The statements in the tables below are about two different chemical equilibria. The symbols have their usual meaning, for example AG Gibbs free energy of reaction and stands for the equilibrium constant. stands for the standard In each table, there may be one statement that is false because it contradicts the other three statements. If you find a false statement, check the box next to It. Otherwise, check the "no false statements box under the table. statement false? statement false? Ink>0
AH° R<1
AG'>0
AG'>0
In > AH">TAS
no false statements

Answer:

2

Explanation:

BIonary bi means 2

i hope this helps :)

Answer:

The option where the atoms are furthest apart.

Explanation:

Gasses are the most energetic of the three basic states of matter, their atoms have more kinetic energy than either solids or liquids and will therefore also have the most spread out atoms.

Answer:

The law of multiple proportions states that whenever the same two elements form more than one compound, the different masses of one element that combine with the same mass of the other element are in the ratio of small whole numbers.

Answer:

0, 0, 0, +1, 0

Explanation:

Because the compound has a +1 charge overall, the sum of the formal charges must equal +1.  The correct answer is the Lewis structure with as many atoms as possible having the formal charge of 0 with just a single atoms having the +1 formal charge.

Answer:

lungs

Explanation:

The standard entropy change for the given reaction is -319.1 J/K.

The standard entropy change (ΔS°) for a reaction can be determined by calculating the difference between the sum of the standard entropies of the products and the sum of the standard entropies of the reactants.

In this reaction, the reactants are CH4 (methane) and O2 (oxygen gas), and the products are CO2 (carbon dioxide gas) and H2O (water vapor).

Using the standard entropy values from the tables, the sum of the standard entropies of the reactants is 186.3 J/K, and the sum of the standard entropies of the products is 505.4 J/K.

ΔS° = (ΣS° products) - (ΣS° reactants)

    = 505.4 J/K - 186.3 J/K

    = -319.1 J/K

Therefore, the standard entropy change for this reaction is -319.1 J/K. A negative value indicates that the reaction results in a decrease in entropy, meaning the products have less disorder or randomness compared to the reactants.

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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:

C: The atomic nucleus contains most of the mass.

Explanation:

the sum of PE and KE is mechanical energy this means energy during motion and position I think the answer is motion and stored or it may be internal

Answer:

a

Explanation:

i did the test

A 12.1 M HCl solution with a density of 1.19 g/mL has a molality of 16.1 m.

Molality is defined as the total moles of a solute contained in a kilogram of a solvent.

Let's suppose we have 1 L of the solution.

The solution is 12.1 M, that is, there are 12.1 moles of HCl in 1 L of solution.

The molar mass of HCl is 36.465 g/mol.

12.1 mol × 36.465 g/mol = 441 g

The density of the solution is 1.19 g/mL.

1000 mL × 1.19 g/mL = 1190 g

In 1190 g of solution, there are 441 g of HCl.

mWater = 1190 g - 441 g = 749 g = 0.749 kg

b = 12.1 mol / 0.749 kg = 16.1 m

A 12.1 M HCl solution with a density of 1.19 g/mL has a molality of 16.1 m.

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Answer:

Wind energy

Explanation:

Answer: Wind energy

Explanation:

Answer:

The energy of each transition is approximately \(1.98\times 10^{-19}\; \rm J\).

The frequency of photons released in such transitions is approximately \(3.00\times 10^{14}\; \rm Hz\).

Explanation:

The Rydberg Equation gives the wavelength (in vacuum) of photons released when the electron of a hydrogen atom transitions from one main energy level to a lower one.

The Rydberg Equation gives the following relation:

\(\displaystyle \frac{1}{\lambda_\text{vac}} = R_\text{H} \cdot \left(\frac{1}{{n_2}^2}} -\frac{1}{{n_1}^2}\right)\).

Rearrange to obtain and expression for \(\lambda_\text{vac}\):

\(\displaystyle \lambda_\text{vac} = \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)}\).

In this question, \(n_1 = 7\) while \(n_2 = 3\). Therefore:

\(\begin{aligned} \lambda_\text{vac} &= \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)} \\ &\approx \frac{1}{\displaystyle 1.09678 \times 10^{7}\; \rm m^{-1} \cdot \left(\frac{1}{3^2} - \frac{1}{7^2}\right)} \approx 1.0 \times 10^{-6}\; \rm m \end{aligned}\).

Note, that \(1.0\times 10^{-6}\; \rm m\) is equivalent to \(1000\; \rm nm\). That is: \(1.0\times 10^{-6}\; \rm m = 1000\; \rm nm\).

Look up the speed of light in vacuum: \(c \approx 3.00\times 10^{8}\; \rm m \cdot s^{-1}\). Calculate the frequency of this photon:

\(\begin{aligned} f &= \frac{c}{\lambda_\text{vac}} \\ &\approx \frac{3.00\times 10^{8}\; \rm m\cdot s^{-1}}{1.0\times 10^{-6}\; \rm m} \approx 3.00 \times 10^{14}\; \rm Hz\end{aligned}\).

Let \(h\) represent Planck constant. The energy of a photon of wavelength \(f\) would be \(E = h \cdot f\).

Look up the Planck constant: \(h \approx 6.62607 \times 10^{-34}\; \rm J \cdot s\). With a frequency of \(3.00\times 10^{14}\; \rm Hz\) (\(1\; \rm Hz = 1\; \rm s^{-1}\),) the energy of each photon released in this transition would be:

\(\begin{aligned}E &= h \cdot f \\ &\approx 6.62607 \times 10^{-34}\; \rm J\cdot s^{-1} \times 3.00 \times 10^{14}\; \rm s^{-1} \\ &\approx 1.98 \times 10^{-19}\; \rm J\end{aligned}\).

The energy of the transition between n = 7 and n = 3  is 1.96 × 10^-19 J while the frequency is 3 × 10^14 Hz.

Using the Rydberg Equation for energy;

ΔE = -RH(1/n^2final - 1/n^2initial)

Given that;

nfinal = 3

ninitial = 7

RH = 2.18 × 10^-18 J

ΔE = - 2.18 × 10^-18(1/3^2 - 1/7^2)

ΔE = - 2.18 × 10^-18(0.11 - 0.02)

ΔE = - 1.96 × 10^-19 J

For the second part;

Since the wavelength is 1000nm, we have;

λ = 1000nm

c = 3 × 10^8 m/s

f = ?

c = λf

f = c/λ

f = 3 × 10^8 m/s/1000 × 10^-9 m

f = 3 × 10^8 m/s/ 1 × 10^-6 m

f = 3 × 10^14 Hz

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Answer:

Explanation: Liquid Chromatography  

I'm sorry if i'm wrong

Thermal Energy is directly proportional to the temperature

Or

\(\\ \sf\longmapsto Thermal\: Energy\propto Temperature \)

So if temperature increases the thermal Energy also increases.

An object which is 20°C warmer than other has more thermal energy

The molar mass of Be is 9.01 g/mol. Atoms in 18.02 g of Be is : D.) 1.204 x 1024 atoms Be

The mass in grams of one mole of the compound is known as molar mass of a substance.

As moles of Be = mass of Be / molar mass of Be

moles of Be = 18.02 g / 9.01 g/mol

moles of Be = 2 mol

number of atoms of Be = moles of Be x Avogadro's number

number of atoms of Be = 2 mol x 6.022 x 10²³

So, number of atoms of Be = 1.204 x 10²⁴

Therefore, the answer is option D, 1.204 x 10²⁴ atoms Be.

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Answer:

32

Explanation:

because the number of heliem is your aswer

Answer:

so the balloon can fill up with air. If the balloon pops, then the air goes away.

The answer responses to  the structures shown in the diagram are:

A. chromosomes

C. They would be the same.

B. They are in pairs because each one comes from a different parent.

The chromosomes are in pairs because humans have a diploid number of chromosomes, meaning they have two sets of chromosomes, one inherited from each parent.

The nucleus is important in eukaryotic cells and has many important parts that help the cell work properly. There are some parts inside cells called the nuclear membrane, nucleoplasm, nucleolus, and chromatin. Chromatin is made up of DNA and other proteins.

Every part of a person's body has the same genes, but the way they are organized can be different in different types of cells. The chromosomes in our skin cells might not be the same as the chromosomes in our muscle cells, even if they come from the same person.

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Identify the structures shown.

A. chromosomes

B. mitochondria

C. nuclei

D. vacuoles

C

Describe how the structures from this cell would compare to the structures in the nucleus of another body cell from the same person.

A. There would be longer.

B. They would be shorter.

C. They would be the same.

D. They would be different.

Describe how the structures from this cell would compare to the structures in the nucleus of another body cell from the same person.

A. There would be longer.

B. They would be shorter.

C. They would be the same.

D. They would be different.

Explain why the structures are in pairs.

A. They aren't in pairs.

B. They are in pairs because each one comes from a different parent.

C. This cell is making a copy of itself.

D. The cell always has 2 copies in case 1 is damaged.

To find the number of moles of a given solution, multiply the volume of the solution (in liters) by the molarity of the solution (in molarity). In this case, the volume of the solution is 1.5 liters and the molarity is 0.77 molarity.

Molarity is a measure of concentration of a solute in a solution expressed as the number of moles of solute per liter of solution (mol/L). It is a commonly used unit of concentration in chemistry. Molarity is often used to measure the amount of a solute in a given volume of solution and is also used to calculate the amount of a solute needed to create a given volume of solution at a given concentration.

Moles of KCl = 1.5 L x 0.77 M = 1.155

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Answer: is is true

Explanation:

Answer: 0.056

Explanation:

The molarity of the solution is the ratio of the number moles of solute to the volume of solution. Hence, the molarity of the given solution is 0.056 M. Thus, option D is correct.

Molarity is a common term used to express the concentration of a solution . It is the ratio of the number of moles of solute to the volume of solution in liters.

Given that,

volume of HCl solution  = 9 L

no. of moles of HCl = 0.50 moles.

Then,

molarity = no.of moles/volume in L

M = 0.50 Moles/9 L

   = 0.056 M.

Therefore, the molarity of the solution is 0.056 M.

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287.13 grams of Al₂O₃ were decomposed. Therefore, option A is correct.

The term molar mass is defined as the mass in grams of one mole of the compound. A mole of any substance is 6.022 × 10²³ molecules. It is called as Avogadro's number. The molar mass is a bulk property, it is not molecular, property of a substance.

2Al₂O₃ ⇒ 4Al + 3O₂

Thus, 4 mol of Al combine with 3 mol of oxygen to form 2 mol of Al₂O₃.

2 mol of Al corresponds to 2 × 27 = 54g

Thus, the weight of Al used in the reaction is 108 g.

Molar mass of Al₂O₃ is 101.96g/mol.

5.63 moles of Al = 5.63 × 54 / 101.96

= 287.13 grams

Thus, 287.13 grams of Al₂0₃ were decomposed, option A is correct.

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The significance of the point where the pressure drops to zero is that at that point the gases would completely show ideal behavior.

The gas laws can be used to describe the behavior of the ideal gases. It is pertinent to note that the ideal gas law can strictly be applied to gases that are at a high temperature and low pressure.

We can be able to find the  temperature when the pressure is equal to zero either be the use of the equation line or by experiment. In that case, we would be able to obtain the point at which the pressure drops to zero.

The significance of the point where the pressure drops to zero is that at that point the gases would completely show ideal behavior.

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Answer:

\(\% m/m=10.1\%\)

Explanation:

Hello,

In this case given the molal solution of sucrose, we can assume there are 0.329 moles of sucrose in 1 kg of solvent, thus, computing both the mass of sucrose and solvent in grams, we obtain:

\(m_{sucrose}=0.329mol*\frac{342g}{1mol}=112.5g\)

\(m_{solvent}=1000g\)

In such a way, we proceed to the calculation of the mass percent as follows:

\(\% m/m=\frac{112.5g}{112.5g+1000g}*100\%\\ \\\% m/m=10.1\%\)

Regards.

Answer:

Mass = 204 g

Explanation:

Given data:

Mass of ammonia produced = ?

Number of moles of nitrogen = 6 moles

Solution:

Chemical equation;

N₂ + 3H₂    →    2NH₃

now we will compare the moles of nitrogen and ammonia.

               N₂        :          NH₃

                1          :            2

               6          :          2/1×6 = 12

Mass of ammonia;

Mass = number of moles × molar mass

Mass = 12 mol × 17 g/mol

Mass = 204 g

From the specific heat capacity of mercury, the amount of heat energy required to raise the temperature of 4.0 g of mercury metal from 9.3 °C to 83.0 °C is 77.792 J.

The specific heat capacity of a substance is the amount of heat required to raise the temperature of a unit mass of the substance by one degree Celsius or kelvin.

The specific heat capacity of a substance is a constant that can be used to calculate the amount of heat required to raise the temperature of a given mass of a substance to any temperature.

The specific heat capacity of mercury is 0.140 J/g/k.

The formula for calculating specific heat capacity is given below:

Specific heat capacity, c = Δq/mΔT

where;

Δq = heat change

m = mass of the substance

ΔT = temperature change

The Heat required, Δq, will then be:

Δq = m * c * ΔT

Heat required, Δq = 4.0 * 0.140 * (83.0 - 9.3)

Heat required, Δq = 77.792 J

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Answer:

thermodynamic deals with the heat and temperature and their relationship with energy

Explanation:

i hope this will help you

Answer:  I Hope It Helps :)

Explanation:

P1 proton in the compound appear most upfield in the 1H NMR spectrum.

What is the main principle of NMR?

The foundation of NMR is the idea that all nuclei are electrically charged and that many of them have spins. Energy can go from the base energy to a higher energy level if an external magnetic field is provided.

The protons are most shielded ones among all the four types of protons as they experience +I effect from adjacent CH2 groups and are also situated far away from carbonyl group. therefore the -I effect is very less compared to the protons 2, 3 and 4. all these factors makes the proton 1 the most shielded and hence they will have very low delta value that is upfield.

Therefore, P1 proton in the compound appear most upfield in the 1H NMR spectrum.

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Answer:

\(\rho =1.96\frac{g}{L}\)

Explanation:

Hello there!

In this case, since this imaginary gas can be modelled as an ideal gas, we can write:

\(PV=nRT\)

Which can be written in terms of density and molar mass as shown below:

\(\frac{P}{RT} =\frac{n}{V} \\\\\frac{P}{RT} =\frac{m}{MM*V}\\\\\frac{P*MM}{RT} =\frac{m}{V}=\rho\)

Thus, by computing the pressure in atmospheres, the resulting density would be:

\(\rho = \frac{165/760 atm * 314.2 g/mol}{0.08206\frac{atm*L}{mol*K}*425K} \\\\\rho =1.96\frac{g}{L}\)

Best regards!

Answer: atomic mass, atomic number

Explanation:

Answer:

Option D: Cations

Explanation:

Do you need only one answer .