A chlorine atom is adsorbed on a small patch of surface (see sketch at right). This patch is known to contain possible adsorption sites. The atom has enough energy to move from site to site, so it could be on any one of them. Suppose a atom also becomes adsorbed onto the surface. Calculate the change in entropy. Round your answer to significant digits, and be sure it has the correct unit symbol.

The question is incomplete. The complete question is:

The half-life for the decay of carbon-14 is 5.73x10^3 years. Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of woodfrom an archeological dig is measured to be 2.8x10^3 Bq. The activity in a similiar-sized sample of fresh wood is measured to be 3.0x10^3 Bq. Calculate the age of the artifact. Round your answer to 2 significant digits.

Answer:

570 years

Explanation:

The activity of the fresh sample is taken as the initial activity of the wood sample while the activity measured at a time t is the present activity of the wood artifact. The time taken for the wood to attain its current activity can be calculated from the formula shown in the image attached. The activity at a time t must always be less than the activity of a fresh wood sample. Detailed solution is found in the image attached.

Answer:

Number of atoms, N = 17,259,259

Explanation:

Given:

Radius of Iridium = 135 pm

distance = 2.33 mm

To determine number of iridium atom that would be laid side by side to span a distance of 2.33 mm, we say let the number of the atom = N

[tex]N =\frac{Distance}{Iridium. Radius} = \frac{2.33X10^{-3}}{135 X10^{-12}} \\\\N = 17259259.26[/tex]

Therefore, If a iridium atom has a radius of  135 pm, then 17,259,259 atoms of Iridium would be laid side by side to span a distance of 2.33 mm.

Number of atoms, N = 17,259,259

17,259,259 atoms of iridium are present in a distance of 2.33 mm

First, to calculate the amount of iridium atoms in a distance of 2.33 mm it is necessary to divide the two values:

                                       [tex]N = \frac{Distance}{Radius} [/tex]

As the element radius value is in pm, it is necessary to transform this unit to suit the other:

                                         [tex]135pm = 135\times 10^{-9} mm[/tex]

Now, we can apply the values in the expression:

                                             [tex]N = \frac{2.33}{135\times10^{-9}}[/tex]

                                        [tex]N = 17,259,259 [/tex] atoms

So, 17,259,259 atoms of iridium are present in a distance of 2.33 mm.

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Answer:

For a: The standard Gibbs free energy of the reaction is -347.4 kJ

For b: The standard Gibbs free energy of the reaction is 746.91 kJ

Explanation:

Relationship between standard Gibbs free energy and standard electrode potential follows:

[tex]\Delta G^o=-nFE^o_{cell}[/tex]           ............(1)

The given chemical equation follows:

[tex]2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)[/tex]

Oxidation half reaction:   [tex]Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-[/tex]       ( × 2)

Reduction half reaction:   [tex]3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)[/tex]

We are given:

[tex]n=2\\E^o_{cell}=+1.08V\\F=96500[/tex]

Putting values in equation 1, we get:

[tex]\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ[/tex]

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

The given chemical equation follows:

[tex]6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)[/tex]

Oxidation half reaction:   [tex]Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-[/tex]       ( × 6)

Reduction half reaction:   [tex]2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)[/tex]

We are given:

[tex]n=6\\E^o_{cell}=-1.29V\\F=96500[/tex]

Putting values in equation 1, we get:

[tex]\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ[/tex]

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

Final answer:

The standard reaction Gibbs free energy for an electrochemical cell reaction is calculated using the formula ΔG° = -nFE°cell, where n is the number of moles of electrons transferred, F is the Faraday constant, and E°cell is the standard cell potential.

Explanation:

To calculate the standard reaction Gibbs free energy (ΔG°) for an electrochemical cell reaction, you can use the following equation that relates ΔG° to the standard cell potential (E°cell):

ΔG° = -nFE°cell

Where:

n is the number of moles of electrons transferred in the reaction,

F is the Faraday constant (96,485 C/mol e⁻), and

E°cell is the standard cell potential.

To perform the calculations for each reaction:

Identify n, the number of electrons transferred in the balanced reaction.

Use the given E°cell values to calculate ΔG° using the formula above.

It's important to remember that these equations should be applied using standard conditions, which means all solutes are at 1 M concentration, all gases are at 1 atm pressure, and the temperature is 298 K (25 °C).

Explanation:

An alpha particles is basically a helium nucleus and it contains 2 protons and 2 neutrons.  

Symbol of an alpha particle is [tex]^{4}_{2}\alpha[/tex]. Whereas a neutron is represented by a symbol [tex]^{1}_{0}n[/tex], that is, it has zero protons and only 1 neutron.

Therefore, reaction equation when an aluminum- nuclide transforms into a phosphorus- nuclide by absorbing an alpha particle and emitting a neutron is as follows.

        [tex]^{27}_{13}Al + ^{1}_{0}n \rightarrow ^{30}_{15}P + ^{1}_{0}n[/tex]

In a nuclear reaction where an aluminum nuclide absorbs an alpha particle and emits a neutron to become a phosphorus nuclide, the balanced equation is ²⁷₁₃Al + ⁴₂α → ³⁰₁₅P + ¹₀n.

When an aluminum nuclide absorbs an alpha particle and emits a neutron to transform into a phosphorus nuclide, we must consider both mass and atomic numbers for conservation in the nuclear reaction equation. Based on atomic numbers, aluminum with Z=13 will turn into phosphorus with Z=15 by absorbing an alpha particle (Helium nucleus) with a mass number of 4 and an atomic number of 2.

In terms of mass numbers, since a neutron (mass number of 1) is emitted from the aluminum, the mass number of the final phosphorus nuclide would be the mass number of aluminum (27) plus the mass number of the alpha particle (4), minus the mass number of the neutron (1), resulting in a phosphorus nuclide with a mass number of 30, which is phosphorus-30. The balanced nuclear equation is:

Answer:

A. m and n are independent from the molar coefficients of the reactants in the balanced chemical equation.

B. m and n must be determined by experiment.

Explanation:

rate = k[H2O2]^m × [I-]^n

The Order of Reaction refers to the power dependence of the rate on the concentration of each reactant.

Either the differential rate law or the integrated rate law can be used to determine the reaction order of reactants from experimental data.

The equilibrium expression (Ka) for the ionization of bicarbonate ion is given by the formula [H3O+][CO32-] / [HCO3-], where A, B, and C are the equilibrium concentrations of H3O+, CO32-, and HCO3- respectively.

To find the equilibrium expression (Ka) for the ionization reaction of the bicarbonate ion (HCO3-(aq)), we must consider the reaction, HCO3-(aq) + H2O(l) ⇌ H3O+(aq) + CO32-(aq). The expression for Ka, known as the acid-ionization constant, follows the general form:

Ka = [H3O+][CO32-] / [HCO3-]

For the reaction given:

The reaction quotient will be Ka = [A][B] / [C]. Utilizing the values provided, assuming that the concentrations of the products [A] and [B] are equivalent as they form in a 1:1 ratio, and given that we're focusing on the ionization equilibrium of bicarbonate ion as a weak acid, the Ka value can be derived.

Equilibrium expression [tex]\( K_a = \frac{[H_3O^+][CO_3^{2-}]}{[HCO_3^-]} \)[/tex]. Concentrations at equilibrium: [tex]\([A] = [HCO_3^-]\), \([B] = [H_3O^+]\), \([C] = [CO_3^{2-}]\)[/tex].

The equilibrium expression (Ka) for the ionization reaction of [tex]HCO_3^-[/tex](aq) in water is given by the following general formula for an acid dissociation constant:

[tex]\[ K_a = \frac{[H_3O^+][CO_3^{2-}]}{[HCO_3^-]} \][/tex]

Here, the concentration of water ([tex]H_2O[/tex]) is not included in the expression because it is the solvent and its concentration remains relatively constant during the reaction.

For the given reaction:

[tex]\[ HCO_3^-(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + CO_3^{2-}(aq) \][/tex]

The expressions for the concentrations of A ([tex]HCO_3^-[/tex]), B ([tex]H_3O^+[/tex]), and C [tex](CO_3^{2-})[/tex] at equilibrium would be:

[tex]\[ [A] = [HCO_3^-] \] \[ [B] = [H_3O^+] \] \[ [C] = [CO_3^{2-}] \][/tex]

These expressions represent the molar concentrations of the species A, B, and C at equilibrium.

The equilibrium constant expression [tex]\( K_a \)[/tex] is then derived from these concentrations, with the product of the concentrations of B and C divided by the concentration of A, as shown in the initial equation.

Answer:

H+  +  OH−  --> H2O

Explanation:

Hydrochloric acid is represented by the chemical formular; HCl. This is an ionic substance so in water it breaks apart into hydrohrn ions; H+ and chloride ions; Cl−. It is a strong acid, hence it completely dissociates.

Potassium Hydroxide is also an ionic substance it also breaks apart in water into potassium ions; K+ and hydroxide ions; OH−. It is a strong base, hence it completely dissociates.

The complete ionic equation for the reaction is given as;

H+  +  Cl−  +  K+  +  OH−  -->  K+  +  Cl−  +  H2O

The Hydrogen ion and the Hydroxide ions combine to form water.

The net ionic equation is given as;

H+  +  OH−  --> H2O

Cl- and K+ ions were cancelled out because they do not undergo any changes therefore are not part of the net ionic equation. They are referred to as spectator ions.

Answer:

1. HNO3

2. HNO2

Explanation:

Nitrogen acid may refer to any of:

1. Nitric acid, HNO3 or

2. Nitrous acid, HNO2

Acid rain is caused by a chemical reaction that begins when compound like nitrogen oxides are released into the air. These substances can rise very high into the atmosphere, where they mix and react with water, oxygen, and other chemicals to form more acidic pollutants, known as acid rain. nitrogen oxides dissolve very easily in water and can be carried very far by the wind. As a result, the compounds can travel long distances where they become part of the rain, sleet, snow, and fog as acids.

Equation of reaction:

NO2 + H2O ==> HNO3

NO + H2O ==> HNO2

Final answer:

Nitrogen oxides (NO and NO2) from automobile exhaust contribute to acid rain primarily through the formation of nitric acid and nitrous acid.

Explanation:

High temperatures in the automobile engine cause nitrogen and oxygen gases from the air to combine to form nitrogen oxides (NO and NO2). These nitrogen oxides in automobile exhaust contribute to acid rain by forming nitric acid and nitrous acid. Nitric acid is formed when nitrogen dioxide (NO2), a highly reactive gas and a major component of nitrogen oxides, reacts with atmospheric water. Nitrous acid forms under similar conditions but is less prevalent. The reaction involving nitrogen oxides, particularly with the presence of water vapor, contributes significantly to the phenomenon of acid rain, which has adverse effects on natural water bodies, soil, and vegetation by altering their chemical composition.

Explanation:

It is known that [tex]Fe(OH)_{3}[/tex] is insoluble in water. As a result, plants are not able to absorb [tex]Fe^{3+}[/tex] readily through osmosis.

Therefore, the [tex]Fe^{3+}[/tex] in [tex]Fe(OH)_{3}[/tex] would be released in acidic environments, using neutralization, Iron(III) ions can be released.

Hence, the easiest way is to add low concentrations of [tex]H_{2}SO_{4}[/tex] to the soil is as follows.

         [tex]Fe(OH)_{3} + H_{2}SO_{4} \rightarrow H_{2}O + Fe_{2}(SO_{4})_{3}[/tex]

Thus, we can conclude that [tex]Fe_{2}(SO_{4})_{3}[/tex] is soluble and is good for plants too.

Plants can be iron deficient even when grown in soils with high iron content because the form of iron present, Fe(OH)3, is insoluble and not easily accessible by plants. Therefore, they may not be able to absorb enough iron to meet their requirements.

The fact that plants can still be iron deficient even when grown in soils with high iron content can be explained by the form of iron present. In this case, the iron is in the form of Fe(OH)3. While plants require iron for proper growth and development, they can only absorb it in a specific form, called Fe²+. Fe(OH)3 is insoluble and cannot be easily accessed by plants. Therefore, even with high iron content, the plants may still not be able to absorb enough iron to meet their requirements, resulting in iron deficiency.

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Answer: Ecell = -0.110volt

Explanation:

Zn--->Zn^+2 + 2e^-.........(1) oxidation

Cu^2+ 2e^- --->Cu........(2)reduction

Zn + Cu^2+ ----> Cu + Zn^+2 (overall

For an electrochemical cell, the reduction potential set up is given by

E(cell) = E(cathode) - E(anode)

E(cell) = E(oxidation) - E(reduction)

E(cathode) = E(oxidation)

E(anode) = E(reduction)

Given that

E(oxidation) = -0.763v

E(reduction) = +0.337v

E(cell) = -0.763 - (+0.337)

E(cell) = -0.763- 0.337

E(cell) = -0.110volt

Explanation:

Expression to calculate the value of for the given reaction is as follows.

And, it is given that

        [CO] = [tex][H_{2}][/tex] = 0.13 mol

        = 0.43 mol

Putting the given values into the above formula as follows.

         [tex]K_{c} = \frac{[CO][H_{2}]}{[H_{2}O]}[/tex]

                  = [tex]\frac{0.13 \times 0.13}{0.43}[/tex]

                   = 0.04

When additional amount of is added then all of has reacted.

So, new = 0 mols

       new = 0.43 + 0.13 = 0.56 mols

The reaction equation is as follows.

                  [tex]C + H_{2}O \rightleftharpoons CO + H_{2}[/tex]

Initial:    -     0.56     0.13     0    

Change:  -      -x         +x         +x

Equilibm.: -    0.56 - x   0.13 + x    x

So,

            0.04 = [tex]\frac{(0.13 + x)(x)}{(0.56-x)}[/tex]

            [tex]0.0224 - 0.04x = x^{2} + 0.13x[/tex]

            [tex]x^{2} + 0.17x - 0.0224[/tex] = 0

                  x = 0.087 mols

Therefore, the amount of [CO] at equilibrium is as follows.

             0.13 + 0.087

            = 0.217 mols

thus, we can conclude that the amount of CO in the flask when the system returns to equilibrium is 0.217 moles.

Answer:

Nickel

Explanation:

This element and its alloys are used in pumps, valves, and other components that are in contact with acid and petroleum solutions

Final answer:

Nickel and its alloys are used in components like pumps and valves that are exposed to acid and petroleum solutions due to their corrosion resistance.

Explanation:

The element and its alloys mentioned in the question that are used in pumps, valves, and other components in contact with acid and petroleum solutions is likely to be Nickel. This is because nickel and its alloys are known for their corrosion resistance and are used in environments that would rapidly degrade other metals. For instance, nickel is used in the desalination of seawater and nickel steel is used for manufacturing armor plates and burglar-proof vaults. Alloys such as brass (copper and zinc) and bronze (copper, tin, and sometimes zinc) are also important but are not specific to the context given in the question, which highlights usage in acid and petroleum solutions where nickel's properties are most relevant.

Answer: The final temperature of the coffee is 43.9°C

Explanation:

To calculate the final temperature, we use the equation:

[tex]q=mC(T_2-T_1)[/tex]

where,

q = heat released = [tex]1.0\times 10^3J=1000J[/tex]

m = mass of water = 10.0 grams

C = specific heat capacity of water = 4.184 J/g°C

[tex]T_2[/tex] = final temperature = ?

[tex]T_1[/tex] = initial temperature = 20°C

Putting values in above equation, we get:

[tex]1000J=10.0g\times 4.184J/g^oC\times (T_2-20)\\\\T_2=43.9^oC[/tex]

Hence, the final temperature of the coffee is 43.9°C

Final answer:

The final temperature of the coffee, after adding 1.0 x 10° J of heat to 10.0 g of water initially at 20°C, is approximately 44°C, calculated using the heat transfer formula in thermodynamics.

Explanation:

The subject of this question is Physics, specifically the concept of energy transfer through heating in thermodynamics. To calculate the final temperature of the prepared coffee, we can use the formula q = mcΔT, where q is the heat added, m is the mass of the water, c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature. By rearranging the formula to solve for ΔT, we get ΔT = q / (mc).

Given that 1.0 x 103 J of heat is added to 10.0 g of water initially at 20°C, we can calculate ΔT as follows:

ΔT = (1.0 x 103 J) / (10.0 g * 4.18 J/g°C) = 23.923°C (approximately)

Therefore, the final temperature of the water (or coffee) is 20°C + 23.923°C = 43.923°C, which when rounded off, gives us approximately 44°C.

Answer:

The overall equilibrium constant K in terms of the equilibrium constants [tex]K_1\& K_2[/tex]:

[tex]K=K_1\times K_2[/tex]

Explanation:

[tex]CH_4(g) H_2O(g)\rightleftharpoons CO(g)+3H_2(g)[/tex]

Equilibrium constant of reaction :

[tex]K_1=\frac{[CO][H_2]^3}{[CH_4][H_2O]}[/tex]

[tex]CO(g) +H_2O(g)\rightleftharpoons CO_2(g) + H_2(g)[/tex]

Equilibrium constant of reaction :

[tex]K_2=\frac{[CO_2][H_2]}{[CO][H_2O]}[/tex]

The net reaction is:

[tex] CH_4(g)+2H_2O(g)\rightleftharpoons CO_2(g)+4H_2(g) [/tex]

Equilibrium constant of reaction :

[tex]K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}[/tex]

[tex]K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}\times \frac{[CO]}{[CO]}[/tex]

Rearranging the equation :

[tex]K=\frac{[CO][H_2]^3}{[CH_4][H_2O]}\times \frac{[CO_2][H_2]}{[CO][H_2O]}[/tex]

[tex]K=K_1\times K_2[/tex]

The overall equilibrium constant K in terms of the equilibrium constants [tex]K_1\& K_2[/tex]:

[tex]K=K_1\times K_2[/tex]

To neutralize the acid in the given equation, you would need 0.1269 grams of K2CO3.

To determine the number of grams of K2CO3 needed to neutralize the acid in the given equation, we need to use stoichiometry. First, we determine the molar ratio between K2CO3 and HBr, which is 1:2 based on the balanced equation. Then, we convert the given mass of HBr to moles using its molar mass. Finally, we use the molar ratio and the molar mass of K2CO3 to calculate the mass of K2CO3 needed.

Here's the step-by-step calculation:

Therefore, you would need 0.1269 grams of K2CO3 to neutralize the acid according to the given equation.

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To neutralize the acid according to the reaction 2HBr(aq) + K₂CO₃(aq) → 2KBr(aq) + CO₂(g) + H₂O(l), calculate the moles of HBr and use stoichiometry to find the required moles of K₂CO₃. Convert these moles to grams to determine the amount needed.

Chemical Reaction and Calculation:

To determine the grams of K₂CO₃ required to neutralize the acid according to the equation:

2HBr(aq) + K₂CO₃(aq) → 2KBr(aq) + CO₂(g) + H₂O(l)

we need to follow these steps:

For a specific example, if you have 0.50 moles of HBr, the balanced equation shows a 2:1 molar ratio between HBr and K₂CO₃. Therefore, you would require 0.25 moles of K₂CO₃. Since the molar mass of K₂CO₃ is 138.205 g/mol, you would need:

0.25 moles × 138.205 g/mol = 34.55 grams of K₂CO₃

Answer:

[HCl] = 0.035 M

Explanation:

COMPLETE QUESTION:

What is the concentration (M) of HCl in a solution that is prepared by dissolving 25.5 g of HCl in 20.0 L  of water? (Assume volume does not change after adding HCl.)

Molarity → moles of solute in 1L of solution

As we assume volume does not change after adding  HCl, solution's volume is 20L

We convert solute's mass to moles → 25.5 g. 1mol / 36.45 g = 0.699 moles

Molarity (mol/L) → 0.699 mol/20L = 0.035 M

To find the concentration of HCl, divide the mass of HCl (25.5 g) by its molar mass to get moles, then divide by the volume of the solution in liters, assuming a volume of 1L for simplicity.

Firstly, the molar mass of HCl must be determined, which is approximately 36.46 g/mol. To find the concentration (Molarity, M) of HCl in the solution, the mass of HCl (25.5 g) must be converted into moles by dividing by the molar mass (moles of HCl = 25.5 g / 36.46 g/mol = 0.699 moles).

Then, the volume of water needs to be converted into liters (assuming it to be 1L for this example, as the exact volume was not provided). Finally, the concentration can be calculated by dividing the moles of HCl by the volume of the solution in liters (Concentration, M = 0.699 moles / 1L = 0.699 M)

Answer:

0.54 g of Pb in 60 m³ of [Pb] = 9 ppb

Explanation:

We need to know, that ppb is a sort of concentration that indicates (in value of volume) ng / mL (ppb = parts per billion)

9 ppb means that 9 ng are contained in 1mL of solution

We know that 1mL = 1 cm³

Let's convert 60 m³ to cm³ → 60 m³ . 1×10⁶ cm³ / 1m³ = 6×10⁷ cm³

Then, we can make a rule of three:

1 cm³ has 9 ng of Pb

Therefore in 6×10⁷ cm³ we must have ( 6×10⁷ cm³ . 9ng) / 1 cm³ =

5.4×10⁸ ng

We convert ng to g → 5.4×10⁸ ng . 1 g / 1×10⁻⁹ ng = 0.54 g

Drinking water with a concentration of Pb of 9.0 ppb, has a molarity of Pb of 4.3 × 10⁻⁸ M. There are 0.54 g of Pb in a 60 m³ pool.

The maximum allowable concentration of lead in drinking water is 9.0 ppb, that is, 9.0 μg of Pb per liter of water.

We want to calculate the molarity (M) of the solution. We will use the following expression, assuming that the volume of water is equal to the volume of the solution.

M = mass Pb / molar mass Pb × liters of solution

M = (9.0 × 10⁻⁶ g) / (207.2 g/mol) × 1 L = 4.3 × 10⁻⁸ M

To calculate the mass of Pb in a 60 m³ pool, we need to consider that 1 m³ = 1000 L.

60 m³ Water × (1000 L Water/ 1 m³ Water) = 60,000 L Water

60,000 L Water × (9.0 × 10⁻⁶ g Pb/1 L Water) = 0.54 g Pb

Drinking water with a concentration of Pb of 9.0 ppb, has a molarity of Pb of 4.3 × 10⁻⁸ M. There are 0.54 g of Pb in a 60 m³ pool.

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Answer: The value of equilibrium constant for new reaction is [tex]1.92\times 10^{-25}[/tex]

Explanation:

The given chemical equation follows:

[tex]O_2(g)\rightarrow \frac{2}{3}O_3(g)+\frac{1}{2}O_2(g)[/tex]  

The equilibrium constant for the above equation is [tex]5.77\times 10^{-9}[/tex]

We need to calculate the equilibrium constant for the equation of 3 times of the above chemical equation, which is:

[tex]3O_2(g)\rightarrow 2O_3(g)[/tex]

The equilibrium constant for this reaction will be the cube of the initial reaction.

If the equation is multiplied by a factor of '3', the equilibrium constant of the new reaction will be the cube of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

[tex]K_{eq}'=(5.77\times 10^{-9})^3=1.92\times 10^{-25}[/tex]

Hence, the value of equilibrium constant for new reaction is [tex]1.92\times 10^{-25}[/tex]

The equilibrium constant for equation (b) is 1.73x10^8.

Therefore, the equilibrium constant (Kc) for equation (b) is 1 / (5.77x10^-9) = 1.73x10^8.

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The question is incomplete, here is the complete question:

A(aq) + B(aq) → 2C(aq)

If the value of Kc for the reaction is 434, what is the concentration of C at equilibrium if initial concentrations of A and B are both 0.500 M. (Hint: Everything is squared after you set-up the equilibrium expression with the values given.

Answer: The concentration of C at equilibrium is 0.912 M

Explanation:

We are given:

Initial concentration of A = 0.500 M

Initial concentration of B = 0.500 M

The given equation follows:

                     [tex]A(aq.)+B(aq.)\rightarrow 2C(aq.)[/tex]

Initial:               0.5   0.5

At eqllm:        0.5-x   0.5-x         2x

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

[tex]K_c=\frac{[C]^2}{[A][B]}[/tex]

We are given:

[tex]K_c=434[/tex]

Putting values in above equation, we get:

[tex]434=\frac{(2x)^2}{(0.5-x)\times (0.5-x)}\\\\x=0.456,0.553[/tex]

Neglecting the value of x = 0.553 M because the equilibrium concentration of A and B will become negative, which is not possible

So, equilibrium concentration of C = 2x = 2(0.456) = 0.912 M

Hence, the concentration of C at equilibrium is 0.912 M

Answer:

A. AN INCREASE IN BLOOD ACIDITY NEAR THE TISSUES

B. AN INCREASE IN BLOOD TEMPERATURE NEAR THE TISSUES.

C. THE PRESENCE OF A PRESSURE GRADIENT FOR OXYGEN.

Explanation:

Metabolically active tissues need more oxygen to carry out theirs functions. They are involved during excercise and other active phsiological conditions.

There is the reduction in the amount of oxygen reaching these tissues resulting in carbon IV oxide build up, lactic acid formation and temperature increases.

The acidity of the blood near the tissues is increased due to the accumulation of carbon IV oxide in the tissues resulting into a decreased pH. This reduces the affinity of heamoglobin to oxygen in the blood near the metabollically active tissues.

There is also the increase in temperature causing rapid offload of oxygen from oxy-heamoglobin molecules.

The partial pressure of oxygen gradient also affects the rate of oxygen offload by the blood. In metabollically active tissues, the partial pressure of oxygen is reduced in the tissues causing a direct offloading of oxygen to the tissues.

Answer:

[tex]C-C> C=C> C\equiv C[/tex]

So, 2 [tex]>[/tex] 1 [tex]>[/tex] 3

Explanation:

Carbon is a  non-metal and has 6 electrons. In the excited state, it contains 4 electrons in its outermost shell and so 4 electrons take place in covalent bond formation. In a molecule , 2 carbon atoms can be bound by a single bond, the double bond and by a triple bond. Since the bond length of [tex]C\equiv C[/tex] is 120 pm, the bond length of [tex]C=C[/tex] is 134 pm and the bond length of [tex]C-C[/tex] is 154 pm. Thus in the given compound, [tex]C= C-C\equiv C[/tex] , in which [tex]C= C[/tex] is labeled as 1,  [tex]C-C[/tex] is labeled as 2 and [tex]C\equiv C[/tex] is labeled as 3. So the bond length of carbon-carbon bonds in the molecule, in decreasing order is [tex]2 > 1 >3[/tex].

Answer:

a) 0.000112 M/s is the average reaction rate between 0.0 seconds and 1500.0 seconds.

b) 0.00011 M/s is the average reaction rate between 200.0 seconds and 1200.0 seconds.

c) Instantaneous rate of the reaction at t=800 s :

Instantaneous rate  : [tex]\frac{0.031 M}{800.0 s}=3.875\times 10^{-5} M/s[/tex]

Explanation:

Average rate of the reaction is given as;

[tex]R_{avg}=-\frac{\Delta A}{\Delta t}=\frac{A_2-A_1}{t_2-t_1}[/tex]

a.) The average reaction rate between 0.0 s and 1500.0 s:

At 0.0 seconds the concentration was  = [tex]A_1=0.184 M[/tex]

[tex]t_1=0.0s[/tex]

At 1500.0 seconds the concentration was  = [tex]A_2=0.016 M[/tex]

[tex]t_2=1500 s[/tex]

[tex]R_{avg]=-\frac{0.016 M-0.184 M}{1500.0 s-0.0 s}=0.000112 M/s[/tex]

0.000112 M/s is the average reaction rate between 0.0 seconds and 1500.0 seconds.

b.) The average reaction rate between 200.0 s and 1200.0 s:

At 0.0 seconds the concentration was  = [tex]A_1=0.129 M[/tex]

[tex]t_1 =200.0 s[/tex]

At 1500.0 seconds the concentration was  = [tex]A_2=0.019M[/tex]

[tex]t_2=1200 s[/tex]

[tex]R_{avg]=-\frac{0.019 M-0.129M}{1200.0s-200.0s}=0.00011 M/s[/tex]

0.00011 M/s is the average reaction rate between 200.0 seconds and 1200.0 seconds.

c.) Instantaneous rate of the reaction at t=800 s :

At 800 seconds the concentration was  = [tex]A=0.031 M[/tex]

[tex]t =800.0 s[/tex]

Instantaneous rate  : [tex]\frac{0.031 M}{800.0 s}=3.875\times 10^{-5} M/s[/tex]

Answer:

7.85 g H₃BO₃

2.92 g NaOH

Explanation:

The strategy for solving this question is to first utilize the Henderson- Hasselbach equation to calculate the ratio of conjugate base concentration to weak acid:

pH = pKa + log [A⁻]/ [HA]

In this case:

pH = pKa + log [H₂BO₃⁻]/[H₃BO₃]

We know pH and indirectly pKa ( = - log Ka ).

9.00 = -log(5.8 x 10⁻¹⁰) + log [H₂BO₃⁻]/[H₃BO₃]

9.00 = 9.24 + log [H₂BO₃⁻]/[H₃BO₃]

log [H₂BO₃⁻]/[H₃BO₃] = - 0.24

taking inverse log function to both sides of the equation:

[H₂BO₃⁻]/[H₃BO₃]  = 10^-0.24 = 0.58

We are also told we want to have a total concentration of boron of 0.200 mol/L, and if we call x the concentration of  H₂BO₃⁻ and y the concentration of H₃BO₃, it follows that:

x + y = 0.200 ( since we have 1 Boron atom per formula of each compound)

and from the Henderson Hasselbach calculation, we have that

x / y = 0.58

So we have a system of 2 equations with two unknowns, which when solved give us that

x = 0.073  and y = 0.127

Because we are told the volume is one liter it follows that the number of moles of boric acid and the salt are the same numbers 0.073 and 0.127

gram boric acid = 0.127 mol x molar mass HBO₃ = 0127 mol x 61.83 g/mol

                          = 7.85 g boric acid

grams NaOH = 0.073 mol x molar mass NaOH = 0.073 x 40 g/mol

                          = 2.92 g NaOH

Answer:

[tex]\large \boxed{\text{88.1 $^{\circ}$C}}[/tex]

Explanation:

There are two heat transfers involved: the heat gained by the spoon and the heat lost by the coffee.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the spoon be Component 1 and the coffee be Component 2.

Data:  

For the spoon:

[tex]m_{1} =\text{100.0 g; }T_{i} = 10.0 ^{\circ}\text{C; }\\C_{1} = 0.235 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}[/tex]

For the coffee:

[tex]m_{2} =\text{240.0 g; }T_{i} = 90.0 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}[/tex]

Calculations

1. The relative temperature changes

[tex]\begin{array}{rcl}\text{Heat gained by spoon + heat lost by coffee} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{100.0 g}\times 0.235 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{240. g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\23.50\Delta T_{1} + 1004\Delta T_{2} & = & 0\\\end{array}[/tex]

[tex]\begin{array}{rcl}1004\Delta T_{2} & = & -23.50\Delta T_{1}\\\Delta T_{2} & = & -0.02340\Delta T_{1}\\\end{array}\\\text{(The temperature change for the coffee is about 1/40 that of the spoon.)}[/tex]

2. Final temperature of coffee

[tex]\Delta T_{1} = T_{\text{f}} - 10.0 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 90.0 ^{\circ}\text{C}[/tex]

[tex]\begin{array}{rcl}\Delta T_{2} & = & -0.02340\Delta T_{1}\\T_{\text{f}} - 90.0 \, ^{\circ}\text{C} & = & -0.02340 (T_{\text{f}} - 10.0 \, ^{\circ}\text{C})\\& = & -0.02340T_{\text{f}} + 0.2340 \, ^{\circ}\text{C}\\T_{\text{f}} & = & -0.02340T_{\text{f}}+ 90.23 \, ^{\circ}\text{C}\\1.02430T_{\text{f}}& = & 90.23 \, ^{\circ}\text{C}\\\end{array}\\[/tex]

[tex]\begin{array}{rcl}T_{\text{f}} & = & \dfrac{ 90.23 \, ^{\circ}\text{C}}{1.02430}\\\\ & = & \mathbf{88.1 \,^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the coffee is $\large \boxed{\textbf{88.1 $\,^{\circ}$C}}$}\\\text{This is hot enough to cause third-degree burns in less than 1 s.}[/tex]

Using the principles of physics and the formula for heat transfer, the final temperature after introducing a cold spoon to a cup of hot coffee ends up being around 89.6°C. It's further noted that spilling coffee at this temperature could potentially cause 3rd-degree burns, although the specific outcome may vary.

This question can be answered through the principles of energy conservation. In this situation, the chilled spoon will absorb heat from the hot coffee until they both reach a thermal equilibrium. This means they will eventually have the same temperature. The formula for this heat transfer can be expressed as Q (heat energy transferred) = mcΔT, where 'm' is the mass, 'c' is the specific heat, and 'ΔT' is the change in temperature.

Applying this to the spoon and coffee respectively, we will end up with an equation: (mass of coffee * specific heat of coffee * (initial coffee temp - final temp)) = -(mass of spoon * specific heat of spoon * (final temp - initial spoon temp)). Considering the specific heat of water (essentially coffee without impurities) is approximately 4.186 J/g °C, the initial math leads us to a final temperature of around 89.6°C.

Mentioning the last thought in your question, it's likely that spilling coffee at this temperature could potentially cause 3rd-degree burns. The exact effect can vary depending on various factors such as exposure duration and skin sensitivity, but typically, exposure to liquid at a temperature above 70°C can cause severe burns in a matter of seconds.

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Answer:

K = 0,00000135 = 1.35 * 10^-6

Explanation:

Step 1: Data given

The equilibrium constant, K, for any reaction is defined as the concentrations of the products raised by their coefficients divided by the concentrations of the reactants raised by their coefficients. In this case, the concentrations are given as partial pressures.

The partial pressures of H2O = 0.0500 atm

The partial pressures of H2 = 0.00150 atm

The partial pressures of O2 = 0.00150 atm

Step 2: The balanced equation

2H2O(g) ⇆ 2H2(g) + O2(g)

Step 3: Calculate K

K = [O2][H2]² / [H2O]²

K = 0.00150 * 0.00150² /  0.0500²

K = 0,00000135 = 1.35 * 10^-6

Answer:

0.04M

Explanation:

The following were obtained from the question:

C1 = 0.8M

V1 = 5mL

V2 = 100mL

C2 =?

The final concentration of the diluted solution can be obtained by doing the following:

C1V1 = C2V2

0.8 x 5 = C2 x 100

Divide both side by 100

C2 = (0.8 x 5) /100

C2 = 0.04M

The concentration of the diluted solution is 0.04M

Answer:

is this just a passage?

Explanation:

The mass of ammonium phosphate produced by the reaction of 4.9 g of phosphoric acid is 7.45 g.

The question asks for the mass of ammonium phosphate produced by the reaction of 4.9 g of phosphoric acid. To determine the mass of ammonium phosphate produced, we need to balance the chemical equation and calculate the molar mass of both reactants and products.

The balanced equation for the reaction is:3H3PO4 + (NH4)OH → (NH4)3PO4 + 3H2O

The molar mass of phosphoric acid (H3PO4) is 97.99 g/mol. The molar mass of ammonium phosphate ((NH4)3PO4) is 149.0 g/mol.

Using the molar mass of phosphoric acid and the ratio of the reactants and products in the balanced equation, we can calculate the mass of ammonium phosphate produced.First, calculate the moles of phosphoric acid:

moles of H3PO4 = mass (g) / molar mass (g/mol)

moles of H3PO4 = 4.9 g / 97.99 g/mol = 0.050 moles

Since the stoichiometry of the reaction is 1:1 between phosphoric acid and ammonium phosphate, the moles of ammonium phosphate produced is also 0.050 moles.

Finally, calculate the mass of ammonium phosphate:mass of (NH4)3PO4 = moles of (NH4)3PO4 × molar mass of (NH4)3PO4

mass of (NH4)3PO4 = 0.050 moles × 149.0 g/mol = 7.45 g

Therefore, 7.45 g of ammonium phosphate is produced by the reaction of 4.9 g of phosphoric acid.

Answer: This will result in the increase of concentration of NO and decrease in the concentrations of nitrogen gas and oxygen gas.

Explanation:

The given chemical equation follows:

[tex]N_2(g)+O_2(g)\rightleftharpoons 2NO(g);\Delta H=182.6kJ[/tex]

As, enthalpy of the reaction is positive. So, it is an endothermic reaction.

For an endothermic reaction, heat is getting absorbed during a chemical reaction and is written on the reactant side.

[tex]A+\text{heat}\rightleftharpoons B[/tex]

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle. This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

If the concentration of oxygen gas is increased, the equilibrium will shift in the direction where the concentration of oxygen gas will be decreased. Thus, the reaction will shift in the right direction or forward direction.

Hence, this will result in the increase of concentration of NO and decrease in the concentrations of nitrogen gas and oxygen gas.

Final answer:

Adding more O2 to the equilibrium mixture of N2 and O2 reacting to form NO causes the equilibrium to shift right, increasing NO concentration while decreasing N2 and O2 concentrations.

Explanation:

When more O2 is added to the equilibrium mixture of the reaction N2(g) + O2(g) <=> 2NO(g), the system responds by shifting the equilibrium to the right, according to Le Châtelier's Principle, to counteract the increase in O2 concentration. This results in an increase in the concentration of NO and a decrease in the concentrations of N2 and O2. This shift is a direct consequence of the system's attempt to re-establish equilibrium by consuming the added O2 and producing more NO.

The question is incomplete, here is the complete question:

At 45°C, Kc = 0.619 for the reaction N₂O₄(g) ⇌ 2 NO₂(g).

If 50.0 g of N₂O₄ is introduced into an empty 2.12 L container, what are the partial pressures of NO₂ and N₂O₄ after equilibrium has been achieved at 45°C?

Answer: The equilibrium partial pressure of [tex]NO_2\text{ and }N_2O_4[/tex] is 7.12 atm and 3.133 atm respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

Or,

[tex]PV=\frac{w}{M}RT[/tex]

where,

P = Pressure of the gas  = ?

V = Volume of the gas  = 2.12 L

w = Weight of the gas  = 50.0 g

M = Molar mass of gas  = 92 g/mol

R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]

T = Temperature of the gas = [tex]45^oC=[45+273]K=318K[/tex]

Putting values in above equation, we get:

[tex]P\times 2.12L=\frac{50.0g}{92g/mol}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 318\\\\P=\frac{50.0\times 0.0821\times 318}{2.12\times 92}=6.693atm[/tex]

Relation of [tex]K_p[/tex] with [tex]K_c[/tex] is given by the formula:

[tex]K_p=K_c(RT)^{\Delta ng}[/tex]

where,

[tex]K_p[/tex] = equilibrium constant in terms of partial pressure

[tex]K_c[/tex] = equilibrium constant in terms of concentration = 0.619

R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature  = [tex]45^oC=[45+273]K=318K[/tex]

[tex]\Delta n_g[/tex] = change in number of moles of gas particles = [tex]n_{products}-n_{reactants}=(2-1)=1[/tex]

Putting values in above equation, we get:

[tex]K_p=0.619\times (0.0821\times 318)^{1}\\\\K_p=16.16[/tex]

For the given chemical equation:

                 [tex]N_2O_4(g)\rightleftharpoons 2NO_2(g)[/tex]

Initial:            6.693

At eqllm:    6.693-x           2x

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

[tex]K_p=\frac{(p_{NO_2})^2}{p_{N_2O_4}}[/tex]

Putting values in above expression, we get:

[tex]16.16=\frac{(2x)^2}{6.693-x}\\\\x=-7.59,3.56[/tex]

So, equilibrium partial pressure of [tex]NO_2=2x=(2\times 3.56)=7.12atm[/tex]

Equilibrium partial pressure of [tex]N_2O_4=(6.693-x)=(6.693-3.56)=3.133atm[/tex]

Hence, the equilibrium partial pressure of [tex]NO_2\text{ and }N_2O_4[/tex] is 7.12 atm and 3.133 atm respectively.