A 48 g piece of ice at 0.0 ∘C is added to a sample of water at 7.4 ∘C. All of the ice melts and the temperature of the water decreases to 0.0 ∘C. How many grams of water were in the sample?

Answer: The net ionic equation is written below.

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which do not get involved in the chemical equation. It is also defined as the ions which are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of ammonium perchlorate and water is given as:

[tex]NH_4ClO_4(aq.)+H_2O(l)\rightarrow NH_4OH(aq.)+HClO_4(aq.)[/tex]

Ionic form of the above equation follows:

[tex]NH_4^+(aq.)+ClO_4^-(aq.)+H_2O(l)\rightarrow NH_4OH(aq.)+H^+(aq.)+ClO_4^-(aq.)[/tex]

Ammonium hydroxide will not dissociate into its ions because it is a weak base.

As, chlorate ions are present on both the sides of the reaction, thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

[tex]NH_4^+(aq.)+H_2O(l)\rightarrow NH_3^+(aq.)+H_3O^+(aq.)[/tex]

Hence, the net ionic equation is given above.

The net ionic equation for the acid-base hydrolysis equilibrium of ammonium perchlorate in water involves the dissociation of ions and the formation of hydronium ions.

The net ionic equation for the acid-base hydrolysis equilibrium when ammonium perchlorate is dissolved in water can be written as follows:

NH₄ClO₄ + H₂O → NH₄⁺ + ClO₄⁻ + H₃O⁺

This equation represents the dissociation of ammonium perchlorate into its ions in water, with the presence of hydronium ions due to the protonation of water.

Answer:

ΔH(Combustion C₆H₆ = -3,262 Kj/mole

Explanation:

Answer:

391.28771 pounds of carbon-dioxide released into the atmosphere.

Explanation:

Density of the gasoline ,d= 0.692 g/mL

Volume of gasoline in an tanks,V = 22.0 gallons = 83,279.02 mL

Let mass of the gasolin be M

[tex]d=\frac{M}{V}[/tex]

M = V × d = 83,279.02 mL × 0.692 g/mL=57,629.081 g

Assume that gasoline is primarily octane (given)

[tex]2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O[/tex]

Mass of octane burnt in the tank =  M = 57,629.081 g

Moles of octane =[tex]\frac{57,629.081 g}{114.08 g/mol}=505.1637 mol[/tex]

According to reaction, 2 moles of octane gives 16 moles of carbon-dioxide.

Then 505.1637 mol of octane will give:

[tex]\frac{16}{2}\times 505.1637 mol=4,041.3100 mol[/tex] of carbon-dioxide

Mass of 4,041.3100 mol of carbon-dioxide:

4,041.3100 mol × 44.01 g/mol = 177,858.05 g

Mass of carbon-dioxide produced in pounds = 391.28771 pounds

391.28771 pounds of carbon-dioxide released into the atmosphe

The number of pounds of CO₂ released into the atmosphere when a 22.0 gallon tank of gasoline is burned in an automobile engine is 392.3 pounds.

We'll begin by converting 22 gallon to mL.

1 gallon = 3785.412 mL

Therefore,

22 gallon = 22 × 3785.412

22 gallon = 83279.064 mL

Next, we shall determine the mass of 83279.064 mL of gasoline (C₈H₁₈).

Density = 0.692 g/mL

Volume = 83279.064 mL

Mass = Density × Volume

Mass of C₈H₁₈ = 0.692 × 83279.064

Next, we shall determine the mass of C₈H₁₈ that reacted and the mass of CO₂ produced from the balanced equation.

2C₈H₁₈ + 25O₂ —> 16CO₂ + 18H₂O

Molar mass of C₈H₁₈ = (12×8) + (1×18)

= 96 + 18

= 114 g/mol

Mass of C₈H₁₈ from the balanced equation = 2 × 114 = 228 g

Molar mass of CO₂ = 12 + (16×2)

= 12 + 32

= 44 g/mol

Mass of CO₂ from the balanced equation = 16 × 44 = 704 g

Thus,

From the balanced equation above,

228 g of C₈H₁₈ reacted to produce 704 g of CO₂

Next, we shall determine the mass of CO₂ produced by the reaction of 57629.11 g of C₈H₁₈. This can be obtained as follow:

From the balanced equation above,

228 g of C₈H₁₈ reacted to produce 704 g of CO₂

Therefore,

57629.11 g of C₈H₁₈ will react to produce = [tex]\frac{57629.11 * 704}{228}[/tex] = 177942.515 g of CO₂

Finally, we shall convert 177942.515 g of CO₂ to pounds.

453.592 g = 1 pound

Therefore,

177942.515 g = [tex]\frac{177942.515}{453.592}[/tex]

177942.515 g = 392.3 pounds

Therefore, the number of pounds of CO₂ released into the atmosphere is 392.3 pounds.

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Answer:

40.0 L.

Explanation:

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

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

(V₁n₂) = (V₂n₁).

V₁ = 50.0 L, n₁ = 2.0 moles,

V₂ = ??? L, ​n₂ = 2.0 mol - 0.4 mol = 1.6 mol.

∴ V₂ = (V₁n₂)/(n₁) = (50.0 L)(1.6 mol)/(2.0 mol) = 40.0 L.

Answer : The temperature in kelvins is, [tex]T>1108.695K[/tex]

Explanation : Given,

[tex]\Delta H[/tex] = 178.5 KJ/mole = 178500 J/mole

[tex]\Delta S[/tex] = 161.0 J/mole.K

Gibbs–Helmholtz equation is :

[tex]\Delta G=\Delta H-T\Delta S[/tex]

As per question the reaction is spontaneous that means the value of [tex]\Delta G[/tex] is negative or we can say that the value of [tex]\Delta G[/tex] is less than zero.

[tex]\Delta <0[/tex]

The above expression will be:

[tex]0>\Delta H-T\Delta S[/tex]

[tex]T\Delta S>\Delta H[/tex]

[tex]T>\frac{\Delta H}{\Delta S}[/tex]

Now put all the given values in this expression, we get :

[tex]T>\frac{178500J/mole}{161.0J/mole.K}[/tex]

[tex]T>1108.695K[/tex]

Therefore, the temperature in kelvins is, [tex]T>1108.695K[/tex]

Answer:

%Yield = 41.3% (w/w)

Explanation:

4NH₃ + 6NO => 5N₂ + 6H₂O

10g NH₃ = (10/17)mole NH₃ = 0.588 mole NH₃ => 5/4(0.588)mole N₂ = 0.735 mole N₂ x 28g N₂/mole N₂ = 20.58g N₂ (Theoretical Yield)

Given Actual Yield = 8.5g N₂

%Yield = (Actual Yield/Theoretical Yield)100% = (8.5g N₂/20.58g N₂)100% =41.3% Yield  (w/w)

The percent yield of N₂ is 41.34%

From the question,

We are to determine the percent yield of N₂

The given balanced chemical equation for the reaction is

4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l)

This means 4 moles of NH₃ reacts with 6 moles of NO to produce 5 moles of N₂ and 6 moles of H₂O

First, we will determine the number of moles of NH₃ that reacted

From the question,

Mass of NH₃ that reacted = 10.0 g

From the formula

[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]

Molar mass of NH₃ = 17.031 g/mol

∴ Number of moles of NH₃ that reacted = [tex]\frac{10.0}{17.031 }[/tex]

Number of moles of NH₃ that reacted = 0.58716 mole

Now, we will determine the theoretical number of moles of N₂ produced

From the balanced chemical equation

Since

4 moles of NH₃ reacts to produce 5 moles of N₂

Then,

0.58716 mole of NH₃ will react to produce [tex]\frac{0.58716 \times 5}{4}[/tex] moles of N₂

[tex]\frac{0.58716 \times 5}{4} = 0.73395[/tex]

∴ The theoretical number of moles of N₂ that would be produced is 0.73395 mole

Now, we will determine the theoretical yield in grams

Using the formula

Mass = Number of moles × Molar mass

Molar mass of N₂ = 28.0134 g/mol

∴ Mass of N₂ that would be produced = 0.73395 × 28.0134

Mass of N₂ that would be produced = 20.56 g

∴ The theoretical yield of N₂ is 20.56 g

Now, for the percent yield

[tex]Percent\ yield = \frac{Actual\ yield}{Theoretical\ yield}\times 100\%[/tex]

From the question,

Actual yield of N₂ = 8.50 g

∴ Percent yield of N₂ = [tex]\frac{8.50}{20.56}\times 100\%[/tex]

Percent yield of N₂ = 41.34 %

Hence, the percent yield of N₂ is 41.34%

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Answer: The correct answer is [tex]NO(g)+O_2(g)\rightarrow NO_2(g)+O(g);Rate=k[NO][O_2][/tex]

Explanation:

Molecularity of the reaction is defined as the number of atoms, ions or molecules that must colloid with one another simultaneously so as to result into a chemical reaction.

Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.

Elementary reactions are defined as the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical reaction.

For the given reactions:

Molecularity of the reaction = 2 + 1 = 3

Order of the reaction = 1 + 1 = 2

This is not considered as an elementary reaction.

Molecularity of the reaction = 1 + 1 = 2

Order of the reaction = [tex]1+\frac{1}{2}=\frac{3}{2}[/tex]

This is not considered as an elementary reaction.

Molecularity of the reaction = 1 + 1 = 2

Order of the reaction = 1 + 1 = 2

This is considered as an elementary reaction.

Molecularity of the reaction = 1 + 1 = 2

Order of the reaction = 2 + 0 = 2

In this equation, the order with respect to each reactant is not equal to its stoichiometric coefficient which is represented in the balanced chemical reaction. Hence, this is not considered as an elementary reaction.

Hence, the correct answer is [tex]NO(g)+O_2(g)\rightarrow NO_2(g)+O(g);Rate=k[NO][O_2][/tex]

An elementary reaction is a reaction that occurs in a single step and involves the collision of reactant molecules. In this case, both of the given reactions could be considered elementary.

An elementary reaction is a reaction that occurs in a single step and involves the collision of reactant molecules. To determine if a reaction is elementary, we need to examine the overall reaction and see if it can be written as a sum of individual elementary reactions.

In this case, the reaction 2 NO2(g) + F2(g) → 2NO2F(g) can be written as a sum of individual elementary reactions:

NO2(g) + F2(g) → NO2F(g)

NO2(g) + F2(g) → NO(g) + FNO(g)

Since the overall reaction can be broken down into individual elementary reactions, both of these reactions could be considered elementary.

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Answer:

2.067 x 10⁻⁷ M/min.

Explanation:

Rate of the reaction = - 1/2 d[NOCl]/dt = 1/2 d[NO]/dt = d[Cl₂]/dt

∵ d[Cl₂] = 3.39 x 10⁻³ M, dt = 1.64 x 10³ min.

∴ Rate of the reaction = d[Cl₂]/dt = (3.39 x 10⁻³ M) / (1.64 x 10³ min) = 2.067 x 10⁻⁷ M/min.

The rate of the reaction for the decomposition of nitrosyl chloride to nitrogen monoxide and chlorine is calculated by dividing the change in concentration of Cl₂ by the change in time, which yields a rate of 2.07×10⁻¶ M/min.

The balanced chemical equation is 2 NOCl → 2 NO + Cl₂. The reaction rate can be calculated using the change in concentration of Cl₂ over the change in time, as the rate of appearance of Cl₂ is directly related to the rate of the reaction. Since Cl₂ starts at 0 M and increases to 3.39×10⁻³ M over a period of 1.64×10³ minutes, you can calculate the average rate as follows:

Rate = Δ[Cl₂] / Δtime = (3.39×10⁻³ M - 0 M) / (1.64×10³ min) = 2.07×10⁻¶ M/min.

Answer:

37.1°C.

Explanation:

∵ ΔHsoln = Q/n

no. of moles (n) of NaOH = mass/molar mass = (2.5 g)/(40 g/mol) = 0.0625 mol.

The negative sign of ΔHsoln indicates that the reaction is exothermic.

∴ Q = (n)(ΔHsoln) = (0.0625 mol)(44.51 kJ/mol) = 2.78 kJ.

Q = m.c.ΔT,

where, Q is the amount of heat released to water (Q = 2781.87 J).

m is the mass of water (m = 55.0 g, suppose density of water = 1.0 g/mL).

c is the specific heat capacity of water (c = 4.18 J/g.°C).

ΔT is the difference in T (ΔT = final temperature - initial temperature = final temperature - 25°C).

∴ (2781.87 J) = (55.0 g)(4.18 J/g.°C)(final temperature - 25°C)

∴ (final temperature - 25°C) = (2781.87 J)/(55.0 g)(4.18 J/g.°C) = 12.1.

∴ final temperature = 25°C + 12.1 = 37.1°C.

Answer:

Wt Avg Atomic Wt = 28.0577 amu

Explanation:

Atomic Mass = ∑Wt Avg contributions of isotopes of an element

Si-28 => 92.21% => 0.9221 decimal fraction => Wt Avg Si-28 = 0.9221(27.9769) = 25.7695 amu

Si-29 => 4.69% => 0.0469 decimal fraction => Wt Avg Si-29 = 0.0469(28.9765) = 1.3590 amu

Si-30 => 3.10% => 0.0310 decimal fraction => Wt Avg Si-30 = 0.0310(29.9737) = 0.9292

Wt Avg Atomic Wt = ∑Wt Avg Contributions = (25.7695 + 1.3590 + 0.9292)amu = 28.0577 amu

Answer :  The average atomic mass of silicon is, 28.08 amu

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]

For isotope Si-28 :

Mass of isotope Si-28  = 27.9769 amu

Fractional abundance of isotope Si-28  = 92.21 % = 0.9221

For isotope Si-29 :

Mass of isotope Si-29 = 28.9765 amu

Fractional abundance of isotope Si-29 = 4.69 % = 0.0469

For isotope Si-30 :

Mass of isotope Si-30 = 29.9737 amu

Fractional abundance of isotope Si-30 = 3.10 % = 0.0310

Putting values in equation 1, we get:

[tex]\text{Average atomic mass }=[(27.9769\times 0.9221)+(28.9765\times 0.0469)+(29.9737\times 0.0310)][/tex]

[tex]\text{Average atomic mass }=28.08amu[/tex]

Thus, the average atomic mass of silicon is, 28.08 amu

Answer:

Ka = 4.9 x 10ˉ⁴

Explanation:

HOAc    ⇄    H⁺    +    OAcˉ

Ka = [H⁺][OAcˉ]/[HOAc]

Given pH = 2.58 => [H⁺] = [OAcˉ] = 10ˉ²∙⁵⁸ = 2.63 x 10ˉ³M

Ka = (2.63 x 10ˉ³)²/(0.0142) = 4.9 x 10ˉ⁴

Answer:

(a) 7.4 x 10⁹ atoms.

(b)

(c)

Explanation:

k = ln(2)/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(10.5 years) = 0.066 year⁻¹.

kt = ln([A₀]/[A]),

where, k is the rate constant of the reaction (k = 0.066 year⁻¹).

t is the time of the reaction.

[A₀] is the initial concentration of (133-Ba) ([A₀] = 1.1 x 10¹⁰ atoms).

[A] is the remaining concentration of (133-Ba) ([A] = ??? g).

(a) 6 years:

t = 6.0 years.

∵ kt = ln([A₀]/[A])

∴ (0.066 year⁻¹)(6.0 year) = ln((1.1 x 10¹⁰ atoms)/[A])

∴ 0.396 = ln((1.1 x 10¹⁰ atoms)/[A]).

Taking exponential for both sides:

∴ 1.486 = ((1.1 x 10¹⁰ atoms)/[A]).

∴ [A] = (1.1 x 10¹⁰ atoms)/(1.486) = 7.4 x 10⁹ atoms.

(b) 10 years

t = 10.0 years.:

∵ kt = ln([A₀]/[A])

∴ (0.066 year⁻¹)(10.0 year) = ln((1.1 x 10¹⁰ atoms)/[A])

∴ 0.66 = ln((1.1 x 10¹⁰ atoms)/[A]).

Taking exponential for both sides:

∴ 1.935 = ((1.1 x 10¹⁰ atoms)/[A]).

∴ [A] = (1.1 x 10¹⁰ atoms)/(1.935) = 5.685 x 10⁹ atoms.

(c) 200 years:

t = 200.0 years.

∵ kt = ln([A₀]/[A])

∴ (0.066 year⁻¹)(200.0 year) = ln((1.1 x 10¹⁰ atoms)/[A])

∴ 13.2 = ln((1.1 x 10¹⁰ atoms)/[A]).

Taking exponential for both sides:

∴ 5.4 x 10⁵ = ((1.1 x 10¹⁰ atoms)/[A]).

∴ [A] = (1.1 x 10¹⁰ atoms)/(5.4 x 10⁵) = 2.035 x 10⁴ atoms.

Final answer:

To determine the number of 133Ba atoms left after a certain time, the half-life formula is used. After 6 years, about 7.78×109 atoms remain; after 10 years, approximately 5.77×109 atoms; and after 200 years, just 2.09×104 atoms remain, showing the exponential decrease due to radioactive decay.

Explanation:

The barium isotope 133Ba has a half-life of 10.5 years. To determine how many atoms are left after a certain time period, we can use the half-life formula. The formula is:

Where N(t) is the number of atoms remaining after time t, N0 is the initial number of atoms, and T is the half-life of the isotope.

We can see that after 200 years, a negligible amount of the original 133Ba atoms remain due to the nature of radioactive decay.

Answer : The percent yield of the reaction is, 91.32 %

Explanation :  Given,

Mass of [tex]C_6H_5CH_3[/tex] = 1 Kg = 1000 g

Molar mass of [tex]C_6H_5CH_3[/tex] = 92.14 g/mole

Molar mass of [tex]C_6H_5COOH[/tex] = 122.12 g/mole

First we have to calculate the moles of [tex]C_6H_5CH_3[/tex].

[tex]\text{Moles of }C_6H_5CH_3=\frac{\text{Mass of }C_6H_5CH_3}{\text{Molar mass of }C_6H_5CH_3}=\frac{1000g}{92.14g/mole}=10.85mole[/tex]

Now we have to calculate the moles of [tex]C_6H_5COOH[/tex].

The balanced chemical reaction will be,

[tex]2C_6H_5CH_3+3O_2\rightarrow 2C_6H_5COOH+2H_2O[/tex]

From the balanced reaction, we conclude that

As, 2 moles of [tex]C_6H_5CH_3[/tex] react to give 2 moles of [tex]C_6H_5COOH[/tex]

So, 10.85 moles of [tex]C_6H_5CH_3[/tex] react to give 10.85 moles of [tex]C_6H_5COOH[/tex]

Now we have to calculate the mass of [tex]C_6H_5COOH[/tex]

[tex]\text{Mass of }C_6H_5COOH=\text{Moles of }C_6H_5COOH\times \text{Molar mass of }C_6H_5COOH[/tex]

[tex]\text{Mass of }C_6H_5COOH=(10.85mole)\times (122.12g/mole)=1325.002g[/tex]

The theoretical yield of [tex]C_6H_5COOH[/tex]  = 1325.002 g

The actual yield of [tex]C_6H_5COOH[/tex]  = 1.21 Kg = 1210 g

Now we have to calculate the percent yield of [tex]C_6H_5COOH[/tex]

[tex]\%\text{ yield of }C_6H_5COOH=\frac{\text{Actual yield of }C_6H_5COOH}{\text{Theoretical yield of }C_6H_5COOH}\times 100=\frac{1210g}{1325.002g}\times 100=91.32\%[/tex]

Therefore, the percent yield of the reaction is, 91.32 %

Answer:

Increases rate of rxn 9X

Explanation:

Given 4NO + O₂ + 2H₂O => 4HNO₂

Rate = k[NO]²[O₂] ; if [O₂] is kept constant => Rate = k[NO]²

For 2nd order reactions Rate = k[A]². If [A] = a, 2a, 3a … then the Rate trend for a 2nd order reaction is …

Rate 1 = k(a)² = ka²

Rate 2 = k(2a)² = 4ka²

Rate 3 = k(3a)² = 9ka²  <= 3X increase in concentration

Rate 4 = k(4a)² = 16ka²

Increasing the concentration of NO by 3 times while keeping concentrations of O₂ and H₂O constant  increases rate of reaction by 9 times.  

The correct answer is that the rate will be 9 times faster if the concentration of NO is tripled.

The rate law for a reaction that is second order in nitric oxide (NO) and first order in oxygen (O2) can be written as:

[tex]\[ \text{Rate} = k[\text{NO}]^2[\text{O2}] \][/tex]

where[tex]\( k \)[/tex] is the rate constant, [tex]\([\text{NO}]\)[/tex] is the concentration of nitric oxide, and \[tex]([\text{O2}]\)[/tex] is the concentration of oxygen.

If the concentration of NO is tripled, the new rate can be calculated by substituting [tex]\( 3[\text{NO}] \)[/tex] into the rate law:

[tex]\[ \text{New Rate} = k(3[\text{NO}])^2[\text{O2}] \][/tex]

[tex]\[ \text{New Rate} = k \cdot 9[\text{NO}]^2 \cdot [\text{O2}] \][/tex]

[tex]\[ \text{New Rate} = 9 \cdot k[\text{NO}]^2[\text{O2}] \][/tex]

Since the original rate is [tex]\( k[\text{NO}]^2[\text{O2}] \),[/tex] the new rate is 9 times the original rate because[tex]\( 9 \cdot k[\text{NO}]^2[\text{O2}] \)[/tex]) is 9 times[tex]\( k[\text{NO}]^2[\text{O2}] \).[/tex]

Therefore, tripling the concentration of NO increases the rate of the reaction by a factor of 9. This is because the rate depends on the square of the concentration of NO, and when a concentration is tripled, its square becomes 9 times larger, resulting in a 9-fold increase in the rate of the reaction.

Answer:

Explanation:

1) Chemical equation (given):

2) Mole ratio:

3) C₈H₁₈ (l) moles:

Molar mass: 114.2285 g/mol (taken from a table or internet)

4) Proportion:

5) Ideal gas equation:

Substitute with:

Solve for V:

Approximately 30.28 liters of CO2 are produced for every gallon of octane burned.

To find the number of liters of CO2 produced for every gallon of octane burned, we need to use the balanced equation and conversion factors. From the equation, we can see that for 2 moles of octane burned, we get 16 moles of CO2. Using the molar volume of gases at STP (22.4 L/mol), we can convert the moles of CO2 to liters. Finally, we can use the given conversion factor to convert from gallons to liters.

First, let's calculate the moles of CO2 produced from 1 gallon of octane burned:

16 moles CO2 / 2 moles octane x 3.785 L / 1 gallon = 30.28 L CO2

Therefore, for every gallon of octane burned, approximately 30.28 liters of CO2 are produced.

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Answer:

11.9 moles of carbon-dioxide will produced.

Explanation:

Given moles of ethane = 5.95 mol

[tex]2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)[/tex]

According to reaction 2 moles of ethane produces 4 moles of carbon-dioxide.

Then, 5.95 moles of ethane will produce:

[tex]\frac{4}{2}\times 5.95 mol=11.9 mol[/tex] of carbon-dioxide

11.9 moles of carbon-dioxide will produced.

First, the stoichiometric relationship in the balanced chemical equation between ethane (C2H6) and carbon dioxide (CO2) is identified. From this it is shown that 1 mole of ethane produces 2 moles of carbon dioxide. Therefore, 5.95 moles of ethane will produce 11.9 moles of CO2.

The combustion of ethane, C2H6, is described by the balanced chemical equation: 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g). This indicates a stoichiometric relationship that 2 moles of ethane produce 4 moles of carbon dioxide, or 1 mole of ethane produces 2 moles of carbon dioxide. Therefore, if 5.95 moles of ethane are combusted in excess oxygen, it would produce twice this amount in moles of carbon dioxide. Hence, the reaction would produce 11.9 moles of CO2.

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Answer:

Explanation:

The Brayton cycle is Joule cycle and is the model for the operation of a gas turbine engine.

The general equation for the efficiency of an engine working between two temperatures is:

In this case:

Thus, substituting the data into the formula, you get:

Answer: the temperature of the air exhausted from the engine is 347 K

Answer:

VP as function of time => VP(Ar) > VP(Ne) > VP(He).

Explanation:

Effusion rate of the lighter particles will be higher than the heavier particles. That is, the lighter particles will leave the container faster than the heavier particles. Over time, the vapor pressure of the greater number of heavier particles will be higher than the vapor pressure of the lighter particles.

=> VP as function of time => VP(Ar) > VP(Ne) > VP(He).

Review Graham's Law => Effusion Rate ∝ 1/√formula mass.

The option that will be true regarding the relative values of the partial pressures of the gases remaining in the vessel after some of the gas mixture has effused is;

A) Pressure He < Pressure Ne < Pressure Ar

The missing options are;

a) Pressure He < Pressure Ne < Pressure Ar

b) Pressure He < Pressure Ar < Pressure Ne

c) Pressure Ar < Pressure Ne < Pressure He

d) Pressure He = Pressure Ar = Pressure Ne

Now, to answer this question, let us first write the formula for graham's law of diffusion/effusion.

r ∝ 1/√M

Where;

r is rate of diffusion/effusion

M is molar mass

In effusion, there is usually a barrier with very small holes that prevents the gas from expanding very fast into a new volume. Thus, we can say that the heavier the gas, the lesser the effusion rate and the lighter the gas, the faster the effusion rate.

Now, the pressure of the heavier gas is usually higher than that of the lighter ones.

The molar mass of the given gases are;

Molar mass of He = 4 g/mol

Molar mass of Ne = 20.18 g/mol

Molar mass of Ar = 39.95 g/mol

From the effusion equation, we see that the rate of effusion is inversely proportional to the molar mass.

Thus, the higher the molar mass, the lesser the effusion rate and as such, the higher the pressure since it is a heavy gas.

Thus, Ar has the highest Molar mass and will have the highest pressure. Next is Neon(Ne), then Helium(He).

Thus, in conclusion;

Pressure He < Pressure Ne < Pressure Ar

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Answer:

the 2nd part

Explanation:

could conversion is called cellular respiration

Glycolysis results in the conversion of one glucose molecule into two pyruvate molecules, a net gain of two ATP molecules, and the reduction of NAD+ to NADH. It does not result in the production of CO2 or the conversion of NADH back to NAD+.

Glycolysis is a metabolic pathway occurring in the cytosol of cells where one glucose molecule is converted into two, three-carbon compounds, specifically, Pyruvate. This process is anaerobic, meaning it doesn't require oxygen. There's also a net production of two ATP molecules per glucose molecule, so the option indicating a net loss is incorrect. Furthermore, in the process of glycolysis, the compound NAD+ is actually reduced to NADH, not the other way around. CO2 production doesn't occur in glycolysis itself but in subsequent processes of cellular respiration namely the Krebs cycle and oxidative phosphorylation.

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The molarity of the KOH solution is calculated by first determining the number of moles of H2SO4 using its volume and molarity. Then, the number of moles of KOH is calculated based on the stoichiometry of the reaction. Finally, the molarity of KOH is computed using the definition of molarity (moles/volume), which results in a molarity of 0.84 M for the KOH solution.

To solve the student's question regarding the molarity of a KOH solution, we'll refer to the balanced chemical equation which is given as 2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l). This equation tells us that the ratio of moles of KOH to H2SO4 is 2:1. This means 1.00 mole of H2SO4 reacts with 2.00 moles of KOH.

First, calculate the number of moles of H2SO4 using its volume and molarity: moles H2SO4 = volume (L) × molarity (M) = 25.2 mL (which is 0.0252 L) × 1.50 M = 0.0378 mol H2SO4.

Then, calculate the number of moles of KOH, knowing that 2 moles of KOH react with each mole of H2SO4: moles KOH = 2 × moles H2SO4 = 2 × 0.0378 mol = 0.0756 mol KOH.

Finally, calculate the molarity of KOH using the definition of molarity = moles/volume(L): Molarity KOH = moles KOH / volume (L) = 0.0756 mol / 0.090 L = 0.84 M.

So, the molarity of the KOH solution is 0.84 M.

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The molarity of the KOH solution, you can use the balanced chemical equation and stoichiometry to calculate the solution concentration as 0.840 M.

Molarity of KOH Solution:

In the given titration, 25.2 ml of 1.50 M H₂SO₄  neutralized 90.0 ml of KOH solution. By using the balanced chemical equation provided, you can determine that 1 mole of H₂SO₄ reacts with 2 moles of KOH. Using this information, you can calculate the molarity of the KOH solution to be 0.840 M.

Due to the conjugate base of the hydrogen atom is aromatic, Hb is regarded as the most acidic. Because the conjugate base of the hydrogen atom Hc is anti-aromatic, it is the least acidic.

The correct options are:

(A) - (a)

(B) - (d)

The hydrogen connected at the heptatriene's tertiary position (at the 7-methyl) would be particularly acidic, as its removal would leave a positive charge that could be transported around the ring via resonance.

The hydrogen connected to the pentadiene (5-methyl) at the tertiary position would not be acidic, as removing it would result in an anti-aromatic structure.

Thus, the least acidic H atom is Hc and the most acidic H atom is Hb.

Learn more about hydrogen atom, here:

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The equilibrium constant, Kc, of the given reaction is calculated by substitifying the equilibrium concentrations into the Kc expression. Considering the chemical equation 2 HI(g) + Cl2(g) ⇌ 2 HCl(g) + I2(s), the Kc expression is [HCl]^2 / ([HI]^2 * [Cl2]). Substituting the given equilibrium concentrations in, Kc equates to 4.67 × 10^33.

The equilibrium constant (Kc) for a chemical reaction is a number that expresses the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of its stoichiometric coefficient in the balanced chemical equation. In the equation given: 2 HI(g) + Cl2(g) ⇌ 2 HCl(g) + I2(s), I2 (s) is a pure solid and according to the rules of equilibrium, pure solids and liquids are not included in the equilibrium expression. Therefore, the equilibrium constant expression does not include I2.

For this reaction, the Kc expression is: Kc = [HCl]^2 / ([HI]^2 * [Cl2])

Substituting the equilibrium concentrations into this expression gives: Kc = (0.13 M)^2 / ((5.6 × 10^-16 M)^2 * 0.0019 M) = 4.67 × 10^33

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Answer:

The solution on this question must be a buffer solution.

Explanation

Buffer solution has a molecule that able to withstand pH changes when a small amount of strong acid is added. In the solution, there is conjugate base that could react with H+. When a strong acid is added, the reaction shifts to the left so the conjugate eats up a bit H+ resulting in a lower amount of expected pH changes.

Answer:

6) (a) 0.499; (b) 31.7 %

7) 0.15

Explanation:

6) (a) Absorbance

Beer's Law is

[tex]A = \epsilon cl\\A = \text{35.9 L&\cdot$mol$^{-1}$cm$^{-1}$} $\times$ 0.0278 mol$\cdot$L$^{-1} \times $ 0.5 cm = \mathbf{0.499}[/tex]

(b) Percent transmission

[tex]A = \log {\left (\dfrac{1}{T}}\right)}\\\\\%T = 100T\\\\T = \dfrac{\%T}{100}\\\\\dfrac{1}{T} = \dfrac{100 }{\%T}\\\\A = \log \left(\dfrac{100 }{\%T} \right ) = 2 - \log \%T\\\\0.499 = 2 - \log \%T\\\\\log \%T = 2 - 0.499 = 1.501\\\\\%T = 10^{1.501} = \mathbf{31.7}[/tex]

7) Absorbance

[tex]A = \log \left (\dfrac{I_{0}}{I} \right ) = \log \left (\dfrac{I_{0}}{0.70I_{0}} \right ) = \log \left (\dfrac{1}{0.70} \right ) = -\log(0.70) = \mathbf{0.15}}[/tex]

The absorbance of the solution with the given molar extinction coefficient, concentration, and path length is approximately 0.5, and the percent transmittance is approximately 31.6%. When the transmitted light is 70% of the initial light beam intensity, the absorbance is approximately 0.523.

The subject matter is related to the concept of absorbance in Chemistry. In this specific case, you are required to know and apply the Beer-Lambert law, which states that the absorbance (A) of a solution is directly proportional to its concentration (c) and the path length (l). The formula used is A = εcl, where ε is the molar extinction coefficient.

For part (a), plug the given values into the above formula to get: A = 35.9 L/(mol cm) * 0.0278 mol/L * 0.5 cm = 0.4987. So, the absorbance of the solution is approximately 0.5.

Now, for part (b), the percent transmittance (%T) can be calculated using the relationship A = 2 - log(%T). Solving for %T gives: %T = 10^(2-A) = 10^(2-0.4987) = 31.6%. So the %T is approximately 31.6%.

For part (7), the decrease in the intensity of light by 70% means the transmitted light is 30% of the initial intensity. The absorbance in this case can be calculated directly from the formula A = -log(I/I0) where I/I0 = 0.30. Hence, A = -log(0.30) = 0.523. So the absorbance, when the transmitted light intensity is 70% of the initial light beam intensity, is approximately 0.523.

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Answer:

n = 5

Explanation:

methane => 1 Carbon => CH₃

ethane => 2 Carbons => C₂H₆

propane => 3 Carbons => C₃H₈

butane => 4 Carbons  => C₄H₁₀

pentane => 5 Carbons => C₅H₁₂

hexane => 6 Carbons => C₆H₁₄

heptane => 7 Carbons => C₇H₁₆ => CH₃(CH₂)₅CH₃

Octante => 8 Carbons => C₈H₁₈

Nonane => 9 Carbons => C₉H₂₀

Decane => 10 Carbons => C₁₀H₂₂

Answer:

3,964 years.

Explanation:

k = ln(2)/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(5,730 years) = 1.21 x 10⁻⁴ year⁻¹.

kt = ln([A₀]/[A]),

where, k is the rate constant of the reaction (k = 1.21 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of the sample ([A₀] = 100%).

[A] is the remaining concentration of the sample ([A] = 61.9%).

∴ t = (1/k) ln([A₀]/[A]) = (1/1.21 x 10⁻⁴ year⁻¹) ln(100%/61.9%) = 3,964 years.

The wooden harpoon handle found in the Inuit archaeological site, with 61.9% of the original carbon-14 content remaining and knowing that the half-life of carbon-14 is 5,730 years, is approximately 4000 years old.

To determine the age of the wooden harpoon handle found in an Inuit archaeological site, we use the concept of carbon-14 dating. Since the half-life of carbon-14 is 5,730 years, this means that after 5,730 years, 50% of the original carbon-14 would have decayed. In our case, the handle has 61.9% of the original carbon-14 content remaining.

We can formulate the decay process mathematically using the equation N(t) = N0 * (1/2)^(t/T), where N(t) is the remaining amount of carbon-14 at time t, N0 is the initial amount of carbon-14, T is the half-life, and t is the time that has passed.

Applying this to our case, we're looking to find 't', given that N(t)/N0 = 0.619 and T = 5730 years. The equation becomes 0.619 = (1/2)^(t/5730).

To solve for 't', we take the natural logarithm of both sides:  ln(0.619) = ln((1/2)^(t/5730)) =>  ln(0.619) = (t/5730)*ln(1/2). Solving for 't', we find that 't' = 5730 * ln(0.619) / ln(1/2). This yields 't' = 5,730 * (-0.4797) / (-0.6931) = approximately 4000 years.

Therefore, the wooden harpoon handle is approximately 4000 years old based on its carbon-14 content.

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Answer:

-720.2 kJ

Explanation:

We have two equations:

(I)   X + ½O₂ ⟶ XO;       ΔH = −549.3 kJ

(II) XCO₃ ⟶ XO + CO₂; ΔH =  +170.9 kJ  

From these, we must devise the target equation:

(III) X +½O₂ +CO₂ ⟶ XCO₃; ΔH = ?

The target equation has X on the left, so you rewrite Equation(I).

(I) X + ½O₂ ⟶ XO; ΔH = −549.3 kJ

Equation (IV) has XO on the right, and that is not in the target equation.

You need an equation with XO on the left, so you reverse Equation (II).  

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

(IV)  XO + CO₂ ⟶ XCO₃ ;  ΔH = -170.9 kJ  

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

When you add equations, you add their ΔH values.

You get the target equation (III):

(I)   X + ½O₂ ⟶ XO;                ΔH = -549.3 kJ

(IV) XO + CO₂ ⟶ XCO₃ ;       ΔH =  -170.9 kJ  

(III) X + ½O₂ + CO₂ ⟶ XCO₃; ΔH = -720.2 kJ

The enthalpy change ΔH for the reaction X(s)+1/2O2(g)+CO2(g)⟶XCO3(s) is +720.2 kJ, according to Hess's law.

The enthalpy change ΔH for the reaction X(s)+1/2O2(g)+CO2(g)⟶XCO3(s) can be deduced using Hess's law. According to Hess's law, the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps into which a reaction can be divided.

The two provided reactions can be manipulated to yield the desired reaction:

1. Reverse the first reaction and change the sign of ΔH:

XO(s)⟶X(s)+1/2O2(g) with ΔH=+549.3 kJ

2. Keep the second reaction as is:

XCO3(s)⟶XO(s)+CO2(g) with ΔH=+170.9 kJ

Adding these reactions gives the desired reaction:

X(s)+1/2O2(g)+CO2(g)⟶XCO3(s)

The enthalpy change ΔH for the overall reaction is the sum of the enthalpy changes for the individual reactions, or ΔH= (+549.3 kJ) + (+170.9 kJ) = +720.2 kJ.

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Answer:

Explanation:

The relevant fact here is:

That means that the precipitate that was thrown away, before evaporating the remaining liquid solution under vacuum, does not count; you must only use the amount of solute that was dissolved after cooling the solution to 21°C.

Then, the amount of solute dissolved in the 600 ml solution at 21°C is the weighed precipitate: 0.084 kg = 84 g.

With that, the solubility can be calculated from the followiing proportion:

      ⇒ y = 84. g solute × 100 ml solution / 600 ml solution = 14. g.

The correct number of significant figures is 2, since the mass 0.084 kg contains two significant figures.

The answer is 14. g of solute per 100 ml of solution.

To calculate the solubility of compound X at 21.° C, you would need to know the number of moles of X in the precipitate, which can be calculated if the molar mass of X is known. The solubility can then be obtained by dividing the number of moles by the volume of solution in liters. Without the molar mass, a precise value can't be obtained.

In the given scenario, we can find the solubility of compound X by determining the number of moles in the precipitate and dividing it by the volume of the solution. The mass of the precipitate given is 0.084 kg or 84 g. Assuming compound X is a simple ionic compound that follows a 1:1 ratio like NaCl, the number of moles of compound X in the solution equals the number of moles of the precipitate. Therefore, the solubility of Compound X at 21.° C can be calculated by dividing the number of moles by the volume of solution in liters. A precise value can't be obtained without the molar mass of the compound. However, if we know the molar mass of the compound, we could obtain the molarity. This molarity would represent the solubility of the compound X at 21.° C.

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