Think of a substance that does not dissolve in water. Any Examples?

Answer:

22.5 Km

Explanation:

45km/hr * 0.5 hr = 22.5km

A single electron from the outermost shell of this element moves to the outermost shell of an atom with seven electrons.

When forming ions elements typically gain or lose the minimum number of electrons required to make a complete octet. For example, fluorine has 7 valence electrons, so it is most likely to gain an electron and form an ion of charge At the outermost energy level he states 8 electrons must be reached for an atom to be stable. Particles smaller than an atom.

Ionization is the process by which ions are formed by gaining or losing electrons from atoms or molecules. When an atom or molecule receives an electron it becomes a negatively charged anion and when it loses an electron it becomes a positively charged cation. Energy can be lost or gained in the formation of ions.

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Answer:

A

Explanation:

Because ethanol consists C H and O formula C2H5OH

Formula of salt is NaCl

Formula of water is H2O

But fórmula of Aluminum is just Al

Answer:

Inorganic chemical reactions involve compounds without carbon atoms. Some of these chemical reactions emit light as the chemical compounds react. When light is emitted as a result of the chemical reaction, the reaction is said to be luminescent.

Explanation:

Answer:

Chemiluminescence is the production of light from a chemical reaction. Two chemicals react to form an excited (high-energy) intermediate, which breaks down releasing some of its energy as photons of light

Okay, let's solve this step-by-step without using quadratics:

1) The equilibrium constant Kc = 0.36 means the equilibrium lies to the left. So there will be more N2O4 than NO2 at equilibrium.

2) The initial concentration of N2O4 is 0.1 mol dm^-3. Let's call this [N2O4]initial.

3) At equilibrium, the concentrations of N2O4 and NO2 will be [N2O4]equil and [NO2]equil respectively.

4) We know the equilibrium constant expression for this reaction is:

Kc = ([NO2]equil)^2 / [N2O4]equil

5) Setting this equal to 0.36 and plugging in 0.1 for [N2O4]initial, we get:

0.36 = ([NO2]equil)^2 / (0.1 - [NO2]equil)

6) Simplifying, we get:

0.036 = [NO2]equil^2

7) Taking the square root of both sides, we get:

[NO2]equil = 0.06 mol dm^-3

So the equilibrium concentration of NO2 is 0.06 mol dm^-3.

Let me know if you have any other questions! I can also provide a more step-by-step explanation if needed.

The substance in the image above would be classified as a mixture of elements (option E).

A compound is a substance formed by chemical bonding of two or more elements in definite proportions by weight.

On the other hand, a mixture is made when two or more substances are combined, but they are not combined chemically.

According to this question, an image is shown with two different substances or elements as distinguished by coloration (white and purple). These elements are combined but not chemically bonded, hence, is a mixture.

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1054.67 grams of calcium phosphate are theoretically produced if we start with 3.40 moles of ca(no3)2 and 2.40 moles of li3po4.

To determine the theoretical yield of calcium phosphate (Ca3(PO4)2) produced from 3.40 moles of Ca(NO3)2 and 2.40 moles of Li3PO4, we need to identify the limiting reactant and use stoichiometry.

First, we need to determine the moles of calcium phosphate produced from each reactant. The balanced equation for the reaction is:

3Ca(NO3)2 + 2Li3PO4 → Ca3(PO4)2 + 6LiNO3

From the equation, we can see that the molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3:1. Therefore, the moles of calcium phosphate produced from Ca(NO3)2 would be 3.40 moles.

Similarly, the molar ratio between Li3PO4 and Ca3(PO4)2 is 2:1. Therefore, the moles of calcium phosphate produced from Li3PO4 would be 2.40/2 = 1.20 moles.

Since the moles of calcium phosphate produced from Ca(NO3)2 (3.40 moles) are higher than those produced from Li3PO4 (1.20 moles), Ca(NO3)2 is the limiting reactant.

To calculate the mass of calcium phosphate, we can use the molar mass of Ca3(PO4)2, which is approximately 310.18 g/mol.

Mass of calcium phosphate = Moles of calcium phosphate × Molar mass

Mass of calcium phosphate = 3.40 moles × 310.18 g/mol

Mass of calcium phosphate ≈ 1054.67 grams

Therefore, theoretically, approximately 1054.67 grams of calcium phosphate would be produced when starting with 3.40 moles of Ca(NO3)2 and 2.40 moles of Li3PO4.

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Answer:

1. Matter is made of tiny particles. 2. There is empty space between the particles. 3. The particles are in constant motion.

Explanation:

Let's see

\(\\ \rm\Rrightarrow C_{12}H_{22}O_{11}\)

\(\\ \rm\Rrightarrow 12(12)+22(1)+11(16)\)

\(\\ \rm\Rrightarrow 144+22+176\)

\(\\ \rm\Rrightarrow 166+176\)

\(\\ \rm\Rrightarrow 342g/mol\)

Exergonic: Reactions with negative delta G°’ values are exergonic reactions that release free energy.Exergonic means that the reaction will occur spontaneously. The value of delta G°’ is less than zero.

Endergonic: When the delta G°’ value is positive, the reaction is endergonic. These are reactions that need to be powered by an external force to occur.

At equilibrium: When delta G°’ = 0, the reaction is at equilibrium. The forward and reverse reactions occur at the same rate, and the concentrations of reactants and products do not change. Thus, the reaction is at equilibrium.

Here is how the K’eq values should be matched with the appropriate delta G°’ values:

K'eq  a. 1 : 0kJ/mol  b. 10⁻⁵ : 22.84 kJ/mol c. 10⁴ : -28.53 kJ/mol  d. 10² : -5.69 kJ/mol  e. 10⁻¹ : 11.42 kJ/mol AG°' (kJ/mol)

Therefore, the correct match for K’eq values with the appropriate delta G°’ values are as follows:

K'eq = 10⁴ corresponds to delta G°’ = -28.53 kJ/mol, which is exergonic.

K'eq = 10⁻⁵ corresponds to delta G°’ = 22.84 kJ/mol, which is endergonic.

K'eq = 10² corresponds to delta G°’ = -5.69 kJ/mol, which is exergonic.

K'eq = 1 corresponds to delta G°’ = 0 kJ/mol, which is at equilibrium.

K'eq = 10⁻¹ corresponds to delta G°’ = 11.42 kJ/mol, which is endergonic.

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Answer:

Explanation:

A) As temperature is increased, Kinetic energy increases. As KE ∝ velocity, velocity increases, and thus, there will be more frequent collisions.

B) As temperature increases, heat energy applied is converted into kinetic energy. Thus, a greater proportion of particles will have sufficient energy to react, upon colliding.

C) Mass remains constant

Answer: (D) Thermal

Explanation: This is Thermal because it is the only one that makes sense  

Cyclops are heterotrophic

None of the above.

The energy difference between two energy levels in the hydrogen atom is given by the equation:

ΔE = E₂ - E₁ = -R_H(1/n₂² - 1/n₁²)

where ΔE is the energy difference, E₂ and E₁ are the energy levels, R_H is the Rydberg constant, and n₂ and n₁ are the principal quantum numbers.

In this case, the photon with a wavelength of 122 nm can excite the electron from n=1 to n=2. To determine if other photons can also excite the electron to the same level, we can use the equation:

λ = c / ν

where λ is the wavelength, c is the speed of light, and ν is the frequency of the photon.

Using the given wavelength values, we can calculate the corresponding frequencies and then determine if the energy difference matches the required energy to excite the electron from n=1 to n=2.

Calculating the frequencies and energy differences for the given wavelengths, we find:

For 244 nm:

ν = c / λ = (3.00 x 10^8 m/s) / (244 x 10^(-9) m) ≈ 1.23 x 10^15 Hz

ΔE = E₂ - E₁ = -R_H(1/n₂² - 1/n₁²) = -R_H(1/2² - 1/1²) = -R_H(1/4 - 1) = 3/4 R_H

For 61 nm:

ν = c / λ = (3.00 x 10^8 m/s) / (61 x 10^(-9) m) ≈ 4.92 x 10^15 Hz

ΔE = E₂ - E₁ = -R_H(1/n₂² - 1/n₁²) = -R_H(1/2² - 1/1²) = -R_H(1/4 - 1) = 3/4 R_H

For 94 nm:

ν = c / λ = (3.00 x 10^8 m/s) / (94 x 10^(-9) m) ≈ 3.19 x 10^15 Hz

ΔE = E₂ - E₁ = -R_H(1/n₂² - 1/n₁²) = -R_H(1/2² - 1/1²) = -R_H(1/4 - 1) = 3/4 R_H

Conclusion:

None of the calculated energy differences match the required energy difference to excite the electron from n=1 to n=2, which is 3/4 R_H. Therefore, none of the given wavelengths can be used to excite the electron from n=1 to n=2.

None of the above statements are true. The given wavelengths cannot be used to excite the electron from n=1 to n=2 in hydrogen.

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To find the concentration of hydroxide ion (OH-) in the aqueous solution of methylamine, we need to use the equilibrium expression for the reaction of methylamine with water:

CH3NH2 + H2O ⇌ CH3NH3+ + OH-

The equilibrium constant expression for this reaction is given by:

Kw = [CH3NH3+][OH-] / [CH3NH2]

We can assume that the concentration of [CH3NH3+] (methylammonium ion) is negligible compared to the initial concentration of CH3NH2. Therefore, we can simplify the equilibrium expression to:

Kw ≈ [OH-][CH3NH2]

Given that Kb (the base dissociation constant) for methylamine is 4.4 x 10^-4, we can write:

Kw = [OH-][CH3NH2] = Kb[CH3NH2]

Plugging in the values:

Kw = [OH-][0.050 M] = (4.4 x 10^-4)[0.050 M]

Now we can solve for [OH-]:

[OH-] = (4.4 x \(10^{-4} ^\))[0.050 M] / [0.050 M]

Canceling out the [0.050 M] terms:

[OH-] = 4.4 x \(10^{-4} ^\)

Therefore, the concentration of hydroxide ion in the solution is 4.4 x \(10^{-4} ^\)

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Answer:

Answers are C, D

Explanation:

Solid particles always stick together no matter what happens unless it is changing into a liquid. If you are talking about vibration for solid particles that only applies to thermal vibratio so not A. They do not slide past each other because they are packed very tightly together. They have a definite shape and volume because they stay together unless facing heat. Hopefully this helps you :)

pls mark brainlest ;)

The statements about the particles present in her model that are true;

C and D

Particles of solids are closely packed and this feature distinguishes a solid from other forms of matter ( liquid and gas ). Solids have a definite volume and shape due to their closely packed particles.

Vibration in solids occur when solids are under thermal pressure and from her skating activity, there was no form of thermal pressure therefore solids vibrating in place would not be included in her model.

Hence we can conclude that The statements about the particles present in her model that are true; they are packed very closely together and  They cause the solid to have a definite volume and definite shape

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Answer:

how are you doing, please kindly help me with my homework

Gold will have the highest final temperature because it has the lowest specific heat.

The lower the specific heat, the less heat that is needed to change the temperature. So, the heat will have cause the greatest temperature change for gold, since it has the smallest specific heat. The temperature change for platinum will be close to that of gold since their specific heats are so similar.

The chemical element gold has the atomic number 79 and the symbol Au (derived from the Latin aurum). As a result, it is among the naturally occurring elements with a higher atomic number. In its purest form, it is a bright, somewhat orange-yellow, dense, soft, malleable, and ductile metal. Gold is a transition metal and a group 11 element in terms of chemistry. One of the least reactive chemical elements, it is solid under normal circumstances.

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In this question, we have:

31.1 grams of O2

84.3 grams of F2

94.9 L of total volume

55.77°C of temperature which is equal to 328.92 K

Now, to find the pressure of this container, we can find the number of moles of each gas, and add both values together making it one value of moles and then we will use the Ideal gas law to find the pressure, so let's start with O2

The molar mass of O2 is 32g/mol and we have 31.1 grams

32g = 1 mol

31.1g = x moles

x = 0.972 moles of O2

Now for F2, the molar mass is 38g/mol, and we have 84.3 grams

38g = 1 mol

84.3g = x moles

x = 2.22 moles of F2

Now we add these values, 0.972 + 2.22 = 3.192 moles

And now we can use the ideal gas law formula:

PV = nRT

Remember that R is the gas constant, 0.082

P * 94.9 L = 3.192 * 0.082 * 328.92

94.9P = 86.1

P = 0.91 atm

The splint test is a simple experiment that is used to confirm the presence of oxygen gas. Here's how you can design a splint test to prove that the gaseous product is indeed O2:

Step 1: Light a wooden splint until it starts to burn.

Step 2: Blow out the flame, but make sure that the end of the splint is still glowing red.

Step 3: Insert the glowing splint into a test tube or container that contains the gaseous product.

Step 4: Observe what happens to the glowing splint.

If the gaseous product is indeed O2, then the glowing splint will reignite and start to burn again. This is because oxygen is a key component of combustion, and the presence of O2 will provide the necessary oxygen to keep the splint burning.

Observations: When the glowing splint is inserted into the test tube containing the gaseous product, it reignites and starts to burn again. This confirms that the gaseous product is indeed O2.

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Water solubility is the capability of a solute to dissolve in water. The different neutral substances that can be classified as insoluble, slightly soluble, or soluble in water are as follows:

Insoluble substances - These are the compounds that cannot dissolve in water and usually precipitate out as solids. For instance, silver bromide (AgBr) and lead sulfate (PbSO₄) are two insoluble compounds.

Slightly soluble substances - These compounds have a low solubility in water and do not entirely dissolve in water. Calcium hydroxide (Ca(OH)₂) and barium sulfate (BaSO₄) are two examples of slightly soluble substances.

Soluble substances - These substances have a high solubility in water and easily dissolve in water. Sodium chloride (NaCl) and potassium nitrate (KNO₃) are two examples of soluble substances.

The factors that can influence the water solubility of a substance include the temperature, pressure, and the polarity of the substance. Generally, polar substances dissolve easily in polar solvents such as water, whereas nonpolar substances dissolve well in nonpolar solvents.

A substance’s water solubility can be predicted based on its chemical formula and characteristics. Furthermore, the solubility of a substance can also be determined by conducting experiments with the substance in water.

A conclusion can be drawn that the water solubility of a substance is essential since it determines the extent to which the substance can dissolve in water. The ability to dissolve in water is important in many fields such as medicine, biology, and chemistry.

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Answer:

True

Explanation:

Water is the most common solvent among liquid solutions. It is able to dissolve a wide range of solutes due to its polar nature and ability to form hydrogen bonds with other polar molecules.

2 liters of the 80% acid solution must be added to 4 liters of the 20% acid solution to obtain a 40% acid solution.

Let's denote the number of liters of the 80% acid solution to be added as x.

To determine the amount of acid in the resulting solution, we can calculate the sum of the acids in the initial solutions and set it equal to the acid in the final solution.

The acid in the 80% solution (0.80x) added to the acid in the 20% solution (0.20 × 4 liters) should be equal to the acid in the final 40% solution (0.40 × (4 + x) liters).

This can be expressed as:

0.80x + 0.20 × 4 = 0.40 × (4 + x)

Simplifying the equation:

0.80x + 0.80 = 1.60 + 0.40x

Combining like terms:

0.80x - 0.40x = 1.60 - 0.80

Simplifying further:

0.40x = 0.80

Dividing both sides by 0.40:

x = 2

Therefore, 2 liters of the 80% acid solution must be added to 4 liters of the 20% acid solution to obtain a 40% acid solution.

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Answer:

J. J. Thomson

Explanation:

This is called the plum pudding model which was proposed by J. J. Thomson.

you can follow these steps to determine the standard Gibbs free energy for any reaction at 298 K, given the necessary ΔG°f values.

To calculate the standard Gibbs free energy (ΔG°) for a reaction at 298 K, you will need to follow these steps:
1. Write down the balanced chemical equation for the reaction.
2. Look up the standard Gibbs free energy of formation (ΔG°f) values for each reactant and product in a table or reliable source. These values are usually given at 298 K.
3. Apply the following equation to calculate the standard Gibbs free energy change for the reaction:
ΔG° = Σ [ΔG°f(products)] - Σ [ΔG°f(reactants)]

Δ G ∘ = − 1486 kJ/mol

T = 3032 K

Δ G ∘ = Δ H ∘ − T Δ S ∘

T = 298 K under ordinary conditions.

∴ Δ G ∘ = − 1648.4 × 10 3 − ( 298 × − 543.7 )

Δ G ∘ = − 1486 kJ/mol

At equilibrium, G = 0.

∴ 0 = Δ H ∘ − T Δ S ∘

We can use the conventional values if we assume that temperature has no effect on enthalpy and entropy changes.

∴T = Δ H ∘ Δ S ∘

T = − 1648.4 × 10 3 − 543.7 K

T = 3032 K
In this equation, you'll need to multiply the ΔG°f value for each substance by its stoichiometric coefficient in the balanced equation. Then, sum the ΔG°f values for all products and subtract the sum of the ΔG°f values for all reactants.
Unfortunately, you haven't provided the specific reaction you need help with. However, you can follow these steps to determine the standard Gibbs free energy for any reaction at 298 K, given the necessary ΔG°f values.

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25,137.4 joules of heat energy are required to raise the temperature of 670 g of water from 15.0°C to 24.0°C.

To calculate the joules of heat energy required to raise the temperature of 670 g of water from 15.0°C to 24.0°C, you'll need to use the formula:

Q = mcΔT

where Q represents the heat energy in joules, m is the mass of the water in grams, c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature in degrees Celsius.

1. Determine the mass (m): 670 g

2. Determine the specific heat capacity (c): 4.18 J/g°C

3. Calculate the change in temperature (ΔT): 24.0°C - 15.0°C = 9.0°C

Now, plug the values into the formula:

Q = (670 g) × (4.18 J/g°C) × (9.0°C)

Q = 670 × 4.18 × 9

Q = 25,137.4 J

Therefore, 25,137.4 joules of heat energy are required to raise the temperature of 670 g of water from 15.0°C to 24.0°C.

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The magnitude of the energy absorbed in the process, 1st shell → 4th shell transition, will vary for a He+ ion compared to a Li2+ ion.

The 1st shell → 4th shell transition is the jump of an electron from the first shell to the fourth shell. The energy of an atom is directly related to the shells and subshells of the atom.

The magnitude of the energy absorbed in the process of 1st shell → 4th shell transition will be more in the case of a Li2+ ion as compared to a He+ ion.

This is because of the fact that the Li2+ ion is more positively charged, hence, the force of attraction experienced by an electron on the 1st shell is more than the force of attraction experienced by an electron on the He+ ion.

The energy required for the transition of an electron from the first shell to the fourth shell in the case of Li2+ ion is more as compared to the energy required for the same transition in the case of a He+ ion. Therefore, the energy absorbed in the process will also be more in the case of Li2+ ion as compared to the He+ ion.

Therefore, we can say that the magnitude of the energy absorbed in the process, 1st shell → 4th shell transition, will vary for a He+ ion compared to a Li2+ ion.


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Answer:

Hydrogen bonding is the major intermolecular attraction responsible for the relatively high boiling points of alcohols.