of the mixture, we can observe two prominent wavelengths, 650 nm and 600 nm. These correspond to the wavelengths of G and J, respectively. Since the spectrum of the mixture does not exhibit the wavelengths specific to L (550 nm) or M (500 nm), we can conclude that only G and J are present in the mixture.

Therefore, based on the given bright-line spectra and the observed wavelengths in the mixture's spectrum, the elements G and J are the ones present in the mixture.

This analysis relies on the principle that each element has characteristic wavelengths at which they emit light. By comparing the observed wavelengths in the mixture's spectrum with those of the individual elements, we can determine the elements present in the mixture.

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Answer::

Explanation::

The sun, the moon, the stars, the earth all are made up of matter.

Matter refers to anything that has mass and occupies space. It is the substance that makes up all physical objects in the universe, including both living and non-living things. Matter can exist in different states, namely solid, liquid, and gas, depending on the arrangement and movement of its particles. Matter is composed of atoms, which are the smallest units of matter that retain the chemical properties of an element. Atoms combine to form molecules, which can be made up of one or more different types of atoms bonded together. These molecules then come together to form different substances.

The properties of matter, such as its density, color, texture, and ability to conduct heat or electricity, are determined by the composition, arrangement, and interactions of its particles. Matter can undergo physical and chemical changes, including phase transitions (such as melting, freezing, and vaporization) and chemical reactions, where substances can be transformed into new substances with different properties. It is important to note that matter also includes forms that are not directly visible to the  eye, such as subatomic particles

The sun, the moon, the stars, and the Earth are all made up of matter. Matter refers to anything that has mass and occupies space. It is composed of atoms and molecules, which are the building blocks of all substances. While symbols can represent or signify various concepts or objects, they are not physical entities made up of matter. A mixture is a combination of two or more substances, but it does not encompass celestial bodies like the sun, moon, stars, or Earth. Material is a more general term that can refer to various physical substances, but it does not specifically indicate the composition or nature of celestial bodies.

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Answer:

the density of the helium gas would be approximately 0.369 g/L when the balloon is placed in the freezer at -10°C and a pressure of 2.0 atm.

Explanation:

To calculate the density of helium gas in the balloon after it is placed in the freezer at -10°C and a pressure of 2.0 atm, we can use the ideal gas law and the relationship between density, molar mass, and molar volume.

First, let's find the initial molar volume of the helium gas using the given conditions:

PV = nRT

Where:

P = pressure = 1.0 atm

V = volume = 0.75 L

n = number of moles = 0.0303 mol

R = ideal gas constant = 0.0821 L·atm/(mol·K)

T = temperature in Kelvin

To convert Celsius to Kelvin, we add 273.15:

T = 30°C + 273.15 = 303.15 K

Using the ideal gas law, we can calculate the initial molar volume:

V_initial = (n * R * T) / P

V_initial = (0.0303 mol * 0.0821 L·atm/(mol·K) * 303.15 K) / 1.0 atm

V_initial ≈ 0.754 L

Next, we can calculate the molar mass of helium (He) using the atomic mass of helium:

Molar mass of He = 4.003 g/mol

Now we can calculate the initial density of the helium gas in the balloon:

Initial density = (mass of helium gas) / (volume of helium gas)

Initial density = (0.0303 mol * 4.003 g/mol) / 0.754 L

Initial density ≈ 0.161 g/L

Now let's find the final density of the helium gas when the balloon is placed in the freezer at -10°C and a pressure of 2.0 atm.

We will use the ideal gas law again with the new conditions:

P_final = 2.0 atm

T_final = -10°C + 273.15 = 263.15 K (converted to Kelvin)

To find the final molar volume, we rearrange the ideal gas law equation:

V_final = (n * R * T_final) / P_final

V_final = (0.0303 mol * 0.0821 L·atm/(mol·K) * 263.15 K) / 2.0 atm

V_final ≈ 0.328 L

Finally, we can calculate the final density of the helium gas:

Final density = (mass of helium gas) / (volume of helium gas)

Final density = (0.0303 mol * 4.003 g/mol) / 0.328 L

Final density ≈ 0.369 g/L

These substances have dispersed particles that are large enough to scatter light, making the beam visible. Therefore, out of the options provided, a mixture that scatters light is an example of a colloid. Option B)

A colloid is a type of mixture in which particles are dispersed throughout a medium, creating a homogeneous appearance. Unlike solutions, where the particles are completely dissolved, and suspensions, where the particles settle out, colloids have particles that are larger than those in solutions but smaller than those in suspensions. One characteristic of colloids is that they can scatter light due to the size of the particles. This scattering of light is known as the Tyndall effect. Examples of colloids include milk, fog, and aerosol sprays. These substances have dispersed particles that are large enough to scatter light, making the beam visible. Therefore, out of the options provided, a mixture that scatters light is an example of a colloid. Therefore option B) is correct

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Note Complete Question

which is an example of a colloid?

a mixture that settles out,

b mixture that scatters light,

c mixture that is separated by filtration,  

d salt and water mixture?

Answer:Electrical Conductivity,soluble

Explanation:

Salt is a non-magnetic solid and is soluble in water. Sand is a non-magnetic solid and is insoluble in water.

Electrical Conductivity: Salt is an electrolyte and conducts electricity when dissolved in water or in a molten state. This is because salt dissociates into ions (Na+ and Cl-) that can carry electric current. In contrast, sand is a covalent compound and does not conduct electricity, as it does not dissociate into ions in the same way as salt. Sand is considered an insulator in terms of electrical conductivity.

The molecular formula for the organic compound is C4H4O.

To determine the molecular formula of the organic compound, we need to calculate the number of moles of carbon (C), hydrogen (H), and oxygen (O) in the compound and find the simplest whole number ratio between them.

Given:

Mass of the organic compound = 11.3 g

Percentage composition:

Carbon (C) = 70.58%

Hydrogen (H) = 5.9%

Oxygen (O) = 23.50%

First, we calculate the mass of each element in the organic compound:

Mass of C = 70.58% of 11.3 g = 7.986 g

Mass of H = 5.9% of 11.3 g = 0.667 g

Mass of O = 23.50% of 11.3 g = 2.655 g

Next, we convert the masses of each element to moles using their respective molar masses:

Molar mass of C = 12.01 g/mol

Molar mass of H = 1.008 g/mol

Molar mass of O = 16.00 g/mol

Moles of C = 7.986 g / 12.01 g/mol ≈ 0.665 mol

Moles of H = 0.667 g / 1.008 g/mol ≈ 0.661 mol

Moles of O = 2.655 g / 16.00 g/mol ≈ 0.166 mol

Now, we divide the moles of each element by the smallest number of moles to find the simplest whole number ratio:

C: 0.665 mol / 0.166 mol ≈ 4

H: 0.661 mol / 0.166 mol ≈ 4

O: 0.166 mol / 0.166 mol = 1

Therefore, the empirical formula of the organic compound is C4H4O.

To find the molecular formula, we need to determine the molecular weight of the compound. Given that the molecular weight of the compound is 11.3 g, which is equal to the empirical formula weight (C4H4O), we can conclude that the molecular formula is the same as the empirical formula.

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