Atoms of Mg combine with atoms of F to form a compound. Atoms of which of the following elements combine with atoms of F in the same ratio?


Al
Li
Cl
Ba

Answer:

1a) 1.67 atm

1b) 68,330 g/mol

Explanation:

1a) For Boyle's Law, when a state change occurs without a change in the temperature, the product of the pressure by the volume remains constant. For Dalton's Law, the total pressure of a gas mixture is the sum of the partial pressure of the components. Then:

P1V1 + P2V2 = PV

Where P1 is the initial pressure of CO₂, V1 is the initial volume of it, P2 is the initial pressure of O₂, V2 is the initial pressure of it, P is the pressure of the mixture and V is the final volume of the mixture (V1 + V2).

1*250 + 2*500 = P*750

750P = 1250

P = 1.67 atm

1b) Let's call hemoglobin by Hem. The stoichiometry reaction is:

Hem + 4O₂ → HemO₂

So, let's calculate the number of moles of oxygen in the reaction, by the ideal gas law, PV = nRT, where P is the pressure, V is the volume (0.0023 L), n is the number of moles, R is the ideal gas constant (62.3637 L.torr/mol.K), and T is the temperature (37°C = 310 K).

743*0.0023 = n*62.3637*310

19,332.747n = 1.7089

n = 8.84x10⁻⁵ mol

For the reaction, the stoichiometry is:

1 mol of Hem -------------------- 4 mol of O₂

x ------------------- 8.84x10⁻⁵ mol of O₂

By a simple direct three rule

4x = 8.84x10⁻⁵

x = 2.21x10⁻⁵ mol of hemoglobin

The molar mass is the mass divided by the number of moles:

M = 1.51/2.21x10⁻⁵

M = 68,330 g/mol

The total pressure of the mixture of CO2 and O2 is 3.00 atm. The molar mass of hemoglobin is approximately 64.7 g/mol.

To determine the total pressure of the mixture of CO2 and O2, we can use Dalton's Law of Partial Pressures. According to this law, the total pressure of a mixture of gases is equal to the sum of the partial pressures of each gas. In this case, the partial pressures of CO2 and O2 are 1.00 atm and 2.00 atm, respectively. Therefore, the total pressure of the mixture is 1.00 atm + 2.00 atm = 3.00 atm.

To find the molar mass of hemoglobin, we can use the ideal gas law. The equation for the ideal gas law is 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. We can rearrange this equation to solve for the molar mass (M).

M = (mRT)/(PV), where m is the mass of the substance in grams. Plugging in the given values, we have M = (1.51 g)(0.0821 L·atm/mol·K)(310 K)/(743 torr)(0.0821 L·atm/mol·K)(2.30 mL/1000 L). Simplifying the calculation gives us a molar mass of hemoglobin approximately equal to 64.7 g/mol.

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Answer:

[tex][I^-]=4.6*10^{-8}M[/tex]

Explanation:

The expression of the Ksp:

[tex]Ksp_{AgCl}=[Ag^{+}][Cl^-][/tex]

[tex]Ksp_{AgI}=[Ag^{+}][I^-][/tex]

When the product of the concentrations of both ions equals the Ksp, the salt starts to precipitate.

For the AgCl:

[tex]1.8*10^{-10}M^{2}=[Ag^{+}]*0.1M[/tex]

[tex][Ag^{+}]=1.8*10^{-9}M[/tex]

Initially the concentration of I- was 0.1 M, due to the lower Ksp than the AgCl's, the AgI will precipite before. So, when AgCl starts to precipitate the concentration of I- will be in equilibrium, following the Ksp equation.

[tex]8.3*10^{-17}M^{2}=1.8*10^{-9}M*[I^-][/tex]

[tex][I^-]=4.6*10^{-8}M[/tex]

Explanation:

There are two type of reaction possible , i.e. , exothermic and endothermic reaction ,

Exothermic reaction -

The type of reaction in which energy in the form of heat or light is released , is known as exothermic reaction ,

The reaction mixture usually get heated after the reaction .( temperature increases ) .

The sign of ΔH of an exothermic reaction is negative .

Similarly ,

Endothermic reaction -

The type of reaction in which energy is absorbed in the form of heat is known as endothermic reaction .

The reaction mixture usually get cooled after the reaction .( temperature decreases ) .

The sign of ΔH of an Endothermic  reaction is positive .

a. dry ice evaporating ,

Endothermic reaction , ΔH = positive .

b. a sparkler burning ,

Exothermic reaction ΔH = negative .

c. the reaction that occurs in a chemical cold pack often used to ice athletic injuries

Endothermic reaction , ΔH = positive .

Dry ice evaporates in an endothermic process with a positive ΔH. A sparkler burning is an exothermic process with a negative ΔH. The reaction in a chemical cold pack is endothermic with a positive ΔH.

In chemistry, understanding whether a process is exothermic or endothermic is crucial. Here’s the breakdown for each scenario:

Answer:

1.08g

Explanation:  

Like all other hydrocarbons, methane burns in oxygen to form carbon iv oxide and water. The chemical equation of this equation is shown below;

  CH4 (g)  +  2O2  (g) ----------->  2H2O (l)   +   CO2 (g)

From the reaction equation, we can see that one mole of methane gave 2  moles of water. This is the theoretical yield. We need to note the actual yield.

To get the actual yield, we get the number of moles of methane reacted. To get this, we divide the mass of methane by  the molar mass of methane. The molar mass of methane is 16g/mol. The number of moles is thus 0.481/16 = 0.03 moles.

Since 1 mole methane gave 2 moles of water, this shows that 0.06 moles of water were produced. The mass of water thus produced is 0.06 multiplied by the molar mass of water. The molar mass of water is 18g/mol. The mass produced is 0.06 * 18 = 1.08g

Now, we do same for the mass of oxygen. From the reaction equation, 2 moles of oxygen produced two moles of water. Hence, we can see from here that the number of moles here are equal. We then proceed to calculate the actual number of moles of oxygen produced. This is the mass of the oxygen divided by the molar mass of molecular oxygen. The molar mass of molecular oxygen is 32g/mol. The number of moles thus produced is 0.54/32 = 0.016875 mole. The number of moles are equal, this means that the number of moles of oxygen produced is also 0.016875

Now, to get the mass of water produced, we multiply the number of moles by the molar mass of water. The molar mass of water is 18g/mol.

The mass is thus, 0.016875 * 18 = 0.30375g

1.08g is higher and thus is the maximum mass

Answer:Answer: The statements that describe amphibolic characteristics of the citric acid cycle is that both catabolic and anabolic processes occur.

Explanation: the metabolic pathways includes; catabolic, anabolic and amphibolic processes.

A catabolic pathway is an exergonic system that produces chemical energy in the form of Adenosine Triphosphate (ATP) usually from energy containing sources such as carbohydrates, fats, and proteins.

An anabolic pathway is a biosynthetic pathway, whereby smaller molecules combine to form larger and more complex ones.[

An amphibolic pathway is one that can be either catabolic or anabolic based on the need for energy. The citric acid pathway contains both energy producing and utilizing pathway.

The citric acid cycle is amphibolic, participating in both catabolic and anabolic processes, producing intermediates for various biosynthetic pathways and involving oxidation and reduction reactions.

The citric acid cycle, also known as the Krebs cycle, is considered amphibolic because it has both catabolic and anabolic functions. It is a central metabolic pathway in the cell, involved in the oxidation of acetyl-CoA into CO2, and is crucial for energy production in the form of ATP. Furthermore, the cycle provides intermediates for various anabolic pathways, making it an integral part of cellular metabolism. For instance, it produces
-ketoglutarate, which is a precursor for amino acid synthesis, and succinyl-CoA, which is required for heme group synthesis. Also, the cycle involves both oxidation and reduction reactions, further asserting its amphibolic nature.

Anaplerotic and cataplerotic pathways are essential for maintaining the function of the citric acid cycle. Anaplerotic pathways replenish cycle intermediates that are removed for biosynthesis, whereas cataplerotic pathways describe the removal of intermediates from the cycle for the synthesis of other compounds. This balance allows the cycle to efficiently meet the fluctuating needs of the cell for both energy production and the synthesis of crucial molecules.

To calculate ΔG for the reaction under given conditions, use the equation ΔG = ΔG° + RT ln(Q), substituting given values and solving. Positive ΔG suggests a non-spontaneous reaction, while negative ΔG indicates a spontaneous reaction.

The student's question pertains to the change in Gibbs Free Energy (ΔG) for the reaction CO2(g) + CCl4(g) ⇌ 2 COCl2(g) under specified conditions. The ΔG of a reaction can be calculated using the equation ΔG = ΔG° + RT ln(Q), where ΔG° is the standard free energy change, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (here, 25°C + 273.15 = 298.15 K), and Q is the reaction quotient. The equation thus becomes ΔG = 46.9 kJ + (8.314 J/(mol·K) * 298.15 K * ln((0.653^2)/(0.459 * 0.984))).

Converting R from J to kJ gives us ΔG = 46.9 kJ + (0.008314 kJ/(mol·K) * 298.15 K * ln((0.653^2)/(0.459 * 0.984))). Solving this equation, we obtain a value for ΔG, which represents the Gibbs free energy under the given conditions. A positive ΔG suggests the reaction is non-spontaneous, while a negative ΔG indicates a spontaneous reaction. An equilibrium will lean towards the side with the lower Gibbs free energy.

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Answer:

The change in temperature = 7.65 °C

Explanation:

Step 1: Data given

10.0 grams of ammonium nitrate dissolves in 100.0 grams of water

Hsoln = 25.7 kJ/mol

Molar mass = 80.04 g/mol

Heat capacity of the solution = 4.2 J

Step 2: Calculate moles  ammonium nitrate

Moles = mass / molar mass

Moles = 10.0 grams / 80.04 g/mol

Moles =0.125 moles

Step 3: Calculate q

q = 25.7 kJ/mol * 0.125 moles

q = 3.2125 kJ = 3212.5 J

Step 4: Calculate change in temperature

q = m*c*ΔT

3212.5 J = 100g *4.2 J * ΔT

ΔT= 7.65

The change in temperature = 7.65 °C

Explanation:

The given data is as follows.

Molar mass of ammonium nitrate = 80.0 g/mol

So, we will calculate the number of moles of ammonium nitrate as follows.

     No. of moles = [tex]\frac{\text{given mass}}{\text{molar mass}}[/tex]

                            = [tex]\frac{10.0 g}{80.0 g/mol}[/tex]

                            = 0.125 mol

Heat released due to solution of ammonium nitrate = [tex]\Delta H \times \text{no. of moles}[/tex]

                    = [tex]25.7 kJ/mol \times 0.125  mol[/tex]

                    = 3.2125 KJ

                    = 3212.5 J              (as 1 kJ = 1000 J)

Therefore, calculate the total mass of solution as follows.

   mass of solution(m) = (10.0 + 100.0 ) g

                                    = 110.0 g

Hence, heat released will be calculated as follows.

                 Q = [tex]m \times C \times \Delta T[/tex]

          3212.5 J = [tex]110.0 \times 4.2 J \times \Delta T[/tex]

           [tex]\Delta T = 6.95^{o}C[/tex]

Thus, we can conclude that the change in temperature of the solution is [tex]6.95^{o}C[/tex].

Answer:

The Kc is 1.36 (but this is not an option, may be the options are wrong, or may be I was .. Thanks!)

Explanation:

Let's think all the situation.

               2 ICl(g)   ⇄   I₂(g)    +    Cl₂(g)

Initially      0.20              -               -

Initially I have only 0.20 moles of reactant, and nothing of products. In the reaction, an x amount of compound has reacted.

React          x              x/2               x/2

Because the ratio is 2:1, in the reaction I have the half of moles.

So in equilibrium I will have

           (0.20 - x)          x/2             x/2

Notice that I have the concentration in equilibrium so:

0.20 - x = 0.060

x = 0.14

So in equilibrium I have formed 0.14/2 moles of I₂ and H₂ (0.07 moles)

Finally, we have to make, the expression for Kc and remember that must to be with concentration in M (mol/L).

As we have a volume of 2L, the values must be /2

Kc = ([I₂]/2 . [H₂]/2) / ([ICl]/2)²

Kc = (0.07/2 . 0.07/2) / (0.060/2)²

Kc = 1.225x10⁻³ / 9x10⁻⁴

Kc = 1.36

Final answer:

The value of Kc for the decomposition of ICl to I2 and Cl2 is calculated using the equilibrium concentrations and the equilibrium expression. By plugging in the given values and solving the expression, we find Kc to be 0.11.

Explanation:

To calculate the equilibrium constant (Kc) for the decomposition of ICl to I2 and Cl2 at a high temperature using the reaction 2 ICl(g) → I2(g) + Cl2(g), we need to apply the concept of the equilibrium constant expression which is based on the molar concentrations of the products raised to the power of their coefficients divided by the molar concentrations of the reactants raised to the power of their coefficients.

At the start, we have 0.20 mol ICl in a 2.0 L flask, which gives us an initial concentration of 0.10 M (0.20 mol / 2.0 L). At equilibrium, the [ICl] is given as 0.060 M. The change in concentration for ICl (reactant) is the initial concentration minus the equilibrium concentration, which is 0.10 M - 0.060 M = 0.040 M. Since the reaction consumes 2 moles of ICl for every 1 mole of I2 and Cl2 produced, each product will have a concentration change of 0.020 M (0.040 M / 2). Therefore, at equilibrium, both I2 and Cl2 have concentrations of 0.020 M.

The equilibrium constant expression is:

Kc = [I2][Cl2] / [ICl]2

Substituting the equilibrium concentrations we get:

Kc = (0.020 M)(0.020 M) / (0.060 M)2 = 4/9 × 10-3

Calculating the value yields:

Kc = 0.11

Answer:

The correct answer is -521.6 KJ/mol

Explanation:

In order to solve the problem, we have to rearrange and to add the reactions with the aim to obtain the requested reaction.

We have:

1) H₂(g) + F₂ (g) → 2HF(g) => ∆H₁ = -546.6 kJ/mol

2) 2H₂(g) + O₂(g) → 2H₂0(l) => ∆H₂ = -571.6 kJ/mol

If we multiply reaction 1 by 2 and add the inverse reaction of 2, we obtain:

2H₂(g) + 2F₂ (g) → 4HF(g) => 4 x ∆H₁ = 4 x (-546.6 kJ/mol)

2H₂0(l) → 2H₂(g) + O₂(g) => (-1) x ∆H₂ = (-1) x (571.6 kJ/mol)

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

2H₂(g) + 2F₂ (g)+  2H₂0(l) → 4HF(g) + 2H₂(g) + O₂(g)

We eliminate 2H₂(g) is repeated in both members of the reaction, and we obtain: 2F₂(g) + 2H₂0(l) -> 4HF(g) + O₂(g)

Finally, the ΔH of the reaction is calculated as follows:

ΔHtotal= (2 x ΔH₁) + ((-1)ΔH₂

ΔHtotal= (2 x (-546.6 KJ/mol)) + ((-1) x (-571.6 KJ/mol)

ΔHtotal= -521.6 KJ/mol

The standard enthalpy change for the combustion of C3H6(g) + H2(g) → C3H8(g) at 25°C is -219.4 kJ/mol.

The standard enthalpy change for the reaction is obtained by calculating the difference between the enthalpy of the products and the enthalpy of the reactants. In this case, the enthalpy change can be determined using the enthalpy of formation values for the compounds involved. The balanced equation for the combustion of propane is:

C3H6(g) + H2(g) → C3H8(g)

The enthalpy change can be calculated as follows:

∆H° = ∑∆H°f(products) - ∑∆H°f(reactants)

∆H° = [2*(∆H°f(CO2(g))) + ∆H°f(H2O(l))] - [∆H°f(C3H6(g)) + ∆H°f(H2(g))]

Substitute the given values for the enthalpies of formation:

∆H° = [2*(-393.5 kJ/mol) + (-285.8 kJ/mol)] - [(-2058.3 kJ/mol) + 0 kJ/mol]

Simplify the equation:

∆H° = -219.4 kJ/mol

Therefore, the standard enthalpy change for the combustion of C3H6(g) + H2(g) → C3H8(g) at 25°C is -219.4 kJ/mol.

Answer: The value of equilibrium constant for reverse reaction is [tex](\frac{1}{K_1})^{1/2}[/tex]

Explanation:

The given chemical equation follows:

[tex]2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)[/tex]

The equilibrium constant for the above equation is [tex]K_1[/tex]

We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:

[tex]SO_3(g)\rightarrow SO_2(g)+\frac{1}{2}O_2(g)[/tex]

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

If the equation is multiplied by a factor of '[tex]\frac{1}{2}[/tex]', the equilibrium constant of the reverse reaction will be the 1/2 power of the equilibrium constant  of initial reaction.

The value of equilibrium constant for reverse reaction is:

[tex]K_{eq}'=(\frac{1}{K_1})^{1/2}[/tex]

Hence, the value of equilibrium constant for reverse reaction is [tex](\frac{1}{K_1})^{1/2}[/tex]

Answer:

The pH will be 4.5

Explanation:

The mixture of benzoic acid (weak acid) and its salt will make a buffer.

The pH of buffer solution can be calculated using Henderson Hassalbalch's equation, which is

[tex]pH=pKa+log\frac{[salt]}{[acid]}[/tex]

pKa = -logKa

pKa = -log([tex]6.3X10^{-5}[/tex])

pKa = 4.2

[tex]pH=4.2 + log\frac{0.41}{0.22}=4.2+0.27=4.47=4.5[/tex]

Answer:

A precipitate will begin to form at [Cu+] = 3.0 *10^-10 M

The precipitate formed is CuI

Explanation:

Step 1: Data given

The solution contains 0.021 M Cl- and 0.017 M I-.

Ksp(CuCl) = 1.0 × 10-6

Ksp(CuI) = 5.1 × 10-12.

Step 2:  Calculate [Cu+]

Ksp(CuCl) = [Cu+] [Cl-]  

1.0 * 10^-6  = [Cu+] [Cl-]  

1.0 * 10^-6  = [Cu+] [0.021]  

[Cu+] = 1.0 * 10^-6 / 0.021

[Cu+] = 4.76 *10^-5 M

Ksp(CuI) = [Cu] [I]  

5.1 * 10^-12  = [Cu+] [I-]  

5.1 * 10^-12 =[Cu+] [0.017]  

[Cu+] = 5.1 * 10^-12 / 0.017

[Cu+] = 3.0 *10^-10 M

[Cu+]from CuI hast the lowest concentration

A precipitate will begin to form at [Cu+] = 3.0 *10^-10 M

The precipitate formed is CuI

Answer:

The half-life of strontium-90 is 28.81 years

Explanation:

Step 1: Data given

Mass of strontium 90 = 1.000 grams

After 2 years there remain 0.953 grams

Step 2: Calculate half-life time

k = (-1/t) * ln (Nt/N0)

⇒ with k = rate constant for the decay

⇒ with N0 = the mass of strontium at the start (t=0) = 1.000 grams

⇒ with Nt =  the mass of strontium after time t (2 years) = 0.953 grams

k = (-1/2) * ln(0.953/1)

k = (-1/2) * (-0.0481) = 0.02405 /yr

t1/2 = 0.693/k= 0.693/ 0.02405 = 28.81  years

The half-life of strontium-90 is 28.81 years

The given data does not provide a clear half-life for strontium-90 since the mass of the substance did not halve in 2 years. Thus, none of the options provided fits the scenario.

The half-life of a radioactive substance is the time taken for half of the atoms in a sample to decay. Here, we have the sample of strontium-90 that has not halved in 2 years since we still have 0.953g out of 1.000g. Given the available options, none of them matches with the provided data because if a substance's mass did not reduce to half in 2 years then its half-life should be greater than 2 years. Possibly, there might be a calculation or data collection error in the problem scenario provided.

The half-life of strontium-90 can be calculated using the given information. If we start with 1.000 g of strontium-90 and after 2.00 years, 0.953 g remains, we can calculate the percentage of strontium-90 remaining: (0.953 g / 1.000 g) x 100% = 95.3%. Since half-life is the time required for half the atoms in a sample to decay, we can conclude that it took 2.00 years for 50% of the strontium-90 to decay. Therefore, the half-life of strontium-90 is 2.00 years.

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In a galvanic cell, oxidation occurs at the anode, the cathode is the positive electrode, cations flow toward the cathode, and electrons flow from the anode to the cathode.

In a galvanic cell, the following processes occur:

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Answer:

The correct answer is A) 1.6 x 10-6

Explanation:

A weak monoprotic acid has the following dissociation equilibrium. At the beggining (t=0), the concentration of the monoprotic acid (HA) is equal to 0.10 M and the concentration of the ions H⁺ and A⁻ is zero (no dissociation). At a time t, dissociation occur and there is x concentration of H⁺ and A⁻ which is given by the dissociation constant Ka.

           HA(aq)    ⇄     H⁺(aq)    +       A⁻(aq)

t=0        0.10 M              0                     0

t               -x                    +x                  +x

eq          0.10 M-x             x                   x

Ka= [tex]\frac{x^{2} }{0.10 - x}[/tex]

As the pH is 3.40, we can calculate the concentration of both H⁺ and A⁻, as follows:

pH= - log (conc H⁺)= -log x

⇒ x = [tex]10^{-3.40}[/tex]= 3.98 x 10⁻⁴

Now we introduce x in the previous equation to calculate Ka:

Ka= [tex]\frac{(3.98 x 10^{-4} )^{2} }{(0.10 - (3.98 x 10^{-4}) }[/tex]

Ka= 1.59 x 10⁻⁶ ≅ 1.60 x 10⁻⁶

Final answer:

To find the acid-ionization constant Ka, the pH is used to calculate the concentration of H+ ions, and this value is squared and divided by the initial concentration of the acid to obtain Ka, which is 1.6 x 10^-6.

Explanation:

The question seeks to determine the acid-ionization constant (Ka) for a weak monoprotic acid with a given molarity and pH. The pH of the solution is 3.40, which means the concentration of hydrogen ions [H+] is 10-3.40 M. Since the acid is weak and monoprotic, its dissociation in water can be represented by HA → H+ + A-. The given concentration of acid (0.10 M) will slightly ionize into H+ and A- ions.

To solve for Ka, the equilibrium expression is Ka = [H+][A-] / [HA]. Given that [H+] = 10-3.40, we can assume that the concentration of A- at equilibrium is also 10-3.40 since the acid donates one proton per molecule (in a 1:1 ratio). The concentration of un-ionized HA will then be approximately ([initial HA] - x) = (0.10 M - 10-3.40 M). However, because the ionization of a weak acid is very small compared to the initial concentration, we can approximate [HA] at equilibrium to be 0.10 M. From this, the Ka can be approximated to be (10-3.40)2 / 0.10 M = 1.6 x 10-6.

Answer:

The answer is: 11759 Hz

Explanation:

Given: Chemical shift: δ = 211.5 ppm, Spectrometer frequency = 556 MHz = 556 × 10⁶ Hz

In NMR spectroscopy, the chemical shift (δ), expressed in ppm, of a given nucleus is given by the equation:

[tex]\delta (ppm) = \frac{Observed\,frequency (Hz)}{Frequency\,\, of\,\,the\,Spectrometer (MHz)} \times 10^{6}[/tex]

[tex]\therefore Observed\,frequency (Hz)= \frac{\delta (ppm)\times Frequency\,\, of\,\,the\,Spectrometer (MHz)}{10^{6}}[/tex]

[tex]Observed\,frequency= \frac{211.5 ppm \times 556 \times 10^{6} Hz}{10^{6}} = 11759 Hz[/tex]

Therefore, the signal is at 11759 Hz from the TMS.

Answer: The value of equilibrium constant for the above equation is 36

Explanation:

We are given:

Initial moles of X = 4.00 moles

Initial moles of Y = 4.00 moles

Equilibrium moles of Z = 6.00 moles

Volume of vessel = 1.00 L

Initial concentration of X = [tex]\frac{4}{1}=4[/tex]

Initial concentration of Y = [tex]\frac{4}{1}=4[/tex]

Equilibrium concentration of Z = [tex]\frac{6}{1}=6[/tex]

The given chemical equation follows:

                  [tex]X(g)+Y(g)\rightarrow 2Z(g)[/tex]

Initial:         4         4

At eqllm:     4-x      4-x        2x

Calculating for 'x', we get:

[tex]\Rightarrow 2x=6\\\\\Rightarrow x=3[/tex]

The expression of [tex]K_c[/tex] for above equation follows:

[tex]K_c=\frac{[Z]^2}{[X]\times [Y]}[/tex]

Putting values in above equation, we get:

[tex]K_c=\frac{6^2}{1\times 1}\\\\K_c=36[/tex]

Hence, the value of equilibrium constant for the above equation is 36

The equilibrium constant, Kc, for the given reaction is 36. This is calculated using the ratio of product concentrations to reactant concentrations at equilibrium, each concentration being raised to the power of its stoichiometric coefficient.

The equilibrium constant, Kc, describes the ratio of product concentrations to reactant concentrations at equilibrium, with each concentration raised to the power of its stoichiometric coefficient. Here, the equation is:

X(g) + Y(g) ---> 2 Z(g)

In terms of moles to start, we have 4.00 mol of X and Y, and 0 mol of Z. At equilibrium, we end up with 6.00 mol of Z, which, since 2 mol of Z are generated for every 1 mol of X and Y, means we have lost 3.00 mol of X and Y.

As the volume is 1L, the molar concentrations are the same as the number of moles. So, the concentrations at equilibrium are: [X] = [Y] = 1.00 M and [Z] = 6.00 M.

Applying this to the Kc expression gives: Kc = ([Z]^2) / ([X] * [Y]) = (6.00^2) / (1.00 * 1.00) = 36. Therefore, the equilibrium constant, Kc, is 36.

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To calculate the mass of dry ice, we need to consider heat transfer and use the formula q = m × c × ΔT. The mass of dry ice can be found using the equation mdry ice = (q / ΔHsublimation). By plugging in the given values, we can find the mass of dry ice that should be added to the water.

In order to calculate the mass of dry ice that should be added to the water, we need to consider the heat transfer that occurs during the process. The heat gained by the water is equal to the heat lost by the dry ice. We can use the formula:

q = m × c × ΔT

where q is the heat gained by the water, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

By rearranging the equation, we can solve for the mass of dry ice:

mdry ice = (q / ΔHsublimation)

where mdry ice is the mass of dry ice, q is the heat gained by the water, and ΔHsublimation is the heat of sublimation of CO2.

In this case, the temperature change is the difference between the initial temperature of the water (90°C) and the final temperature (10°C). The specific heat capacity of water is approximately 4.18 J/g°C, and the heat of sublimation of CO2 is 571 J/g. By plugging in the values and solving the equation, we can find the mass of dry ice that should be added to the water.

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1. The standard free-energy change for the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 25 ∘C is -358,000 J (to three significant figures).

2. The standard cell potential for the reaction X(s)+2Y+(aq)→X2+(aq)+2Y(s) at 25 ∘C is 2.90 V (to three significant figures).

1. To calculate the standard free-energy change at 25 ∘C for the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s), we can use the following equation:

ΔG∘=−nFE∘

where:

* ΔG∘ is the standard free-energy change (in J)

* n is the number of moles of electrons transferred (2 in this case)

* F is the Faraday constant (96,500 C/(mol e⁻))

* E∘ is the standard cell potential (in V)

The standard cell potential for the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) is 1.83 V. Therefore, the standard free-energy change is:

ΔG∘=−(2 mol e⁻)(96,500 C/(mol e⁻))(1.83 V)

ΔG∘=−358,000 J

Therefore, the standard free-energy change for the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 25 ∘C is -358,000 J (to three significant figures).

2. To calculate the standard cell potential at 25 ∘C for the reaction X(s)+2Y+(aq)→X2+(aq)+2Y(s), we can use the following equation:

E∘=−ΔG∘nF

where:

* E∘ is the standard cell potential (in V)

* ΔG∘ is the standard free-energy change (in J)

* n is the number of moles of electrons transferred (2 in this case)

* F is the Faraday constant (96,500 C/(mol e⁻))

We are given that ΔH∘ = -675 kJ and ΔS∘ = -357 J/K. We can use these values to calculate ΔG∘ using the following equation:

ΔG∘=ΔH∘−TΔS∘

where T is the temperature in Kelvin (298 K in this case).

ΔG∘=−675 kJ−(298 K)(−357 J/K)

ΔG∘=−559,890 J

Now we can calculate the standard cell potential using the equation above:

E∘=−ΔG∘nF

E∘=−(−559,890 J)/(2 mol e⁻)(96,500 C/(mol e⁻))

E∘=2.90 V

Therefore, the standard cell potential for the reaction X(s)+2Y+(aq)→X2+(aq)+2Y(s) at 25 ∘C is 2.90 V (to three significant figures).

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The standard free-energy change for the reaction Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s) is -372 kJ at 25 °C. The standard cell potential for the reaction X(s) + 2Y⁺(aq) → X²⁺(aq) + 2Y(s) is 2.95 V at 25 °C. These values were determined using proper thermodynamic equations and constants.

1. Calculate the Standard Free-Energy Change:

The given reaction is: Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s).

To find the standard free-energy change (ΔG°), we use the given equation ΔG° = -nFE°.

2. Calculate the Standard Cell Potential:

Given the reaction: X(s) + 2Y⁺(aq) →  X²⁺(aq) + 2Y(s)

And ΔH° = -675 kJ and ΔS° = -357 J/K.

The standard cell potential for this reaction at 25 °C is 2.95 V.

Final answer:

The chemical equation representing the decomposition of hydrogen peroxide into water and oxygen gas is 2 H2O2 → 2 H2O + O2. This represents a stoichiometric relationship with a 2:1 ratio of hydrogen peroxide to oxygen and a 1:1 ratio of hydrogen peroxide to water.

Explanation:

The stoichiometry of the chemical reaction where hydrogen peroxide (H2O2) decomposes into water (H2O) and oxygen gas (O2) can be expressed by the following balanced chemical equation:

2 H2O2 → 2 H2O + O2

According to the equation given by the student, 0.0800 mol of H2O2 decomposes to produce 0.0800 mol of H2O and 0.0400 mol of O2. However, based on stoichiometric coefficients, we should expect a 1:1 ratio between H2O2 and H2O, and a 2:1 ratio between H2O2 and O2. Therefore, the decomposition of 0.0800 mol H2O2 should yield 0.0400 mol O2 according to the equation, which aligns with what was provided by the chemist.

Final answer:

The enthalpy change (ΔHreaction) for the given reaction is calculated using Hess's law and is found to be 122 kJ/mol.

Explanation:

To calculate the enthalpy change (ΔHreaction) for the reaction 3NO₂(g) + H₂O(l) → 2HNO3(aq) + NO(g), we will apply Hess's law and use the standard enthalpies of formation (ΔHf) for each compound.

The formula to calculate ΔHreaction is:

ΔHreaction = ∑(ΔHf products) - ∑(ΔHf reactants)

Using the provided ΔHf values:

We can calculate ΔHreaction as follows:

ΔHreaction = [2 × ΔHf(HNO3) + ΔHf(NO)] - [3 × ΔHf(NO2) + ΔHf(H2O)]

ΔHreaction = [(2 × 207 kJ/mol) + (90 kJ/mol)] - [(3 × 32 kJ/mol) + (286 kJ/mol)]

ΔHreaction = (414 kJ/mol + 90 kJ/mol) - (96 kJ/mol + 286 kJ/mol)

ΔHreaction = 504 kJ/mol - 382 kJ/mol

ΔHreaction = 122 kJ/mol

AgI precipitates first due to its smaller Ksp. The I⁻ concentration for the second compound (PbI₂) to begin precipitating is 9.3 x 10⁻⁵ M.

The compound that will precipitate first will be the one with the smallest solubility product constant, Ksp. The given Ksp values for AgI and PbI₂ are 1.5 x 10⁻¹⁶ and 8.7 x 10⁻⁹ respectively; thus, AgI has a smaller Ksp, indicating it is less soluble and precipitates first.

To find the I⁻ concentration for the second compound to precipitate, we set up the Ksp expression for PbI₂ (assuming AgI has already precipitated out, AgNO₃ is gone and the remaining AgI does not affect this calculation): [Pb²⁺][I⁻]² = Ksp. Pb(NO₃)₂ fully dissociates, so [Pb²⁺] is assumed to be 0.10 M, and we can substitute this and the Ksp into our expression, then solve for [I-]: (0.10 M)([I⁻])² = 8.7 x 10⁻⁹ M, from which we find [I-] = sqrt((8.7 x 10⁻⁹ M) / 0.10 M) = 9.3 x 10⁻⁵ M.

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Answer:

1) Exothermic.

2) It goes up.

3) It moves out.

4) It does work.

5) 94 kJ.

Explanation:

1) An endothermic reaction is a reaction that absorbs heat from the surroundings, and an exothermic reaction is a reaction that releases heat to the surroundings. So the reaction placed is exothermic.

2) Because the heat is flowing to the water, its temperature will go up.

3) By the first law of the thermodynamics:

ΔU = Q - W

Where ΔU is the total energy, Q is the heat, and W is the work. Because the energy and the heat are being released, they are both negative:

-244 = -150 - W

W = 94 kJ

The work is positive, so it's being doing by the system, it means that the system is expanding, and the piston moves out.

4) As explained above, the gas mixture (the system) does work.

5) As shown above, W = 94 kJ

a) Approximately 3.125% of 254/102 No remains 5.0 minutes after it forms.

b) About 0.391% of 254/102 No is left 1.0 hour after it forms.

In nuclear decay the percentage of a substance remaining can be calculated using the formula

[tex]\[ \text{Final amount} = (\frac{1}{2})^{\frac{\text{time elapsed}}{\text{half-life}}} \times 100 \][/tex]

a) For 5.0 minutes after formation we convert the time to seconds (5.0 minutes = 300 seconds) and apply the formula

[tex]\[ \text{Final amount} = (\frac{1}{2})^{\frac{300 \, \text{seconds}}{55 \, \text{seconds}}} \times 100 \approx 3.125\% \][/tex]

b) For 1.0 hour after formation we convert the time to seconds (1.0 hour = 3600 seconds) and use the formula

[tex]\[ \text{Final amount} = (\frac{1}{2})^{\frac{3600 \, \text{seconds}}{55 \, \text{seconds}}} \times 100 \approx 0.391\% \][/tex]

Nuclear decay and half-life involve complex mathematical models that govern the decay of radioactive substances. Understanding these processes is crucial in various scientific fields including nuclear physics medicine and environmental science.

Answer:

a. -206,4kJ

b. Surroundings will gain heat.

c. -115kJ are given off.

Explanation:

It is possible to obtain ΔH of a reaction using Hess's law that consist in sum the different ΔH's of other reactions until obtain the reaction you need.

Using:

(1) C₆H₄(OH)₂(l) → C₆H₄O₂(l) + H₂(g) ΔH: +177.4 kJ

(2) H₂(g) + O₂(g) → H₂O₂ (l) ΔH: -187.8 kJ

(3) H₂(g) + 1/2O₂(g) → H₂O(l) ΔH: -285.8 kJ

It is possible to obtain:

C₆H₄(OH)₂(l) + H₂O₂(l) → C₆H₄O₂(l) + 2H₂O(l)

From (1)-(2)+2×(3). That is:

(1) C₆H₄(OH)₂(l) → C₆H₄O₂(l) + H₂(g) ΔH: +177.4 kJ

-(2) H₂O₂(l) → H₂(g) + O₂(g) ΔH: +187.8 kJ

2x(3) 2H₂(g) + O₂(g) → 2H₂O(l) ΔH: 2×-285.8 kJ

The ΔH you obtain is:

+177,4kJ + 187,8kJ - 2×285.8 kJ = -206,4kJ

b. When ΔH of a reaction is <0, the reaction is exothermic, that means that the reaction produce heat and the surroundings will gain this heat.

c. 20,0g of H₂O are:

20,0g×[tex]\frac{1mol}{18,01g}[/tex] = 1,11 mol H₂O

As 2 moles of H₂O are produced when -206,4kJ are given off, when 1,11mol of H₂O are produced, there are given off:

1,11mol H₂O×[tex]\frac{-206,4kJ}{2mol}[/tex] = -115kJ

I hope it helps!

Final answer:

To calculate the enthalpy change (ΔH°) for the given reaction, use Hess's Law and the enthalpy changes provided. Heat would be gained by the surroundings as the reaction occurs, based on the positive enthalpy change. When 20.0 g of H2O is produced, approximately -317 kJ of heat would be given off.

Explanation:

To calculate the enthalpy change (ΔH°) for the given reaction, we can use the Hess's Law and the enthalpy changes provided. We need to find a combination of reactions whose sum gives us the desired reaction. Here's an approach:

1. Reverse the first equation and multiply it by 2 to get 2CO(g) → 2C(s) + O2(g) with ΔH° = -442.0 kJ.

2. Use the second equation as it is: C(s) + O2(g) → CO2(g) with ΔH° = -393.5 kJ.

3. Add the two equations together: 2CO(g) + C(s) + O2(g) → 2CO2(g) with ΔH° = -835.5 kJ.

This combined equation is equivalent to the given reaction, but we need to multiply it by -2 to match the coefficients. Therefore, the enthalpy change for the given reaction is ΔH° = -2*(-835.5 kJ) = 1671 kJ.

For part (b), based on the positive enthalpy change, heat would be gained by the surroundings as the reaction occurs. In part (c), we can use the enthalpy change value from part (a) to calculate the heat produced. Since the reaction produces 2 moles of H2O, we can use the provided enthalpy change for H2O(l) to find the heat produced when 20.0 g of H2O is produced.

Using the given enthalpy change values:

ΔH° C6H4(OH)2(l) → C6H4O2(l) + H2(g) = +177.4 kJ

ΔH° H2(g) + O2(g) → H2O(l) = -285.8 kJ

We can set up the following equation to find the heat produced:

(20.0 g H2O) x (1 mol H2O / 18.015 g H2O) x (-285.8 kJ / 1 mol H2O) = -317 kJ.

Answer:

The major product of the reaction is (2S,3R)-3-methoxy - 3-methylpentan-2-ol.

Explanation:

Epoxide reacts with methanol in the presence of sulfuric acid.

In this chemicl reaction methanol act as nucleophile and sulfuric acid act as catalyst.

The chemical reaction is as follows.

An epoxide is a ring compound in which the oxygen atom closes the ring.

An epoxide is a ring compound in which the oxygen atom closes the ring. The reaction of the epoxide with methanol in the presence of acid is an example of a ring opening reaction.

The product of this reaction is shown in the image attached to this answer. The acid attacks the epoxide and opens the ring then te methanol attacks subsequently to yield the product.

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Answer:

Approximately [tex]\rm -249.4\; kJ \cdot mol^{-1}[/tex].

Explanation:

[tex]\rm CO\; (g) + 3\; H_2\; (g) \to CH_4\; (g) + H_2 O\; (g)[/tex].

Note that hydrogen gas [tex]\rm H_2\; (g)[/tex] is the most stable allotrope of hydrogen. Since [tex]\rm H_2[/tex] is naturally a gas under standard conditions, the standard enthalpy of formation of [tex]\rm H_2\; (g)[/tex] would be equal to zero. That is:

Look up the standard enthalpy of formation for the other species:

(Source: CRC Handbook of Chemistry and Physics, 84th Edition (2004).)

[tex]\displaystyle \Delta H^{\circ}_\text{reaction} = \sum \Delta H^{\circ}_f(\text{products}) - \sum \Delta H^{\circ}_f(\text{reactants})[/tex].

In other words, the standard enthalpy change of a reaction is equal to:

In this case,

[tex]\begin{aligned} & \sum \Delta H^{\circ}_f(\text{products}) \\ =& \Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) + \Delta H^{\circ}_f(\mathrm{H_2O\;(g)})\end{aligned}[/tex].

[tex]\begin{aligned} & \sum \Delta H^{\circ}_f(\text{reactants}) \\ =& \Delta H^{\circ}_f(\mathrm{CO\;(g)}) + 3\times \Delta H^{\circ}_f(\mathrm{H_2\;(g)})\end{aligned}[/tex].

Note that the number [tex]3[/tex] in front of [tex]\Delta H^{\circ}_f(\mathrm{H_2\;(g)})[/tex] corresponds to the coefficient of [tex]\rm H_2[/tex] in the chemical equation.

[tex]\begin{aligned}&\Delta H^{\circ}_\text{reaction} \\ =& \sum \Delta H^{\circ}_f(\text{products}) - \sum \Delta H^{\circ}_f(\text{reactants})\\ =& \left(\Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) + \Delta H^{\circ}_f(\mathrm{H_2O\;(g)})\right) \\ &- \left(\Delta H^{\circ}_f(\mathrm{CO\;(g)}) + 3\times \Delta H^{\circ}_f(\mathrm{H_2\;(g)})\right) \\ =& \Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) + (-74.6) - (3 \times 0 -110.5)\end{aligned}[/tex].

In other words,

[tex]\begin{aligned} & \Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) + (-74.6) - (3 \times 0 -110.5) \\=& \Delta H^{\circ}_\text{reaction} = -213.5\; \rm kJ\cdot mol^{-1} \end{aligned}[/tex].

Therefore,

[tex]\begin{aligned}& \Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) \\ =& -213.5 - ((-74.6) - (3 \times 0 -110.5)) \\=& -249.4\; \rm kJ\cdot mol^{-1} \end{aligned}[/tex].

Answer:

2.64 × 10⁶ g

Explanation:

We can find the mass of air using the ideal gas equation.

[tex]P.V=n.R.T=\frac{m}{M} .R.T[/tex]

where,

P is the pressure (P = 1.00 atm)

V is the volume (V = 2.95 × 10⁶ L)

n is the number of moles

R is the ideal gas constant (0.08206atm.L/mol.K)

T is the absolute temperature (121°C + 273 = 394 K)

m is the mass

M is the molar mass (28.09 g/mol)

[tex]m=\frac{P.V.M}{R.T} =\frac{1.00atm \times 2.95 \times 10^{6} L \times 28.97g/mol}{(0.08206atm.L/mol.K) \times 394 K } =2.64 \times 10^{6} g[/tex]

Answer:

Chlorine gas leaked = 20 g        

Explanation:

Given: Molar mass of chlorine gas: m = 70.9 g/mol

Initial Mass of chlorine gas: w₁ = 65.5 g, Initial Absolute pressure: P₁ = 6 × 10⁵ Pa, Initial temperature: T₁ = 73 °C = 73 + 273 = 346 K      (∵ 0°C = 273 K)

Final Absolute pressure: P₂ = 3.70 × 10⁵ Pa, Final temperature: T₂ = 34 °C = 34 + 273 = 307 K

Volume is constant

Final mass of chlorine gas: w₂ = ? g

Chlorine gas have leaked = w₁ - w₂ = ? g

Initial number of moles of chlorine gas: [tex]n_{1}= \frac{w_{1}}{m_{1}} =\frac{65.5 g}{70.9 g/mol} = 0.924 mole[/tex]

According to the Ideal gas law: P.V = n.R.T

∴ at constant volume,

[tex]\frac{n_{1}\times T_{1}}{P_{1}} = \frac{n_{2}\times T_{2}}{P_{2}}[/tex]

[tex]n_{2} = \frac{n_{1}\times T_{1} \times P_{2}}{P_{1} \times T_{2}}[/tex]

[tex]n_{2} = \frac{(0.924)\times (346 K) \times (3.70 \times 10^{5} Pa)}{(6 \times 10^{5} Pa) \times (307 K)} [/tex]

Final number of moles of chlorine gas: [tex]n_{2} = 0.642 = \frac{w_{2}}{m} [/tex]

⇒ Final mass of chlorine: [tex]w_{2} = 0.642 mol \times m = (0.642 mol) \times (70.9 g/mol) = 45.5 g[/tex]

Therefore, Chlorine gas leaked = w₁ - w₂ = 65.5 g - 45.5 g = 20 g

No chlorine gas has leaked out of the tank.

To calculate the amount of chlorine gas leaked out of the tank, we need to use the Ideal Gas Law equation: 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.

First, we calculate the initial number of moles using the given mass of chlorine gas (65.5 g) and its molar mass (70.9 g/mol):

65.5 g / 70.9 g/mol = 0.9237 mol

Next, we use the initial conditions (temperature = 73°C = 346 K, pressure = 6.00 × 10^5 Pa) to calculate the initial volume:

V = nRT / P = (0.9237 mol) * (8.314 J/mol K) * (346 K) / (6.00 × 10^5 Pa) ≈ 0.175 m^3

Similarly, we use the final temperature (34°C = 307 K) and pressure (3.70 × 10^5 Pa) to calculate the final volume:

V = nRT / P = (0.9237 mol) * (8.314 J/mol K) * (307 K) / (3.70 × 10^5 Pa) ≈ 0.227 m^3

Finally, we calculate the difference in the volumes to determine the amount of leaked chlorine gas:

Volume leaked = Initial volume - Final volume = 0.175 m^3 - 0.227 m^3 ≈ -0.052 m^3

Since volume cannot be negative, we conclude that no chlorine gas has leaked out of the tank.

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