An "empty" container is not really empty if it contains air. How may moles of nitrogen are in an "empty" two-liter cola bottle at atmospheric pressure and room temperature (25∘C)? Assume ideal behavior.What is the partial pressure of oxygen in air at atmospheric pressure (1 atm)? Assume ideal behavior

Explanation of pH changes in a titration between NH₃ and HCl.

The pH in a titration is determined by the amount of acid and base before and after the equivalence point. In the given scenario, titrating NH₃ with HCl, the initial pH is determined by the NH₃ concentration before any HCl is added.

(a) 0.00 mL: Before any HCl is added, NH₃ is in excess, making the solution basic with a pH >7.

(b) 25.00 mL: At the equivalence point, the pH is determined by the salt formed during the reaction. For NH₃ titrated with HCl, the salt formed is NH₄Cl, making the pH acidic.

Answer:

3.46*10⁻²² kJ

Explanation:

By the Avogadro's number, 1 mole of electrons at 1 mole of atoms correspond to 6.02x10²³ electrons. So it's necessary 208.4 kJ to remove 6.02x10²³ electrons. To remove a single electron:

6.02*10²³ electrons ---------------- 208.4 kJ

1 electron                  ---------------- x

By a simple direct three rule:

6.02*10²³ x = 208.4

x = 3.46*10⁻²² kJ

Answer: The concentration of [tex]Au^{3+}[/tex] is [tex]1.096\times 10^{-6}[/tex]

Explanation:

The given chemical cell follows:

[tex]Pt(s)|H_2(g,1atm)|H^+(aq,1.0M)||Au^{3+}(aq,?M)|Au(s)[/tex]

Oxidation half reaction: [tex]H_2(g,1atm)\rightarrow 2H^{+}(aq,1.0M)+2e^-;E^o_{2H^{+}/H_2}=0.0V[/tex]       ( × 3)

Reduction half reaction: [tex]Au^{3+}(aq,?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V[/tex]       ( × 2)

Net cell reaction: [tex]3H_2(g,1atm)+2Au^{3+}(aq,?M)\rightarrow 6H^{+}(aq,1.0M)+2Au(s)[/tex]

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:

[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]

Putting values in above equation, we get:

[tex]E^o_{cell}=1.50-(0.0)=1.50V[/tex]

To calculate the EMF of the cell, we use the Nernst equation, which is:

[tex]E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}[/tex]

where,

[tex]E_{cell}[/tex] = electrode potential of the cell = 1.23 V

[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

[tex][H^{+}]=1.0M[/tex]

[tex][Au^{3+}]=?M[/tex]

Putting values in above equation, we get:

[tex]1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})[/tex]

[tex][Au^{3+}]=-1.0906\times 10^{-6},1.096\times 10^{-6}[/tex]

Neglecting the negative value because concentration cannot be negative.

Hence, the concentration of [tex]Au^{3+}[/tex] is [tex]1.096\times 10^{-6}[/tex]

Answer:

72 kJ of heat is removed.

Explanation:

First, the ethanol vapor will reduce its temperature until the temperature of the boiling point, then it will occur a phase change from vapor to liquid, and then the temperature of the liquid will decrease. The total heat will be:

Q = Q1 + Q2 + Q3

Q1 = n*cv*ΔT1, Q2 = m*Hl, and Q3 = n*cl*ΔT2

Where n is the number of moles, cv is the specific heat of the vapor (65.44 J/K.mol, cl is the specific heat of the liquid (111.46 J/K.mol), Hl is the heat of liquefaction (-836.8 J/g), m is the mass, and ΔT is the temperature variation (final - initial).

Q = n*cv*ΔT1 + m*Hl + n*cl*ΔT2

The molar mass of ethanol is 46 g/mol, and the number of moles is the mass divided by the molar mass:

n = 74.2/46 = 1.613 moles

Q = 1.613*65.44*(78.37 - 83) + 74.2*(-836.8) + 1.613*111.46*(26 - 78.37)

Q = -72000 J

Q = -72 kJ (because it is negative, it is removed)

Answer:

The amount of air is 11.1 mmol

Explanation:

Density shows the relation between mass and volume, so you can know the mass which has been sealed in the Erlenmayer.

Density = mass/ volume

Density . volume = mass

0.001185 g/ml . 272.2 ml = 0.322 g

We have the mass now, so the molar weight determinates the mols

Mass/molar weight = mol

0.322g/ 28.96 g/m = 0.0111 mol

To get a better number we can inform on milimols

Mols . 1000 = milimols

The amount of air sealed in the 272.2 mL Erlenmeyer flask is approximately 0.0111 mol, calculated using the given density and molar mass of air at 25°C.

To determine the amount of air sealed in the flask, we need to use the density of the air (given as 0.001185 g/mL at 25°C), the volume of the flask and the molar mass of air.

First, using the density and the volume of the flask, we can calculate the mass of the air in the flask with the formula density = mass/volume. We multiply the given density of 0.001185 g/mL by the volume of 272.2 mL to find the mass of the air to be around 0.3224 g.

Next, to convert this mass to moles, we use the given molar mass of air (28.96 g/mol). Moles = mass / molar mass, hence the air in the flask is approximately 0.0111 mol. Thus, there is approximately 0.0111 mol of air sealed in the Erlenmeyer flask.

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With every addition of NaOH, the concentration of HC2H3O2 in the solution decreases due to the neutralization reaction. Thus, the quantity of HC2H3O2 after adding 5.0 mL of NaOH will be less than the amount after adding 1.0 mL of NaOH. This is due to the stoichiometry of the reaction, where one mole of added NaOH neutralizes one mole of acetic acid.

In this titration analysis, the neutralization reaction involves acetic acid (HC2H3O2) and Sodium Hydroxide (NaOH), creating sodium acetate and water as products. When 1.0 mL of NaOH is added, it reacts with the acetic acid present in the flask, reducing its quantity. When 5.0 mL of NaOH is added, more acetic acid reacts, and therefore, the amount of HC2H3O2 in the flask decreases more significantly compared to when only 1 mL was added. The principle of this result is based upon the stoichiometry of the neutralization reaction, indicating that for each mole of NaOH added, one mole of acetic acid is neutralized. Therefore, as more NaOH is added, more HC2H3O2 is neutralized, hence contributing to a lower concentration of HC2H3O2 in the solution.

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The amount of HC2H3O2(aq) in the flask will be less after the addition of 5.0mL NaOH than after the addition of 1.0mL NaOH. This happens because the NaOH neutralizes the acid directly proportional to its quantity added. More the NaOH added, more the acidic substance HC2H3O2 gets neutralized.

In a titration process involving HC2H3O2(aq) and NaOH(aq), following the neutralization equation, each acid molecule reacts with one base molecule. Hence, the quantity of HC2H3O2 will reduce in direct proportion to the amount of NaOH added. So if you add 5.0mL of NaOH, you will neutralize 5 times as much HC2H3O2 as when you add 1.0mL of NaOH.

Therefore, the quantity of HC2H3O2(aq) remaining in the flask after the addition of 5.0mL NaOH will be less than that after the addition of 1.0mL NaOH. The reason is that more NaOH means more HC2H3O2(aq) has been neutralized into water and the salt NaC2H3O2.

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Explanation:

The given data is as follows.

      50 ml of [tex]Cl_{2}[/tex],       50 ml of [tex]C_{2}H_{4}[/tex]

And, it is known that at STP 1 mole of a gas occupies 22.4 L. Hence, moles present in 50 ml of gas are as follows.

          [tex]\frac{50}{22.4 \times 1000}[/tex]      (As 1 L = 1000 ml)

          = [tex]2.23 \times 10^{-3}[/tex] moles

So, according to the given equation [tex]2.23 \times 10^{-3}[/tex] moles of [tex]Cl_{2}[/tex] reacts with [tex]2.23 \times 10^{-3}[/tex] moles of [tex]C_{2}H_{4}[/tex].

Hence, moles of [tex]C_{2}H_{4}Cl_{2}[/tex] is equal to the moles of [tex]C_{2}H_{4}[/tex] and [tex]Cl_{2}[/tex].

Therefore, moles of [tex]C_{2}H_{4}Cl_{2}[/tex] = [tex]2.23 \times 10^{-3}[/tex] moles

           1 mole of [tex]C_{2}H_{4}Cl_{2}[/tex] = 22.4 L

   [tex]2.23 \times 10^{-3}[/tex] moles = [tex]22.4 \times 2.23 \times 10^{-3} moles[/tex]        

                                = 50 ml of product

Thus, we can conclude that 50 ml of products if pressure and temperature are kept constant.

Answer:

-87.4 kJ/mol

Explanation:

We can rewrite Arrhenius' equation as:

ln(k₂/k₁) = -Ea/R * (1/T₂ - 1/T₁)

Where k₂ and k₁ are the rates of the reaction at temperatures T₂ and T₁, Ea is the activation energy and R is the universal gas constant.

For this problem:

k₂ = 5

k₁ = 17

T₂ = 15°C = 288.16 K

T₁ = 25°C = 298.16 K

R = 8.314 J/mol·K

We put the data in the equation and solve for Ea:

ln(5/17) = -Ea/8.314 J/mol·K * (1/288.16K - 1/298.16K)

-1.224 = -Ea/8.314 J/mol·K * 1.1639x10⁻⁴K⁻¹

Ea = -87416.78 J/mol ≅ -87.4 kJ/mol

Answer:

A

Explanation:

The slow step of the reaction is the rate determining step. This means to get the rate law for a chemical activity with slow and fast steps, we have to consider the slow step if we are to write the rate law successfully.

Since the compound NO is reacting with itself, we have to raise the value of the concentration in the square bracket by 2. That is why we have the concentration squared.

Answer: it’s c

Explanation:

Answer:

1.40 atm

Explanation:

To answer this question we can use Gay-Lussac's law, which states:

When volume and number of moles remain constant.

We put the known data in the equation and solve for P₂:

Using the ideal gas law, the pressure inside an aerosol can increases proportionately with temperature. If the can's temperature is raised to the boiling point of water (100°C or 373 K), the pressure increases from 1.11 atm to 1.40 atm.

The pressure inside an aerosol can, like this one which has a residual propellant, can be explained by the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature when the volume and amount of gas are held constant. If the can is heated to the boiling point of water (100°C or 373 K), we can expect the pressure to increase proportionately.

To find out the new pressure, we need to convert the initial temperature to Kelvin (23°C = 296 K), and then use the formula P1/T1 = P2/T2 where P1 is the initial pressure, T1 is the initial temperature, P2 is the pressure at elevated temperatures, and T2 is the temperature at the boiling point of water.

So, P2 = P1×T2/T1 = 1.11 atm × 373 K / 296 K = 1.40 atm. Therefore, the pressure inside the can at the boiling point of water is 1.40 atm.

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The question deals with the photoelectric effect, where light ejects electrons from a metal's surface. To identify the metal, one must calculate the work function using the kinetic energy of the emitted electrons and the frequency of the incident light.

The question concerns the photoelectric effect, a phenomenon in physics where electrons (called photoelectrons) are ejected from a metal surface when it is exposed to light of a certain frequency or wavelength that is above the metal's threshold frequency. You've given a light with a wavelength of 190 nm incident on an unknown metal, and the most energetic electrons emitted have 4.0 eV of kinetic energy.

To identify which metal this could be, one needs to calculate the work function of the metal. This involves using the equation for the photoelectric effect: KE = hf - Φ, where KE is the kinetic energy of the ejected electrons, h is Planck's constant, f is the frequency of the incident light, and Φ represents the work function (also known as the binding energy) of the metal.

First, we need to convert the wavelength of 190 nm into frequency using the equation c = λf, where c is the speed of light, λ is the wavelength, and f is the frequency. Once we have the frequency, we can plug it into Planck's equation to find the energy of the photons. Subsequently, we'll use the initial energy of the photons to determine the work function. By comparing this calculated work function with the known work functions of various metals, we can identify the most likely metal of the surface.

Answer:

C(graphite) → C(diamond), ΔH = -0.45

CH4 + 2O2 → CO2 + 2H2O + 212,800 cal

Explanation:

NH3(g) + 12.0 kcal → ½N2(g) + 3/2H2(g) in this reaction we see that energy is added to the reactants so this is an endorthermic process. It takes in energy

C(graphite) → C(diamond), ΔH = -0.45 kcal/mole The change in enthalpy for this reaction is listed and it is negative, telling us that this is an exothermic reaction.

C + 2S → CS2, ΔH = 27,550 cal  The change of enthalpy for this reaction is listed and it is positive which tells us that the reaction is endothermic reaction

CH4 + 2O2 → CO2 + 2H2O + 212,800 cal In this reaction, the energy is part of the equation on the products side. It is given out. this tells us that the reaction is exothermic

2H2 O → 2H2 + O2, ΔH = +58 kcal/mole H2O In this reaction, the energy is part of the equation on the products side. It is given out. this tells us that the reaction is exothermic

Answer:

a. 5 0 0 +1/2

5 0 0 -1/2

Valid

b. 4 1 -1 +1/2

4 1 0 +1/2

4 1 +1 +1/2

Valid

c. 3 2 -1 +1/2

3 2 0 +1/2

3 2 +1 +1/2

3 2 0 +1/2

3 2 +2 +1/2

Pauli violation

d. 3 1 -1 +1/2

3 1 0 +1/2

3 3 +1 +1/2

other violation

Explanation:

The four quantum number and possible values are:

n = 1,2,3.....

l = 0 , (n-1), (n-2).....

m = +l , 0 , -l

s = [tex]+\frac{1}{2}[/tex] or [tex]-\frac{1}{2}[/tex]

Pauli's exclusion principle: No two electrons in an atom can have all the four quantum numbers same.

Let us check each case:

a. 5 0 0 +1/2

5 0 0 -1/2

Valid

b. 4 1 -1 +1/2

4 1 0 +1/2

4 1 +1 +1/2

Valid

c. 3 2 -1 +1/2

3 2 0 +1/2

3 2 +1 +1/2

3 2 0 +1/2

3 2 +2 +1/2

Pauli violation

The two electrons have same four quantum numbers

3 2 0 +1/2

3 2 0 +1/2

d. 3 1 -1 +1/2

3 1 0 +1/2

3 3 +1 +1/2

other violation

As mentioned above in the condition the value of "l" can be only less than "n"

So for 3 3 +1 +1/2 : n = 3 and l= 3, which is not valid.

The arrangement of electrons in orbitals are showed by four sets of quantum numbers.

According to the Pauli exclusion theory, no two electrons in an atom should have all the four quantum numbers as the same. According to this principle, the spin quantum number of electrons in an atom must differ even if they are in the same orbital.

For the first set;

5 0 0 +1/2

5 0 0 -1/2

This set, correctly corresponds to the 5s  orbital so it is valid.

For the second set;

4  1  -1  +1/2

4  1  0  +1/2

4  1  +1  +1/2

This should have corresponding to the 4p orbital so it is valid.

For the third set;

3  2 -1  +1/2

3  2  0  +1/2

3  2  +1  +1/2

3  2  0  +1/2

3  2 +2  +1/2

This set should correspond to a 3d orbital but we can see that are two electrons in the set that has exactly the same quantum numbers of 3 2 0 +1/2. This violates the Pauli exclusion theory so we should mark "Pauli violation".

For the fourth set:

3  1  -1  +1/2

3   1   0  +1/2

3  3  +1  +1/2

This set should have corresponded to a 3p orbital but remember that the values of l only range from 0 to (n - 1). This means that we can not have n =3, l=3 so the arrangement 3  3  +1  +1/2 is not possible. We should mark  "other violation".

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Answer:

elemental form,  readily form compounds

Explanation:

Copper, silver, and gold have all been know since ancient times because they appear in nature in_elemental form___ and were thus discovered thousands of years ago. But the majority of elements__readily form compounds_and, consequently, are hard to find in nature.

Answer:

A, B, & C

Explanation:

got it right on edu

The thermal energy of the bathtub can be increased by adding hot water, heating the water, and adding more water to the tub at the same temperature. Thus, options A, B, and C are correct.

Thermal energy is the increase in the collision and the kinetic energy of particles of matter due to a rise in the temperature of the substance. This movement produces energy called thermal energy.

The thermal energy of the substance can be increased by providing them with the same energy molecules. The addition of the hot water will increase the energy due to increased collision of the particles of the hot water.

Heating the water will increase the kinetic energy and the rate of a collision resulting in increased thermal energy. Also, water with the same temperature can be added to increase energy.

Therefore, in options A, B, and C adding hot water and heating water increases the thermal energy.

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Answer: D.) A method to determine the sequence of bases in a gene.

Explanation:Polymerase chain reaction (PCR) is a method widely used in molecular biology to make several copies of a specific DNA segment. Using PCR, copies of DNA sequences are exponentially amplified to generate thousands to millions of more copies of that particular DNA segment

Answer:

Ammonia acts as an Arrhenius base because it increases the concentration of OH⁻ in aqueous solution.

Explanation:

The acid-base theory of Arrhenius explains that in aqueous solutions both acid and base dissociate, releasing ions in the solution. The acid release the ion H⁺ and some anion, and the base release the ion OH⁻ and some cation.

In water, the reaction of ammonia is:

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

Because of that, ammonia is an Arrhenius base.

Ammonia acts as an Arrhenius base because it increases the concentration of hydroxide ions in solution through a reaction with water molecules where it accepts a proton, forming NH4+ and OH-. The Arrhenius definition is focused on hydroxide ion production; however, broader definitions like Brønsted-Lowry and Lewis provide a more general understanding of acid-base chemistry.

Ammonia, NH3, acts as an Arrhenius base because it increases the concentration of hydroxide ion, OH-, in aqueous solution by accepting a proton from water. This reaction can be represented by:

NH3(aq) + H2O(l) ⇒ NH4+(aq) + OH-(aq)

In this reaction, ammonia takes a hydrogen ion (H+) from a water molecule to produce an ammonium ion (NH4+) and a hydroxide ion (OH-). Unlike what might be initially assumed, ammonia does not release hydroxide ions into a solution directly; rather, the concentration of hydroxide ions increases due to this proton transfer.

Additionally, more fundamental definitions of acids and bases, beyond the Arrhenius model, include the Brønsted-Lowry theory, where bases are substances that accept hydrogen ions (H+), and the Lewis model, where a base is an electron pair donor.

Overall, while the Arrhenius model was originally seen as limited, it helps to explain how substances like ammonia function as bases without containing hydroxide ions in their formula.

Answer:

SO₂

Explanation:

The dipole-dipole force is not only determined by the electron density around each atoms in the molecule (dependent of electronegativity difference), but also how the atoms in the molecules are arranged. In general, the more symmetric a molecule is, the less dipole force it exerts as each dipole moments cancels each other out.

Now let's examine each answer

Answer:

1.[tex]S^{-}(g)+e^{-} \rightarrow S^{2-}(g)[/tex]

2.[tex]Ti^{2+}(g) \rightarrow Ti^{3+}(g) \,\, IE = 2652.5\,kJ.mol^{-1}[/tex]

3.The electron affinity of  [tex]Mg^{2+}[/tex] is zero.

4.[tex]O^{2-}(g) \rightarrow O^{-}(g)+e^{-}[/tex]

Explanation:

1.

Electron affinity:

It is defined as the amount of energy change when an electron is added to atom in the gaseous phase.

The electron affinity of [tex]S^{-}[/tex] is as follows.

[tex]S^{-}(g)+e^{-} \rightarrow S^{2-}(g)[/tex]

2.

Ionization energy:

Amount of energy required to removal of an electron from an isolated gaseous atom.

The third ionization energy of Titanium is as follows.

[tex]Ti^{2+}(g) \rightarrow Ti^{3+}(g) \,\, IE = 2652.5\,kJ.mol^{-1}[/tex]

3.

The electronic configuration of Mg: [tex]1s^{2}2s^{2}2p^{6}3s^{2}[/tex]

By the removal of two electrons from a magnesium element we get [tex]Mg^{2+}[/tex] ion.

[tex]Mg^{2+}[/tex] has inert gas configuration i.e,[tex]1s^{2}2s^{2}2p^{6}[/tex]

Hence, it does not require more electrons to get stability.

Therefore,the electron affinity of  [tex]Mg^{2+}[/tex] is zero.

4.

The ionization energy of [tex]O^{2-}[/tex] is follows.

[tex]O^{2-}(g) \rightarrow O^{-}(g)+e^{-}[/tex]

The electron affinity equation for S⁻ is S⁻(g) + e⁻ → S²⁻(g). The third ionization energy equation for titanium is Ti²⁺(g) → Ti³⁺(g) + e⁻. The equation representing the electron affinity of Mg²⁺ is Mg²⁺(g) + e⁻ → Mg³⁺(g). The ionization energy equation for O²⁻ is O²⁻(g) → O³⁻(g) + e⁻.

1. The electron affinity of S⁻: The electron affinity represents the energy change that occurs when an electron is added to a neutral atom or ion. For sulfur (S⁻), the equation representing this process is: S⁻(g) + e⁻ → S²⁻(g). Here, the physical state of S⁻ and S²⁻ is gas.

2. The third ionization energy of titanium: Ionization energy is the energy required to remove an electron from an atom or ion. For titanium, the equation representing the third ionization energy is: Ti²⁺(g) → Ti³⁺(g) + e⁻. The physical state of Ti²⁺ and Ti³⁺ is gas.

3. The electron affinity of Mg²⁺: The electron affinity of a cation, such as Mg²⁺, is the energy change when an electron is added to the cation. Here, the equation representing this process is: Mg²⁺(g) + e⁻ → Mg³⁺(g). The physical state of Mg²⁺ and Mg³⁺ is gas.

4. The ionization energy of O²⁻: Ionization energy is the energy required to remove an electron from an atom or ion. For oxygen, the equation representing the ionization energy of O²⁻ is: O²⁻(g) → O³⁻(g) + e⁻. The physical state of O²⁻ and O³⁻ is gas.

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Explanation:

As the given reaction equation is as follows.

         [tex]N_{2}(g) + O_{2}(g) \rightleftharpoons 2NO(g)[/tex]

Hence, expression for [tex]K_{p}[/tex] of this reaction is as follows.

        [tex]K_{p} = \frac{P^{2}_{NO}}{P_{N_{2}} \times P_{O_{2}}}[/tex]

Now, putting the given values into the above expression as follows.

         [tex]K_{p} = \frac{P^{2}_{NO}}{P_{N_{2}} \times P_{O_{2}}}[/tex]

                 = [tex]\frac{(0.050)^{2} atm}{(0.15 atm) \times (0.33 atm)}[/tex]  

                 = [tex]5.05 \times 10^{-2}[/tex]

Thus, we can conclude that the value of [tex]K_{p}[/tex] is [tex]5.05 \times 10^{-2}[/tex].

Final answer:

The equilibrium constant, Kp, for the reaction N2(g) + O2(g) = 2 NO(g) at the given conditions is approximately 0.1010 when calculated using the given equilibrium partial pressures.

Explanation:

The question is asking us to calculate the equilibrium constant, KP, for the reaction given the equilibrium partial pressures of the reactants and products at a certain temperature. Using the reaction N2(g) + O2(g) = 2 NO(g), we can express the equilibrium constant KP in terms of the partial pressures of the gases:

KP = (PNO)2 / (PN2 * PO2)

Substituting in the given partial pressures:

KP = (0.050 ATM)2 / (0.15 ATM * 0.33 ATM) = 0.0050 / 0.0495 = 0.1010

So, the equilibrium constant KP at 2200°C for this reaction is approximately 0.1010.

Answer:

C

Explanation:

A nuclear reaction is one in which 2 atoms combine or an atom divides to give rise to entirely different atoms. This is basically the difference it has with a chemical reaction. In a chemical reaction, only new compounds are formed but not new elements. In the case of a nuclear reaction however, even if new elements are not formed, a different atom would be formed.

Answer:

162 kJ

Explanation:

The reaction given by the problem is:

If we turn it around, we have:

If we think now of HOBr and NH₃ as our reactants, then now we need to find out which one will be the limiting reactant when we have 9 moles of HOBr and 2 moles of NH₃:

-81 kJ is our energy change when there's one mol of NH₃ reacting, so we multiply that value by two when there's two moles of NH₃ reacting. The answer is 81*2 = 162 kJ.

Final answer:

The energy change for the reverse reaction of NBr3 and H2O to form HOBr and NH3 is -81 kJ. When considering nine moles of HOBr and two moles of NH3, thrice the stoichiometric amounts, the expected energy change would be three times as much, resulting in -243 kJ.

Explanation:

The energy change, ΔH, for a reaction indicates how much energy is absorbed or released during the reaction. In the given reaction, NBr3(g) + 3 H2O(g) → 3 HOBr(g) + NH3(g), the energy change is +81 kJ, meaning 81 kJ of energy is absorbed in the forward reaction. Since energy is a state function and the magnitude of energy change is the same for a reaction whether it goes in the forward or reverse direction, the energy change for the reverse reaction would be -81 kJ. The reaction in reverse would release 81 kJ of energy.

However, the question asks for the expected energy change for reversing the original reaction with nine moles of HOBr and two moles of NH3. The energy change provided (+81 kJ) is for the stoichiometric amounts as per the original equation. For nine moles of HOBr and two moles of NH3, which are thrice the stoichiometric amounts indicated by the original reaction, the energy released would be three times as much. Therefore, the expected energy change for the reverse reaction under these conditions would be 3 × (-81 kJ) = -243 kJ.

Answer:

[tex]Age=2.52*10^9 years[/tex]

Explanation:

To determine the age, we need to know how much U have become Pb, so first we have to calculate the Pb moles present in a sample:

[tex]n_{Pb}=\frac{0.337g}{206g/mol}=0.00164 mol[/tex]

[tex]n_{U}=\frac{1 g}{238g/mol}=0.0042 mol[/tex]

The percentage of U degradation:

[tex]P=\frac{n_{Pb}}{n_{Pb}+n_{U}}[/tex]

[tex]P=\frac{0.00164}{0.00164+0.0042}=0.28[/tex]

Assuming that the life time is linear:

[tex]Age=\frac{4.5*10^9 years}{0.5 life time}*0.28 life time[/tex]

[tex]Age=2.52*10^9 years[/tex]

Answer:

7.1 x 10^9 years

Explanation:

This is the answer according to Mastering Chemistry.

Water, Carbon Dioxide, Acids, and Oxygen are the inorganic chemicals common in cells.

Thus, inorganic chemicals like Water, Carbon Dioxide, Acids, and Oxygen

are essentail for cells.

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Answer:

The product of the reaction is 1,2 di-bromo-1-methylcyclopentane.

Explanation:

In the reaction of 1-methylcyclopentene with Br2 in CCl4, a bromonium ion intermediate forms, followed by nucleophilic attack by Cl-, resulting in the product 1-chloromethylcyclopentane.

In the reaction of 1-methylcyclopentene with Br2 in CCl4, a bromonium ion intermediate is formed. The bromonium ion is a three-membered ring with a bromine atom attached to two adjacent carbon atoms. Due to the lack of stereochemistry information, the structure of the intermediate is a general representation without specifying the stereochemical arrangement.

Following the formation of the bromonium ion, it undergoes a nucleophilic attack by the chloride ion (Cl-) from the solvent CCl4. The chloride ion opens the bromonium ring, leading to the substitution of bromine by chloride on one of the adjacent carbons. The product of this reaction is 1-chloromethylcyclopentane. The chlorine atom is now attached to the carbon that originally carried the methyl group.

This reaction mechanism is a classic example of halogenation of alkenes, where the bromine adds to the alkene in a concerted manner, forming a cyclic intermediate. The subsequent nucleophilic attack by the chloride ion results in the replacement of bromine with chlorine, leading to the final halogenated product.

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Answer:

(a) ΔU = 7.2x10²

(b) W = -5.1x10²

(c) q = 5.2x10²

Explanation:

From the definition of power (p), we have:

[tex] p = \frac {\Delta W}{\Delta t} = \frac {\Delta U}{\Delta t} [/tex] (1)

where, p: is power (J/s = W (watt)) W: is work = ΔU (J) and t: is time (s)  

(a) We can calculate the energy (ΔU) using equation (1):

[tex] \Delta U = p \cdot \Delta t = 100 \frac{J}{s} \cdot 2.0\cdot 10^{-2} h \cdot \frac{3600s}{1h} = 7.2 \cdot 10^{2} J [/tex]  

(b) The work is related to pressure and volume by:

[tex] \Delta W = -p \Delta V [/tex]

where p: pressure and ΔV: change in volume = V final - V initial      

[tex] \Delta W = - p \cdot (V_{fin} - V_{ini}) = - 1.0 atm (5.88L - 0.85L) = - 5.03 L \cdot atm \cdot \frac{101.33J}{1 L\cdot atm} = -5.1 \cdot 10^{2} J [/tex]

(c) By the definition of Energy, we can calculate q:

[tex] \Delta U = \Delta W + \Delta q [/tex]

where Δq: is the heat transfer

[tex] \Delta q = \Delta U - \Delta W = 7.2 J - (-5.1 \cdot 10^{2} J) = 5.2 \cdot 10^{2} J [/tex]    

I hope it helps you!  

Answer:

Final temperature T is 26.197 degrees celcius.

Explanation:

As the system is in thermal equilibrium, the heat lost by aluminium is the heat gained by water.

specific heat capacity of aluminium = 0.9 J/g per degree celcius

specific heat capacity of water = 4.186 J/g per degree celcius

Let the final common temperature attained be "T".

Heat lost by aluminium = [tex](m)(c)(65.6-T)[/tex]

Heat gained by water = [tex](m)(c)(T-22)[/tex]

by law of conservation of energy,

Heat lost by hot body = Heat gained by cold body

[tex](m)(c)(65.6-T)[/tex] = [tex](m)(c)(T-22)[/tex]

[tex](27.5)(0.9)(65.6 - T) = (55.5)(4.186)(T - 22)[/tex]

calculating the value of T comes out as 26.197 degrees celcius.

The question is incomplete. The complete question is attached below.

Answer : The correct option is, (A) 7-chloro-3-ethyl-4-methyl-3-heptene

Explanation :

The rules for naming of alkene are :

First select the longest possible carbon chain.

The longest possible carbon chain should include the carbons of double bonds.

The naming of alkene by adding the suffix -ene.

The numbering is done in such a way that first carbon of double bond gets the lowest number.

The carbon atoms of the double bond get the preference over the other substituents present in the parent chain.

If two or more similar alkyl groups are present in a compound, the words di-, tri-, tetra- and so on are used to specify the number of times of the alkyl groups in the chain.

The given compound name will be, 7-chloro-3-ethyl-4-methyl-3-heptene.

The structure of given compound is shown below.

The structure given can be named as trans-7-chloro-3-ethyl-4-methyl-3-heptene.

The structure given can be named as trans-7-chloro-3-ethyl-4-methyl-3-heptene. The name is determined by identifying the longest carbon chain, the substituents attached to it, and their positions. In this case, the longest carbon chain has 7 carbons, with a chlorine atom attached at position 7. There is an ethyl group at position 3 and a methyl group at position 4. The presence of double bonds is indicated by the -ene ending.

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Answer:

Phosphorus cycle

Explanation:

Biogeochemical cycle, any of the natural pathways by which essential elements of living matter are circlated.

There are four types of biogeochemical cycle, they are ; water cycle,carbon cycle,nitrogen cycle and phosphorous cycle

Carbon cycle is the cycle in which photosynthesis and cellular respiration take place.

Water cycle involves transpiration.

Nitrogen cycle Is the cycle that is dependent upon bacteria for nitrogen fixation and denitrification.

Phosphorus cycle is one of the slowest biogeochemical cycle. It does not stay in the atmosphere, because it is normally in a liquid state at room temperature. It does not include the atmosphere.

Final answer:

The phosphorus cycle is the biogeochemical cycle that does not include an atmospheric component, differentiating it from cycles like water, carbon, and nitrogen.

Explanation:

The biogeochemical cycle that does not include the atmosphere is the phosphorus cycle. Unlike the water, carbon, and nitrogen cycles, the phosphorus cycle does not involve the atmospheric pathway. Phosphorus mainly cycles through rocks, soil, water, and living organisms. It is essential for life as part of DNA, RNA, and ATP molecules, which are vital for genetic information and energy transfer in cells. However, phosphorus doesn't enter the atmosphere in a gaseous phase; hence, it is distinct amongst biogeochemical cycles.

Ionic solids are more likely to dissolve in water and can become better conductors of electricity via chemical substitution.

Both covalent-network solids and ionic solids have different properties. Ionic solids dissolve readily in water and can conduct electricity when melted or dissolved because their ions are free to move. Covalent-network solids, on the other hand, are insoluble in water and are poor conductors of electricity in any state because their particles are electrically neutral.

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