Nitrogen is generated by a chemical reaction and 1.80 liters is collected over water at 35.0°C when the pressure in the laboratory is 775 mm Hg. What is the partial pressure (in mm Hg) of nitrogen? The partial pressure of water at this temperature

42.2 mm Hg.
5.45 mm Hg
817 mm Hg
733 mm Hg
18.4 mm Hg

The molar mass of the compound N₂O₅ is approximately 108.02 g/mol, and when rounded to the nearest whole number, it is approximately 108 g.

To calculate the mass of a compound, we need to determine the molar mass of each element in the compound and then sum them together based on the molecular formula. In this case, we have the compound N₂O₅, which consists of two nitrogen atoms (N) and five oxygen atoms (O).

To calculate the molar mass, we look up the atomic masses of nitrogen (N) and oxygen (O) from the periodic table:

Nitrogen (N) atomic mass = 14.01 g/mol

Oxygen (O) atomic mass = 16.00 g/mol

Now, we calculate the molar mass of each element in the compound:

Molar mass of nitrogen (N) = 14.01 g/mol × 2 = 28.02 g/mol

Molar mass of oxygen (O) = 16.00 g/mol × 5 = 80.00 g/mol

Finally, we sum the molar masses of nitrogen and oxygen to get the molar mass of the compound:

Molar mass of N₂O₅ = 28.02 g/mol + 80.00 g/mol = 108.02 g/mol

Therefore, the molar mass of N₂O₅ is approximately 108.02 g/mol.

To round the answer to the nearest whole number, we obtain:

Mass of N₂O₅ ≈ 108 g

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I really wish I could help but this is all I know

Answer:

Millimeter can be used to describe the width of a small piece of metal

Explanation:

The most appropriate unit to describe the width of a small piece of metal would be millimeters (mm). Millimeters are commonly used to measure small lengths or dimensions, such as the width of objects. This unit provides a precise and accurate measurement for small-scale items.

Ounces (oz) are typically used to measure weight or volume and may not be suitable for describing the width of an object.

Moles (mol) are used in chemistry to quantify the amount of a substance, so it is not relevant for measuring dimensions.

Milliliters (ml) are used to measure volume, specifically the capacity of liquids, and are not typically used to describe the width of a solid object.

Therefore, when discussing the width of a small piece of metal, millimeters (mm) would be the appropriate unit as it specifically measures small lengths or dimensions.

AI said C. Millimeter.

I hope this helped! :)

The molar mass of saccharin is 0.900 g / [Sac].

Saccharin is a weak organic base with a K​b of 4.80 × 10-3 and we are given that a 0.900-g sample of saccharin dissolved in 45.0mL of water has a pH of 12.310.

We are supposed to calculate the molar mass of saccharin.

The formula for finding the molar mass of a substance is:Molar mass = (mass of substance) / (number of moles).

We are given that the Kb of saccharin is 4.80 × 10-3.

Since it is a base, it reacts with water to form the conjugate acid of saccharin (HSac) and hydroxide ions (OH-).

The balanced chemical equation for this reaction is: C7H4NO3S + H2O → HSac + OH-.

We can use the Kb value to find the concentration of the hydroxide ions produced:Kb = [HSac][OH-] / [Sac].

Initial concentration of saccharin, [Sac] = (0.900 g) / (Molar mass) = 0.900 / M Molar mass = 0.900 / [Sac].

Now we can use the given pH value to find the concentration of the hydroxide ions using the expression:pH = 14 - pOHpOH = 14 - pH[OH-] = 10^-pOH.

Substitute these values in the expression for Kb and solve for [HSac]:Kb = [HSac][OH-] / [Sac][HSac] = (Kb x [Sac]) / [OH-]

Now we can substitute the values we have into the expression for the molar mass:Molar mass = (mass of substance) / (number of moles)Number of moles = [Sac]Molar mass = 0.900 / [Sac].

Therefore, the molar mass of saccharin can be calculated by first finding the concentration of hydroxide ions produced using the pH value, then using the Kb value to find the concentration of the conjugate acid of saccharin, and finally using these values to find the molar mass of saccharin using the formula Molar mass = (mass of substance) / (number of moles).

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The maximum amount of product that can be obtained from given amounts of reactants in a chemical reaction. The correct answer is C.

A chemical reaction is a process that causes one group of chemical components to change chemically into another. Reactants are substances that interact to make new substances, whereas products are those substances that result from the interaction.

Chemical Reaction Types

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0.595 moles of water can be made at 112.6 grams of [tex]HNO_{3}[/tex] are consumed

To determine the number of moles of water produced when 112.6 grams of [tex]HNO_{3}[/tex] are consumed, we use the equation's stoichiometry and molar masses.

To determine the number of moles of water produced when 112.6 grams of [tex]HNO_{3}[/tex] are consumed, we need to use the molar mass of [tex]HNO_{3}[/tex] and the stoichiometric coefficients of the balanced chemical equation.

The molar mass of [tex]HNO_{3}[/tex] is calculated as follows:

1 mole of hydrogen (H) = 1 g/mol

1 mole of nitrogen (N) = 14 g/mol

3 moles of oxygen (O) = 3 × 16 g/mol = 48 g/mol

Adding these together, the molar mass of [tex]HNO_{3}[/tex] is 1 + 14 + 48 = 63 g/mol.

Now, we can set up a conversion factor using the stoichiometry of the balanced equation:

From the equation: 5 + 6 [tex]HNO_{3}[/tex] -> [tex]H_{2}SO_{4}[/tex] + 6 [tex]NO_{2}[/tex] + 2 [tex]H_{2}O[/tex]

From the coefficients: 6 moles of [tex]HNO_{3}[/tex] produce 2 moles of [tex]H_{2}O[/tex]

To find the moles of water produced, we use the following calculation:

112.6 g [tex]HNO_{3}[/tex] × (1 mol [tex]H_{2}O[/tex] / 63 g [tex]HNO_{3}[/tex]) × (2 mol [tex]H_{2}O[/tex] / 6 mol [tex]HNO_{3}[/tex]) = 0.595 mol [tex]H_{2}O[/tex]

Therefore, when 112.6 grams of [tex]HNO_{3}[/tex] are consumed, approximately 0.595 moles of water can be produced according to the given balanced equation and molar masses.

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Taking into account the reaction stoichiometry, 128.4 grams of C are required to produce 10.7 moles of methane.

In first place, the balanced reaction is:

C + 2 H₂ → CH₄

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The molar mass of the compounds is:

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

The following rule of three can be applied: If by reaction stoichiometry 1 mole of CH₄ is produced by 12 grams of C, 10.7 moles of CH₄ are produced by how much mass of C?

mass of C= (10.7 moles of CH₄×12 grams of C)÷1 mole of CH₄

mass of C= 128.4 grams

Finally, 128.4 grams of C are required.

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128.4 grams of carbon are required to produce 10.7 moles of methane.

The chemical reaction of carbon and hydrogen gas to produce methane gas can be written as:C + 2H2 → CH4The given reaction can be interpreted as: 1 mole of Carbon reacts with 2 moles of Hydrogen to produce 1 mole of Methane.

The number of moles of methane produced from a given number of moles of carbon can be calculated using the stoichiometric coefficients of the balanced chemical equation.

Here, the number of moles of methane produced can be calculated as follows:Number of moles of methane produced = (Number of moles of Carbon) / 1 * 1 = Number of moles of Carbon.

Therefore,Number of moles of Carbon = Number of moles of methane producedSince the number of moles of methane produced is given as 10.7 moles, the number of moles of carbon required can be calculated as:

Number of moles of Carbon = Number of moles of methane produced= 10.7 moles

Using the molar mass of carbon, the number of grams of carbon required can be calculated as follows:

Mass of carbon = Number of moles of carbon × Molar mass of carbon= 10.7 moles × 12 g/mol= 128.4 g.

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Phosphatidylserine is a type of phospholipid that is mainly found in cell membranes. Its structure is made up of two fatty acid chains, a phosphate group, a serine molecule, and a glycerol molecule.

The fatty acid chains are hydrophobic, meaning they repel water, while the phosphate group and serine molecule are hydrophilic, meaning they attract water.

The glycerol molecule acts as a bridge that connects the two fatty acid chains to the phosphate group and serine molecule.
The structure of phosphatidylserine is important for its function in the cell membrane.

Because of the hydrophobic and hydrophilic components of its structure, phosphatidylserine is able to form a lipid bilayer, which is a barrier that separates the inside of the cell from the outside environment.

The hydrophilic heads of the phosphatidylserine molecules face outward and interact with water, while the hydrophobic tails face inward and repel water.
Phosphatidylserine also plays a role in cell signaling and apoptosis, which is programmed cell death.

It acts as a signaling molecule by binding to proteins that are involved in cellular pathways.

In addition, phosphatidylserine is translocated to the outer leaflet of the cell membrane during apoptosis, which signals to immune cells that the cell is ready to be removed.
In conclusion, the structure of phosphatidylserine is made up of two fatty acid chains, a phosphate group, a serine molecule, and a glycerol molecule. Its hydrophobic and hydrophilic components allow it to form a lipid bilayer in cell membranes, and it also plays a role in cell signaling and apoptosis.

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In most experiments, the decay of a large number of radioactive isotopes over time is measured in order to calculate the half-life of a radioactive isotope. They compute the quantity of radioactive atoms still present at various time points and graph the data. Time is represented on the x-axis, while the quantity of radioactive atoms left is shown on the y-axis.

In an exponential decay curve, the remaining number of atoms declines quickly at initially before gradually levelling out on the graph. The amount of time it takes for half of the radioactive atoms in a sample to decay is known as the half-life. Scientists seek for the point on the graph where the number of atoms left is half of what it was to establish the half-life. They pick out this location on the graph and calculate the associated time. The radioactive isotope's half-life is represented by this period of time.

Similar to your example with the dimes, you could display this data on a graph if you were to run an experiment in which you threw a lot of dimes and counted how many were still there after each toss. The x-axis would show the number of tosses, and the y-axis would show how many dime are left.

If the surviving dimes follow a pattern similar to radioactive decay, you would notice a steady decline in their quantity over time. Even if the graph isn't a perfect exponential decay curve, you can still locate a point where the number of dimes left is half of what they were initially. The "half-life" of the dimes in your experiment would be represented by the corresponding number of throws at this stage.

You can comprehend the general idea of finding the time or number of events necessary for a quantity to reduce to half of its initial value by drawing a comparison between figuring out the half-life of radioactive isotopes and the fictitious scenario of figuring out the "half-life" of dimes.

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- In reactions 1 and 2, the highlighted atom (oxygen) is reduced.

- In reaction 3, the highlighted atom (sulfur) is neither oxidized nor reduced.

- In reaction 4, the highlighted atom (oxygen) is reduced.

In the given chemical reactions, we need to identify whether the highlighted atom is being oxidized or reduced. Let's analyze each reaction individually:

Reaction 1: 4 HF (g) + SiO₂ (s) → SiF₄ (g) + 2 H₂O (g)

In this reaction, the highlighted atom is oxygen (O). Oxygen in SiO₂ undergoes a change in oxidation state from -2 to 0 in SiF₄. Therefore, the highlighted atom (oxygen) is reduced.

Reaction 2: 2 Cl (aq) + 2 H₂O (l) → 2 OH (aq) + H₂ (g) + Cl₂ (g)

In this reaction, the highlighted atom is oxygen (O). Oxygen in H₂O undergoes a change in oxidation state from -2 to -1 in OH. Therefore, the highlighted atom (oxygen) is reduced.

Reaction 3: H₂S (aq) + 2 NaOH (aq) → Na₂S (aq) + 2 H₂O (l)

In this reaction, the highlighted atom is sulfur (S). Sulfur in H₂S undergoes a change in oxidation state from -2 to -2 in Na₂S. Therefore, the highlighted atom (sulfur) is neither oxidized nor reduced.

Reaction 4: 2 H₂ (g) + O₂ (g) → 2 H₂O (g)

In this reaction, the highlighted atom is oxygen (O). Oxygen in O₂ undergoes a change in oxidation state from 0 to -2 in H₂O. Therefore, the highlighted atom (oxygen) is reduced.

To summarize:

- In reactions 1 and 2, the highlighted atom (oxygen) is reduced.

- In reaction 3, the highlighted atom (sulfur) is neither oxidized nor reduced.

- In reaction 4, the highlighted atom (oxygen) is reduced.

It's important to note that oxidation and reduction involve changes in the oxidation state of atoms, indicating the gain or loss of electrons. The analysis above is based on the change in oxidation state of the highlighted atom in each reaction.

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Answer:

Like charges repel each other

Explanation:

two objects that carry the same type of electric charge will exert a force of repulsion on each other.

for example, if 2 north poles of a magnet are put together they will repel each other

hope this helps!

(i) A mole is the amount of a substance containing 6.022 × 10²³ particles.

(ii) Compounds are formed by the chemical bonding of different elements.

(iii) Molecular mass is the sum of atomic masses in a molecule.

(iv) There are two types of mix homogeneous and heterogeneous mixtures.

(v) Free radicals are highly reactive species with unpaired electrons.

(vi) Gram formula mass is the sum of atomic masses expressed in grams per mole.

(i) Mole: A mole is defined as the quantity of a substance that has an equal number of particles that are in 12 grams of carbon-12.

It is denoted by mol and is used in chemistry to measure quantities of atoms or molecules.

Example:1 mol of oxygen gas contains 6.022 × 10²³ oxygen molecules.

(ii) Compounds: Compounds are substances made up of two or more different elements that are chemically bonded together.

They can be broken down into simpler substances through chemical reactions.

Example:Water (H2O) is a compound made up of two hydrogen atoms and one oxygen atom that are chemically bonded together.

(iii) Molecular Mass: The molecular mass of a compound is the sum of the atomic masses of all the atoms in the molecule.

It is expressed in atomic mass units (amu) or grams per mole.

Example:The molecular mass of water (H2O) is 18.015 amu.

(iv) Types of Mixture: A mixture is a combination of two or more substances that are not chemically combined.

There are two types of mixtures - homogeneous and heterogeneous.

Example:A homogeneous mixture is a solution, such as saltwater, where the solute (salt) is evenly distributed in the solvent (water).

A heterogeneous mixture is a mixture that is not evenly distributed, such as oil and water.

(v) Free Radical: A free radical is an atom or molecule that has an unpaired electron in its outer shell and is highly reactive. They can be both harmful and helpful to the human body.

Example:A common free radical is the hydroxyl radical (OH·) that is formed by the body during metabolism.

(vi) Gram formula mass: The gram formula mass is the sum of the atomic masses of all the atoms in a compound.

It is expressed in grams per mole and is used to determine the mass of a certain number of molecules or atoms.

Example:The gram formula mass of water (H2O) is 18.015 g/mol.

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Answer:

ok, here is your answer

Explanation:

To find the number of moles of water that can be produced when 112.6 grams of HNO3 are consumed, we need to first convert the given mass of HNO3 to the number of moles using its molar mass:

Molar mass of HNO3 = 1(1) + 1(14) + 3(16) = 63 g/mol

Number of moles of HNO3 = mass/molar mass = 112.6 g/63 g/mol = 1.79 mol

From the balanced chemical equation, we can see that for every 6 moles of HNO3, 2 moles of H2O are produced. Therefore, we can use a mole ratio to find the number of moles of water produced when 1.79 moles of HNO3 are consumed:

6 moles HNO3 : 2 moles H2O

1.79 moles HNO3 : x moles H2O

x moles H2O = (2 moles H2O/6 moles HNO3) * 1.79 moles HNO3

x moles H2O = 0.5963 moles H2O

Finally, we can convert the number of moles of water to the mass of water using its molar mass:

Molar mass of H2O = 2(1) + 1(16) = 18 g/mol

Mass of water = number of moles * molar mass = 0.5963 mol * 18 g/mol = 10.73 g

Therefore, 10.73 grams of water can be produced when 112.6 grams of HNO3 are consumed.

When 112.6 grams of HNO3 are consumed, approximately 0.5957 moles of water can be produced.

To determine the number of moles of water produced when 112.6 grams of HNO3 are consumed, we need to use the molar mass of HNO3 and the stoichiometric coefficients of the balanced equation.

The molar mass of HNO3 is calculated as follows:

H = 1 g/mol

N = 14 g/mol

O = 16 g/mol (three oxygen atoms in HNO3)

Molar mass of HNO3 = (1 * 1) + 14 + (16 * 3) = 63 g/mol

Now, let's calculate the number of moles of HNO3:

moles of HNO3 = mass of HNO3 / molar mass of HNO3

moles of HNO3 = 112.6 g / 63 g/mol ≈ 1.787 moles

From the balanced equation: 5 + 6 HNO3 -> H2SO4 + 6 NO2 + 2 H2O

we can see that for every 6 moles of HNO3 consumed, 2 moles of water are produced.

Therefore, we can use a ratio to find the number of moles of water:

moles of water = (moles of HNO3 * 2) / 6

moles of water = (1.787 moles * 2) / 6 ≈ 0.5957 moles

Thus, when 112.6 grams of HNO3 are consumed, approximately 0.5957 moles of water can be produced.

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Answer:

It is not possible to determine the half-life of the cubes solely from the data in Table 9.1.

To determine the half-life, we would need data showing the decay of the cubes over time. The second data set and graph for the trials that started with only cubes were necessary to observe the decay and calculate the half-life. By analyzing the decay curve, the time it takes for half of the cubes to decay can be determined.

The half-life of the cubes cannot be determined from the data in Table 9.1 alone. The second data set and graph were crucial to measure the decay and calculate the half-life accurately. This additional information provides a clearer understanding of the decay process and allows for the determination of the time it takes for the cubes to decay by half.

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Answer: troposphere.

Explanation: the troposphere is 3/4 of the atmosphere and ozone layer acts as greenhouse gas and affects global warming

Answer:

troposphere

Explanation:

In the troposphere, greenhouse gases, such as carbon dioxide (CO2), methane (CH4), and water vapor (H2O), absorb and re-emit infrared radiation, effectively trapping heat near the Earth's surface. This trapped heat contributes to the warming of the troposphere and is responsible for the greenhouse effect.

The researcher initially added approximately 13 g of KF to the water.

First, you calculate the amount of heat absorbed by the water when the KF dissolves.

We can use the equation:

q = mCΔT

We can assume that the KF has a negligible heat capacity and does not contribute to the heat absorption.

Using the given values, we have:

q = mCΔT

q = (238 g - mKf)(4.184 J/g·°C)(31.51°C - 18.98°C)

q = (238 g - mKf)(86.05 J/g)

Next, we can use the fact that the amount of KF added is equal to the amount of KF dissolved in the water.

We can also assume that the KF fully dissociates into K+ and F- ions when it dissolves:

KF(s) → K+(aq) + F-(aq)

Therefore, the moles of K+ ions are equal to the moles of KF added to the water.

We can use the following equation to calculate the moles of K+ ions:

moles K+ = mass KF / molar mass KF

where the molar mass of KF is 58.10 g/mol.

Since the solution is electrically neutral, the moles of F- ions are also equal to the moles of KF added.

Next, we can use the following equation to calculate the concentration of the ions in the solution:

C = moles / volume

Assuming that the volume of the solution is equal to the mass of the solution (since the density of water is close to 1 g/mL), we have:

C K+ = moles K+ / (238 g - mKf)

C F- = moles F- / (238 g - mKf)

Finally, we can use the fact that KF is a strong electrolyte and dissociates completely in water to write the following equation for the heat absorbed during the dissolution:

q = (moles K+) ΔHfus + (moles K+)ΔHsol + (moles F-)ΔHsol

The heat of fusion of KF is 10.8 kJ/mol, and the heat of solution is -155.2 kJ/mol.

Substituting all the equations and values above, we get:

(238 g - mKf)(4.184 J/g·°C)(31.51°C - 18.98°C) =

((mKf / 58.10 g/mol) ΔHfus + (mKf / 58.10 g/mol)ΔHsol + (mKf / 58.10 g/mol)ΔHsol) + ((238 g - mKf) (mKf / 58.10 g/mol) (-155.2 kJ/mol))

Simplifying this equation gives:

mKf ≈ 13 g

Therefore, the researcher initially added approximately 13 g of KF to the water.

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Answer:

B - convergent destructive (subduction)

Explanation:

option A is not a plate boundary. A constructive plate boundary moves away from eachother and a convergent collision boundary would suggest no slipping whereas a convergent destructive boundary has subduction take place which is when a denser plate slips under a less dense plate.

Explanation:

To solve this problem, we can use the equation:

ΔG° = -RT ln(K)

where ΔG° is the standard Gibbs free energy change, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and K is the equilibrium constant.

First, we need to find the value of K. We can use the equation:

K = (P_B × P_C) / (P_A)^2

where P_B and P_C are the partial pressures of B and C, respectively, and P_A is the partial pressure of A.

We are given that the total pressure is 9.72 atm and the partial pressure of A is 5.71 atm. Therefore, we can calculate the partial pressures of B and C:

P_B = (2 × P_A) / (2 + 1) = 3.81 atm

P_C = (1 × P_A) / (2 + 1) = 1.90 atm

Substituting these values into the equation for K, we get:

K = (3.81 × 1.90) / (5.71)^2 = 0.246

Now we can use this value of K to calculate ΔG°:

ΔG° = -RT ln(K)

ΔG° = -(8.314 J/mol·K)(298 K) ln(0.246)

ΔG° = -8,870 J/mol or -8.87 kJ/mol

Therefore, the value of the standard Gibbs free energy change for this reaction at 25°C is -8.87 kJ/mol.

The law that relates to the ideal gas law is P₁V₁/T₁ = P₂V₂/T₂.

This is known as Boyle's law, and it states that the pressure of a gas is inversely proportional to its volume at a constant temperature.

The other laws you mentioned are also gas laws, but they do not relate to the ideal gas law. Charles' law states that the volume of a gas is directly proportional to its temperature at a constant pressure. Avogadro's law states that the volume of a gas is directly proportional to the number of moles of gas at a constant pressure and temperature. Gay-Lussac's law states that the pressure of a gas is directly proportional to its temperature at a constant volume.

The ideal gas law is a combination of Boyle's law, Charles' law, Avogadro's law, and Gay-Lussac's law. It can be expressed as follows:

PV = nRT

where:

P = pressure of the gas (in Pa)

V = volume of the gas (in m³)

n = number of moles of gas

R = universal gas constant (8.314 J/mol K)

T = temperature of the gas (in K)

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The value of Kc for the given reaction is 2.70.

The value of Kc for the given reaction can be determined by using the equilibrium concentrations of the reactants and products.

Before that, let us understand what Kc is.

Kc is the equilibrium constant for a chemical reaction in terms of concentration.

It is calculated by dividing the product of the equilibrium concentrations of the products raised to their stoichiometric coefficients by the product of the equilibrium concentrations of the reactants raised to their stoichiometric coefficients.

Kc = [products]^coefficients/[reactants]^coefficients

The given reaction is:N2(g) + 3 H2(g) ⇌ 2 NH3(g).

The balanced chemical equation tells us that two moles of NH3 are produced from one mole of N2 and three moles of H2, so the stoichiometric coefficients are 1 for N2, 3 for H2, and 2 for NH3.

The equilibrium concentrations of N2, H2, and NH3 are [N2]eq = 2.66 M, [H2]eq = 0.64 M, and [NH3]eq = 3.34 M respectively.

Substituting these values in the expression for Kc, we get:Kc = [NH3]^2/[N2][H2]^3Kc = (3.34 M)^2/((2.66 M)(0.64 M)^3)Kc = 2.70

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Types of luminous flames:

1. Yellow Luminous Flame

2. Smoky Luminous Flame

3. Orange Luminous Flame

4. Blue Luminous Flame

Luminous flames are characterized by their visible glow, which is caused by the incomplete combustion of fuel. The presence of soot particles in the flame causes the emission of light. There are different types of luminous flames, which can be classified based on their fuel composition and burning conditions. Here are some common types of luminous flames:

1. Yellow Luminous Flame: This is the most common type of luminous flame, often seen in open fires, candles, and gas stoves. It appears yellow due to the presence of soot particles in the flame. Yellow flames indicate incomplete combustion of hydrocarbon fuels, such as methane, propane, or natural gas. The high carbon content in these fuels leads to the formation of soot, which emits visible light.

2. Smoky Luminous Flame: This type of flame is characterized by a significant amount of black smoke and soot production. It is commonly observed in poorly adjusted or malfunctioning burners or engines. The excessive presence of unburned fuel in the flame results in incomplete combustion and the emission of dark smoke particles.

3. Orange Luminous Flame: An orange flame indicates a higher combustion temperature compared to a yellow flame. It is often seen in more efficient burners or when burning fuels with a higher carbon content, such as oil or diesel. The higher temperature helps in burning more of the carbon particles, reducing the amount of soot and making the flame appear less yellow.

4. Blue Luminous Flame: A blue flame is typically associated with complete combustion. It indicates efficient burning of fuel, resulting in minimal soot formation. Blue flames are commonly observed in gas burners or Bunsen burners. The blue color is a result of the combustion of gases, such as methane, in the presence of sufficient oxygen.

It's important to note that the luminosity of a flame can vary depending on factors such as fuel-air mixture, combustion temperature, and the presence of impurities. Achieving complete combustion and minimizing the production of soot is desirable for efficient and cleaner burning processes.

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Answer:

[Xe]: 6s²

Explanation:

that's the noble gas config for Barium.

The electron configuration of barium (Ba) in noble gas notation is [Xe] 6s². This shorthand notation signifies that barium has chemically stable electron configuration similar to that of the noble gas xenon (Xe), plus two additional electrons in the 6s orbital.

The electron configuration of an element represents the distribution of electrons in an atom's electron shells. Barium (Ba) is a chemical element with atomic number 56 in the periodic table. Its electron configuration can be written out in full, but the noble-gas notation provides a convenient shorthand. To write the electron configuration of barium in noble gas notation, you first locate the nearest noble gas that precedes barium in the periodic table. In this case, the nearest noble gas is xenon (Xe), which has an atomic number of 54. This leaves two more electrons, which go into the 6s orbital. So, the electron configuration of barium in noble gas notation is [Xe] 6s².

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Answer:

To find the percentage of SiO2 in the sample mixture, we need to compare the mass of SiO2 to the total mass of the sample mixture and then calculate the percentage.

Given:

Mass of SiO2 recovered = 1.20 g

Total mass of the sample mixture = 2.0 g

To find the percentage of SiO2, we can use the following formula:

Percentage of SiO2 = (Mass of SiO2 recovered / Total mass of the sample mixture) * 100

Substituting the given values:

Percentage of SiO2 = (1.20 g / 2.0 g) * 100

Calculating:

Percentage of SiO2 = 0.60 * 100

Percentage of SiO2 = 60%

Therefore, the percentage of SiO2 in the 2.0 g sample mixture is 60%.

128.4 grams of carbon are required to produce 10.7 moles of methane.

The balanced chemical equation for the reaction C + 2H2 - CH4 is as follows:C + 2H2 ⟶ CH4.

In this equation, we can see that for one mole of methane (CH4), one mole of carbon (C) and two moles of hydrogen (H2) are required.

The molar mass of carbon is 12 g/mol.

The molar mass of methane is the sum of the molar masses of its constituent atoms, which is 12 g/mol (for carbon) + 4(1 g/mol) (for hydrogen) = 16 g/mol.

To find the amount of carbon required to produce 10.7 moles of methane, we can use the following proportion:1 mole CH4 : 1 mole C

Therefore,10.7 moles CH4 : x moles C.

Thus,x = 10.7 moles C.

Since we know the molar mass of carbon (12 g/mol), we can convert the moles of carbon to grams:mass = moles × molar massmass of carbon required = 10.7 moles × 12 g/mol= 128.4 g.

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In a combustion reaction, a substance reacts with oxygen to produce carbon dioxide and water. In this case, the hydrocarbon C10H8 (often known as naphthalene) is reacting with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O).

Therefore, the correct answer is: Combustion.

The options "synthesis" and "decomposition" do not apply to this particular reaction since synthesis involves the combination of simpler substances to form a more complex one, and decomposition involves the breakdown of a compound into simpler substances.

The balanced chemical equation for the reaction C + 2H2 → CH4 tells us that one mole of carbon produces one mole of methane. So, if 10.7 moles of methane are produced, then 10.7 moles of carbon are required to produce it.

To find the mass of carbon required to produce 10.7 moles of methane, we can use the molar mass of carbon (12 g/mol) and the following calculation:

Mass of carbon = Number of moles of carbon × Molar mass of carbon

Mass of carbon = 10.7 mol × 12 g/mol

Mass of carbon = 128.4 g

Therefore, 128.4 grams of carbon are required to produce 10.7 moles of methane, CH4.

Answer:

128.40g

Explanation:

Grams of carbon = moles of carbon * molar mass of carbon

Grams of carbon = 10.7 moles * 12 g/mol

3.574 moles of water can be made when 112.6 grams of HNO3 are consumed.

The balanced equation is5 + 6 HNO3 → H2SO4 + 6 NO2 + 2H2OOne mole of HNO3 weighs 63.02 grams.

Therefore, 112.6 grams of HNO3 would be:

112.6 g ÷ 63.02 g/mol = 1.787 moles HNO3

According to the balanced equation, 2 moles of H2O are produced for every mole of HNO3.

Therefore, the moles of H2O that can be produced from 1.787 moles of HNO3 are:

1.787 moles HNO3 × 2 moles H2O / 1 mole HNO3 = 3.574 moles H2O

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Answer:

We need more? What else is in the question? This is unanswerable.

Explanation: