The equilibrium constant is given for one of the reactions below. Determine the value of the missing equilibrium constant. Deuterium, D, is an isotope of hydrogen. 2 HD(g) ⇌ H2(g) + D2(g) Kc = 0.28 5 H2(g) + 5 D2(g) ⇌ 10 HD(g) Kc = ? The equilibrium constant is given for one of the reactions below. Determine the value of the missing equilibrium constant. Deuterium, D, is an isotope of hydrogen. 2 HD(g) ⇌ H2(g) + D2(g) Kc = 0.28 5 H2(g) + 5 D2(g) ⇌ 10 HD(g) Kc = ? 581 0.73 1.3 0.0017 2.06

Answer:

Explanation:

From the data it is clear that after each .80881 hours , the concentration becomes half . so its decomposition appears to be first order .

k = 1 / t ln A₀ / A

= (1/.80881) ln 2

= .857

for the given case

A = .0161 , A₀ = .479

t = ?

t = 1/k ln( .479 / .0161)

= (1 / .857 )ln (.479 / .0161 )

= (1 / .857 ) x ln 29.75

= 3.959 hours

Answer:

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Answer:

the speed of which they are running.

Explanation:

Answer:

Your question is half unfinished, regarding the chromium III oxide the correct option that expresses the inorganic formula of said compound is "A"

Explanation:

In the reaction an initial salt reacts giving as product water vapor, nitrogen gas and an oxide that is chromium oxide.

Chromium oxide is an oxide that adopts the structure of corundum, compact hexagonal. It consists of an anion oxide matrix with 2/3 of the octahedral holes occupied by chromium. Like corundum, Cr2O3 is a tough, brittle material.

It is used as a pigment, green in color.

Answer:

Valency electrons is the number outermost electron present in atom of an element . The valency electron of an atom determines the kind of bonding an element undergoes. The valency electrons determines whether an atom of element will either donate electrons or receive electrons.

Explanation:

Valency electrons is the number outermost electron present in atom of an element . The valency electron of an atom determines the kind of bonding an element undergoes. The valency electrons determines whether an atom of element will either donate electrons or receive electrons.

Elements donates  or receive electron to attain or fulfill the octet rule and this account for the stability. Most metallic elements donates electrons and because they have few valency electrons , this also contributes to their ability to loose electrons easily . Metallic elements like sodium, magnesium, calcium etc loose electrons during bonding with other non metals. When metals loose electrons they form cations

Non metallic element have high number of valency electron and they tend to receive electrons from metallic elements to attain stability . When they gain electron they form anions.  

An example of bonding that involves donating and receiving are bonding between Na and Cl atoms to form NaCl . The metallic sodium loose one electron while the chlorine atom gains one electron to attain the octet rule.

Answer:

The absolute pressure inside the ball is 1.25atm.

Explanation:

Given-

Volume, V = 5..27L

Temperature, T = 27°C

Gauge Pressure, Pg = 0.25 atm

Atmospheric pressure, Patm = 1 atm

Absolute pressure, Pabs = ?

We know,

Therefore, absolute pressure inside the ball is 1.25atm.

Answer:

The absolute pressure inside the ball is 1.25 atmospheres.

The ball contains 0.267 moles of air

Explanation:

Answer:

Explanation:

1.The solution contains 20 mL of methanol and 100 mL of water.

2.The solution contains 20 mL of methanol and 80 mL of water.

3.The solution contains 20 g of methanol and 100 mL of water.

4.The solution contains 20 g of methanol and 80 mL of water.

Answer:

20 % volume/volume methanol in water contains 20 mL of methanol and 80 mL of water.

Explanation:

Total volume of the solution= 20 + 80 = 100 mL

Answer:

The grams of rust present in the bicycle  frame is  [tex]x = 8.50 g[/tex]

Explanation:

From the question we are told that

      The number of oxygen atom contained is  [tex]n = 9.62 *10^{22} \ atoms[/tex]

The molar mass of the compound is [tex]M_{Fe_3O_3} = 159.69 g[/tex]

At standard temperature and pressure the number of oxygen atom in one mole of  iron(III) oxide is  mathematically evaluated as

        [tex]N_o = 3 * Ne[/tex]

Where Ne is the avogadro's constant with a value [tex]N_e = 6.023 *10^{23} \ atoms[/tex]

      So

            [tex]N_o= 3 * 6.023*10^{23} \ atoms[/tex]

            [tex]N_o= 1.8069*10^{24} \ atoms[/tex]

So

   [tex]1.8069*10^{24} \ atoms[/tex] is  contained in [tex]M_{Fe_3O_3} = 159.69 g[/tex]

   [tex]9.62 *10^{22} \ atoms[/tex]   is contained in x

So

     [tex]x = \frac{159.69 * 9.62 *10^{22}}{1.8069 *10^{24}}[/tex]

    [tex]x = 8.50 g[/tex]

Final answer:

To calculate the grams of rust on the bicycle frame, we need to use stoichiometry. By converting the number of oxygen atoms to moles and then to grams using the molar mass of iron(III) oxide, we find that there are approximately 5.37x10^22 grams of rust on the bicycle frame.

Explanation:

To determine the grams of rust present on the bicycle frame, we need to calculate the molar mass of iron(III) oxide and then use stoichiometry to convert the number of oxygen atoms to grams of rust. The molar mass of Fe2O3 is 159.69 g/mol. Given that 1 mole of Fe2O3 contains 3 moles of oxygen atoms, we can calculate the moles of Fe2O3 using the given number of oxygen atoms and then convert it to grams using the molar mass.

9.62x10^22 oxygen atoms x (1 mol Fe2O3 / 3 mol O) x (159.69 g Fe2O3 / 1 mol Fe2O3) = 5.37x10^22 g Fe2O3

Therefore, there are approximately 5.37x10^22 grams of rust present on the bicycle frame.

Answer:

Explanation:

CHECK THE FILE BELOW FOR THE ANSWER

Answer:

a) Warmer

b) Exothermic

c) -10.71 kJ

Explanation:

The reaction:

KOH(s) → KOH(aq) + 43 kJ/mol

It is an exothermic reaction since the reaction liberates 43 kJ per mol of KOH dissolved.

Hence, the dissolution of potassium hydroxide pellets to water provokes that the beaker gets warmer for being an exothermic reaction.

The enthalpy change for the dissolution of 14 g of KOH is:

[tex] n = \frac{m}{M} [/tex]

Where:

m: is the mass of KOH = 14 g

M: is the molar mass = 56.1056 g/mol

[tex] n = \frac{m}{M} = \frac{14 g}{56.1056 g/mol} = 0.249 mol [/tex]

The enthalpy change is:

[tex] \Delta H = -43 \frac{kJ}{mol}*0.249 mol = -10.71 kJ [/tex]

The minus sign of 43 is because the reaction is exothermic.

I hope it helps you!

Answer:

See explanation below

Explanation:

A claisen reaction, is often used between two esters (One of them, usually having alpha hydrogens atoms) to form Beta Keto esters. Depending of the reagents, this reaction is taking place with base where one ester acts as nucleophile and the other as electrophile.

Now, we have here ethyl propanoate and ethyl formate. The ethyl propanoate has two alpha hydrogens, while ethyl formate do not have that. So, the base will react with the ethyl propanoate first, and then, it will react as nucleophile with the formate.

Now, the base to be used in this case will have to be a relatively strong base such sodium ethanoate, because if we use a strong base such NaOH, this will cause the saponification of the ester and not the condensation. So, in order to promove the claisen condensation, we need to use a base not too strong.

In the picture attached you have the mechanism and structure of the final product.

Concentration of C, [C] will not appear in the equilibrium constant expression for the reaction.

Ratio of product of concentration of the products to the product of concentration of reactants.

Here,

H₂O(g) + C(s) ⇆ H₂ + CO

The expression for equilibrium constant for the reaction is given by,

K = [H₂] [CO]/ [H₂O] [C]  

where K is the equilibrium constant of the reaction.

We know, the concentration of solids does not change in a reaction and therefore the concentration is taken as unity.

In the given reaction, since the reactant C is in solid form, its concentration is taken as 1.

So, K = [H₂] [CO]/ [H₂O]

As a result, the concentration of C, [C] will not be included in the expression for equilibrium constant.

Hence,

Concentration of C, [C] will not appear in the equilibrium constant expression for the reaction.

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In the reaction H2O(g) + C(s) → H2(g) + CO(g), the component that does not appear in the equilibrium constant expression is C(s), carbon in its solid form. This is because pure solids do not change concentration and do not affect the state of equilibrium.

To answer which component will not appear in the equilibrium constant expression for the reaction H2O(g) + C(s) → H2(g) + CO(g), we need to understand how equilibrium expressions are formed. The equilibrium constant expression for a reaction is based on the balanced chemical equation and includes the concentrations of the gaseous and aqueous species, raised to the power of their coefficients in the balanced equation. Pure solids and pure liquids are not included in the expression because their concentrations don't change during the reaction.

In the given reaction, C(s), which is carbon in its solid form, does not appear in the equilibrium constant expression because as a pure solid, its concentration is constant and does not affect the state of equilibrium. Therefore, the equilibrium constant expression for this reaction would be K = [CO][H2], excluding both the water (H2O) because it is in gaseous form and included in the expression, and the carbon (C) because it is a solid.

The anion that would bond with K+ in a 1:1 ratio to form a neutral ionic compound is F- (fluoride ion)

The anion that would bond with K+ in a 1:1 ratio to form a neutral ionic compound is B. F- (fluoride ion).

Potassium (K+) is a cation with a +1 charge, and to form a neutral ionic compound, it needs to bond with an anion with a -1 charge. The fluoride ion (F-) has a -1 charge, and therefore, it can form a 1:1 ratio with K+ to create a neutral compound.

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Answer:

54 L

Explanation:

Given data

We have a concentrated NaCl solution and we will add water to it to obtain a diluted NaCl solution. We can find the volume of the final solution using the dilution rule.

[tex]C_1 \times V_1 = C_2 \times V_2\\V_2 = \frac{C_1 \times V_1}{C_2} = \frac{9.0M \times 450mL}{0.075M} = 5.4 \times 10^{4} mL = 54 L[/tex]

The volume of the final solution should be 54L.

Since

Initial concentration (C₁): 9.0 M

Initial volume (V₁): 450 mL

Final concentration (C₂): 0.075 M

We know that

C1 * V1 = C2 * V2

V2 = C1 * V1/ C2

= 9.0M * 450 mL/ 0.075 M

= 54 L

hence, The volume of the final solution should be 54L.

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Answer:

The figures and parameters to answer this question are not clearly stated, so I will answer it in the closest best way I can.

Explanation:

Molality of alanine = mole / weight of solvent( kg)

= (170×1000)/(89×600)

= 3.18 molal

We know ∆ T​​​​​​f​​​​​ = K​​​​f × m

K​​​​​​f​​​​​ = ∆ T​​​​f / molality

= 7.9/3.18 = 2.48 °c.kg.mol-1

Now using the value of cryoscopic constant we calculate can't Hoff factor for Ammonium chloride.

i = ∆ T​​​​​​f /( K​​​​f × molality of NH​​​​​4​​​​Cl)

= (24.7 × 53.5 × 600) /(2.48 × 170 × 1000)

= 1.8

Answer:

b.one that results in lighter flower petal colors without changing the plant’s ability to reproduce

Explanation:

on edge

Beneficial mutations improve the survival and reproductive chances of an organism. Examples include mosquitos' resistance to insecticides, the color change in peppered moths during the Industrial Revolution, and the Antennapedia mutation in Drosophila.

A beneficial mutation can be understood as the type of mutation that improves an organism's chances of survival and reproduction. For instance, consider the case of mosquitoes. The mutation that has provided them with resistance to some insecticides is an example of a beneficial mutation. This resistance has allowed them to survive even when faced with these chemicals, thus enhancing their chances of propagation.

Similarly, during the Industrial Revolution, peppered moths underwent a beneficial mutation where their coloration turned from light to dark. As industrial pollution darkened the environment, the dark coloration assisted in camouflage, aiding their survival.

The mutant allele can also interfere with the normal gene's function or its distribution in the body, which can result in beneficial mutations. For example, the Antennapedia mutation in Drosophila that allowed for the development of legs in place of antennae. However, whether a mutation is beneficial or not primarily depends on its effects on the organism's ability to mature sexually and reproduce successfully.

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Answer:

Cathode: O2 + 4H+ +4e--------> 2H20

Anode: 2H2 -4e- ---------> 4H+

Overall: 2H2 + O2 → 2H2O

Explanation:

A hydrogen-oxygen fuel cell is an alternative cell to rechargeable cells and batteries. In this cell, hydrogen and oxygen is used to produce voltage and water is the only byproduct.

At the cathode (positive electrode):

O2 + 4H+ +4e--------> 2H2O

At the anode (negative electrode);

2H2 -4e- ---------> 4H+

Adding the two half reactions we have:

2H2 + O2 + 4H+ + 4e-  ----------->  2H2O + 4H + 4e-

The overall reaction after cancelling out the like terms in the reaction is:

2H2 (g) + O2 (g) --------> 2H2O (l)

The hydrogen fuel cell is an electrochemical cell in which hydrogen is oxidized and oxygen is reduced.

A fuel cell is an electrochemical cell in which oxygen enters into the positive cathode and hydrogen flows into the negative cathode. Hydrogen is oxidized and oxygen is reduced during the electrochemical reaction.

At the negative electrode;

2H2(g)  ----->  4H^+(aq)+ 4e

At the positive electrode;

O2(g) + 4H^+(aq) + 4e -----> 2H2O(l)

The overall reaction is;

2H2(g)  + O2(g)  -----> 4H^+(aq) + 2H2O(l)

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Answer:

See explaination

Explanation:

Chlorine’s disinfection properties have helped improve the lives of billions of people around the world. Chlorine also is an essential chemical building block, used to make many products that contribute to public health and safety, advanced technology, nutrition, security and transportation.

Please kindly check attachment for the step by step solution of the given problem.

The threshold for unpleasant taste and order in the respective units are;

A) 0.005 mg/L

A) 0.005 mg/LB) 5 μg/L

A) 0.005 mg/LB) 5 μg/LC) 0.005 ppm

A) 0.005 mg/LB) 5 μg/LC) 0.005 ppmD) 5 ppb

We are given the concentration of chlorophenols as; M = 5 mg/m³

A) Converting mg/m³ to mg/L means that;

1 mg/m³ = 0.001 mg/L. Thus;

5 mg/m³ = (5 × 0.001)/1

>> 0.005 mg/L

B) Converting mg/m³ to μg/L means that;

1 mg/m³ = 1 μg/L

Thus; 5 mg/m³ = 5 × 1/1 μg/L

>> 5 μg/L

C) Converting mg/m³ to ppm gives;

1 mg/m³ = 0.001 ppm

Thus;

5 mg/m³ = (5 × 0.001)/1 ppm

>> 0.005 ppm

D) Converting mg/m³ to ppb gives;

1 mg/m³ = 1 ppm

Thus;

5 mg/m³ = 5 × 1/1 ppb

>> 5 ppb

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Explanation:

(a) The given data is as follows.

            B = [tex]2.180 \times 10^{-18} J[/tex]

            Z = 4 for Be

Now, for the first excited state [tex]n_{f}[/tex] = 2; and [tex]n_{i} = \infinity[/tex] if it is ionized.

Therefore, ionization energy will be calculated as follows.

         I.E = [tex]\frac{-Bz^{2}}{\infinity^{2}} - (\frac{-2.180 \times 10^{-18} J /times (4)^{2}}{(2)^{2}})[/tex]

              = [tex]8.72 \times 10^{-18} J[/tex]

Converting this energy into kJ/mol as follows.

           [tex]8.72 \times 10^{-18} J \times 6.02 \times 10^{23} mol[/tex]  

           = 5249 kJ/mol

Therefore, the ionization energy of the [tex]Be^{3+}[/tex] ion in its first excited state in kilojoules per mole is 5249 kJ/mol.

(b) Change in ionization energy is as follows.

         [tex]\Delta E = -Bz^{2}(\frac{1}{(4)^{2}} - {1}{(2)^{2}}) = \frac{hc}{\lambda}[/tex]

   [tex]\frac{hc}{\lambda} = 0.1875 \times 2.180 \times 10^{-18} J \times (4)^{2}[/tex]                

        [tex]\lambda = \frac{6.626 \times 10^{-34} \times 2.998 \times 10^{8} m/s}{0.1875 \times 2.180 \times 10^{-18} J \times 16}[/tex]

                     = [tex]303.7 \times 10^{-10} m[/tex]

or,                 = [tex]303.7^{o}A[/tex]

Therefore, wavelength of light given off from the [tex]Be^{3+}[/tex] ion by electrons dropping from the fourth (n = 4) to the second (n = 2) energy levels [tex]303.7^{o}A[/tex].

The ionization energy of the Be³+ ion in its first excited state is approximately 5250 kJ/mol. The wavelength of light emitted when an electron transitions from n=4 to n=2 in a Be³+ ion is approximately 75.9 nm.

Ionization Energy and Wavelength Calculation for Be³+

The problem involves calculating the ionization energy of a Be³+ ion in its first excited state and the wavelength of light emitted during an electron transition.

a) Ionization Energy

For a one-electron species like Be³+, the energy in the nth state is given by:

E = –BZ²/n², where Z is the charge on the nucleus and B = 2.180 × 10⁻¹⁸ J.

In the first excited state (n = 2) for Be³+ (Z = 4), we get:

En = –B(4)²/2² = –2.180 × 10⁻¹⁸ J × 16/4 = –8.72 × 10⁻¹⁸ J.

To find the ionization energy, we need the energy to liberate 1 mole of Be³+ ions:

Ionization Energy = |En| × Avogadro's Number = 8.72 × 10⁻¹⁸ J × 6.022 × 10²³ mol⁻¹ = 5.25 × 10⁶ J/mol = 5250 kJ/mol.

b) Wavelength of Light

The energy difference between the fourth (n=4) and second (n=2) energy levels is calculated using:

ΔE = E₄ - E₂ = -BZ² / 4² - (-BZ² / 2²) = BZ² (1/4² - 1/2²)

ΔE = 2.180 × 10⁻¹⁸ J × 16 × (1/16 - 1/4) = 2.180 × 10⁻¹⁸ J × 16 × (-¾) = -2.616 × 10⁻¹⁷ J

The emitted photon's wavelength is given by:

λ = hc /ΔE = (6.626 × 10⁻³⁴ Js × 3.00 × 10⁸ m/s) / 2.616 × 10⁻¹⁷ J ≈ 7.59 × 10⁻⁸ m = 75.9 nm

The expected bond angle between bromine and oxygen, based on hybridization and molecular geometry, would likely be slightly less than the ideal tetrahedral angle of 109.5°, due to electron repulsion effects.

The bond angle between bromine and oxygen can be predicted using the concepts of hybridization and molecular geometry. If bromine and oxygen were to form a molecule similar to water (H₂O), one might expect a tetrahedral geometry because the valence orbitals of oxygen consist of one 2s orbital and three 2p orbitals. These combine during hybridization to form four hybrid orbitals that point toward the corners of a tetrahedron. The ideal bond angle for a tetrahedron is 109.5°. However, due to the effects of lone pair repulsion, the actual bond angle would likely be less, similar to how the H-O-H bond angle in water is experimentally found to be 104.5°.

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The actual angle between the bromine-oxygen bonds in BrO3⁻ is expected to be less than 120° due to the presence of the lone pair on the central bromine atom.

The ideal bond angle for a sp³ hybridized central atom is 109.5°. However, the presence of lone pairs on the central atom can distort the bond angles. Lone pairs are more electron-rich than bonding pairs, so they repel the bonding pairs more strongly. This repulsion causes the bond angles to decrease.

In the case of BrO3⁻, the central bromine atom is sp³ hybridized and has one lone pair. The lone pair will repel the two bonding pairs between the bromine and oxygen atoms. This repulsion will cause the bond angles between the bromine and oxygen atoms to decrease from the ideal angle of 120°.

The actual bond angle between the bromine-oxygen bonds in BrO3⁻ has been measured to be 107.1°. This is slightly less than the ideal angle of 120°, as expected.

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The question probable may be:

Compared to the ideal angle  120 ,the central atom Br lone pair 1 How you would expect the actual angle between the bromine-oxygen bonds to be  

Answer:

the rate constant of the reaction at a temperature of 741 °C is [tex]0.22858 \ s^{-1}[/tex]

it will take 3.0313 s to consume half of the reactant if an identical experiment is performed at 741 °C

Explanation:

Given that :

[tex]k_1 = 7.21*10^{-3} s^{-1} \\ \\ E_a = 247 kJ/mol \ \ \ = 247*10^3 \ J/mol \\ \\ T_1 = 634 ^ {^0} C= (273 + 634) K = 907 \ K \\ \\ T_2 = 741^{^0 } C = (273+ 741) K = 1014 \ K \\ \\ R =8.314 \ \ J/mol/K[/tex]

a)

According to Arrhenius Equation ;

[tex]In\frac{k_2}{k_1} = -\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )[/tex]

[tex]In\frac{k_2}{7.21*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{1}{1014} -\frac{1}{907} )[/tex]

[tex]In\frac{k_2}{7.21*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{907-1014}{907*1014} )[/tex]

[tex]In\frac{k_2}{7.21*10^{-3}} = \frac{-247*10^3*-107}{8.314*907*1014}[/tex]

[tex]In\frac{k_2}{7.21*10^{-3}} = 3.456412[/tex]

[tex]\frac{k_2}{7.21*10^{-3}} =e^{ 3.456412[/tex]

[tex]k_2 = 31.70302 * 7.21*10^{-3}[/tex]

[tex]k_2 = 0.22858 \ s^{-1}[/tex]

Therefore ,  the rate constant of the reaction at a temperature of 741 °C is [tex]0.22858 \ s^{-1}[/tex]

b) Given that  :

[tex]t_{(1/2)} = 96.1 \ s[/tex]

[tex]k_1 = \frac{0.693}{ t_{(1/2)}}[/tex]

[tex]k_1 = \frac{0.693}{ 96.1 \ s}[/tex]

[tex]k_1 = 7.211*10^{-3} \ s^{-1}[/tex]

[tex]In\frac{k_2}{7.211*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{1}{1014} -\frac{1}{907} )[/tex]

[tex]In\frac{k_2}{7.211*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{907-1014}{907*1014} )[/tex]

[tex]In\frac{k_2}{7.211*10^{-3}} = \frac{-247*10^3*-107}{8.314*907*1014}[/tex]

[tex]In\frac{k_2}{7.211*10^{-3}} = 3.456412[/tex]

[tex]\frac{k_2}{7.211*10^{-3}} =e^{ 3.456412[/tex]

[tex]k_2 =e^{ 3.456412 * 7.21*10^{-3}[/tex]

[tex]k_2 = 0.22862 \ s^{-1}[/tex]

[tex]t_{(1/2)_2}} = \frac{0.693}{k_2}[/tex]

[tex]t_{(1/2)_2}} = \frac{0.693}{0.22862}[/tex]

[tex]t_{(1/2)_2}} =3.0313 \ s[/tex]

it will take 3.0313 s to consume half of the reactant if an identical experiment is performed at 741 °C

Answer:

ok

Explanation:

I have the same question

Answer:

Burning a match is a chemical change while the rest are physical changes.

Explanation:

Burning a match represents a chemical change, where a substance transforms into new substances with distinct chemical properties. In contrast, boiling water, melting ice, and breaking glass are examples of physical changes, altering the substance's form but not its identity.

Among the given options, burning a match is a chemical change. A chemical change is a process that involves a substance changing into a new substance with different chemical properties. In this case, when a match is burned, it undergoes a chemical reaction leading to the formation of new substances such as heat, light, water vapor, and carbon dioxide, which were not there before the match was lit. On the other hand, boiling water, melting ice, and breaking glass are all examples of physical changes, which involve changes in the form of a substance but do not change the identity of the substance itself.

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Answer: The value of Keq for the reaction expressed in scientific notation is  [tex]1.6\times 10^{-5}[/tex]

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.

For the given chemical reaction:

[tex]2NOCl(g)\rightarrow 2NO(g)+Cl_2(g)[/tex]

The expression for [tex]K_{eq}[/tex] is written as:

[tex]K_{eq}=\frac{[NO]^2\times [Cl_2]}{[NOCl]^2}[/tex]

[tex]K_{eq}=\frac{(1.2\times 10^{-3})^2\times (2.2\times 10^{-3})}{(1.4\times 10^{-2})^2}[/tex]

[tex]K_{eq}=1.6\times 10^{-5}[/tex]

The value of Keq for the reaction expressed in scientific notation is  [tex]1.6\times 10^{-5}[/tex]

Answer:

its A

Explanation:

Complete Question

Despite the destruction from Hurricane Katrina in Sept 2005,the lowest pressure for a hurricane in the Atlantic Ocean was measured several weeks after Katrina. Hurricane Wilma registered at atmospheric pressure of 88.2 kPa on Oct 19, 2005,about 2 kPa lower than Hurricane Katrinia. What was the difference between the two hurricanes in:

millimeters of Hg? ___ mm Hg

atmospheres? ___ atm

millibars? ___ mb

Answer:

The difference in pressure in mm of  Hg is  [tex]Pg = 15.05 \ mm\ Hg[/tex]

The difference in pressure in atm is   [tex]P_{atm} = 0.0198 \ atm[/tex]

The difference in pressure in millibar is   [tex]P_{bar} = 20 \ mbar[/tex]

Explanation:

From the question we are told that

      The pressure of Hurricane Wilma is  [tex]P = 88 kPa[/tex]

      The difference is pressure is  [tex]\Delta P =2\ k Pa[/tex]

Generally

            [tex]1\ atm = 1.01 *10^5 \ Pa[/tex]

            [tex]1 \ atm = 760\ mmHg[/tex]

The pressure difference in mm Hg is mathematically  evaluated as

             [tex]Pg = \Delta P * \frac{1000 Pa}{1kPa} * \frac{1\ atm}{1.01 *10^5} * \frac{760 mm Hg}{1 \ atm}[/tex]

Substituting the value

            [tex]Pg = 2 kPa * \frac{1000 Pa}{1kPa} * \frac{1\ atm}{1.01 *10^5} * \frac{760 mm Hg}{1 \ atm}[/tex]

            [tex]Pg = 15.05 \ mm\ Hg[/tex]

  The pressure difference  in atm is  

                                      [tex]P_{atm } = 2kPa * \frac{1000}{1\ kPa} *\frac{1 \ atm}{1.01 *10^5 Pa}[/tex]

                                               [tex]= 0.0198 \ atm[/tex]

  The pressure in millibars is  

                                            [tex]= 2kPa * \frac{1000Pa}{1 kPa} * \frac{1\ bar }{1*10^5 Pa} * \frac{1000 \ milli bar}{1 bar}[/tex]

                                            [tex]= 20 \ m bar[/tex]

Final answer:

The pressure difference between Hurricane Wilma and Hurricane Katrina was 2.0 kPa, with Wilma's pressure being 88.2 kPa and Katrina's being 90.2 kPa. The pressure drop during a hurricane on the northeast U.S. coast in torr is 50.8 torr. A nonvolatile liquid is used in barometers and manometers because volatility would affect the accuracy of pressure measurements.

Explanation:

The difference in atmospheric pressure between Hurricane Wilma and Hurricane Katrina can be calculated using the information provided. If Wilma's pressure was 88.2 kPa and that was 2.0 kPa lower than Katrina's, then Katrina's pressure was 88.2 kPa + 2.0 kPa = 90.2 kPa. Therefore, the difference in pressure between the two hurricanes was 2.0 kPa.

To calculate the drop in pressure in torr during a hurricane on the northeastern United States coast: the pressure usually at 30.0 in. Hg can drop to 28.0 in. Hg. Since 1 in. Hg is equivalent to 25.4 mm Hg, and 1 mm Hg is the same as 1 torr, the drop in pressure is (30.0 in. Hg - 28.0 in. Hg) x 25.4 mm Hg/in. Hg = 50.8 torr.

It is necessary to use a nonvolatile liquid in a barometer or manometer because a volatile liquid would evaporate and form a vapor that would affect the pressure measurement, leading to inaccurate readings.

The heats of combustion for methanol, ethanol, and n-propanol, when expressed in kJ/mol, are -0.710 × 10³ kJ/mol, -0.644 × 10³ kJ/mol, and -0.56 × 10³ kJ/mol, respectively. This shows that the amount of heat liberated decreases as the molecular weight of the alcohol increases.

The heat of combustion of a substance is defined as the amount of heat energy released when 1 mole of a substance is completely burned in oxygen. Here, the heats of combustion for methanol (CH3OH), ethanol (C2H5OH), and n−propanol (C3H7OH) are given in kJ/g, and we need to convert them into kJ/mol.

To do this conversion, you first need to find the molar mass of each alcohol. You can do this by adding up the atomic masses of each element in the molecule.

Using the atomic masses in g/mol from the periodic table, we have: Methanol (CH3OH): 12.01 + 3(1.01) + 16 + 1.01 = 32.05 g/mol; Ethanol (C2H5OH): 2(12.01) + 5(1.01) +16 + 1.01=46.07 g/mol; n-Propanol (C3H7OH): 3(12.01) + 7(1.01) + 16 + 1.01 = 60.1 g/mol.

Next, divide the heat of combustion given in kJ/g by the molar mass calculated in g/mol to obtain the heat of combustion in kJ/mol:

Methanol: -22.6 kJ/g ÷ 32.05 g/mol = -0.710 × 10³ kJ/mol (scientific notation)

Ethanol: -29.7 kJ/g ÷ 46.07 g/mol = -0.644 × 10³ kJ/mol (scientific notation)

n-Propanol: -33.4 kJ/g ÷ 60.1 g/mol = -0.56 × 10³ kJ/mol (scientific notation)

So, the heats of combustion for the alcohols are: Methanol: -0.710 × 10³ kJ/mol, Ethanol: -0.644 × 10³ kJ/mol, and n-Propanol: -0.56 × 10³ kJ/mol. As you can see, the amount of heat liberated per mole decreases as the molecular mass of the alcohol increases.

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Answer:

A

Explanation:

Just thinking logically.

Answer:

4585 KJ

Explanation:

Data obtained from the question. This includes following:

Number of mole (n) of glycerin = 50 moles

Heat of fusion glycerin (Hf) = 91.7 KJ/mol

Heat (Q) released =?

The heat released can be obtained as follow:

Q = n•Hf

Q = 50 x 91.7

Q = 4585 KJ

Therefore, 4585 KJ of heat is released.

Answer:

The new volume of the balloon is 539 L

Explanation:

As the volume increases, the gas particles (atoms or molecules) take longer to reach the walls of the container and therefore collide less times per unit time against them. This means that the pressure will be less because it represents the frequency of gas strikes against the walls. In this way, pressure and volume are related, determining Boyle's law that says:

"The volume occupied by a given gas mass at constant temperature is inversely proportional to the pressure"

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

o P * V = k

Having an initial state 1 and an final state 2 will be fulfilled:

P1 * V1 = P2 * V2

So, in this case, you know:

Replacing:

760 mmHg*200 L= 282 mmHg*V2

Solving:

[tex]V2=\frac{760 mmHg*200 L}{282 mmHg}[/tex]

V2=539 L

The new volume of the balloon is 539 L

Using Boyle's Law, we calculated the new volume of the weather balloon as approximately 538.7 L when the pressure decreases from 760 mm Hg to 282 mm Hg at constant temperature.

To find the new volume of the weather balloon when the pressure changes, we can use Boyle's Law, which states that the product of the initial pressure and volume is equal to the product of the final pressure and volume, assuming constant temperature.

The formula is:

P₁V₁= P₂V₂

Given: P₁ = 760 mm Hg, V₁ = 200.0 L, P₂ = 282 mm Hg

We need to find V₂.

Rearranging the formula to solve for V₂:

V₂= (P₁V1₁ / P₂)

Substituting the given values:

V₂ = (760 mm Hg * 200.0 L) / 282 mm Hg

V₂ = 152000 mm Hg L / 282 mm Hg

V₂ ≈ 538.7 L

Therefore, the new volume of the balloon is approximately 538.7 L.

Answer: C2H2

Explanation: Because each of the lines represent one bond, and because there are three lines (bonds) between the carbons, it means that they are bonded by three bonds, also known as a triple bond.

The triple bond that we have is shown by image D.

Two atoms share three pairs of electrons in a triple bond, making a total of six electrons present in the bond. Multiple bonds are formed between the atoms as a result of the sharing of electrons, offering a high degree of stability.

Two pi (π) bonds and a sigma (σ) bond define a triple bond. Atomic orbitals overlap head-on to form the sigma bond, whereas side-to-side overlap produces the two pi bonds.

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