This type of nucleus is formed during a solar nuclear reaction by the fusion of four nuclei.

The velocity of the marble can be determined using the de Broglie equation which relates a particle's wavelength to its momentum. Given the marble's mass and wavelength, plug these values into the de Broglie equation and solve for the velocity.

To determine the velocity of the marble, we can use the de Broglie equation, which relates the wavelength of a particle to its momentum. This equation is λ = h/mv, where λ is the wavelength, h is Planck's constant (6.626 x 10^-34 m^2 kg / s), m is the mass of the particle, and v is the speed (velocity) of the particle.

In this problem, we are given: the mass (m) of the marble as 8.66 g (which we convert to kg by dividing by 1000 to get 0.00866 kg), the wavelength (λ) of the marble is given as 3.46 x 10^-33 m, and we are asked to solve for velocity (v). Inserting these numbers into the de Broglie equation, we get: v = h/(mλ) = (6.626 x 10^-34 m^2 kg / s) / (0.00866 kg x 3.46 x 10^-33 m) (solve this on your calculator to get the answer). Due to the given options, the calculated answer must be one of them.

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A gaseous compound is 30.4% nitrogen and 69.6% oxygen by mass. A 5.25-g sample of the gas occupies a volume of 1.00 L and exerts a pressure of 1.26 atm at -4.0°C. Which of the following is its molecular formula?  

1) NO2  

2) N3O6  

3) N2O5  

4) N2O4  

5) NO

Answer:

A. vibrating objects

Explanation:

Final answer:

The pH of a 0.160 M ammonia solution can be calculated using the pOH method. The concentration of hydroxide ions in the solution is approximately 1.125 x 10^-4 M, resulting in a pOH of approximately 3.95. The pH of the solution is approximately 10.05.

Explanation:

The pH of a 0.160 M ammonia solution can be calculated using the pOH method. Ammonia is a weak base, so we can use the equation:
pOH = -log[OH-]
pH + pOH = 14

First, we need to find the concentration of hydroxide ions ([OH-]) in the solution. The Kb value for ammonia (NH3) is 1.8 x 10-5.
Given that [NH3] = 0.160 M, we can calculate [OH-] using the Kb expression:
Kb = [NH3][OH-] / [NH4+]
Since [NH4+] can be ignored in this dilute ammonia solution, the equation simplifies to:
Kb = [NH3][OH-].

Now, rearranging the equation to solve for [OH-], we get:
[OH-] = Kb / [NH3]
Substituting the given values into the equation:
[OH-] = (1.8 x 10-5) / (0.160)
[OH-] = 1.125 x 10-4 M

Next, we can calculate the pOH of the solution using the equation:
pOH = -log[OH-]
pOH = -log(1.125 x 10-4)
pOH ≈ 3.95

Finally, we can find the pH using the equation:
pH = 14 - pOH
pH = 14 - 3.95
pH ≈ 10.05

An electrolyte is a compound that conducts an electric current in an aqueous solution or in the molten state. This happens when the substance dissolves in water and forms ions. The dissolved or molten ions can move freely, allowing the solution to conduct electricity.

A compound that conducts an electric current in an aqueous solution or in the molten state is known as an electrolyte. Electrolytes are formed when certain substances dissolve in water and either undergo a physical or a chemical change that yields ions in solution. These substances are important as they conduct electric current in a solution due to the presence of ions.

For example, ionic compounds like NaCl (table salt) disassociate completely in water forming Na+ and Cl- ions, giving the solution the ability to conduct electricity. In another example, covalent compounds like HCl (hydrochloric acid) also conduct electricity in water. This happens because when HCl is dissolved in water, it reacts with water molecules to form ions of H+ and Cl-.

However, the ability to conduct electricity is not limited to aqueous solutions; compounds can also conduct electricity in the molten state. For instance, when table salt is melted, it’s able to conduct electric current because the ionic bonds are broken, freeing the ions and enabling them to move and carry the current.

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Answer:

A

Explanation:

This kind of bond is known as an ionic bond and is based on electrostatic forces between two charged molecules/atoms. To achieve stable configuration, sodium (2.8.1) transfers its valence electron to chlorine that also needs one electron to achieve stable configuration (2.8.8), and combine in an ionic bond.

Answer: The correct answer is Option A.

Explanation:

Sodium is the 11th element of the periodic table having electronic configuration of [tex][Ne]3s^1[/tex]

This element will loose 1 electron in order to attain stability and will form [tex]Na^+[/tex] ion.

Chlorine is the 17th element of the periodic table having electronic configuration of [tex][Ne]3s^23p^5[/tex]

This element will gain 1 electron in order to attain stability and will form [tex]Cl^-[/tex] ion.

These two elements form ionic compound by transferring of electrons from sodium atom to chlorine atom.

Hence, the correct answer is Option A.

If you don't wash the thermometer, residual NaOH will react with the HCl solution. This is a highly exothermic reaction and will change the temperature of the solution, and thus throw off your measurement.

Not rinsing and drying the thermometer between measurements in an experiment involving HCl and NaOH solutions could skew results by unintentionally introducing acid or base into the other solution, altering the pH and temperature readings, and potentially the titration results.

In the experimental procedure where you measure the temperature of an HCl solution and then a NaOH solution without rinsing and drying the thermometer, your results may be compromised. This is because there could be a carry-over of acid (HCl) or base (NaOH) from one solution to the other which could potentially affect the pH of the solutions being measured. The pH of a solution is dependent on its hydrogen ion concentration, so any residual acid or base on the thermometer could interact with the other solution, altering its pH and therefore its temperature.

Titration is a technique that uses a solution of known concentration to find the concentration of an unknown solution. In your experiment, if you measured the temperature of the HCl solution and then immediately measured the temperature of the NaOH solution without rinsing and drying the thermometer, you might unintentionally introduce some HCl into the NaOH solution. This could potentially skew your results as the introduction of acid could decrease the pH of the NaOH solution, making it less basic than it actually is, and this could affect the titration results.

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Let the mass be X g

percentage = X/ 6.50 * 100 =2.2%

X= 0.143 g

The mass is 0.143 g

The mass is X g

============================

percentage is X/6.50*100 =2.2%

-----------------------------------------------

X=0.143g

-------------------------------

| The mass is 0.143g |

-------------------------------

0.002512 moles of H2O

The reaction between acetic acid ( CH3COOH) and NaOH is given by the equation;

CH3COOH + NaOH ------> CH3COONa + H2O

Number of moles of CH3COOH = molarity × volume in litres

                                                     = 0.08 × 31.4/1000

                                                     = 2.512 × 10^-3  

Similarly number of moles of NaOH = 1 × 24.3/1000

                                                      = 0.0243

From the reaction the mole ratio of CH3COOH : NaOH

Therefore; 0.0243 moles of NaOH will react with 0.0243 moles of CH3COOH but no.of moles of CH3COOH given in the question are 0.002512  moles, which is less than what is required.

Thus; CH3COOH is the limiting reagent and amount of products produced will depend on amount of CH3COOH only.

Since; 1 mole of CH3COOH gives 1 mole of water.

Then; 0.002512 moles of CH3COOH will give 0.002512 moles of H2O

The number of moles produced from the reaction of 24.3 mL of 0.100 M NaOH with 31.4 mL of 0.08M acetic acid is 2.43x10⁻³ moles.

The acid-base reaction between NaOH and CH₃COOH is the following:

NaOH + CH₃COOH ⇄ H₂O + CH₃COONa

To find the number of moles of water produced we need to calculate the number of moles of NaOH and CH₃COOH

[tex] n = C*V [/tex]

Where:  

n: is the number of moles

C: is the concentration

V: is the volume

So, the number of moles of NaOH and CH₃COOH is:

[tex] n_{NaOH_{i}} = C_{NaOH}*V_{NaOH} = 0.100 M*0.0243 L = 2.43 \cdot 10^{-3} moles [/tex]

[tex] n_{{CH_{3}COOH}_{i}} = C_{CH_{3}COOH}*V_{CH_{3}COOH} = 0.08 M*0.0314 L = 2.51 \cdot 10^{-3} moles [/tex]

Now, we need to find the limiting reactant knowing that 1 mol of NaOH reacts with 1 mol of CH₃COOH:

[tex] n_{NaOH} = \frac{1\: mol NaOH}{1\: mol CH_{3}COOH}*2.51 \cdot 10^{-3} \: moles\: CH_{3}COOH = 2.51 \cdot 10^{-3} \: moles [/tex]

We can see that we need 2.51x10⁻³ moles of NaOH to react with CH₃COOH, and initially, we have 2.43x10⁻³ moles of NaOH, so the limiting reactant is NaOH.  

Knowing that 1 mol of NaOH reacts with 1 mol of CH₃COOH to produce 1 mol of H₂O and 1 mol of CH₃COONa, the number of moles of water produced is:

[tex] n_{H_{2}O} = n_{NaOH} = 2.43 \cdot 10^{-3} moles [/tex]

Therefore, the number of moles of water produced is 2.43x10⁻³ moles.

You can find more about the acid-base reaction here:

I hope it helps you!

Answer:

5. ΔH = 238.7 kJ; 6 ΔH, = -1504.6 kJ

Step-by-step explanation:

Question 5:

We have three equations:  

(I)  CH₃OH + ³/₂O₂ ⟶ CO₂ + 2H₂O;  ΔH = -726.4 kJ

(II)  C + O₂ ⟶ CO₂;                              ΔH = -393.5 kJ

(III) H₂ + ½O₂ ⟶ H₂O;                         ΔH = -285.8 kJ

From these, we must devise the target equation:  

(IV) C + 2H₂ + ½ O₂ → CH₃OH; ΔH = ?  

The target equation has 1C on the left, so you rewrite Equation (II)..  

(V) C + O₂ ⟶ CO₂;                               ΔH = -393.5 kJ

Equation (V) has 1CO₂ on the right, and that is not in the target equation.  

You need an equation with ½O₂ on the left, so you reverse Equation (I).  

When you reverse an equation, you reverse the sign of its ΔH.  

(VI) CO₂ + 2H₂O ⟶ CH₃OH + ³/₂O₂;  ΔH = 726.4 kJ

Equation (VI) has 2H₂O on the left, and that is not in the target equation.

You need an equation with 2H₂O on the right. Double Equation (III).

When you double an equation, you double its ΔH.  

(VII) 2H₂ + O₂ ⟶ 2H₂O;                        ΔH = -571.6 kJ

Now, you add equations (V), (VI), and (VII), cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

We get the target equation (IV)  

(V)   C + O₂ ⟶ CO₂;                              ΔH = -393.5 kJ

(VI)  CO₂ + 2H₂O ⟶ CH₃OH + ³/₂O₂ ;  ΔH =  726.4 kJ

(VII) 2H₂ + O₂ ⟶ 2H₂O;                        ΔH = -571.6 kJ

(IV)  C + 2H₂ + ½ O₂ → CH₃OH;             ΔH = -238.7 kJ  

Question 6

We have three equations:  

(I) Mg + ½O₂ → MgO;  ΔH = -601.7 kJ

(II) Mg + S ⟶ MgS;      ΔH = -598.0 kJ

(III) S + O₂ ⟶ SO₂;       ΔH = -296.8 kJ

From these, we must devise the target equation:  

(IV) 3Mg + SO₂ → MgS + 2MgO; ΔH = ?  

The target equation has 1SO₂ on the left, so you reverse Equation(III).

(V) SO₂ ⟶ S + O₂;         ΔH = 296.8 kJ

Equation (V) has 1S on the right, and that is not in the target equation.  

You need an equation with 1S on the left, so you rewrite Equation (II).  

(VI) Mg + S ⟶ MgS;        ΔH = -598.0 kJ

Equation (V) has 1O₂ on the right, and that is not in the target equation.

You need an equation with 1O₂ on the left. Double Equation (I).

(VII) 2Mg + O₂⟶ 2MgO ;  ΔH = -1203.4 kJ  

Now, you add equations (V), (VI), and (VII), cancelling species that appear on opposite sides of the reaction arrows.

We get the target equation (IV)  

(V)   SO₂ ⟶ S + O₂;                    ΔH =    296.8 kJ

(VI)   Mg + S ⟶ MgS;                  ΔH = -  598.0 kJ

(VII) 2Mg + O₂ ⟶ 2MgO;            ΔH = -1203.4 kJ

(IV)  3Mg + SO₂ → MgS + 2MgO; ΔH = -1504.6 kJ

Answer:

26. 42.3 kJ; 27. 143 kJ  

Explanation:

26. Formation of water gas

We know that we will need the balanced equation with masses, molar masses, and enthalpies, so let’s gather all the information in one place.

A_r:        12.01

                 C + H₂O ⟶ CO + H₂; ΔH_r = +113 kJ

Mass/g:  4.50

(a) Calculate the moles of C

Moles of C = 4.50 × 1/12.01 = 0.375 mol C

(b) Calculate the energy absorbed

The conversion factor is 113 kJ/1 mol Al

Heat = 0.375 × 113/1 = 42.3 kJ

27. Oxidation of glucose

A_r:         180.16

               C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O; ΔH_r =  -2803 kJ

Mass/g:     9.22

(a) Calculate the moles of C₆H₁₂O₆

Moles of C₆H₁₂O₆ = 9.22 × 1/180.16 = 0.051 07 mol C₆H₁₂O₆

(b) Calculate the energy absorbed

The conversion factor is (-2803 kJ/1 mol C₆H₁₂O₆)

Heat = 0.051 07 × (-2803)/1 = -143 kJ

The oxidation of glucose releases 143 kJ .

Answer: Adding Hydrogen Gas

Explanation:

The Reduction Reaction that occurs during the refinement of Iron Ore is as follows:

[tex]Fe_3O_4 + H_2  -> 3FeO + H_2O\\\\FeO + H_2   -> Fe + H_2O[/tex]

Since Hydrogen Gas is a reactant in the equation with Iron ore , adding more hydrogen gas will shift the reaction toward the right side, increase the amount of  refined iron produced.

Answer:

The correct answer is increase in temperature.

Explanation:

During heating of the iron ore with the hydrogen gas at high temperature, the production of water and element iron takes place. High temperature is one of the essential condition for this chemical reaction to take place.  

The more the temperature is, the more will be the amount of the production of iron, this is due to the presence of high temperature, which is one of the mediating force for the production of iron. Therefore, an increase in temperature will produce a greater amount of iron.  

Answer:

0.957 atm.

Explanation:

PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the container in L.

n is the no. of moles of the gas.

R is the genral gas constant.

T is the temperature of the gas in K.

∵ n = mass/molar mass.

∴ PV = (mass/molar mass)RT

∴ P = (mass/V)(RT/molar mass)

∵ d = mass/V

∴ P = dRT/molar mass

∵ d = 2.51 g/L, R = 0.082 L.atm/mol.K, T = 25°C + 273 = 298 K, molar mass of SO₂ = 64.06 g/mol.

∴ P = dRT/molar mass = (2.51 g/L)(0.082 L.atm/mol.K)(298.0 K)/(64.06 g/mol) = 0.957 atm.

The pressure in the flask is approximately 0.993 atmospheres.

The correct answer is that the pressure in the flask containing SO2 gas at a density of 2.51 g/L at room temperature (25.0°C) can be determined using the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to calculate the molar mass of SO2. The molar mass of sulfur (S) is approximately 32.07 g/mol and that of oxygen (O) is approximately 16.00 g/mol. Since SO2 has one sulfur atom and two oxygen atoms, its molar mass is:

[tex]Molar mass of SO2 = 32.07 g/mol (S) + 2 \times 16.00 g/mol (O) = 64.07 g/mol[/tex]

Next, we can use the density (d) of the gas to find the mass (m) of the gas in a container of known volume (V). The density is given as 2.51 g/L, so for a volume of 1 liter, the mass would be:

[tex]m = d \times V = 2.51 g/L \times 1 L = 2.51 g[/tex]

 Now, we can find the number of moles (n) of SO2 using the molar mass (M):

[tex]n = m / M = 2.51 g / 64.07 g/mol = 0.0392 mol[/tex]

The ideal gas constant (R) is 0.0821 L·atm/mol·K. The temperature (T) in Kelvin is room temperature (25.0°C) plus 273.15, which gives:

[tex]T = 25.0 + 273.15 = 298.15 K[/tex]

Now we can rearrange the ideal gas law to solve for pressure (P):

[tex]P = nRT / V[/tex]

Since we are considering a volume of 1 liter, the equation simplifies to:

[tex]P = (0.0392 mol)(0.0821 L\times atm/mol\times K)(298.15 K) / 1 L[/tex]

[tex]P =0.993 atm[/tex]

 Therefore, the pressure in the flask is approximately 0.993 atmospheres.

Answer:

Explanation:

The spontaneity of a reaction can be determined by the thermodynamic property named Gibb's free energy or simply free energy (G).

The change in the free energy of a system is defined as the difference between the increase in enthalpy (ΔH) and the product of the temperature (T) times the increase in entropy (ΔS):

The sign of ΔG tells if a reaction is spontaneous according to this:

The question states that a reaction is exothermic, and its entropy change is positive. That means:

Remember that the temperature is stated in absolute scale, so T is always positive.

Hence, ΔG = ΔH - T ΔS = (negative) - T (positive) = (negative) + (negative) = negative.

Conclusion: since ΔG is negative, regardless the temperature, you conclude that the reaction is spontaneous at all temperatures, which is the option c.

The answer is C. This reaction is spontaneous at all temperatures.

Heat is an energy that is easily received and easily released so that it can change the temperature of these substances into rising or falling. Heat can also move from one substance to another through a medium or intermediary. For example, two substances that have different temperatures are mixed in a container. Then the temperature of the two objects will be the same.

An endothermic reaction is a reaction that absorbs heat (there is heat transfer from the environment to the system). Endothermic reaction (endo-is a prefix meaning "inward"), where heat must be supplied to the system by the environment. In this reaction, there is heat transfer from the environment to the system so that the ambient temperature drops and becomes colder. Endothermic reactions absorb a certain amount of energy so that the energy of the system increases. Because enthalpy increases, the enthalpy change is positive. In endothermic reactions, the system absorbs energy. Therefore, the enthalpy of the system will increase. That is, the product enthalpy (Hp) is greater than the reactant enthalpy (Hr). As a result, the enthalpy change is the difference between the product enthalpy and the reactant enthalpy (Hp-Hr) which is positive. And the enthalpy changes for endothermic reactions can be stated as follows:

∆H = Hp-Hr> 0 (Positive)

Exothermic reactions are chemical reactions that produce heat. In this reaction, there is heat transfer from the system to the environment so that the environment becomes hotter. The exothermic reaction sign is a negative enthalpy change (∆H = -). This is because the exothermic reaction will release energy so that the enthalpy of the system decreases. Examples of exothermic reactions include combustion reactions, such as wood combustion, methane, propane, and the reaction between aluminum powder and iron oxide. Another example of an exothermic reaction in the manufacture of ethanol from glucose fermentation.

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Grade: High School

Subject: Chemistry

keywords: Exothermic reactions, Heat

Given, final veocity = 47m/s, initial velocity= 0, time = 5s

V = u+at

a = v-u/t

   = 47- 0/5

   = 9.4 ms^-2

There are not too many things that are more insoluble than AgCl. When you actually do this experiment, virtually all of the AgCl sinks to the bottom of the test tube (or whatever container you are doing it in). The CuCl2 is very soluble. It is a gorgeous Forest Green with a very slightly blue tinge so you can tell that you are successful.  

CuCl2 is very soluble.

That makes A incorrect.

AgCl is really insoluble

Looks like B is the answer.

C is just wrong. No complex is formed. The equation is a double replacement.

AgNO3 + HCl ====> AgCl(s) + HNO3 Not a good example but it is correct.

Cu(NO3)2 + 2HCl ===> CuCl2 + 2HNO3

D is not true. Ag will react with the HCl That's how you get AgCl going to the bottom.

E A double replacement need not transfer electrons. In this case it does not. E is incorrect.

thanks for the answers ッ. (btw they’re on the bottom of the question if anyone doesn’t see it.

Alpha particles consist of two protons and two neutrons, it is produced during alpha decay and release energy.

Alpha particles are the small part of a matter which carry positive charge in it.

From the definition of the alpha particles, it is clear that it is positively charged particle and no electron is present in it. It mainly consist 2 protons and 2 neutrons in it. It is produced by the decay of very heavy radioactive elements which are unstable and releases energy with alpha particle to gain stability, and this decay is known as alpha - decay.

Hence, option (1), (3) and (5) is correct.

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Answers:

1. 282 g/mol

2. (a) pH = 10.0; (b) pKₐ = 5.5; Kₐ = 3 × 10⁻⁶; (c) thymolphthalein

Step-by-step explanation:

1. Molar mass of unknown acid

The equation for the reaction is

HA + NaOH ⟶ NaA + H₂O

(a) Calculate the moles of NaOH

Moles of NaOH = 0.023 64 L NaOH × (0.1 mol NaOH/1 L NaOH)

= 2.364 × 10⁻³ mol NaOH

(b) Moles of HA

Now, you use the molar ratio from the balanced chemical equation to find the moles of unknown acid.

Moles of HA = 2.364 × 10⁻³ mol NaOH × (1 mol HA/1 mol NaOH)

= 2.364 × 10⁻³ mol HA

(c) Molar mass of HA

You know now that 0.5632 g of HA .

MM = mass/moles = 0.5632 g/2.364 × 10⁻³ mol = 238.2 g/mol

2. Titration of acetic acid

(a) Equivalence point

The equivalence point is the pH at the steepest point of the titration curve.

In your titration of vinegar, the equivalence point appears to be at about

pH 10.0.

(b) Half-way point

At the half-way point, you have neutralized half the acid HA and converted it into the sane amount of A⁻, so [HA] = [A⁻].

Kₐ = [H₃O⁺] × [A⁻]/[HA] = [H₃O⁺]

At the half-way point, Kₐ = [H₃O⁺] and pKₐ = pH.

Your equivalence point is at about 21.3 mL, so the half-way point is at 10.7 mL.

The pH at 10.6 mL is about 5.5.

pH = pKₐ = 5.5

Kₐ = 10^(-pKₐ) = 10^(-5.5)  = 3 × 10⁻⁶

(c) Appropriate indicator

An indicator is a weak acid in which the acid form HA is a different colour than the base form A⁻.

The colour change occurs when [HA] = [A⁻] and pH = pKₐ.

Your indicator should have pKₐ ≈ 10 (Kₐ≈ 10⁻¹⁰), so it changes colour at the equivalence point (pH 10).

The best indicator is thymolphthalein, because it has Kₐ = 10⁻¹⁰.

I believe it’s called Wisteria

1.0875 x 10-2 atm

2O3(g) → 3O2(g)

rate = -(1/2)∆[O3]/∆t = +(1/3)∆[O2)/∆t  

The average rate of disappearance of ozone ... is found to  

be 7.25 × 10–3 atm over a certain interval of time.

This means (ignoring time)

∆[O3]/∆t = -7.25 × 10^–3 atm  

(it is disappearing, thus the negative sign)

rate = -(1/2)∆[O3]/∆t  

rate = -(1/2)*(-7.25 × 10^–3 atm)

      = 3.625 × 10^–3 atm  

Now use the other part of the expression:  

rate = +(1/3)∆[O2)∆t  

3.625 × 10–3 atm = +(1/3)∆[O2)/t  

∆[O2)/∆t = (3)*(3.625× 10^–3 atm)

              = 1.0875 x 10-2 atm over the same time interval

From the parameters given, the rate of appearance of O2 is 1.1 * 10^-2.

The equation of the reaction is;

2O3(g) → 3O2(g)

We can see from the equation that; [tex]-\frac{1}{2} \frac{d[O3]}{dt} = \frac{1}{3} \frac{d[O2]}{dt}[/tex]

Hence it follows that;

[tex]-\frac{3}{2} \frac{d[O3]}{dt} = \frac{d[O2]}{dt}[/tex]

Since we already have the rate of disappearance of O3 from the problem as 7.25 × 10^-3, the rate of appearance of O2 is now given by;

[tex]\frac{d[O2]}{dt} = \frac{3}{2} * 7.25 * 10^-3[/tex]

[tex]\frac{d[O2]}{dt} = 1.1 * 10^-2[/tex]

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Answer:

A. 11

Explanation:

atoms are made of three types of sub atomic particles - protons, neutrons and electrons

Protons are positively charged, electrons are negatively charged and neutrons are neutral.

Protons and neutrons are located inside the nucleus and electrons orbit around the nucleus in energy shells.

Atoms make up an element and atoms are neutral charged with an equal number of protons and electrons. Neutral atom of sodium would therefore have an equal number of positive charges and negative charges.

Since Na has 11 protons, for it to be neutral it would have 11 electrons

Answer:

Methane(CH₄).

Explanation:

∨ ∝ 1/√M.

where, ∨ is the rate of diffusion of the gas.

M is the molar mass of the gas.

∨₁/∨₂ = √(M₂/M₁)

Carbon dioxide: 44.01 g/mol.

Water vapor : 18.0 g/mol.

Ammonia (NH₃): 17.0 g/mol.

Methane(CH₄): 16.0 g.mol.

Since, methane has the lowest molar mass. So, it will diffuse faster within a system.

In a hydrogen atom, the 3s → 1s transition would result in the emission of the photon with the highest energy, according to the Bohr model. Energy released in photon form is highest when the transition ends at the ground state (n = 1), from the highest available initial energy level.

For a hydrogen atom, the electronic transition that would result in the emission of the photon with the highest energy would be the transition from the highest initial energy level to the lowest final energy level, which is the ground state (n = 1). Comparing the given transitions, 2p → 1s is not listed; hence, we consider the next highest, which would be 3s → 1s.

The energy of a photon emitted during an electron transition is directly proportional to the difference in energy between the two energy levels according to the equation E = hν, where E is the energy of the photon, h is Planck's constant, and ν is the frequency of the photon. Since the ground state has the lowest energy, any transition to it from a higher energy level will release more energy as a photon compared to transitions between higher levels.

The Bohr model is used to predict the energies involved in transitions between energy levels in a hydrogen atom. Emission of photons occurs when an electron falls from a higher energy level (n > m) to a lower one (n < m) and the energy of the emitted photon is equal to the difference in energy between those levels. The transitions that release the most energy are those that end at the n = 1 level, known as the Lyman series.

Specifically, a transition from n = 3 to n = 1 would result in a higher energy photon than those from n = 5 or 6 to levels other than n = 1, as indicated in the choices provided.

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The statement above is False;

Answer:

Explanation:

1) Since the two rooms are connected by an open door, you assume pressure equilibrium: the pressure on the two rooms is the same.

2) Since you consider two equal size rooms, both volumes are equal.

3) Assuming ideal gas behavior, pressure (P), temperature (T), volume (V) and number of moles (n) are related by the equation PV = nRT

4) Naming T₁ the lower temperature, T₂ the higher temperature, n₁ the number of moles of air in the room at lower temperature, and n₂ the number of moles of air in the room at higher temperature, you get:

5) That means that the amount of molecules (number of moles) is inversely related to the temperature: the higher the temperature the lower the number of moles, and the lower the temperature the greater the number of moles.

Hence, the answer is that the room that contains more air molecules is the room mantained at a lower temperature.

The room that has the lower temperature is expected to have the greater number of gas molecules since the molecules

Ideal gases are the gases that obey the ideal gas equation. We know that for ideal gases, the temperature of a given mass of gas is inversely proportional to the number of moles of gas hence we can write; n₁ / n₂ = T₂ / T₁.

Since the above is true, it the follows that the room that has the lower temperature is expected to have the greater number of gas molecules since the molecules posses less kinetic energy to escape.

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The pKa of the indicator in this scenario is 6. The color of the solution will be yellow when the pH > 6 and blue when the pH < 6. At pH = 6, the solution will display a mixture of both colors.

The pKa of an indicator is the negative logarithm (base 10) of the Ka value. In this case, given the Ka value of 1.0Ã10â6, the pKa is -log(1.0Ã10â6), which equals 6. The colors the indicator presents depend on the pH of the solution. When the pH is greater than pKa, the solution tends to have the color of the conjugate base, which in this case is yellow. Conversely, when the pH is less than pKa, the solution takes on the color of the protonated form (the acid form), which in this case is blue. At a pH equal to the pka, you would see a mixed color, as appreciable amounts of both the conjugate acid and base are present. This information can be applied when considering the use of indicators in a titration scenario or when attempting to determine the approximate pH of an unknown solution.

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The pKa of the indicator ha with a ka value of 1.0x10^(-6) is 6.

The pKa of an indicator can be calculated using the Henderson-Hasselbalch equation, which relates the ratio of the concentrations of the conjugate base and conjugate acid forms of the indicator to the pH of the solution:

pKa = -log (ka)

For the given indicator, ha, with a ka value of 1.0x10^(-6), the pKa can be calculated as follows:

pKa = -log (1.0x10^(-6)) = 6

Therefore, the pKa of the indicator ha is 6.

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Answer:

        c. 2N₂O₅(g) ⇄ 4 NO₂(g)+ O₂(g), and

        d. N₂O₄(g) ⇄ 2 NO₂(g)

Explanation:

Volume and pressure are inversely related (Boyle's law).

Volume and number of molecules are directly related (Avogadro's principle).

As per Le Chatelier's principle, the reaction will shift toward the side that permits to overcome or minimize the force that disturbs the equilibrium.

With those concepts you can predict which equilibria will shift toward formation of more products if the volume of a reaction mixture at equilibrium increases by a factor of 2

Let's dig into each option.

a. 2 SO₂(g)+ O₂(g) ⇄ 2 SO3(g)

Incorrect.

There are three molecules in the reactant side and two in the product side. so the increase in volume will favor the reverse reaction. This is, the equilibrium will shift to the formation of more reactants, and this is an incorrect choice.

b. NO(g) + O₃(g) ⇄ NO₂(g) + O₂(g)

Incorrect.

There are the same number of molecules in the reactant side and the product side. Hence, the increase of volume will not produce a change in the equlibrium.

c. 2N₂O₅(g) ⇄ 4 NO₂(g)+ O₂(g)

Correct.

There are 2 molecules in the reactant side (left) and four molecules in the product side (right). So, the increase in volume in this system will produce a shift toward the product side.

d. N₂O₄(g) ⇄ 2 NO₂(g)

Correct.

There are more molecules in the product side than in the reactant side, so you predict that the equilibrium will shift toward the formation of more product to overcome the increase of volume.

Equilibrium shift is the shift of the reaction towards the stressed conditions in the reactions. The shift towards the formation of the products will occur in [tex]\rm 2N_{2}O_{5}(g) \rightleftharpoons 4 NO_{2}(g)+ O_{2}(g)[/tex] and [tex]\rm N_{2}O_{4}(g) \rightleftharpoons 2 NO_{2}(g)[/tex].

According to the Le principle the reaction shift towards the site or the reactants and products where the disturbance or the stress is present so that it can be overcome.

In the third equation reaction, [tex]\rm 2N_{2}O_{5}(g) \rightleftharpoons 4 NO_{2}(g)+ O_{2}(g)[/tex]  the number of the molecules on the left side is 2 and on the right side of the product have 4 molecules.  When the volume is increased then the reaction shifts towards the right or the product side.

In the fourth reaction, [tex]\rm N_{2}O_{4}(g) \rightleftharpoons 2 NO_{2}(g)[/tex] the number of the products are more compared to the reactants and hence the increase in the volume will shift the reaction towards the formation of the product.

Therefore, option c. [tex]\rm 2N_{2}O_{5}(g) \rightleftharpoons 4 NO_{2}(g)+ O_{2}(g)[/tex] and option d. [tex]\rm N_{2}O_{4}(g) \rightleftharpoons 2 NO_{2}(g)[/tex] are the reactions that shift towards product formation.

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That's just an error in the source.

There are 3H₂O on each side of the reaction arrows, so all the water molecules must cancel.

The approximate value of the volume of the gas is 958.3L

Using the ideal gas equation which is a combination of pressure law, Boyle's law and Charles's law.

Mathematically

[tex]PV=nRT[/tex]

we convert the value of temperature from degree Celsius to Kelvin.

To do this, simply add 273.15 to the °C value.

[tex]T=15 + 273.15 = 288.15K[/tex]

From the data given, we can substitute into the equation and solve for v

[tex]PV = nRT\\3.75*V=1.5*8.314*288.15\\3.75V=3593.52\\V=\frac{3593.52}{3.75}\\V=958.27L[/tex]

The approximate value of the volume is 958.3L

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