Lines in one spectral series can overlap lines in another.
(d) What does this overlap imply about the H atom line spectrum at longer wavelengths?

Answer:

Give any two application for isotopes

Isotopes of iodine are used for radiotherapy in treatment of hyperthyroidism, cancer, etc.

Uranium, Radium, Polonium isotopes are used in atomic reactors.

Cobalt isotopes are used for irradiation of food products.

✅1) Isotopes of iodine are used for radiotherapy in treatment of hyperthyroidism, cancer, etc.

✅2) Uranium, Radium, Polonium isotopes are used in atomic reactors.

IamSugarBee

1. The molar mass of a compound is the sum of the atomic masses of all the atoms present in one molecule of the compound. To calculate the relative molecular mass (also called the molecular weight or molar mass) of a compound, we add up the atomic masses of all the atoms in the compound.

1a) H₂SO₄

The atomic masses of H, S, and O are 1.008, 32.06, and 16.00, respectively.

Relative molecular mass of H₂SO₄ = (2 x atomic mass of H) + atomic mass of S + (4 x atomic mass of O)

= (2 x 1.008) + 32.06 + (4 x 16.00)

= 98.08

1b) Ca(NO₃)₂

The atomic masses of Ca, N, and O are 40.08, 14.01, and 16.00, respectively.

Relative molecular mass of Ca(NO₃)₂ = atomic mass of Ca + (2 x atomic mass of N) + (6 x atomic mass of O)

= 40.08 + (2 x 14.01) + (6 x 16.00)

= 164.09

1c) CuSO₄.7H₂O

The atomic masses of Cu, S, O, and H are 63.55, 32.06, 16.00, and 1.008, respectively.

Relative molecular mass of CuSO₄.7H₂O = atomic mass of Cu + atomic mass of S + (4 x atomic mass of O) + (14 x atomic mass of H)

= 63.55 + 32.06 + (4 x 16.00) + (14 x 1.008)

= 249.69

2. An oxide of nitrogen contains only one nitrogen atom per molecule and has a relative molecular mass = 30. Determine its molecular formula.

The relative molecular mass of the oxide of nitrogen (NOx) is 30.

The atomic mass of N is 14.01, and the atomic mass of O is 16.00.

We can set up the equation: 14.01 + 16.00x = 30

Solving for x, we get x = 1.

Therefore, the molecular formula of the oxide of nitrogen is NO.

3. An oxide of nitrogen contains only one oxygen atom per molecule and has a relative molecular mass = 44. Determine its molecular formula.

The relative molecular mass of the oxide of nitrogen (NOx) is 44.

The atomic mass of N is 14.01, and the atomic mass of O is 16.00.

We can set up the equation: 14.01x + 16.00 = 44

Solving for x, we get x = 2.

Therefore, the molecular formula of the oxide of nitrogen is N₂O.

4. An oxide of sulfur contains only one sulfur atom per molecule and has a relative molecular mass = 80. Determine its molecular formula.

The relative molecular mass of the oxide of sulfur (SOx) is 80.

5. The atomic mass of S is 32.06, and the atomic mass of O is 16.00.

We can set up the equation: 32.06 + 16.00x = 80

Solving for x, we get x = 3.

Therefore, the molecular formula of the oxide of sulfur is SO₃.

6. The molar mass of the oxide of silicon (SiOx) is 60 g/mol.

The atomic mass of Si is 28.09, and the atomic mass of O is 16.00.

We can set up the equation: 28.09 + 16.00x = 60

Solving for x, we get x = 2.

Therefore, the molecular formula of the oxide of silicon is SiO₂.

1. The molar mass of Ar is 39.95 g/mol. Therefore, the number of moles of Ar in 8 g is:

moles = mass / molar mass = 8 g / 39.95 g/mol = 0.200 moles

2. The molar mass of H₂SO₄ is 98.08 g/mol (2 + 32 + 32 + 4x16). Therefore, the number of moles of H₂SO₄ in 1.25 kg is:

moles = mass / molar mass = 1250 g / 98.08 g/mol = 12.73 moles

3. The molar mass of Ca(OH)₂ is 74.09 g/mol (40.08 + 2x1.01 + 2x16.00). Therefore, the mass of 0.04 moles of Ca(OH)₂ is:

mass = moles x molar mass = 0.04 moles x 74.09 g/mol = 2.96 g

4. The molar mass of CuSO₄·5H₂O is 249.68 g/mol (63.55 + 32.07 + 4x16.00 + 5x18.02). Therefore, the mass of 0.001 moles of CuSO₄·5H₂O is:

mass = moles x molar mass = 0.001 moles x 249.68 g/mol = 0.250 g

5. The mass of the hydrocarbon is given as 6 g, and the number of moles is 0.2 moles. Therefore, the molar mass of the hydrocarbon is:

molar mass = mass / moles = 6 g / 0.2 moles = 30 g/mol

6. The mass of the oxide of nitrogen is given as 115 g, and the number of moles is 2.5 moles. Therefore, the molar mass of the oxide of nitrogen is:

molar mass = mass / moles = 115 g / 2.5 moles = 46 g/mol.

1.a) In 1 molecule of HCl, there is 1 hydrogen atom. Therefore, in 0.5 moles of HCl, there are:

0.5 moles HCl x 1 mole H per 1 mole HCl = 0.5 moles H

b) In 1 molecule of H₂O, there are 2 hydrogen atoms. Therefore, in 0.5 moles of H₂O, there are:

0.5 moles H₂O x 2 moles H per 1 mole H₂O = 1 mole H

c) In 1 molecule of Na₂CO₃, there are 3 oxygen atoms. Therefore, in 0.2 moles of Na₂CO₃, there are:

0.2 moles Na₂CO₃ x 3 moles O per 1 mole Na₂CO₃ = 0.6 moles O

d) In 1 molecule of MgCl₂, there are 2 chloride ions. Therefore, in 1.2 moles of MgCl₂, there are:

1.2 moles MgCl₂ x 2 moles Cl⁻ per 1 mole MgCl₂ = 2.4 moles Cl⁻

This is a stoichiometry question, and will be answered in two steps, but the first thing for every stoichiometry question is to have the reaction properly balanced, which the question already provided us:

S8 + 8 O2 -> 8 SO2

First step is to find the limiting and excess reactant, this will tell us which element is in excess and by how much, let's gather the data for the reactants first:

30.14 grams of S8, molar mass of S8 is 256.5g/mol

287.43 grams of O2, molar mass of O2 is 32g/mol

Now we will focus on S8, and see how many moles we have of it in this reaction:

256.5 g = 1 mol

30.14 g = x moles

x = 0.12 moles of S8 in this reaction

According to the molar ratio, 1:8, we need 1 mol of S8 and 8 moles of O2 in order to have the reaction to proceed

1 S8 = 8 O2

0.12 S8 = x O2

x = 0.96 moles of O2 if we have 0.12 moles of S8

We only need 0.96, if we have more than that, is excess, if we have less than that, this means that O2 is the limiting and S8 is in excess, so let's check how much of O2 we do have

32g = 1 mol

287.43 g = x moles

x = 8.98 moles of O2, we have a big excess of O2, we only need 0.96 moles and we have almost 9 times more

Now that we know which element is the limiting, we can use this information to find the theoretical yield of SO2

The molar ratio will also be 1:8, and since we have 0.12 moles of S8, we will have 8 times that, which is 0.96 moles of SO2, using its molar mass, 64.1g/mol, we will have the theoretical yield

64.1g = 1 mol

x grams = 0.96

x = 61.5 grams is the theoretical yield

55.59 grams is the actual yield

Now we use these two values in the percent yield formula:

%yield = actual yield/theoretical yield

%yield = 55.59/61.5

%yield = 0.90

Limiting reactant = S8

Excess reactant = O2

Percent yield = 90%

Answer:

I think B

tell me if I'm wrong

Answer:

the electron has a negative charge and is found outside the nucleus

Answer:

a lot but im pretty sure if u just take avogadros number and multiply it by 9.8 u get the answer

Explanation:

There are 5.89 × 10²⁴ atoms in 9.8 moles on NaCl. Details about number of atoms can be found below.

The number of atoms of a compound can be calculated by multiplying the number of moles of the substance by Avogadro's number.

no of atoms = no of moles × 6.02 × 10²³

According to this question, there are 9.8 moles of NaCl.

no of atoms = 9.8 × 6.02 × 10²³

no of atoms = 5.896 × 10²⁴

Therefore, there are 5.89 × 10²⁴ atoms in 9.8 moles on NaCl.

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Answer:

the movement of large amounts of soil and rock debris down a slope

Explanation:

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Using an ice bath to cool the solution is most likely to reduce the concentration of sodium chloride in the solution. Option D is correct.

When a solution is cooled, the solubility of most solids decreases. As a result, some of the sodium chloride may precipitate out of the solution, reducing the concentration of the solute. The other options listed would not reduce the concentration of sodium chloride in the solution.

Heating the solution on a hot plate could potentially increase the solubility of sodium chloride and lead to more dissolving, whereas adding more sodium chloride would only increase the concentration. Removing some solution with a pipette would not change the concentration, as the amount of solute would remain the same in the remaining solution. Hence Option D is correct.

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Final concentrations of acsch and tacrine in that reaction is 600x10⁻¹²mm.

The symptoms for mild to moderate Hypertension are managed by tacrine. Alzheimer's disease cannot be cured or prevented from worsening with tacrine. However, some Alzheimer's sufferers may benefit from tacrine since it helps them think more clearly.

The behavior, procedure, or event of responding. B: opposition or resistance to a power, influence, or movement, particularly: a propensity for an old-fashioned and typically obsolete social or political order or policy.

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Answer:

element.

.it is pure substance composed of only one type of atoms.

.the number of element is more than 118.

compound.

.a compound is composed of atoms of two or more elements.

.the number of compound is very large in comparison of elements.

hope it help

The chemical must break down within 283 seconds.

The fact that the rate constant "k" is expressed in units of "s1" indicates that the reaction is a first-order one.

We also have the following for the first-order process from chemical kinetics:

equation is in attached image

Where,

K is the reaction's rate constant.

R0 = starting reactant concentration

R=remaining reactant concentration at time t.

t is the length of time needed for the change to take place.

Considering the query we have,

\(K=\frac{2.303}{t} log \frac{R_{0}}{R}\\K=9.05 {*} 10^{-4}S^{-1}\)

R₀=100g

R=100-13=87g

By changing the necessary variables in equation (1), we obtain;

\(t=\frac{2.303} k *log \frac{100}{87}\\\\\\t=\frac{2.303}{9.05*10^{-4}} (log100-log87)\\\\t=\frac{2.303} {9.05*10^{-4}} *1.113\\\\t=\frac{2.563}{9.05}*10^{4} sec\\\\t=0.283*10^{4} sec\\\\t=283sec\)

Therefore, 283 seconds are needed for the chemical to break down.

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H2SO4 + 2RbOH -> Rb2SO4 + 2H2O

If you want an explanation, keep reading.

In the first portion, there are two hydrogen ions and four sulfate ions.

The second portion has one rubidium ions and one hydroxide ion.

On the other side of the equation, in order to keep those two rubidiums balanced, you'll need to add a two at the beginning of the second portion, but in that process you are giving a second hydroxide value.

Back to the right side, there is there is water (H2O).

On the first portion, there were two hydrogen ions. The second portion also has two hydroxides because of the value change (adding the two to the front).

So on the fourth portion, you'd have to add another two so you could balance the four hydrogen ions (H2 and 2OH) and the two oxygen ions (2OH).

I hope this was easy to understand.

Answer: C

Explanation:

The density of the material which is the ratio of the mass and volume of the material is 10g/cm³

The density of a material is obtained using the relation ::

The density of the material can be obtained by obtaining the slope of the graph :

Therefore, the density of the materials is :

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The combined water's ultimate temperature is 45.6 0C.

The final temperature of the mixture is calculated as follows;

Based on the principle of conservation of energy;

Heat lost by hot water = heat gained by the cold water

mcΔθ2  = mcΔθ1

75 × 4.184 × (75.7-T) = 100 × 4.184 ×  (T-23)

75 ×  (75.7-T) = 100×  (T-23)

(75.7 - T) = 1.333(T - 23)

simplifying the above equation, then we get

75.7-T = 1.333 T - 30.659

75.7 + 30.659 = 1.333 T + T

106.359 = 2.333 T

T = 106.359 / 2.333

T = 45.6 °C

Thus, the final temperature of the mixed water is 45.6 °C

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The electronic configuration of atoms and ions is based on the arrangement of electrons in various energy levels and sublevels. The configuration is typically denoted by the principle energy level (n), the type of orbital, and the number of electrons in that orbital.

A. Iron (Fe)

i. In its neutral state, Iron (atomic number 26) has the electron configuration:

    1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶

 ii. In its ion state, it's common for Iron to lose two or three electrons to form Fe²⁺ or Fe³⁺, respectively.

    - For Fe²⁺, the electron configuration would be: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶

    - For Fe³⁺, the electron configuration would be: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

B. Chromium (Cr)

 i. In its neutral state, Chromium (atomic number 24) has a slightly unusual electron configuration due to electron configurations being more stable when half-filled or fully-filled. This results in the electron configuration:

    1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵

ii. In its ion state, Chromium typically forms Cr²⁺ or Cr³⁺ ions.

    - For Cr²⁺, the electron configuration would be: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴

    - For Cr³⁺, the electron configuration would be: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³

The CEC for this soil sample is 675.2 cmol(+)/kg.

The % Base Saturation for this soil sample is approximately 136.62%.

The % Acid Saturation for this soil sample is approximately 60.55%.

To    calculate the CEC, % Base Saturation (B.S.), and % Acid Saturation for the given soil sample:

a. Calculation of CEC (Cation Exchange Capacity):

CEC is the sum of exchangeable cations in the soil. From the given results, we have:

CEC = Ca2+ + Mg2+ + K+ + Al3+

CEC = (100 mg/L + 45 mg/L + 85.5 mg/L + 94.2 mg/L) / (25 g / 1000)

CEC = 168.7 mg / (25 g / 1000)

CEC = 675.2 cmol(+)/kg

b. Calculation of % Base Saturation (B.S.):

% B.S. represents the percentage of CEC occupied by base cations. In this case, we consider Ca2+, Mg2+, and K+ as base cations. The formula to calculate % B.S. is:

% B.S. = (Ca2+ + Mg2+ + K+) / CEC * 100

% B.S. = (100 mg/L + 45 mg/L + 85.5 mg/L) / (168.7 cmol(+)/kg) * 100

% B.S. = 230.5 mg / (168.7 cmol(+)/kg) * 100

% B.S. = 136.62%

c. Calculation of % Acid Saturation:

% Acid Saturation represents the percentage of CEC occupied by acid cations, in this case, H+ and Al3+. The formula to calculate % Acid Saturation is:

% Acid Saturation = (H+ + Al3+) / CEC * 100

% Acid Saturation = (8.0 mg/L + 94.2 mg/L) / (168.7 cmol(+)/kg) * 100

% Acid Saturation = 102.2 mg / (168.7 cmol(+)/kg) * 100

% Acid Saturation = 60.55%

Please note that the given values were in milligrams per liter (mg/L), and the CEC and % Saturation values were calculated assuming a conversion from mg/L to cmol(+)/kg using the mass of the soil sample (25 g).

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16.32 has the least amount of decimal places

Answer:

16.32

Explanation:

Yes its a larger number however, If you were to move the decimal point to the end of each number then 16.32 would be moved the least.

Answer:

hlo

how r u ?

hindi-kya hal ?

1:The energy involved in a phase change depends on two major factors: the number and strength of bonds or force pairs. The number of bonds is proportional to the number of molecules and thus to the mass of the sample. The strength of forces depends on the type of molecules. The heat Q required to change the phase of a sample of mass m is given by

2:According to the law of conservation of energy, energy can neither be created nor destroyed but can be transformed from one form to another. Conversion of electric energy into sound energy is an example of the conservation of energy.

Explanation:

Hope it helps UvU

When 75.0 mL of 0.20 M CaSO4 solution is mixed with 45.0 mL of 0.35 M Li3PO4 solution, we can determine the theoretical yield of Ca3(PO4)2 and identify the limiting reagent using the principles of stoichiometry and the concept of limiting reactants.

To determine the theoretical yield of Ca3(PO4)2 and identify the limiting reagent, we need to compare the number of moles of CaSO4 and Li3PO4 present in the given solutions. First, we can calculate the number of moles for each reactant by multiplying the volume (in liters) of each solution by its respective molarity. Next, we can write the balanced chemical equation for the reaction between CaSO4 and Li3PO4 to determine the stoichiometric ratio between the reactants and the product.

By comparing the moles of CaSO4 and Li3PO4 and considering the stoichiometric ratio, we can identify the limiting reagent. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed. The reactant that yields the smaller number of moles is the limiting reagent. Once the limiting reagent is determined, we can use its stoichiometric ratio to calculate the theoretical yield of Ca3(PO4)2, which represents the maximum amount of product that can be obtained from the given reactants.

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100.0 g of oxygen gas at a pressure of 1.5 atm and a temperature of 25°C would occupy a volume of 49.2 L.

To solve this problem, we can use the Ideal Gas Law, which states:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

We need to rearrange this equation to solve for the volume V:

V = (nRT) / P

where n is the number of moles of the gas, which we can calculate using the molar mass of oxygen gas:

n = m / M

where m is the mass of the gas and M is the molar mass of oxygen gas (32 g/mol).

n = 100.0 g / 32 g/mol = 3.125 mol

Now we can substitute the given values into the equation to find the volume:

V = (nRT) / P

V = (3.125 mol)(0.0821 L·atm/mol·K)(298 K) / 1.5 atm

V = 49.2 L

Therefore, 100.0 g of oxygen gas at a pressure of 1.5 atm and a temperature of 25°C would occupy a volume of 49.2 L.

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Answer: Predominant tropical easterly winds sweep across the central and southern portions of the state, keeping the temperatures mild

Explanation:

bc

Answer:

Hunds rule is violated here.

Hunds rule : pairing of electrons takes place only if all the orbitals are filled with single electrons each.

The correct answer is d. 1,3-cyclohexadiene undergoes dehydrogenation readily because it would gain considerable stability by becoming benzene. Benzene is a highly stable and aromatic compound that possesses resonance energy due to its delocalized pi-electrons.

Dehydrogenation is a chemical reaction that involves the removal of hydrogen from a molecule. In the case of 1,3-cyclohexadiene, the removal of two hydrogen atoms would result in the formation of benzene. This transformation would result in the formation of a highly stable compound, which has much lower energy than its precursor.
Moreover, 1,3-cyclohexadiene is an unsaturated compound that possesses a double bond between two carbon atoms. This double bond makes the molecule reactive towards dehydrogenation. During dehydrogenation, the double bond is broken, and the two hydrogen atoms that were attached to the carbon atoms are removed. As a result, the molecule undergoes a structural change, and a highly stable compound, benzene, is formed.
In conclusion, 1,3-cyclohexadiene undergoes dehydrogenation readily because it would gain considerable stability by becoming benzene. This transformation is a result of the removal of two hydrogen atoms from the molecule, and it occurs due to the reactivity of the double bond that the molecule possesses.

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The expected Edman degradation product of Val-Ala-Cys after one cycle is Ala-Cys. In the case of Val-Ala-Cys, the first amino acid to be removed is Valine, leaving Ala-Cys as the product of the first cycle.

Edman degradation is a chemical process that is used to sequence amino acids in a peptide or protein. In this process, the N-terminal amino acid is selectively removed and identified. In the case of Val-Ala-Cys, the first amino acid to be removed is Valine, leaving Ala-Cys as the product of the first cycle. This product can then be subjected to subsequent cycles of Edman degradation to identify the remaining amino acids in the sequence.

After one cycle of Edman degradation, the N-terminal amino acid (Val) is cleaved and converted into a phenylthiohydantoin derivative (PTH-Val). The remaining peptide will be shortened by one amino acid, leaving the dipeptide Ala-Cys. So, the expected product is PTH-Val and Ala-Cys.

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The formula contains : C. 1 nitrogen atom, 4 hydrogen atoms, and 1 chlorine atom

Given

NH₄Cl compound

Required

The number of atoms

Solution

The empirical formula is the smallest comparison of atoms of compound forming elements.  

A molecular formula is a formula that shows the number of atomic elements that make up a compound.

A molecular formula is composed of two or more elements formed of the same or different types.

A subscript indicating the number of atoms forming a molecule

So for NH₄Cl , there are 3 different elements, N, H and Cl, and the number is shown from the subscript