Which of the following is one way that nuclear reactions differ from chemical reactions?

A. Atoms are not involved in nuclear reactions.
B. Nuclear reactions are not affected by temperature.
C. Nuclear reactions are slowed by increasing pressure.
D. Nuclear reactions do not occur in a predictable manner.

Answer:

2.

Explanation:

2Al(s) + 3FeCl₂(aq) → 2AlCl₃(aq) + 3Fe(s).

It is clear that 2 mol of Al react with 3 mol of FeCl₂ to produce 2 mol of AlCl₃ and 3 mol of Fe.

The stoichiometric coefficient for aluminum, Al when the chemical equation is balanced is 2.

To obtain the stoichiometric coefficient for aluminum, we shall write and balance the equation for the reaction. This is illustrated below:

Aluminum => Al

Iron (II) chloride => FeCl₂

Aluminum chloride => AlCl₃

Iron => Fe

Aluminum + Iron (II) chloride —> Aluminum chloride + Iron

Al + FeCl₂ –> AlCl₃ + Fe

There are 3 atoms of Cl on the right side and 2 atoms on the left side. It can be balance by writing 3 before FeCl₂ and 2 before AlCl₃ as shown below:

Al + 3FeCl₂ –> 2AlCl₃ + Fe

There are 2 atoms of Al on the right side and 1 atom on the left. It can be balance by writing 2 before Al as shown below:

2Al + 3FeCl₂ –> 2AlCl₃ + Fe

There are 3 atoms of Fe on the left side and 1 atom on the right side. It can be balance by writing 3 before Fe as shown below:

2Al + 3FeCl₂ –> 2AlCl₃ + 3Fe

Now, the equation is balanced.

The coefficient of Aluminum, Al in the balanced equation is 2.

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Answer:

[tex]\boxed{\text{c) NH$_{3}$; hydrogen bonding}}[/tex]

Explanation:

For each of these molecules, you must determine their VSEPR structure and then identify the strongest intermolecular forces.

Remember that water is a highly polar molecule.

a) CH₄

  Electron geometry: tetrahedral

Molecular geometry: tetrahedral

          Bond polarity: C-H bond nonpolar

  Molecular polarity: nonpolar

        Strongest IMF: London dispersion forces

 Solubility in water: low

A nonpolar molecule is insoluble in a polar solvent.

b) CCl₄

  Electron geometry: tetrahedral

Molecular geometry: tetrahedral

          Bond polarity: C-Cl bond nonpolar

  Molecular polarity: nonpolar (symmetrical molecule. All bond dipoles cancel)

        Strongest IMF: London dispersion forces

 Solubility in water: low

A nonpolar molecule is insoluble in a polar solvent.

d) PH₃

  Electron geometry: tetrahedral

Molecular geometry: trigonal pyramidal

          Bond polarity: P-H bonds are polar

  Molecular polarity: polar (all P-H bond dipoles point towards P)

         Strongest IMF: dipole-dipole

  Solubility in water: soluble

A polar molecule is soluble in a polar solvent.

c) NH₃

  Electron geometry: tetrahedral

Molecular geometry: trigonal pyramidal

          Bond polarity: N-H bonds are highly polar

  Molecular polarity:  highly polar (all N-H bond dipoles point towards N)

         Strongest IMF: hydrogen bonding

  Solubility in water: highly soluble

NH₃ is so polar that it can form hydrogen bonds with water.

[tex]\boxed{\textbf{The compound with the greatest solubility in water is NH$_{3}$}}[/tex]

Answer:

The correct answer is  [tex]NH_{3}[/tex].

Explanation:

The electron geometry of the [tex]NH_{3}[/tex] is tetrahedral and the molecular geometry is a trigonal pyramid.

The [tex]NH_{3}[/tex] has the strongest intramolecular hydrogen bond, which makes them a highly polar molecule.

The polarity is directly proportional to the solubility of the compound in the water.

Therefore,[tex]NH_{3}[/tex]  has the greatest solubility.

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A the nucleus emits partials and or energy

When 10 g of diethyl ether is converted to vapor at its boiling point, approximately 2.12 kJ of heat is absorbed. This is calculated by first determining the number of moles in 10g of diethyl ether and then multiplying that by the given heat of vaporization.

To calculate the heat absorbed during the vaporization of diethyl ether, we first need to know the number of moles of diethyl ether. The molar mass of diethyl ether (C4H10O) is approximately 74 g/mol. So, 10 g of diethyl ether equates to roughly 0.135 moles (10g / 74g/mol).

Given that the heat of vaporization (δHvap) for diethyl ether is 15.7 kJ/mol, the total heat absorbed can be calculated by multiplying the number of moles by the heat of vaporization. Therefore, total heat absorbed would be approximately 2.12 kJ (0.135 moles * 15.7 kJ/mol).

So, when 10 g of diethyl ether is converted to vapor at its boiling point, approximately 2.12 kJ of heat is absorbed.

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To vaporize 10 g of diethyl ether at its boiling point, approximately 2.12 kJ of heat is absorbed. Thus, option C is correct answer.

To find the amount of heat absorbed when 10 g of diethyl ether (C₄H₁₀O) is vaporized, we need to use the enthalpy of vaporization ([tex]\Delta H_{vap}[/tex]) and the molar mass of diethyl ether.

First, calculate the molar mass of diethyl ether (C₄H₁₀O):

Next, determine the number of moles of diethyl ether in 10 g:

Given the enthalpy of vaporization, [tex]\Delta H_{vap}[/tex], is 15.7 kJ/mol, calculate the total heat absorbed using the formula:

Therefore, option c) about 2.12 kJ of heat is absorbed when 10 g of diethyl ether is vaporized at its boiling point of 34.6°C.

The complete question is as follows:

When 10 g of diethyl ether is converted to vapor at its boiling point, how much heat is absorbed? (C₄H₁₀O, [tex]\Delta H_{vap}[/tex] = 15.7 kJ/mol, boiling point: 34.6°C)

A. 20 KJ

B. 0.2 KJ

С. 2.12 kJ

D. 200 KJ

Answer:

Produces energy that are useful in nuclear power plants

Explanation:

Nuclear fission is a the radioactive disintegration of a heavy nucleus into simpler ones. The fission process which initiates a chain reaction releases a lot of neutrons and a large amount of energy.

The energy released in a fission process is very useful in nuclear power plants for producing electricity.

Answer:

D; constructive; mountains

Answer:

D.) Constructive , Mountains

Explanation:

I got it right in study island

Answer:

[tex]\boxed{\text{C. 11.2 L}}[/tex]

Explanation:

The pressure is constant, so we can use Charles' Law to calculate the volume.

[tex]\dfrac{V_{1}}{T_{1}} = \dfrac{V_{2}}{T_{2}}[/tex]

Data:

V₁ = 22.4 L; T₁ = 273.15 K

V₂ = ?;         T₂ = 136.58 K

Calculations:

[tex]\dfrac{ 22.4}{273.15} = \dfrac{ V_{2}}{136.58}\\\\0.082 00 = \dfrac{ V_{2}}{136.58}\\\\V_{2} =0.082 00 \times 136.58 = \boxed{\textbf{11.2 L}}[/tex]

The value of q when e = 0 at the given temperature in the question is :

Q =  1.3 * 10⁻²⁶

Given that

E = 0,  ΔG = -nFE,

therefore ΔG = 0

Also

Given that

ΔG = ΔG° + RTIn q

ΔG° = -  RTIn q

Hence ; Q = e^ (nFE°cell / RT) -- ( 1 )

where : n = 2, F = 96500, E°cell = -0.87 volt, R = 8.314, T = 339 k

insert  values into equation ( 1 )

Q =  1.3 * 10⁻²⁶

Note :  E°cell = reduction half reaction + oxidation half reaction

          = 0.20 volt - 1.07 volt  = -0.87 volt.

Hence we can conclude that The value of q when e = 0 at the given temperature in the question is : Q =  1.3 * 10⁻²⁶

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Answer:

Explanation:

Water molecule is H₂O, which means that there is one oxygen atom per each water molecule.

The balanced chemical equation that represents this is:

The stoichiometric coefficents 1 for O₂ (g) and 2 for H₂O (g) means that two molecules of oxygen are required to produce two molecules of water.

STP stands for standard temperature and pressure. Those conditions are 273.15 K (0 °C, 32 °F) and 100 KPa of absolute pressure.

That means that the reaction is carried out at constant temperature and pressure.

Then, since the  ideal gas law states that the at constant pressure and temperature the volume occupied by the gases is proportional to the number of particles (atoms or molecules), the molecular stoichiometric ratio of 1 molecule of O₂ (g) to 2 molecules for H₂O (g) is equivalent to the volumetric ratio 1 liter of O₂ to 2 liters of H₂O:

Hence, you conclude that 1 liter of oxygen is required to produce 2 liters of water, at STP.

Answer:

KIKOKEN

Explanation:

Alpha particles are respective to the helium-4 ion. Therefore it has a mass of four and a positive charge of two. The correct answer is the mass number increases by 4.

If a nucleus emits an alpha particle, the mass number decreases by 4

An alpha particle is essentially identical to a helium nucleus. An alpha particle has a mass of four units and a positive charge of two units just as the helium nucleus.

Hence, if a nucleus emits an alpha particle, the mass number decreases by 4 while the atomic number decreases by two.

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The boiling point of a 0.1 mole glucose in 200 ml water solution is approximately 100.255°C at 1 atm, after calculating the molality as 0.5 m and applying the molal boiling point elevation constant for water of 0.51°C/m.

The question asks: What is the boiling point of a solution of .1 mole of glucose in 200 ml of water? To calculate this, we need to first determine the molality (m) of the glucose solution since we have the molal boiling point elevation constant (Kb) for water, which is 0.51°C/m.

The molality (m) is calculated by the number of moles of solute per kilogram of solvent (water in this case). Given that 0.1 mole of glucose is dissolved in 200 ml (or 0.2 kg) of water, the solution's molality would be 0.5 m (0.1 mole / 0.2 kg). Since the constant Kb is 0.51°C/m, the boiling point elevation would be 0.5 m x 0.51°C/m = 0.255°C.

The normal boiling point of water is 100°C at 1 atm. We add the boiling point elevation to this to get the boiling point of the glucose solution: 100°C + 0.255°C = 100.255°C. Therefore, the boiling point of the given glucose solution would be approximately 100.255°C at 1 atm.

Answer:

344.02 g.

Explanation:

no. of moles (n) = mass/molar mass

no. of moles Al(OH)₃ = 4.41 mol & molar mass of Al(OH)₃ = 78.01 g/mol.

∴ mass = no. of moles * molar mass = (4.41 mol)*(78.01 g/mol) = 344.02 g.

Answer:

0.643 mol.

Explanation:

where, P is the pressure of the gas in atm (P = 4.0 atm).  

V is the volume of the gas in L (V = 4000 mL = 4.0 L).  

n is the no. of moles of the gas in mol (n = ??? mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),  

T is the temperature of the gas in K (T = 30ºC + 273 = 303 K).

∴ n = PV/RT = (4.0 atm)(4.0 L)/(0.0821 L.atm/mol.K)(303 K) = 0.643 mol.

Answer:

[tex]\boxed{\text{18.2 kJ}}[/tex]

Explanation:

The formula for the heat involved is  

[tex]q = m\Delta_{\text{f}}\text{H}[/tex]

Data:

m = 115 g

[tex]\Delta_{\text{f}}\text{H} = \text{158.3 J/g}[/tex]

Calculation:

[tex]q = \text{115 g} \times \dfrac{\text{158.3 J}}{\text{1 g}}\\\\q = \text{18 200 J} = \textbf{18.2 kJ}}\\\\\text{It takes }\boxed{\textbf{18.2 kJ}} \text{ to melt the p-xylene}[/tex]

Answer:

D) 1/2

Explanation:

Using Ideal gas equation for same mole of gas as

[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]

Given,

P₂ = 4P₁

T₂ = 2T₁

Using above equation as:

[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]

[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{4\times P_1}\times {V_2}}{2\times T_1}[/tex]

[tex]V_2=\frac{1}{2}\times V_1[/tex]

The volume change by half of the original.

¹/₂

Given:

Question:

By what factor does the volume of the sample change?

The Process:

We use an equation of state for an ideal gas:  

[tex]\boxed{\boxed{ \ \frac{pV}{T} = constant \ }}[/tex]

For the same amount of substances in two states, the equations for state-1 and state-2 are as follows:

[tex]\boxed{ \ \frac{p_2V_2}{T_2} = \frac{p_1V_1}{T_1} \ }[/tex]

Let us use the equation above to see the relationship between volumes. Enter all the information in the equation.

[tex]\boxed{ \ \frac{4p_1V_2}{2T_1} = \frac{p_1V_1}{T_1} \ }[/tex]

[tex]\boxed{ \ \frac{4V_2}{2} = V_1} \ }[/tex]

[tex]\boxed{ \ 2V_2 = V_1} \ }[/tex]

[tex]\boxed{ \ V_2 = \frac{1}{2}V_1} \ }[/tex]

Thus by factor ¹/₂, the volume of the sample will change.

- - - - - - - - - -

Notes

[tex]\boxed{ \ \frac{pV}{nT} = R \ } \rightarrow \boxed{ \ pV = nRT \ }[/tex]

n = moles of ideal gas

R = the molar gas constant (in J mol⁻¹ K⁻¹)

Keywords: the pressure of a gas sample, an ideal gas, volume, constant, moles, equation of state , quadrupled, the absolute temperature, doubled, by what factor, change

Answer:

Explanation:

Stoichiometry is the cuantitative study of the chemical reactions.

It is like algebra applied to chemical equations.

The cuantitative relations between the amount of reactants and products is determined by the law of conservation of mass: the number of each kind of atoms in the reactants must equal the number of the same kind of atoms in the products.

Once that relation has been established, as mole ratios, then it can be determined the amount of reactant neeed to obtain a certain amount of product, or vice versa, determine the amount of product that can be obtained from a given amount of reactants.

That is why, after you know the reactants and products in a chemical equation you must balance to assure that the relative amounts are properly established.

Final answer:

The first step in most stoichiometry problems is to plan the problem by writing and balancing the chemical equation correctly, which is foundational for accurate stoichiometry calculations.

Explanation:

The first step in most stoichiometry problems is to plan the problem. This typically involves writing and balancing the chemical equation. Ensuring that all formulas are correct and balanced is crucial as it lays the foundation for all subsequent calculations in the stoichiometry process. Once the equation is balanced, you can proceed to write the ionic and net ionic equations if necessary, assign oxidation numbers, or derive stoichiometric factors to relate the amounts of substances involved. It is also important to identify the 'given' information and what the problem is asking you to 'find,' as well as list other known quantities.

A water molecule consists of two hydrogen atoms bonded to an oxygen atom, and its overall structureis bent. This is because the oxygen atom, in addition to forming bonds with the hydrogen atoms, also carries two pairs of unshared electrons.

Answer:

Explanation:

The structure of water molecule is very simple . it's has a central oxygen atom ( which has a valency of two ) . Since it's valency is two , it can two hydrogen atoms to both of it's sides ( which have a valency of one )

Structure :

H-O-H

Water is known as amphoteric substance , as it has the ability to act as either a base or an acid ( depending upon the substance it's reacting with ) . On top of that a very electro negative oxygen atom reacts with an electro positive hydrogen gives the molecule a strong bonding force ( which results in a molecule that is held by hydrogen bonding , which is a very strong attraction) . And when its bonded with some strong electro forces of attraction , Guess what happens ? It's boiling point increases !!!! ( It means you have to heat it up more to boil water ) . Which also means it has high heat of vaporization ( to get water into it's vapour state ) .all of that just because it has high forces if attraction between one another ...

In the reaction N2 + 3 H2 → 2 NH3, each nitrogen atom is reduced from 0 to -3, and each hydrogen atom is oxidized from 0 to +1. This is an oxidation-reduction reaction where nitrogen gains electrons (reduction), and hydrogen loses electrons (oxidation).

The correct statement for the provided reaction N2 + 3 H2 → 2 NH3 is: 'Each N atom is reduced from 0 to -3. Each H atom is oxidized from 0 to +1.' This reaction is an oxidation-reduction reaction, wherein the nitrogen atom is reduced (its oxidation number decreases from 0 to -3), and each hydrogen atom is oxidized (its oxidation number increases from 0 to +1).

In the process, nitrogen is gaining electrons, thus being reduced. On the other hand, hydrogen is losing electrons, thus being oxidized. This principle is aligned with the redox reactions wherein one element loses electrons (oxidation) and another element gains electrons (reduction). The reaction equation also follows the law of conservation of mass stating that matter cannot be created or destroyed.

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Answer: i really       dont know srry

Explanation:

Answer:

S₂(s) + C(s) → CS₂(s).

Explanation:

S₂(s) + C(s) → CS₂(s).

1 mol of S₂(s) reacts with 1 mol of charcoal (C(s)) to produce 1 mol of CS₂(s).

Answer:

CoCl₃ > Li₂SO₄ > NH₄I.

Explanation:

ΔTf = i.Kf.m,

where, ΔTf is the depression in freezing point.

i is the van 't Hoff factor.

Kf is the molal depression constant of water.

m is the molality of the solution.

(1) Li₂SO₄:

i for Li₂SO₄ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

∴ ΔTb for (Li₂SO₄) = i.Kb.m = (3)(Kf)(m) = 3(Kf)(m).

(2) NH₄I:

i for NH₄I = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (NH₄I) = i.Kb.m = (2)(Kf)(m) = 2(Kf)(m).

(3) CoCl₃:

i for CoCl₃ = no. of particles produced when the substance is dissolved/no. of original particle = 4/1 = 4.

∴ ΔTb for (CoCl₃) = i.Kb.m = (4)(Kf)(m) = 4(Kf)(m).

CoCl₃ > Li₂SO₄ > NH₄I.

Answer:

1. Percentage composition O in CO₂ is 72.7%

2.Percentage composition of O in N₂O₅ is 74.1%

Explanation:

How to calculate percentage composition

I. Calculate the molar mass of the compound by summing up the atomic masses of the elements that makes up the compound

II. The percentage composition of the element is derieved by dividing the atomic mass of the atoms by the molar mass of the compound

III. Now express this ratio as a percentage.

1. Percentage composition of Oxygen in CO₂:

Molar mass of CO₂

Atomic mass of C = 12gmol⁻¹

Atomic mass of O = 16gmol⁻¹

Note: We have two atoms of Oxygen

Molar mass = [12 + (2x16)]gmol⁻¹

= (12 + 32)gmol⁻¹

= 44gmol⁻¹

Percentage composition O in CO₂

= (2x16)/44 x 100

= 32/44 x 100

= 0.727 x 100

= 72.7%

Percentage composition O in CO₂ is 72.7%

2. The percentage composition of O in N₂O₅

Atomic mass of N = 14gmol⁻¹

Atomic mass of O = 16gmol⁻¹

Molar mass of N₂O₅ = [(2x14) + (5x16)]gmol⁻¹

= (28 + 80)gmol⁻¹

= 108gmol⁻¹

Percentage composition of O in N₂O₅

= (5x16)/108 x 100

= 80/108 x 100

= 0.741 x 100

= 74.1%

Percentage composition of O in N₂O₅ is 74.1%

Note: Percentage composition is expressed as a percentage.

Final answer:

The percent composition of oxygen in CO2 is approximately 72.7%, and in dinitrogen pentoxide, it is approximately 74.1%. This is calculated by dividing the mass of oxygen in each molecule by the total molar mass of the molecule and then multiplying by 100%.

Explanation:

Percent Composition of Oxygen in Compounds

The percent composition of an element in a compound represents the mass percentage of that element in the total mass of the compound. To calculate the percent composition of oxygen in CO2 (carbon dioxide) and dinitrogen pentoxide, we need to look at the molar masses of these compounds and the elements within them.

CO2:

Carbon dioxide is composed of one carbon atom and two oxygen atoms. The molar mass of carbon is 12.01 g/mol and oxygen is 16.00 g/mol. Hence, the molar mass of CO2 equals 44.01 g/mol (12.01 + (16.00 × 2)). The mass of oxygen in CO2 is 32.00 g/mol, which we get from (16.00 × 2). To find the percent composition of oxygen in CO2, we divide the mass of oxygen by the molar mass of CO2 and multiply by 100%.

Percent composition of O in CO2 = (32.00 g/mol / 44.01 g/mol) × 100% ≈ 72.7%

Dinitrogen Pentoxide:

Dinitrogen pentoxide consists of two nitrogen atoms and five oxygen atoms. To calculate the percent composition of oxygen, we first find the molar mass of N2O5, which is 108.01 g/mol (14.01 × 2 + 16.00 × 5). The mass of oxygen in N2O5 is 80.00 g/mol, from (16.00 × 5). The percent composition of oxygen is then calculated as follows:

Percent composition of O in N2O5 = (80.00 g/mol / 108.01 g/mol) × 100% ≈ 74.1%

Answer:

818.2 g.

Explanation:

M = (no. of moles of NaCl)/(Volume of the solution (L))

M = 2.0 M.

no. of moles of NaCl = ??? mol,

Volume of the solution = 7.0 L.

∴ (2.0 M) = (no. of moles of NaCl)/(7.0 L)

∴ (no. of moles of NaCl) = (2.0 M)*(7.0 L) = 14.0 mol.

no. of moles of NaCl = mass/molar mass

∴ mass of NaCl = (no. of moles of NaCl)*(molar mass) = (14.0 mol)*(58.44 g/mol) = 818.2 g.

Answer:

False

Explanation:

In chemistry, trigonal planar is a molecular geometry model with one atom at the center and three atoms at the corners of an equilateral triangle, called peripheral atoms, all in one plane. In an ideal trigonal planar species, all three ligands are identical and all bond angles are 120°.

Meaning there shouldn't be any lone pair.

Look up "Structure of a trigonal planar molecule" for a visual

So it is false.

Answer:

Explanation:

The innermost electron shell is the lowest principal energy level, i.e n = 1.

For n = 1 there is only one orbital, the 1s orbital.

As stated by the Pauli's exculsion principle an orbital may have a maximum of two electrons, and they have opposed spins.

Then, the innermost electron shell has just one orbital and, in consequence, can hold up to 2 elecrons.

The innermost electron shell of an atom can hold up to two electrons.

The innermost electron shell of an atom, also known as the first shell or K-shell, can indeed hold a maximum of 2 electrons. This is often based on the quantum mechanical model of the atom, where the electrons are organized into various shells and subshells.

In addition to this, the first shell consists of only one subshell, called the 1s subshell, which accommodates a maximum of 2 electrons. The distribution of electrons in shells and subshells is a fundamental aspect of atomic structure and determines the chemical properties of elements, as well as their interactions in chemical reactions and bonding.

Limiting reactant is the answer

(1/2)^3 is the fraction of a sample after 3 half lives

The heavy particles all passed straight through the foil, because the atoms are mostly empty space.

Answer is third choice

Answer:  Some of the heavy particles bounced off the foil, because there is a dense, positive area in the atom.

Explanation:

In Rutherford's experiment, he took a gold foil and bombarded it with alpha particles which carry positive charge. He thought that the alpha particles will pass straight through the foil, but to his surprise, many of them passed through, some of them deflected their path and a few of them bounced back.

From this he concluded that in an atom, there exist a small positive charge in the center. Due to this positive charge, the alpha particles deflected their path and some of them bounced straight back their path.

Thus he concluded that there is a dense, positive area in the atom.