Trend of atomic number and atomic size of the elements when we move from left to right in different periods of periodic table​

The wavelength of 1212.7 Å corresponds to an electronic transition from n=1 to n=2, while the wavelength of 6562.8 Å does not have enough information to determine the value of n.

In atomic spectra, the wavelength of spectral lines corresponds to specific electronic transitions within an atom. Each line is associated with a particular energy level transition, which is characterized by the principal quantum number (n) of the electronic orbitals involved.

Let's analyze the given wavelengths:

Wavelength of 1212.7 Å: This wavelength corresponds to a transition involving the principal quantum numbers n and n-2. We need to find the values of n for this transition.

We know that the Balmer series in the hydrogen atom is defined by transitions from higher energy levels to the n=2 level. Therefore, the wavelength of 1212.7 Å corresponds to an electronic transition within the Balmer series. From the Balmer series formula,

1/λ = R_H * [tex](1/2^2 - 1/n^2[/tex]), where R_H is the Rydberg constant for hydrogen, we can solve for n. Rearranging the formula, we get:

1/λ = R_H * (1/4 - 1/[tex]n^2[/tex])

1/λ = R_H/n^2 - R_H/4

1/λ + R_H/4 = R_H[tex]/n^2[/tex]

1/λ + 109677 cm^-1 = 109677 cm^-1/n^2

Substituting the value of the wavelength (1212.7 Å = 121270 nm = 12127000 Å) into the equation, we can solve for n:

1/12127000 + 109677 cm^-1 = 109677 cm^-1/n^2

[tex]n^2[/tex] = 109677 cm^-1 / (1/12127000 + 109677 cm^-1)

[tex]n^2[/tex]≈ 109677 cm^-1 / 109678 cm^-1

n ≈ sqrt(1) ≈ 1

Therefore, the electronic transition corresponding to the wavelength of 1212.7 Å is from n=1 to n=2.

Wavelength of 6562.8 Å: This wavelength corresponds to a transition involving the principal quantum numbers n and n-2. To determine the value of n for this transition, we can use a similar approach as before. However, since the wavelength is not specified further, we cannot determine the exact series or atom.

Different atoms or series can have transitions that produce the same wavelength. Without additional information, we cannot pinpoint the specific value of n for this wavelength.

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As combustion always result into the formation of water, the other type of reactions that may result into formation of water are Acid-Base Neutralization Reactions and Hydrogen and Oxygen Reaction.

Acid-Base Neutralization Reactions:

A neutralisation reaction is a chemical process in which an acid and a base combine to produce salt and water as the end products.

H⁺ ions and OH⁻ ions combine to generate water during a neutralisation reaction. Acid-base neutralisation is the most common type of neutralisation reaction.

Example: Formation of Sodium Chloride (Common Salt):

HCl + NaOH → NaCl + H₂O

Hydrogen and Oxygen Reaction:

Water vapour is created when hydrogen gas (H₂) and oxygen gas (O₂) are combined directly. This reaction produces a lot of heat and releases a lot of energy.

Example: 2 H₂ + O₂ → 2 H₂O

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The most reactive element based on the given data among the given options is option c) Element J.  

This can be determined based on their placement on the periodic table. The reactivity of an element is dependent on its position on the periodic table, particularly its electron configuration and the number of valence electrons it has. For instance, elements located in the top left corner of the periodic table are typically the most reactive.

They have fewer electrons in their outermost shell and have a tendency to lose them or combine with other elements in order to obtain a full outer shell or achieve stability.In this case, Element J is most likely located in the far left of the periodic table, most likely in the alkali metals group, which contains some of the most reactive metals.

Alkali metals are highly reactive because they only have one valence electron, making it easy for them to give it up and form positive ions. As a result, Element J is the most reactive among the given elements.The correct answer is c.

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Allosteric enzymes change shape upon binding an effector molecule, displaying a sigmoidal substrate concentration vs. reaction rate curve. The reaction rate increases until saturation, characterized by the enzyme's Km.

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The latent heat of fusion measures " The energy required to melt a substance" option (A).

The latent heat of fusion refers to the amount of energy required to change a substance from a solid state to a liquid state at its melting point while keeping the temperature constant. It is a specific type of latent heat that measures the energy needed for the phase transition of a substance.

When a substance is in a solid state, its particles are tightly packed and have a regular arrangement. As heat is added to the substance, its temperature gradually rises until it reaches the melting point. At this point, further addition of heat does not increase the temperature but instead causes the substance to undergo a phase change and transform into a liquid state. The energy absorbed during this process is known as the latent heat of fusion.

This energy is used to overcome the attractive forces between the particles in the solid and allow them to break free and move more freely in the liquid state. The latent heat of fusion is crucial in various practical applications, such as melting ice, changing solid metals into liquid form for casting, or utilizing phase change materials for thermal energy storage.

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The correct option that represents a precipitation reaction is:

B. K2CO3 + PbCl2 -> 2KCl + PbCO3

In a precipitation reaction, two aqueous solutions are mixed, resulting in the formation of an insoluble solid called a precipitate. This solid is formed due to the combination of certain ions that are no longer soluble in the solution.

In option B, when potassium carbonate (K2CO3) reacts with lead chloride (PbCl2), it produces potassium chloride (2KCl) and lead carbonate (PbCO3) as the products. Lead carbonate is an insoluble compound and forms a precipitate, which indicates a precipitation reaction.

Options A, C, and D do not represent precipitation reactions:

- Option A represents a double displacement reaction between magnesium bromide (MgBr2) and hydrochloric acid (HCl), resulting in the formation of magnesium chloride (MgCl2) and hydrogen bromide (HBr).

- Option C represents a substitution reaction between lithium acetate (LiC2H3O2) and tetrabromotitanium (IV) (TiBr4), forming lithium bromide (LiBr) and tetrakis(acetato) titanium (IV) (Ti(C2H3O2)4).

- Option D represents a double displacement reaction between ammonium nitrate (NH4NO3) and copper chloride (CuCl2), resulting in the formation of ammonium chloride (NH4Cl) and copper nitrate (Cu(NO3)2).

Therefore, option B is the correct representation of a precipitation reaction.

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The compound has an empirical formula of [tex]K_2H_2P_2O_8[/tex] and a molecular formula of [tex]K_2HPO_4[/tex].

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The correct question would be as

The composition of a compound is 28.73% K. 1.48% H, 22.76% P, and 47.03% O. The molar mass of the compound is 136.1 g/mol. What is the Molecular Formula of the compound?

[tex]KH_2PO_4\\KH_3PO_4\\K_2H_4P_20_{12}\\K_2H_3PO_6[/tex]

Atoms of different substances differ by weight. Option D

A) Atoms are real even though they're invisible: Dalton proposed that atoms are fundamental, indivisible particles that make up all matter. While atoms themselves cannot be observed directly, their existence and behavior can be inferred through their effects on matter.

B) The atom could be divided into smaller parts: Initially, Dalton believed that atoms were indivisible and the ultimate building blocks of matter. However, subsequent scientific discoveries, such as the discovery of subatomic particles like protons, neutrons, and electrons, revealed that atoms could be further divided into smaller components.

C) All atoms of a single substance are identical: Dalton postulated that atoms of the same element are identical in size, mass, and chemical properties. According to his atomic theory, different elements are composed of unique atoms, and atoms of the same element are identical to one another.

D) Atoms of different substances differ by weight: Dalton recognized that atoms have different masses and proposed that the differences in atomic weight account for the distinct properties of different elements. He formulated the law of multiple proportions, which states that elements combine in fixed ratios of masses to form compounds.

Option D

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At equilibrium, the number of moles of Cl2(g) present is approximately 347.37 mol.

To determine the number of moles of Cl2(g) at equilibrium, we need to use the given equilibrium constant (Kc) and set up an ICE table to track the changes in the reactants and products.

The balanced equation for the reaction is:

CO(g) + Cl2(g) ⇌ COCl2(g)

Let's set up the ICE table:

  CO(g)     +     Cl2(g)     ⇌     COCl2(g)

Initial: 0.3500 0.05500 0

Change: -x -x +x

Equilibrium: 0.3500 - x 0.05500 - x x

Using the equilibrium concentrations in the ICE table, we can write the expression for the equilibrium constant (Kc) as:

Kc = [COCl2(g)] / [CO(g)][Cl2(g)]

Substituting the values into the equation, we have:

1.2 × 10^3 = (0.05500 - x) / [(0.3500 - x)(0.05500 - x)]

Simplifying the equation, we can cross-multiply and rearrange:

1.2 × 10^3 × (0.3500 - x)(0.05500 - x) = 0.05500 - x

Expanding and rearranging, we get:

0 = (1.2 × 10^3 × 0.05500 - 1.2 × 10^3x + 0.05500x) - x

Simplifying further:

0 = 66 - 1.245x + 0.05500x - x

0 = 66 - 0.19x

0.19x = 66

x = 66 / 0.19

x ≈ 347.37

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Answer:

To find the volume of the thallium, we can use the formula:

density = mass/volume

Rearranging this formula, we get:

volume = mass/density

Plugging in the given values, we get:

Volume = 3.85g / 11.9 cm^-3

Using a calculator, we can solve for the volume:

Volume = 0.3235 cm^3

Therefore, the volume of the thallium is 0.3235 cm^3.

Explanation:

The original pressure[P₁] is approximately 0.0848 atm

We can use the combined gas law equation, which relates the initial and final conditions of a gas sample. The combined gas law equation is as follows:

(P₁ × V₁) / (T₁) = (P₂ × V₂) / (T₂)

Given:

V₁ = 33 liters

T₁ = 24 °C = 24 + 273.15 = 297.15 K (converted to Kelvin)

V₂ = 41,000 milliliters = 41 liters (converted to liters)

T₂ = 13 °C = 13 + 273.15 = 286.15 K (converted to Kelvin)

P₂ = 2.7 atm

We need to find P₁, the original pressure.

Plugging in the values into the combined gas law equation:

(P₁ × 33) / (297.15) = (2.7 × 41) / (286.15)

Simplifying the equation:

33P₁ = (2.7 × 41 × 297.15) / (286.15)

33P₁ ≈ 2.804

Dividing both sides by 33:

P₁ ≈ 2.804 / 33

P₁ ≈ 0.0848 atm

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For the solution with a pH of 7.9:

pH = 7.9

pOH = 14 - pH = 14 - 7.9 = 6.1

[H+] = 10^(-pH) = 10^(-7.9) (in mol/L)

[OH-] = 10^(-pOH) = 10^(-6.1) (in mol/L)

The pH of a solution is a measure of its acidity, while pOH is a measure of its alkalinity. The pH and pOH values are related through the equation pH + pOH = 14.

For the solution with a pH of 4.5:

pH = 4.5

pOH = 14 - pH = 14 - 4.5 = 9.5

[H+] = 10^(-pH) = 10^(-4.5) (in mol/L)

[OH-] = 10^(-pOH) = 10^(-9.5) (in mol/L)

For the solution with a pH of 7.9:

pH = 7.9

pOH = 14 - pH = 14 - 7.9 = 6.1

[H+] = 10^(-pH) = 10^(-7.9) (in mol/L)

[OH-] = 10^(-pOH) = 10^(-6.1) (in mol/L)

Note: The [H+] and [OH-] concentrations can also be calculated using the equation [H+][OH-] = 1 x 10^(-14) at 25°C.

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Answer: 100 ohms.

Explanation:

The circuit is composed of two parallel branches (upper and lower), with one resistor in the upper branch (150) and two resistors in the lower branch (250 and 50).

The lower branch resistors are in series, so the lower branch's resistance is:

250 + 50 = 300.

Now, the upper branch (150) and total lower branch (300) are in parallel, so:

[tex]\frac{1}{R} = \frac{1}{150} + \frac{1}{300}[/tex]

That is,

[tex]\frac{1}{R} = \frac{3}{300} = \frac{1}{100}[/tex],

Solving for R, we find R = 100.

The equivalent resistance of this circuit is 100 ohms.

The experimental energy in J and the n initial and n final by applying the equation in [tex]E= -4.21 * 10^{-19} J[/tex]

The given formula is[tex]E=-2.18*10^-18J(1/n^2final - 1/n^2initial)Z^2[/tex]

The formula to calculate the energy of a photon is given by:E= hc / λwhere:E = energy of a photonh = Planck's constantc = speed of lightλ = wavelength of the photon.

Given values are:

λ = 671 nmh = [tex]6.626 * 10-^{34}J.sc = 3.0 * 10^8 m/s[/tex]

By using the formulaE= hc / λE

= [tex]6.626 * 10^{-34} J.s * 3.0 * 10^{8} m/s / (671 * 10^{-9} m)E[/tex]

= [tex]2.96 * 10^{-19[/tex]J

Now, the energy of a photon in joules is found to be 2.96 × 10^-19 J. We will now find the n final and n initial. We need to find out the principle quantum numbers of n initial and n final. Let us apply the Rydberg formula to find out n initial and n final.

We know that:

λ = [tex]R [1/n^2final - 1/n^2initial][/tex]where:λ = 671 nm

n final  = ?n initial  = ?R = Rydberg constantR = [tex]1.097 * 10^7 m^{-1[/tex]

By substituting the given values, we get:

671 nm =[tex](1.097 * 107 m-1) [1/n^2final - 1/n^2initial][/tex]

On solving this, we get:n initial = 2n final = 1

By substituting the obtained values in the energy formula, we get:

[tex]E=-2.18*10^-18J(1/n^2final - 1/n^2initial)Z^2E=-2.18*10^-18J(1/1^2 - 1/2^2)(3^2)[/tex]

[tex]E= -4.21 * 10^{-19} J[/tex]

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The equation for the reaction between silver nitrate (AgNO3) and sodium chloride (NaCl) is: AgNO3 + NaCl → AgCl + NaNO3.

In this reaction, silver nitrate (AgNO3) reacts with sodium chloride (NaCl) to produce silver chloride (AgCl) and sodium nitrate (NaNO3).

When the two solutions are mixed, the silver ions (Ag+) from silver nitrate combine with chloride ions (Cl-) from sodium chloride to form silver chloride, which is a white, insoluble precipitate. The sodium ions (Na+) from sodium chloride combine with nitrate ions (NO3-) from silver nitrate to form sodium nitrate, which remains in solution.

The reaction is a double displacement reaction, also known as a precipitation reaction, as a solid precipitate (silver chloride) is formed. This reaction occurs due to the exchange of ions between the two reactants.

Silver chloride is sparingly soluble in water and precipitates out of the solution as a solid due to its low solubility. Sodium nitrate, being a soluble ionic compound, remains dissolved in the solution as individual ions.

This reaction is commonly used in the laboratory to test for the presence of chloride ions. The formation of the white precipitate of silver chloride confirms the presence of chloride ions in the solution.

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There are approximately 1.203 × 10^24 atoms of oxygen in 60 grams of acetic acid.

To determine the number of atoms of oxygen in 60 grams of acetic acid (CH3COOH), we need to consider the molar mass and the molecular formula of acetic acid.

The molar mass of acetic acid can be calculated by summing the atomic masses of each element in its molecular formula. The atomic masses of carbon (C), hydrogen (H), and oxygen (O) are approximately 12.01 g/mol, 1.01 g/mol, and 16.00 g/mol, respectively.

Molar mass of CH3COOH = (1 × 12.01 g/mol) + (4 × 1.01 g/mol) + (2 × 16.00 g/mol) + 1.01 g/mol

= 60.05 g/mol

Now, we can calculate the number of moles of acetic acid in 60 grams using the molar mass:

Number of moles = Mass / Molar mass

= 60 g / 60.05 g/mol

≈ 0.999 moles

From the molecular formula of acetic acid, we can see that there are two atoms of oxygen in each molecule.

Therefore, the number of atoms of oxygen in 60 grams of acetic acid can be calculated by multiplying the number of moles by the Avogadro's number, which represents the number of particles (atoms, molecules, or ions) in one mole of a substance. Avogadro's number is approximately 6.022 × 10^23 particles/mol.

Number of atoms of oxygen = Number of moles × Avogadro's number × Number of oxygen atoms in one molecule

= 0.999 moles × 6.022 × 10^23 particles/mol × 2

≈ 1.203 × 10^24 atoms

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Answer:105.99 g/mol

Explanation:

Sodium carbonate is the inorganic compound with the formula Na₂CO₃ and its various hydrates. All forms are white, odourless, water-soluble salts that yield alkaline solutions in water.

Answer:

Percent composition tells you the relative amounts of each element in a molecule by mass. It can be used to determine the empirical formula of a compound, as well as to compare the composition of different molecules.

For example, the percent composition of water (H2O) is 11.19% hydrogen and 88.81% oxygen by mass. This tells us that there are two hydrogen atoms for every one oxygen atom in the molecule.

Explanation:

Brainliest Plsss

C. 2, hydrogen atoms could bong with oxygen in this illustration of an oxygen atom.

In the given illustration of an oxygen atom, there are two unpaired electrons in the outermost electron shell. Each oxygen atom can form a covalent bond by sharing one electron with another atom. In the case of oxygen, it has a valence of 2, which means it can form up to two covalent bonds. Each hydrogen atom has one electron, and it requires one additional electron to complete its outermost electron shell.

Therefore, in the given illustration, the oxygen atom can form two covalent bonds with hydrogen atoms. This is represented by the formula H2O, where one oxygen atom is bonded to two hydrogen atoms.

Hence, the correct answer is C. 2. Two hydrogen atoms can bond with one oxygen atom to form a stable molecule of water. The sharing of electrons in covalent bonds allows atoms to achieve a more stable electron configuration and form compounds with different properties.

The question was incomplete. find the full content below:

How many hydrogen atoms could bong with oxygen in this illustration of an oxygen atom?

A. 0

B. 1

C. 2

D. 6.

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Water in coal-fired power stations is used for cooling, steam generation, and pollution control, including capturing sulfur dioxide and cooling exhaust gases. Efficient water recycling helps minimize environmental impact.

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The ice is used to regulate and control the temperature during the dehydration of [tex]hydrated copper(II) sulfate[/tex], ensuring a safer and more controlled process.

When [tex]hydrated copper(II) sulfate[/tex] [tex](CuSO_ {4} .H_{4} O)[/tex] is heated, the purpose of the ice is to provide a cooling effect during the process. The hydrated copper(II) sulfate contains water molecules (H2O) that are chemically bonded to the copper sulfate compound. The formula [tex]CuSO_{4} .H_{2} O[/tex] indicates that there are x moles of water molecules per mole of copper(II) sulfate.

As the [tex]hydrated copper(II) sulfate[/tex] is heated, the heat energy causes the water molecules to undergo a physical change and turn into steam. This process is known as dehydration. The water molecules break their chemical bonds with the copper sulfate compound and are released in the form of steam.

The presence of ice during the heating process helps maintain a lower temperature in the reaction vessel. The ice absorbs the heat energy from the surroundings, allowing for a controlled and gradual increase in temperature. This controlled heating prevents sudden temperature changes and potential hazards, such as splattering or overheating.

In summary, the ice is used to regulate and control the temperature during the dehydration of [tex]hydrated copper(II) sulfate[/tex], ensuring a safer and more controlled process.

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The molar mass of ZnI2 is approximately 319.18 grams per mole.

To determine the molar mass of ZnI2 (zinc iodide), we need to know the atomic masses of zinc (Zn) and iodine (I) and their respective subscripts in the chemical formula.

The atomic mass of zinc (Zn) is approximately 65.38 grams per mole (g/mol), as found on the periodic table. The atomic mass of iodine (I) is approximately 126.90 g/mol.

Since the chemical formula of zinc iodide is ZnI2, it means there are two iodine atoms for every one zinc atom. Therefore, we multiply the atomic mass of iodine by 2.

Molar mass of ZnI2 = (atomic mass of Zn) + 2 × (atomic mass of I)

                 = 65.38 g/mol + 2 × 126.90 g/mol

                 = 65.38 g/mol + 253.80 g/mol

                 = 319.18 g/mol

Hence, the molar mass of ZnI2 is approximately 319.18 grams per mole.

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Taking 1500 mg of vitamin C daily amounts to approximately 1.2045 pounds per year.

Dr. Linus Pauling suggested that taking 1500 mg of vitamin C daily could result in milder and fewer colds. To determine the weight in pounds per year, we'll first convert milligrams to pounds and then multiply by the number of days in a year.

To convert milligrams to pounds, we need to know that there are 453,592.37 milligrams in a pound. Therefore, 1500 mg is equal to 0.0033 pounds (1500 mg / 453,592.37 mg/lb).

Now, to calculate the weight in pounds per year, we'll multiply 0.0033 pounds by the number of days in a year (365).

Weight in pounds per year = 0.0033 pounds/day * 365 days/year = 1.2045 pounds/year.

Therefore, taking 1500 mg of vitamin C daily amounts to approximately 1.2045 pounds per year.

It's important to note that while this calculation provides the weight equivalent, the effectiveness and recommended dosage of vitamin C for preventing colds should be discussed with a healthcare professional, as individual needs may vary.

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