Consider the following energy-level diagram for a particular electron in an atom.
AE
Based on this diagram, which of the following statements is incorrect?
The wavelength of a photon emitted by the electron jumping from level 2 to level 1 is given by
2==
he
|ΔΕ|
If the electron is in level 1, it may jump to level 2 by absorbing a photon with energy of AE.
If the electron is in level 1, it may jump to level 2 by absorbing any photon having energy of at least
ΔΕ.
We would observe an electron jumping from level 2 to level 1 as a single line in a line spectrum.
If the electron is in level 2, it may jump to level 1 by emitting a photon with energy of AE.

A beam of light has a wavelength of 280 nanometers. The frequency of the light is  1.07 × 10¹⁵ Hz.

the information in the question is given as :

wavelength of beam of light = 280 nm

the relation between the frequency and the wavelength is given as :

F = c / λ

where,

F = frequency of the light

c = speed of light

λ = wavelength of light

speed of light , c is = 3 × 10⁸ m/s

substituting all the value in the formula for the frequency, we get:

F = c / λ

F =  3 × 10⁸  / 280 × 10⁻⁹

F = 0.0107 × 10¹⁷ Hz

F = 1.07 × 10¹⁵ Hz

Thus, A beam of light has a wavelength of 280 nanometers. The frequency of the light is  1.07 × 10¹⁵ Hz.

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Answer:

computer is a device used to make work easy and comfort

approximately 0.27 grams of BaF2 will be dissolved in 125 mL of the saturated solution of BaF2.

How to solve the problem?

The solubility product constant (Ksp) for BaF₂ is given as 1.50 x 10⁻⁶. This means that at equilibrium, the product of the concentration of Ba₂₊ and F- ions will be equal to Ksp.

The balanced equation for the dissolution of BaF₂ is BaF₂(s) ⇔ Ba₂₊(aq) + 2F-(aq).

We can use the Ksp value to determine the concentration of Ba₂₊ and F- ions in a saturated solution of BaF₂

Let x be the molar solubility of BaF₂. Then, the concentration of Ba₂₊ ions in the saturated solution will be x mol/L, and the concentration of F- ions will be 2x mol/L, according to the stoichiometry of the balanced equation.

The Ksp expression can be written as:

Ksp = [Ba₂₊][F-]²

Substituting the expressions for the concentrations of Ba₂₊ and F- ions, we get:

Ksp = x (2x)⅝ = 4x³

Solving for x, we get:

x³= Ksp / 4 = 1.50 x 10⁻⁶/ 4 = 3.75 x 10⁻⁷

x = (3.75 x 10⁻⁷)¹/³ = 0.0077 mol/L

This means that the molar solubility of BaF₂ is 0.0077 mol/L in water.

To determine the mass of BaF₂ that will be dissolved in 125 mL of this saturated solution, we can use the following formula:

mass = concentration x volume x molar mass

The molar mass of BaF₂ is 175.32 g/mol.

The concentration of BaF₂ Is equal to its molar solubility, which is 0.0077 mol/L.

The volume of the solution is 125 mL, which is equal to 0.125 L.

Substituting these values in the formula, we get:

mass = 0.0077 mol/L x 0.125 L x 175.32 g/mol

mass = 0.27 g (approx.)

Therefore, approximately 0.27 grams of BaF₂will be dissolved in 125 mL of the saturated solution of BaF₂

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Therefore, approximately 22.61 grams of ammonium carbonate should be added to 438 mL of 0.18 M ammonium nitrate solution to produce an aqueous 0.67 M solution of ammonium ions.

The balanced equation for the reaction between ammonium carbonate (NH4)2CO3 and ammonium nitrate NH4NO3 is:

(NH4)2CO3 + NH4NO3 -> 2NH4+ + CO3^2- + NO3^-

From the balanced equation, we can see that one mole of (NH4)2CO3 produces 2 moles of NH4+ ions.

Given:

Volume of ammonium nitrate solution = 438 mL = 0.438 L

Molarity of ammonium nitrate solution = 0.18 M

Desired molarity of ammonium ions = 0.67 M

Molar mass of ammonium carbonate = 96.09 g/mol

Calculate the moles of ammonium nitrate:

Moles of NH4NO3 = Molarity × Volume

Moles of NH4NO3 = 0.18 M × 0.438 L

Calculate the moles of ammonium ions:

Moles of NH4+ = Moles of NH4NO3 × 2

Calculate the volume of ammonium carbonate solution required:

Volume of (NH4)2CO3 solution = Moles of NH4+ / Desired molarity of NH4+

Calculate the mass of ammonium carbonate:

Mass of (NH4)2CO3 = Volume of (NH4)2CO3 solution × Molarity × Molar mass

Let's perform the calculations:

Moles of NH4NO3 = 0.18 M × 0.438 L = 0.07884 mol NH4NO3

Moles of NH4+ = 0.07884 mol NH4NO3 × 2 = 0.15768 mol NH4+

Volume of (NH4)2CO3 solution = 0.15768 mol NH4+ / 0.67 M = 0.23546 L

Mass of (NH4)2CO3 = 0.23546 L × 96.09 g/mol = 22.61 g

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The volume of water in the container is 292.2 mL.

The molecules in water must be able to move more quickly in order for the temperature to rise, and in order to do this, the hydrogen bonds that bind them must be severed. These intermolecular interactions must be broken by the heat that water absorbs. Before the temperature of the water can rise.

We can used the formula:

Q = m * c * ΔT

Q = amount of heat absorbed by the water

m = mass of water

c = specific heat capacity of water

ΔT = change in temperature of the water

Given;

Q = 64.4 kJ

ΔT = (73.4 - 22.0) °C = 51.4 °C

c = 4.18 J/(g·°C)

Converting the units of Q to Joules:

Q = 64.4 kJ * 1000 J/kJ = 64400 J

Now:

m = Q / (c * ΔT)

m = 64400 J / (4.18 J/(g·°C) * 51.4 °C)

m = 292.2 g

The density of water:

Density = mass / volume

volume = mass / density

volume = 292.2 g / 1 g/mL = 292.2 mL

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Answer:

B.Oceans

Explanation:

Carbon is stored on our planet in the following major sinks (1) as organic molecules in living and dead organisms found in the biosphere; (2) as the gas carbon dioxide in the atmosphere; (3) as organic matter in soils; (4) in the lithosphere as fossil fuels and sedimentary rock deposits such as limestone, dolomite and ...

The ocean soil and also forests are one of the worlds largest carbon sinks

so the answer is B.Oceans

The equation is not a transmutation because there is conservation of the number of atoms as well as the number of nucleons; option C.

Transmutation refers to the artificial synthesis of new elements by a nuclear reaction of the original element and nuclear particles.

In transmutation, a new element is formed and the atomic number changes.

Based on the above, it can be concluded that the given equation is not a transmutation because there is no change in atomic and nucleon number.

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The amount of water removed from every kilogram of commercial sea salt would be 27.57 kg.

Seawater contains 3.5% sodium chloride by mass. This means every 1 kg of seawater contains 0.035 kg of sodium chloride.

Sea salt contains about 10% water. This means every 1 kg of sea salt contains 0.1 kg of water.

In order to calculate the amount of water removed from every kilogram of commercial sea salt produced, we need to know the amount of seawater that will give 1 kilogram of sea salt.

We said every 1 kg of seawater contains 0.035 kg of salt. How many 0.035 kg can be found in 1kg?

1/0.035 = 28.57

This means 28.57 kg of seawater would need to be processed in order to have 1 kg of salt. Thus, the amount of water removed can be calculated as:

28.57 - 1 = 27.57 kg

In other words, about 27.57 kg of water is removed for every kilogram of commercial salt produced.

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Answer:

10 must be added

Explanation:

Answer:

Boiling point of water increases wben salt is added to the water this propertiy is known as elevation of boiling points

Explanation:

the law of conservation of mass applies.....?

Answer:

it is A as both are always charged

Explanation:

Answer: A the particle is always charged

Explanation:

The reaction between pottasium metal and chlorine gas is an example of combination reaction and the balanced equation is as follows: 2K + Cl₂ → 2KCl

A chemical equation is a symbolic representation of a chemical reaction where reactants are represented on the left, and products on the right.

A chemical equation is said to be balanced when the number of atoms of each element on both sides of the equation are the same.

According to this question, potassium metal reacts with chlorine gas to form solid potassium chloride. The balanced equation is as follows:

2K + Cl₂ → 2KCl

Based on the above equation, pottasium combines with chlorine chemically to form pottasium chloride compound, hence, it is an example of combination reaction.

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Pesticide B was starting to be used since the pests were beginning to adapt and learn how to become immune to Pesticide A.

Due to farmers applying pesticides to safeguard their crops and the possibility that insects may develop resistance to pesticides, the number of pesticides used increased exponentially.

In the production of agricultural products, pesticides are essential. Farmers have employed them to manage weeds and insects, and it has been noted that they have led to notable gains in agricultural output. Without a corresponding rise in food production, the growth in the global population throughout the 20th century would not have been conceivable.

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Plasma is a state of matter in which a gas has been ionized to the point where it contains a significant number of free electrons and positive ions.

In a plasma, the electrons are separated from their parent atoms or molecules and are free to move about. This creates a mixture of positively charged ions and negatively charged electrons, which collectively behave like a fluid rather than individual particles.

Plasmas can be created by heating a gas to high temperatures, subjecting it to a strong electromagnetic field, or by passing an electric current through it. Examples of natural plasmas include lightning, the aurora borealis, and the sun. Plasmas have unique properties and are used in a variety of applications, including fluorescent lighting, plasma cutting and welding, and in plasma TVs.

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Answer:

Follows are the solution to the given question:

Explanation:

Following are the balanced redox equation:

\(Cr_2O_7^{2-} + 6 Fe^{2+} + 14H^{+} \to 2Cr^{3+} + 6 Fe^{3+} + 7H_2O\)

Calculating the average volume of the dichromate:

\((i)\ 13.8 - 2.3 = 11.5 \ ml\\\\(ii)\ 24.4 - 13.8 = 10.6\ ml\\\\(ii)\ 35.2 - 24.4 = 10.8 \ ml\\\\(iv)\ 45.9 - 35.2 = 10.7\ ml\)

\(Mean\ volume = 10.9\ ml = 0.0109\ L\\\\moles\ Fe^{2+} \ titrated = 0.01500 \ L \times 0.07\ \frac{mol}{L} = 0.00105 \\\\moles\ Cr_2O_7^{2-}\ required = 0.00105\ mol\ Fe^{2+} \times 1 \ mol\ \frac{Cr_2O_7^{2-}}{6 \mol\ Fe^{2+} } = 0.000175 \ moles\ Cr_2O_7^{2-}\\\\Concentration \ of\ Cr_2O_7^{2-} = \frac{0.000175\ moles}{0.0109\ L} = 0.01606\ M\\\\\)

The liquid-liquid extraction (LLE) used to extract the free fatty acids from the saponification reaction mixture into heptane is a type of partition chromatography. Partition chromatography involves the separation of components based on their partition coefficient between two immiscible phases, in this case, the aqueous and organic phases.

The free fatty acids preferentially partition into the organic phase, allowing for their isolation and purification.

In the process you described, liquid-liquid extraction (LLE) was used to extract free fatty acids from the saponification reaction mixture into heptane. The type of chromatography associated with this organic extraction is partition chromatography, where the distribution mechanism involves the partitioning of analytes between two immiscible liquid phases.

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When 485 junior college students were surveyed,195 said that they have previously owned a motorcycle. This point estimate suggests that out of the surveyed junior college students, around 40.2% have owned a motorcycle

To find a point estimate for the population proportion (p) of students who have previously owned a motorcycle, we divide the number of students who have owned a motorcycle (195) by the total number of students surveyed (485).

Point Estimate of p = Number of students who have owned a motorcycle / Total number of students surveyed

Point Estimate of p = 195 / 485 ≈ 0.402

Therefore, the point estimate for p, the population proportion of students who have previously owned a motorcycle, is approximately 0.402 or 40.2%.

This point estimate suggests that out of the surveyed junior college students, around 40.2% have owned a motorcycle. It is important to note that this is just an estimate based on the sample data, and the true population proportion can only be determined with complete certainty if the entire population is surveyed. The point estimate provides us with a rough indication of the proportion, but it is subject to some level of sampling variability.

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Mass of NaCl : 5.85 g

Given

5.3 g Sodium carbonate (Na₂CO₃)

Required

Mass of NaCl

Solution

Reaction

Na₂CO₃ + 2HCl ⇒ 2NaCl + H₂O + CO₂

mol of Na₂CO₃ (MW=106 g/mol) :

= mass : MW

= 5.3 g : 106 g/mol

= 0.05

From the equation, mol ratio Na₂CO₃ : NaCl = 1 : 2, so mol NaCl :

= 2/1 x mol Na₂CO₃

= 2/1 x 0.05

= 0.1

Mass of NaCl (MW=58.5 g/mol) :

= mol x MW

= 0.1 x 58.5

= 5.85 g

Answer:

C2H3O2^-(aq) + H^+ —> HC2H3O2

Explanation:

The following equation was given in the question:

Na^+ + C2H3O2^-(aq) + H^+ Cl^- —> Na^+ + Cl^- + HC2H3O2

Now, to obtain the net ionic equation, we simply cancel out the ions common to both side of the equation.

A careful observation of the equation above, shows that sodium ion, Na^+ and chloride ion, Cl^- are common to both side of the equation.

Therefore, to obtain the net ionic equation, we simply cancel out Na^+ and Cl^- from both side of the equation. This is illustrated below:

Na^+ + C2H3O2^-(aq) + H^+ Cl^- —> Na^+ + Cl^- + HC2H3O2

C2H3O2^-(aq) + H^+ —> HC2H3O2

Therefore, the net ionic equation is

C2H3O2^-(aq) + H^+ —> HC2H3O2

A galvanic anode that would NOT be used to provide CP to steel is:.D) Chromium is not commonly used as a galvanic anode for the cathodic protection (CP) of steel. Magnesium, aluminum, and zinc are commonly used galvanic anodes for the CP of steel.

A galvanic anode is a type of sacrificial anode that is used to protect metal structures from corrosion. It is made from a more active metal than the metal being protected, such as zinc, aluminum, or magnesium. When the anode is electrically connected to the metal being protected and immersed in an electrolyte, such as seawater, a galvanic cell is created. This results in the anode corroding instead of the protected metal. As the anode corrodes, it releases electrons that flow through the electrolyte to the metal being protected, preventing it from corroding. Galvanic anodes are commonly used in pipelines, ships, and offshore structures to prevent corrosion.

Galvanic anodes are commonly used as a form of cathodic protection (CP) to protect metallic structures from corrosion. The anode material is more reactive than the metal being protected, and when connected to the structure through a conductive medium, it corrodes preferentially to the protected metal, thereby providing CP.

Magnesium, aluminum, and zinc are all commonly used as galvanic anodes for CP because they are more reactive than steel and corrode preferentially to it. However, chromium is not typically used as a galvanic anode for CP because it is less reactive than steel and would not provide sufficient protection. Instead, chromium is often used as a passive protective coating on steel, as it forms a thin, stable oxide layer that helps to prevent corrosion.

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Answer:

Polypropylene, also known as polypropene, is a thermoplastic polymer used in a wide variety of applications. It is produced via chain-growth polymerization from the monomer propylene. Polypropylene belongs to the group of polyolefins and is partially crystalline and non-polar.

Explanation:

The temperature of a plasma is often higher compared to the temperatures of gases, liquids, or solids.

Plasma is a state of matter that exists at very high temperatures, typically in the range of thousands to millions of degrees Celsius.

At such high temperatures, the atoms and molecules in the plasma gain enough energy to ionize, meaning they lose or gain electrons, resulting in a mixture of charged particles.

This ionization gives plasma its unique properties and behavior.

Plasma is commonly found in phenomena such as lightning, stars, and certain laboratory conditions. Its high temperature is necessary for sustaining the ionization and allowing the plasma to exhibit characteristics such as electrical conductivity and the ability to generate magnetic fields.

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Answer:

Explanation:

good

Round off .00042557 to three significant digits is = 0.000426

Given

The digit of 00042557

Three significant digits is 0.000426

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If mercury (ii) oxide is heated and decomposes, the product of the reaction be 4 Hg.

Mercury is a chemical element with the atomic number 80 and the symbol Hg. Formerly known as hydrargyrum and also known as quicksilver, In order to fragment anything into smaller pieces or to break it into smaller pieces, mercury is used in thermometers, barometers, manometers, sphygmomanometers, float valves, mercury switches, mercury relays, fluorescent lights, and other devices: Methane and carbon dioxide are produced when microbes break down organic waste. staying fresh while decomposes The process of recycling nutrients that an organism (plant or animal) has utilized to develop its body begins with decomposes. It is the process by which the decomposing dead tissues transform into more basic organic forms. For a large number of the species at the base of ecosystems, these serve as a food supply.

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The mass of gold salt that forms when a 73.5g mixture of equal masses of all three reactants is prepared is 18.7924 g.

The balanced chemical equation for the reaction between gold metal, BrF3 and KF is given below:2Au + 3BrF3 + 6KF → 2K[AuF4] + 3Br2The oxidizing agent in the above reaction is BrF3 while the reducing agent is Au. The reaction between gold metal and BrF3 is an oxidation reaction, where BrF3 acts as an oxidizing agent and gold metal acts as a reducing agent.According to the equation, 2 moles of gold produce 2 moles of K[AuF4]. The molar mass of K[AuF4] is 304.12 g/mol.Mass of the mixture of all reactants = 73.5 g

Mass of each reactant = 73.5/3 = 24.5 gNumber of moles of Au in 24.5 g of Au = 24.5/197.0 = 0.1240 mol

Number of moles of BrF3 in 24.5 g of BrF3 = 24.5/136.9 = 0.1788 mol

Number of moles of KF in 24.5 g of KF = 24.5/58.1 = 0.4218 molFrom the balanced chemical equation, we can see that 2 moles of gold produce 1 mole of K[AuF4].Therefore, 0.1240 mol of Au will produce 0.0620 mol of K[AuF4].Mass of K[AuF4] produced = 0.0620 × 304.12 = 18.7924 g

So, the mass of gold salt that forms when a 73.5g mixture of equal masses of all three reactants is prepared is 18.7924 g.

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The mass of C in the 1.2359 g sample of the compound is 0.6116 g, the mass of N is 0.3564 g and the mass of O is calculated to be 0.2031 g.

First, we determine the mass of carbon C as follows;

mass of carbon C = (2.241 g) × (1 mole Carbon dioxide/44.01 g Carbon dioxide) × (1 mole C/ 1 mole Carbon dioxide) × (12.011 C/1 C)

mass of carbon C = 0.6116 g

Now the mass of nitrogen (N) can be determined as follows;

As the mass percent of N is found to be 28.84 % therefore;

Mass of nitrogen (N) = 1.2359 × 0.2884 = 0.3564 g

Now the mass of oxygen (O) can be calculated as follows;

Since the compound contains only C, H, N, and O therefore;

mass of O = g of sample − ( g of H + g of C + g of N )

mass of O = 1.2359 − (0.0648 + 0.6116 + 0.3564)

Mass of O = 0.2031 g

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