A stock solution contains a mixture of ~100 ppm chloride, fluoride, nitrite, bromide, nitrate and phosphate anions. In order to prepare 1 L of 100 ppm nitrite stock solution, you weigh out 150.0 mg of NaNO2. The actual concentration of nitrite would be:______.

Answer:

Fatty acids can be branched or unbranched

Triacylglycerols are a major energy storage form

Mass spectrometry can be used to identify individual lipids in complex mixtures.

Membrane sphingolipids can be used to produce intracellular messengers.

Explanation:

branched fatty acid (BCFA) usually contain one or more methyl group on the carbon chain , mostly found in bacteria as component of membrane lipids while unbrached are straight chain saturated or unsaturated fatty acids.

Triacylglycerol is made up of glcerol esterified to three molecules of same or different fatty acids. it serve as the storageform of fatty acid in the adipose tissues. it is usually mobilized for usage through an hormone triggered process

Mass spectrometry is a seperation technique used to quantify and identify unknown compounds within a sample by differentiating gaseous ion in an electric field and magnetic field according to their mass to charge ratios

sphingolipids are lipids containing the alcohol; sphingosine. They serve as intracellular second messengers and extracellular mediator

Answer:

E) 2.38

Explanation:

The pH of any solution , helps to determine the acidic strength of the solution ,

i.e. ,

and ,

pH is given as the negative log of the concentration of H⁺ ions ,

hence ,

pH = - log H⁺

From the question ,

the concentration of the solution is 0.0042 M , and being it a strong acid , dissociates completely to its respective ions ,

Therefore , the concentration of H⁺ = 0.0042 M .

Hence , using the above equation , the value of pH can be calculated as follows -

pH = - log H⁺

pH = - log ( 0.0042 M )

pH =  2.38 .

Final answer:

HCl is a strong acid, and its dissociation yields an equal concentration of hydrogen ions. Thus, the correct options is E) 2.38.

Explanation:

The pH of a 0.0042 M hydrochloric acid solution can be calculated using the formula [tex]pH = -log[H+][/tex], where [tex][H+][/tex] is the concentration of hydrogen ions.

In a solution of hydrochloric acid, which is a strong acid, the concentration of hydrogen ions [tex][H+][/tex] is equal to the concentration of the acid itself because [tex]HCl[/tex]dissociates completely in water.

Thus, the pH of the 0.0042 [tex]M HCl[/tex] solution is calculated as follows:

[tex]pH = -log(0.0042)[/tex]

= 2.38.

Therefore, the correct answer is E) 2.38.

Answer: b. Catalysis occurs at the active site, which usually consists of a crevice on the surface of the enzyme.

c. Generally, an enzyme is specific for a particular substrate. For example, thrombin catalyzes the hydrolysis of the peptide bond between Arg and Gly.

Explanation:

The role of the enzyme is to produce a specific product, whereas the non-biological catalyst produces more than one product. Thus they have a different degree of reaction specificity.

The active site of the enzyme is the catalytic site of the enzyme. It is the region where the substrate molecule bind and undergoes a chemical reaction. The enzyme exhibit a special crevice or opening at the active site which facilitates the binding with the substrate.

Enzymes usually bind to a specific substrate as they have an active site that allows a particular substrate to bind to the active site of the enzyme. This is because of the shape of the active site of the enzyme which has a binding affinity with a particular substrate any other substrate cannot bind to the active site.

The balanced chemical equation for the combustion of butane is correct. Converting the mass of butane to moles and using the mole ratio, the volume of carbon dioxide produced is approximately 596 liters, as calculated using the ideal gas law.

Combustion of Butane and its Volume Calculation:

1. Balanced Chemical Equation:

The provided equation is correct:

C₄H₁₀ + ¹³/₂O₂ → 4CO₂ + 5H₂O

It is balanced because it has the same number of atoms of each element on both sides of the equation.

2. Volume of Carbon Dioxide Produced:

a) Converting Mass of Butane to Moles:

The mass of butane (0.360 kg) to moles (6.19 mol) using the following formula:

moles = mass / molar mass

The molar mass of butane is approximately 58.12 g/mol.

b) Mole Ratio and CO₂ Production:

The balanced equation tells us that 1 mole of butane reacts with 13/2 moles of oxygen to produce 4 moles of carbon dioxide. Therefore, the 6.19 moles of butane will produce:

6.19 mol butane * (4 mol CO₂ / 1 mol butane) = 24.76 mol CO₂

c) Ideal Gas Law and Volume Calculation:

The ideal gas law relates the volume (V), pressure (P), temperature (T), and number of moles (n) of a gas:

V = nRT/P

where:

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

T is the temperature in Kelvin (293.15 K)

P is the pressure in atm (1 atm)

Plugging in the values:

V = 24.76 mol * 0.08206 L atm/mol K * 293.15 K / 1 atm ≈ 596 L

Therefore, the volume of carbon dioxide produced is approximately 596 liters.

Final answer:

To calculate the heat energy released by burning a lump of coal, find the volume, multiply by density to get mass, convert mass to moles, and then multiply by the energy released per mole of carbon. The final calculated heat energy (q) is approximately -6451.75 kJ.

Explanation:

The question asks us to calculate the heat energy (q) released when a lump of coal is burned. To find this, we first need to determine the mass of the coal using its volume and density. The volume of the coal lump can be determined by multiplying its dimensions: 5.6 cm × 5.1 cm × 4.6 cm. Once we have the volume, we multiply it by the coal's density to get the mass. The combustion of carbon releases -394 kJ of energy per mole of carbon according to the given chemical equation. Since the molar mass of carbon (C) is 12.01 g/mol, we convert the mass of coal to moles and then multiply by the energy released per mole to find the total energy (q).

Step-by-step calculation:

Therefore, the value of q when the lump of coal is burned is approximately -6451.75 kJ.

Answer:

We need 247 mL of NaOH

Explanation:

Step 1: Data given

Molarity of NaOH = 0.587 M

Volume of 0.445 M NiBr2 solution = 163 mL = 0.163 L

Step 2: The balanced equation

NiBr2(aq) + 2NaOH(aq) Ni(OH)2(s) + 2NaBr(aq)

Step 3: Calculate moles of NiBr2

Moles NiBR2 = Molarity NiBR2 * volume

Moles NiBR2 = 0.445 M * 0.163 L

Moles NiBR2 = 0.0725 moles

Step 3: Calculate moles of NaOH

For 1 mol NiBr2 consumed, we need 2 moles NaOH

For 0.0725 moles NiBR2, we need 2* 0.0725 = 0.145 moles NaOH

Step 4: Calculate volume of NaOH

Volume = moles NaOH / Molarity NaOH

Volume = 0.145 moles / 0.587 M

volume = 0.247L = 247 mL

We need 247 mL of NaOH

Answer: The balloon will burst as it occupies higher volume than  maximum inflated volume.

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.  

The equation given by this law is:

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas= 1.00 atm = 760.0 mm Hg  

[tex]V_1[/tex] =  initial volume of gas = 2.50 L

[tex]P_2[/tex] = final pressure of gas= 450.0 mm Hg

[tex]V_2[/tex] = final volume of gas = ?

Putting values in above equation, we get:

[tex]760.0\times 2.50=450.0\times V_2[/tex]

[tex]V_2=4.22L[/tex]

Thus the final volume of the gas is 4.22 L and as the maximum inflated volume is 3.00 L, the balloon will burst.

Answers and Explanation:

a)- The chemical equation for the corresponden equilibrium of Ka1 is:

2. HNO2(aq)⇌H+(aq)+NO−2

Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.

b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:

ΔG= ΔGº + RT ln Q

Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)

At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:

⇒ 0 = ΔGº + RT ln Ka

   ΔGº= - RT ln Ka

   ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)

  ΔGº= 19092.8 J/mol

c)- According to the previous demonstation, at equilibrium ΔG= 0.

d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:

ΔG= ΔGº + RT ln Q

Q= ((H⁺) (NO₂⁻))/(HNO₂)

Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)

Q= 1.88 10⁻⁴

We know that   ΔGº= 19092.8 J/mol, so:

ΔG= ΔGº + RT ln Q

ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)

ΔG= -2162.4 J/mol

Notice that ΔG<0, so the process is spontaneous in that direction.

HNO2 is a weak acid whose dissociation occurs as follows;  HNO2 (aq)    ⇋       H+ (aq)  +      NO2- (aq)

The equation for the dissociation of HNO2 is;

           HNO2 (aq)    ⇋       H+ (aq)  +      NO2- (aq)

We can use the value of Ka to find the  ΔG∘ for the dissociation of nitrous acid in aqueous solution as follows;

ΔG∘ = -RTlnKa

Where;

R = Gas constant = 8.314 J/K. mol

Ka = Acid dissociation constant = 4.5×10^−4

T = temperature = 25 ∘C or 298 K

Substituting values;

ΔG∘ = -(8.314 J/K. mol × 298 K × ln 4.5×10^−4)

ΔG∘ = 19.1 KJ/mol

Given that;

Q = [ H+] [NO2- ]/[HNO2]

[ H+] =  5.9×10−2 M

[NO2- ] = 6.7×10−4 M

[HNO2] = 0.21 M

Q = [5.9×10−2 M] [6.7×10−4 M]/[0.21 M]

Q = 1.88 × 10^−4

From the formula;

ΔG = ΔG∘ + RTlnQ

ΔG =  19.1  KJ/mol + (8.314 J/K. mol  ×  298 K ×  ln 1.88 × 10^−4)

ΔG = -2.15 KJ/mol

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Answer:

[tex]{\rho_{143\ ^0C}}=1.118\times 10^4\ kg/m^3}[/tex]

Explanation:

The expression for the volume expansion is:-

[tex]V_2=V_1\times [1+3\times \alpha\times \Delta T][/tex]

Where,

[tex]V_2\ and\ V_1[/tex] are the volume values

[tex]\alpha[/tex] is the coefficient of linear expansion = [tex]29\times 10^{-6}\ (^0C)^{-1}[/tex]

Also,

Density is defined as:-

[tex]\rho=\frac{Mass}{Volume}[/tex]

or,

[tex]Volume=\frac{Mass}{\rho}[/tex]

Applying in the above equation, we get that:-

[tex]\frac{M}{\rho_2}=\frac{M}{\rho_1}\times [1+3\times \alpha\times \Delta T][/tex]

Or,

[tex]{\rho_2}=\frac{\rho_1}{[1+3\times \alpha\times \Delta T]}[/tex]

So, From the question,

[tex]\Delta T=143-20\ ^0C=123\ ^0C[/tex]

[tex]\rho_1=1.13\times 10^4\ kg/m^3[/tex]

Thus,

[tex]{\rho_2}=\frac{1.13\times 10^4\ kg/m^3}{[1+3\times (29\times 10^{-6}\ (^0C)^{-1})\times \Delta (123\ ^0C)]}[/tex]

[tex]{\rho_2}=1.118\times 10^4\ kg/m^3}[/tex]

The density of lead at 143°C is 1.130418 × 10⁴ kg/m³.

Density of lead changes with temperature, which is directly proportional to the coefficient of linear expansion. We can use the formula given below to find the density of lead at 143°C if we know its density at 20°C and the coefficient of linear expansion.

Δρ = αρΔTwhere,

Δρ = change in density

α = coefficient of linear expansion

ρ = initial density

ΔT = change in temperature

Let's substitute the given values in the above formula and solve for Δρ:

α = 29 × 10⁻⁶ /°C

ρ = 1.13 × 10⁴ kg/m³ (density at 20°C)

ΔT = 143°C - 20°C = 123°C

Δρ = αρΔT

Δρ = 29 × 10⁻⁶ /°C × 1.13 × 10⁴ kg/m³ × 123°C

Δρ = 41.80 kg/m³

Therefore, the density of lead at 143°C is:

ρ = ρ₀ + Δρ

ρ = 1.13 × 10⁴ kg/m³ + 41.80 kg/m³

ρ = 1.130418 × 10⁴ kg/m³

Thus, the density of lead at 143°C is 1.130418 × 10⁴ kg/m³ (to at least four significant figures).

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Answer:

The exception here is

C2 and B2

both the bonds are pi bond in the above cases

Explanation:

The fault here is the unforeseen energy order of molecular orbitals. Since if filled, sigma 2px will be firmly pushed away by the electron density of orbitals already filled with sigma 2s (both electron densities lie on the inter-nuclear axis). Because of this, sigma 2px's energy is found shockingly greater than pi 2py and pi 2pz.

The electrons therefore occupy pi 2py and pi 2pz orbitals as opposed to normal sigma 2px (which remains vacant) while filling.

The existence of these 4 electrons in orbitals pi 2py and pi 2pz results in two pi bonds being formed.

While most species follow the general rule that a single bond is a sigma bond and a double bond comprises a sigma and pi bond, B2 and C2 are exceptions, with their sigma and pi bond energy levels differing from the norm.

The generalization that a single bond is a sigma bond, and a double bond is made up of a sigma bond and a pi bond, with a triple bond consisting of one sigma bond and two pi bonds, is questioned in the given species. In most cases, this generalization holds true, with N2 possessing a triple bond that contains one sigma bond and two pi bonds, O2 containing a double bond with one sigma and one pi bond, and F2 having a single bond that, by definition, is a sigma bond.

However, according to the provided information, there is an exception among the listed species where the energy levels of the sigma and pi bonds differ from the norm. For B2, C2, and N2, the sigma bonds (2p) have higher energy than the pi bonds (2p). This anomalous behavior suggests that in these species, the generalization about bonding is violated.

Therefore, among the species listed, B2 and C2 are exceptions to the rule, with B2 and C2 having differences in their bond energies that do not align with the typical characteristics of sigma and pi bonds.

Answer : The value of [tex]\Delta G_{rxn}[/tex] is, -27.0kJ/mole

Explanation :

The formula used for [tex]\Delta G_{rxn}[/tex] is:

[tex]\Delta G_{rxn}=\Delta G^o+RT\ln K_p[/tex]   ............(1)

where,

[tex]\Delta G_{rxn}[/tex] = Gibbs free energy for the reaction

[tex]\Delta G_^o[/tex] =  standard Gibbs free energy  = -32.8 kJ

/mol

R = gas constant = 8.314 J/mole.K

T = temperature = 298 K

[tex]K_p[/tex] = equilibrium constant

First we have to calculate the value of [tex]K_p[/tex].

The given balanced chemical reaction is,

[tex]2NO(g)+O_2(g)\rightarrow 2NO_2(g)[/tex]

The expression for equilibrium constant will be :

[tex]K_p=\frac{(p_{NO_2})^2}{(p_{NO})^2\times (p_{O_2})}[/tex]

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

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

[tex]K_p=\frac{(0.800)^2}{(0.500)^2\times (0.250)}[/tex]

[tex]K_p=10.24[/tex]

Now we have to calculate the value of [tex]\Delta G_{rxn}[/tex] by using relation (1).

[tex]\Delta G_{rxn}=\Delta G^o+RT\ln K_p[/tex]

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

[tex]\Delta G_{rxn}=-32.8kJ/mol+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\ln (10.24)[/tex]

[tex]\Delta G_{rxn}=-27.0kJ/mol[/tex]

Therefore, the value of [tex]\Delta G_{rxn}[/tex] is, -27.0kJ/mole

Answer: The boiling point of water is an absolute constant

Explanation:

The boiling point of a compound is NOT an absolute constant. This is because at certain conditions such as a change in altitude, the boiling point of a compound changes.

As temperature increases, evaporation increases and vapour pressure increases. Compounds boils when vapour pressure equal to the atmospheric pressure.

At higher altitudes, atmospheric pressure is decreses.

When atmospheric pressure is decresed, the vapour pressure of the compound is lowered to reach boiling point. Therefore, the temperature needed for vapour pressure to equal atmospheric pressure is lower. The boiling point is lower at higher altitude.

Hence boiling point depends on atmospheric pressure

The boiling point of a compound is not an absolute constant.

The statement "The boiling point of a compound is an absolute constant" is FALSE.

The boiling point of a compound is affected by several factors, including intermolecular forces and molecular structure. Compounds with stronger intermolecular forces, such as hydrogen bonds, will generally have higher boiling points. Viscosity, which is the resistance of a liquid to flow, is not directly related to boiling point. Nonvolatile compounds, which do not easily evaporate, may have lower boiling points depending on their molecular structure.

Therefore, the correct statement is that the boiling point is not an absolute constant.

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The equilibrium constant for the reaction is 32.2

The equilibrium constant (K꜀) for a given reaction is simply defined as the ratio of the concentration of the products raised to their coefficients to the concentration of the reactants raised to their coefficients

With the above information, we can obtain the equilibrium constant for the reaction given above as follow:

8H₂S(g) <=> 8H₂(g) + S₈(g)

Concentration of Hydrogen sulphide [H₂S] = 0.250 M

Concentration of Hydrogen [H₂] = 0.400 M

Concentration of sulphur [S₈] = 0.750 M

[tex]K_{C} = \frac{[H_{2}]^8[S_{8}]}{[H_{2}S]^{8}} \\\\K_{C} = \frac{0.4^{8} * 0.750}{0.25^{8}}\\\\[/tex]

Therefore, the equilibrium constant for the reaction is 32.2

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The equilibrium constant Kc for the given reaction is determined from the ratio of the product concentrations to the reactant concentrations, each raised to the power of their stoichiometric coefficients.

The equilibrium constant Kc for a chemical reaction is calculated from the concentrations of the products divided by the concentrations of the reactants, each raised to the power of their stoichiometric coefficients.

Given the reaction equation: 8H2S(g) ⇌ 8H2(g) + S8(g)

And the equilibrium concentrations: [H2S] = 0.250 M, [H2] = 0.400 M, [S8] = 0.750 M

The equilibrium constant Kc is therefore: Kc = ([H2]^8 * [S8]) / ([H2S]^8) = (0.400^8 * 0.750) / (0.250^8).

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Answer : The enthalpy of the reaction is, -2552 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given enthalpy of reaction is,

[tex]4B(s)+3O_2(g)\rightarrow 2B_2O_3(s)[/tex]    [tex]\Delta H=?[/tex]

The intermediate balanced chemical reactions are:

(1) [tex]B_2O_3(s)+3H_2O(g)\rightarrow 3O_2(g)+B_2H_6(g)[/tex]     [tex]\Delta H_A=+2035kJ[/tex]

(2) [tex]2B(s)+3H_2(g)\rightarrow B_2H_6(g)[/tex]    [tex]\Delta H_B=+36kJ[/tex]

(3) [tex]H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)[/tex]    [tex]\Delta H_C=-285kJ[/tex]

(4) [tex]H_2O(l)\rightarrow H_2O(g)[/tex]    [tex]\Delta H_D=+44kJ[/tex]

Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :

(when we are reversing the reaction then the sign of the enthalpy change will be change.)

The expression for enthalpy of the reaction will be,

[tex]\Delta H=-2\times \Delta H_A+2\times \Delta H_B-6\times \Delta H_C-6\times \Delta H_D[/tex]

[tex]\Delta H=-2(+2035kJ)+2(+36kJ)-6(-285kJ)-6(+44)[/tex]

[tex]\Delta H=-2552kJ[/tex]

Therefore, the enthalpy of the reaction is, -2552 kJ/mole

Answer:

that results in an alcohol and a carboxylic acid.

Explanation:

Saponification is a chemical reaction process of alkaline hydrolysis of esters(R'COOR group) by which soap is obtained.

For Example, when a base such as sodium hydroxide [NaOH]  is used to hydrolyze an ester, the products are a carboxylate salt and an alcohol. Because soaps are prepared by the alkaline hydrolysis of fats and oils.

In a saponification reaction, alkaline hydrolysis of fats and oils with sodium hydroxide  yields propane-1,2,3-triol and the corresponding sodium salts of the component fatty acids.

i.e  Fat or oil + caustic alkali ⇒ Soap + propane-1,2,3-triol

As a specific example, ethyl acetate and NaOH react to form sodium acetate and ethanol:

The reaction goes to completion in the image below:

Answer:

ΔG° = -133,1 kJ

Explanation:

For the reactions:

(1) V₂O₃(s) + 3CO(g) → 2V(s) + 3CO₂(g); ΔH° = 369,8 kJ; ΔS° = 8,3 J/K

(2) V₂O₅(s) + 2CO(g) → V₂O₃(s) + 2CO₂(g); ΔH° = –234,2 kJ; ΔS° = 0,2 J/K

By Hess's law it is possible to obtain the ΔH° and ΔS° of:

2V(s) + 5CO₂(g) → V₂O₅(s) + 5CO(g)

Substracting -(1)-(2), that means:

ΔH° = -369,8 kJ - (-234,2 kJ) = -135,6 kJ

ΔS° = - 8,3 J/K - 0,2 J/K = -8,5 J/K

Using: ΔG° = ΔH° - TΔS° at 298K

ΔG° = -135,6 kJ - 298K×-8,5x10⁻³kJ/K

ΔG° = -133,1 kJ

I hope it helps!

Final answer:

To calculate the standard Gibbs free energy change (ΔG°) for the desired reaction at 298 K, reverse the first reaction, multiply the second by 2, sum their ΔH° and ΔS° values, and use the Gibbs free energy equation.

Explanation:

To calculate the standard Gibbs free energy change (ΔG°) for the reaction 2V(s) + 5CO2(g) → V2O5(s) + 5CO(g) at 298 K, we need to use the given information from two different reactions and apply Hess's law. First, we need to reverse the first reaction and multiply the second reaction by 2 to get the formation of V2O5(s) from V(s) and CO2(g). The ΔH° and ΔS° values of the first reaction would become -369.8 kJ and -8.3 J/K, respectively, after being reversed. For the second reaction, these values are -468.4 kJ and 0.4 J/K after being multiplied by 2. Adding these values provides the ΔH° and ΔS° for the overall reaction.

The ΔH° and ΔS° are then used to calculate ΔG° using the Gibbs free energy equation: ΔG° = ΔH° - TΔS°. Substituting the values, we find ΔG° for the overall reaction. This will tell us whether the reaction is spontaneous at 298 K, with a negative ΔG° indicating a spontaneous process.

Answer:

The expression will be given as:

[tex][CH_3OH]=K_c\times [CO]\times [H_2]^2[/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. It is expressed as [tex]K_{c}[/tex]

[tex]aA+bB\rightleftharpoons cC+dD[/tex]

[tex]K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]

[tex]CO(g)+2H_2(g)[/tex] ⇌ [tex]CH_3OH(g)[/tex]

The equilibrium-constant expression for the given reaction  is given by:

[tex]K_{c}=\frac{[CH_3OH]}{[CO][H_2]^2}[/tex]

If we are given with equilibrium constant and equilibrium concentration of carbon monoxide and hydrogen gas we can determine the concentration of methanol at equilibrium.

The expression will be given as:

[tex][CH_3OH]=K_c\times [CO]\times [H_2]^2[/tex]

Answer: The [tex]K_{sp}[/tex] of gold (III) chloride is [tex]3.2\times 10^{-25}[/tex]

Explanation:

We are given:

Solubility of gold (III) chloride = [tex]1.00\times 10^{-4}g/L[/tex]

Molar mass of gold (III) chloride = 303.3 g/mol

To calculate the solubility in mol/L, we divide the given solubility (in g/L) with molar mass, we get:

[tex]\text{Solubility (in mol/L)}=\frac{\text{Solubilty (in g/L)}}{\text{Molar mass}}[/tex]

Putting values in above equation, we get:

[tex]\text{Solubility of gold (III) chloride (in mol/L)}=\frac{1.00\times 10^{-4}g/L}{303.3g/mol}\\\\\text{Solubility of gold (III) chloride (in mol/L)}=3.29\times 10^{-7}mol/L[/tex]

The balanced equilibrium reaction for the ionization of gold (III) chloride follows:

[tex]AuCl_3\rightleftharpoons Au^{3+}+3Cl^-[/tex]

                s       3s

The expression for solubility constant for this reaction will be:

[tex]K_{sp}=[Au^{3+}][Cl^-]^3[/tex]

We are given:

[tex]s=3.29\times 10^{-7}mol/L[/tex]

Putting values in above equation, we get:

[tex]K_{sp}=(s)\times (3s)^3\\\\K_{sp}=9s^4\\\\K_{sp}=9\times (3.29\times 10^{-7})^4[/tex][tex]K_{sp}=(s)\times (3s)^3\\\\K_{sp}=27s^4\\\\K_{sp}=27\times (3.29\times 10^{-7})^4=3.2\times 10^{-25}[/tex]

Hence, the [tex]K_{sp}[/tex] of gold (III) chloride is [tex]3.2\times 10^{-25}[/tex]

The Ksp of AuCl3, calculated from its solubility and molar mass, is 1.2x10^-26.

In order to calculate the Ksp for AuCl3, we first need to convert the solubility from g/L to mol/L. We do this by dividing the given solubility (1.00x10^-4 g/L) by the molar mass of AuCl3 (303.3 g/mol). This gives us a molar solubility of 3.3x10^-7 mol/L. The dissolution reaction of AuCl3 can be represented as follows: AuCl3(s) -> Au3+(aq) + 3Cl-(aq). The solubility product (Ksp) is then obtained by multiplying the concentration of each ion raised to the power of its stoichiometric coefficient in the balanced equation. This gives Ksp=Au3+ (aq) * Cl-(aq)^3 = (3.3x10^-7)^4 = 1.2x10^-26, hence, the answer is option C.) 1.2x10^-26.

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Answer:

Explanation:

Please check the graph attached.

In a Lineweaver-Burk plot, the x-intercept (-1/Km) represents the Michaelis constant, the y-intercept (1/Vmax) shows the maximum enzyme reaction rate, and the slope represents the ratio Km/Vmax.

In a Lineweaver-Burk plot, the x-intercept, the y-intercept, and the slope provide vital information about specific parameters of an enzyme reaction. The x-intercept (-1/Km) gives a view of the Michaelis constant Km, describing the enzyme's affinity for its substrate, while the y-intercept (1/Vmax) gives the maximum reaction rate Vmax. The slope in this plot corresponds to the ratio Km/Vmax, which helps understand the dependency of the enzyme reaction speed on the substrate concentration and maximum reaction speed. Understanding these parameters through the Lineweaver-Burk plot plays a vital role in the analysis of enzyme kinetics.

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Answer:

[tex]e(\%)=0.17\%[/tex]

Explanation:

Volume of a drop:

[tex]V_{drop}=\frac{1 mL}{20 drop}[/tex]

[tex]V_{drop}=0.05 mL[/tex]

To estimate the error:

[tex]e(\%)=\frac{V_{real}-V{theorerical}}{V{theoretical}}*100\%[/tex]

[tex]e(\%)=\frac{(30 mL+0.05mL)-30mL}{30mL}*100\%[/tex]

[tex]e(\%)=0.17\%[/tex]

Answer: The standard enthalpy change of the reaction is coming out to be -16.3 kJ

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H[/tex]

The equation used to calculate enthalpy change is of a reaction is:  

[tex]\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]

For the given chemical reaction:

[tex]Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)[/tex]

The equation for the enthalpy change of the above reaction is:

[tex]\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})][/tex]

We are given:

[tex]\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol[/tex]

Putting values in above equation, we get:

[tex]\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ[/tex]

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

To calculate the standard enthalpy change for the given reaction, use the standard enthalpy of formation values for the reactants and products. Calculate the enthalpy change for the formation of each compound using their respective standard enthalpy of formation values. Then, calculate the overall enthalpy change for the reaction by subtracting the sum of the enthalpy changes for the reactants from the sum of the enthalpy changes for the products. Round the result to the appropriate number of significant figures.

The standard enthalpy change for a reaction can be calculated using the standard enthalpy of formation values. For the given reaction, Mg(OH)2(s) + 2 HCl(g) ⟶ MgCl2(s) + 2 H2O(g), we need to use the standard enthalpy of formation values for the reactants and products.

Step 1: Calculate the enthalpy change for the formation of Mg(OH)2(s) using its standard enthalpy of formation value.

Step 2: Calculate the enthalpy change for the formation of MgCl2(s) using its standard enthalpy of formation value.

Step 3: Calculate the enthalpy change for the formation of H2O(g) using its standard enthalpy of formation value.

Step 4: Calculate the overall enthalpy change for the reaction by subtracting the sum of the enthalpy changes for the reactants from the sum of the enthalpy changes for the products.

Step 5: Round the result to the appropriate number of significant figures.

Answer: The standard enthalpy change for the given reaction at 25°C can be calculated using the standard enthalpy of formation values for the reactants and products.

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The above question is incomplete, here is the complete question:

Calculate the standard molar enthalpy of formation of NO(g) from the following data at 298 K:

[tex]N_2(g) + 2O_2 \rightarrow 2NO_2(g), \Delta H^o = 66.4 kJ[/tex]

[tex]2NO(g) + O_2\rightarrow 2NO_2(g),\Delta H^o= -114.1 kJ[/tex]

Answer:

The standard molar enthalpy of formation of NO is 90.25 kJ/mol.

Explanation:

[tex]N_2(g) + 2O_2 \rightarrow 2NO_2(g), \Delta H^o_{1} = 66.4 kJ[/tex]

[tex]2NO(g) + O_2\rightarrow 2NO_2(g),\Delta H^o_{2} = -114.1 kJ[/tex]

To calculate the standard molar enthalpy of formation

[tex]N_2+O_2\rightarrow 2NO(g),\Delta H^o_{3} = ?[/tex]...[3]

Using Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

[1] - [2] = [3]

[tex]N_2+O_2\rightarrow 2NO(g),\Delta H^o_{3} = ?[/tex]

[tex]\Delta H^o_{3} =\Delta H^o_{1} - \Delta H^o_{2} [/tex]

[tex]\Delta H^o_{3}=66.4 kJ - [ -114.1 kJ] = 180.5 kJ[/tex]

According to reaction [3], 1 mole of nitrogen gas and 1 mole of oxygen gas gives 2 mole of nitrogen monoxide, So, the standard molar enthalpy of formation of 1 mole of NO gas :

=[tex]\frac{\Delta H^o_{3}}{2 mol}[/tex]

[tex]=\frac{180.5 kJ}{2 mol}=90.25 kJ/mol[/tex]

Answer:

b C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻

e HC₂H₃O₂ + H₂O ⇄ H₃O⁺ + C₂H₃O₂⁻

f ka<<1

g. Weak acid molecules and water molecules

Explanation:

The water molecule could act as a base and as an acid, a molecule that have this property is called as amphoteric.

b The salt NaC₂H₃O₂ is dissolved in water as Na⁺ and C₂H₃O₂⁻. The reaction of the anion with water is:

C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻

Where the C₂H₃O₂⁻ is the base and water is the acid.

e. The reaction of HC₂H₃O₂ (acid) with water (base), produce:

HC₂H₃O₂ + H₂O ⇄ H₃O⁺ + C₂H₃O₂⁻

f. As the acetic acid (HC₂H₃O₂) is a week acid, the dissociation in C₂H₃O₂⁻ is not complete, that means that ka<<1

g. The ka for this reaction is 1,8x10⁻⁵, that means that there are more weak acid molecules (HC₂H₃O₂) than conjugate base ions. Also, the water molecules will be in higher proportion than hydronium ions.

I hope it helps!

HC2H3O2 is a weak acid so it dissociates to a small extent in solution.

When NaC2H3O2 is dissolved in water, the following reaction occurs;

[tex]NaC2H3O2(aq) ---->Na^+(aq) + C2H3O2-(aq)[/tex]

The ion combines with water as follows;

[tex]C2H3O2-(aq) + H2O(l) -----> HC2H3O2(aq) + OH-(aq)[/tex]

When  HC2H3O2 is added to water, the following reaction occurs;

[tex]HC2H3O2(aq) + H2O(l) -----> H3O+(aq) + C2H3O2-(aq)[/tex]

We must note that HC2H3O2 is a weak acid. Weak acids only dissociate to a small extent in water therefore Ka << 1.

Since Ka << 1, it the follows that weak acid molecules and water molecules are present in the greatest concentration in solution because HC2H3O2 only dissociates to a very small extent.

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Answer: The unbalanced chemical equations are written below.

Explanation:

An unbalanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side is not equal to the total number of individual atoms on the product side. These equations does not follow law of conservation of mass.

The chemical equation for the reaction of nitrogen gas and oxygen gas follows:

[tex]N_2+O_2\rightarrow NO_2[/tex]

The product formed is nitrogen dioxide.

The chemical equation for the decomposition of dinitrogen pentaoxide follows:

[tex]N_2O_5\rightarrow N_2O_4+O_2[/tex]

The product formed is dinitrogen tetroxide and oxygen gas.

The chemical equation for the reaction of ozone to oxygen gas follows:

[tex]O_3\rightarrow O_2[/tex]

The product formed is oxygen gas.

The chemical equation for the reaction of chlorine and sodium iodide follows:

[tex]Cl_2+NaI\rightarrow NaCl+I_2[/tex]

The product formed is sodium chloride and iodine gas

The chemical equation for the reaction of magnesium and oxygen gas follows:

[tex]Mg+O_2\rightarrow MgO[/tex]

The product formed is magnesium oxide

Final answer:

Unbalanced equations are provided for the chemical reactions described, including the formation of nitrogen dioxide, the decomposition of dinitrogen pentoxide, the decomposition of ozone, and reactions involving chlorine with sodium iodide and magnesium with oxygen.

Explanation:

The unbalanced chemical equations for the reactions described are:

(a) N2 + O2 → NO2

(b) N2O5 → N2O4 + O2

(c) O3 → O2

(d) Cl2 + NaI → I2 + NaCl

(e) Mg + O2 → MgO

Balancing chemical equations is an essential part of understanding chemical reactions. While the equations provided are unbalanced, they represent the initial step in the process of determining the correct stoichiometry to satisfy the law of conservation of mass in a chemical reaction.

Answer:

The reactions are shown in the explanation

Explanation:

The net bronsted equation will include the acid / base and its dissciated product. (conjugate base or acid and hydrogen or hydroxide ions)

a) Vinegar

[tex]CH{3}COOH+H_{2}O--->CH{3}COO^{-}+H_{3}O^{+}[/tex]

b) bleach

[tex]NaOCl+H_{2}O--->HOCl+Na^{+}+OH^{-}[/tex]

c) ammonia

[tex]NH_{3}+H_{2}O--->NH_{4}^{+}+OH^{-}[/tex]

d)Vitamin C

[tex]H_{2}C_{6}H_{6}O_{6}+H_{2}O--->HC_{6}H_{6}O_{6}^{-}+H_{3}O^{+}[/tex]

e)Citric acid

[tex]H_{3}C_{6}H_{6}O_{7}+H_{2}O--->H_{2}C_{6}H_{6}O_{7}^{-}+H_{3}O^{+}[/tex]

f) Washing soda

[tex]Na_{2}CO_{3} +H_{2}O--->NaHCO_{3}^{-}+OH^{-}[/tex]

The net Brønsted equations that show the acidic or basic nature of the following solutions in water are:

(a) HC₂H₃O₂ + H₂O ⇒ C₂H₃O₂⁻ + H₃O⁺

(b) OCl⁻ + H₂O ⇒ HClO + OH⁻

(c) NH₃ + H₂O ⇒ NH₄⁺ + OH⁻

(d) H₂C₆H₆O₆ + H₂O ⇒ HC₆H₆O₆⁻ + H₃O⁺

(e) H₃C₆H₅O₇ + H₂O ⇒ H₂C₆H₅O₇⁻ + H₃O⁺

(f) CO₃²⁻ + H₂O ⇒ HCO₃⁻ + OH⁻

According to Brønsted-Lowry acid-base theory:

Let's consider the net Brønsted equations for the following species.

HC₂H₃O₂ is an acid according to the following equation.

HC₂H₃O₂ + H₂O ⇒ C₂H₃O₂⁻ + H₃O⁺

NaOCl is a base according to the following equation.

OCl⁻ + H₂O ⇒ HClO + OH⁻

NH₃ is a base according to the following equation.

NH₃ + H₂O ⇒ NH₄⁺ + OH⁻

H₂C₆H₆O₆ is an acid according to the following equation.

H₂C₆H₆O₆ + H₂O ⇒ HC₆H₆O₆⁻ + H₃O⁺

H₃C₆H₅O₇ is an acid according to the following equation.

H₃C₆H₅O₇ + H₂O ⇒ H₂C₆H₅O₇⁻ + H₃O⁺

Na₂CO₃ is a base according to the following equation.

CO₃²⁻ + H₂O ⇒ HCO₃⁻ + OH⁻

The net Brønsted equations that show the acidic or basic nature of the following solutions in water are:

(a) HC₂H₃O₂ + H₂O ⇒ C₂H₃O₂⁻ + H₃O⁺

(b) OCl⁻ + H₂O ⇒ HClO + OH⁻

(c) NH₃ + H₂O ⇒ NH₄⁺ + OH⁻

(d) H₂C₆H₆O₆ + H₂O ⇒ HC₆H₆O₆⁻ + H₃O⁺

(e) H₃C₆H₅O₇ + H₂O ⇒ H₂C₆H₅O₇⁻ + H₃O⁺

(f) CO₃²⁻ + H₂O ⇒ HCO₃⁻ + OH⁻

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Answer:

The correct option is: B.) M^-1 s^-1

Explanation:

According to the rate law, the rate of a given chemical reaction is equal to the product of the reactant concentrations. Where, the exponents a and b are the partial orders for the concentration of X and Y, respectively.

The rate equation for the chemical reaction is:

Rate = k [X]ᵃ[Y]ᵇ

Here, k is the rate constant.

The concentration of X and Y reactants is given in mol.L⁻¹ or M (molarity).

The unit of rate of a chemical reaction is mol.L⁻¹.s⁻¹ or molarity per second M.s⁻¹

Given: rate = k [X][Y]              

In the given rate equation, the exponents of concentrations of reactants X and Y is 1. Thus the overall order of the reaction is 2.

Therefore, it is a second order reaction.

Since the unit of rate is M.s⁻¹ and units of [X] and [Y] is M.

∴ [tex]Rate (M.s^{-1}) = k [X (M)] [Y (M)][/tex]

⇒ [tex]k = \frac{Rate (M.s^{-1})}{[X (M)] [Y (M)]} [/tex]

⇒ the unit of k in the given rate law equation = M⁻¹.s⁻¹

The units of k in the rate law equation rate=k[X][Y] are M^-1 s^-1, which correspond to a second-order reaction. These units arise because the rate constant unit must counteract the reactant concentrations unit, providing the reaction rate (usually M s^-1).

The given reaction rate law is rate=k[X][Y] where k is the rate constant, and [X] and [Y] are the molar concentrations of substances X and Y. The units of k, the rate constant, are determined by the reaction's overall order, which, in this case, is a sum total of the individual reaction orders of X and Y. In this scenario, the reaction is second-order (one for [X] and one for [Y]). Therefore, the units of k would be M-1 s-1 (Molarity inverse second inverse) as given by option B. It is inferred as M-1 s-1 because the rate constant unit must cancel out the units of the reactant concentrations to give the rate of the reaction (which is typically in M s-1).

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Final answer:

In an isothermal reversible expansion, ΔG and ΔA do not differ and can be calculated using the natural log of the ratio of final to initial volumes. During an isothermal expansion against a constant external pressure, while ΔG can vary due to the external work being done, ΔA remains more reflective of internal work under constant conditions.

Explanation:

The question asks for the calculation of ΔG (Gibbs Free Energy change) and ΔA (Helmholtz Free Energy change) for the isothermal expansion of an ideal gas under two scenarios: (a) a reversible path and (b) an expansion against a constant external pressure. Both ΔG and ΔA can be insightful in understanding the spontaneity and energy changes of a chemical process.

(a) Isothermal Reversible Expansion

For an isothermal reversible expansion, ΔG and ΔA do not differ because both depend on the same variables in an ideal condition. Since the temperature is constant and the process is reversible, the free energy changes can be calculated using the formulas:

ΔG = -nRT ln(Vf/Vi) ΔA = -nRT ln(Vf/Vi)

Given: n = 2.50 mol, T = 298 K, Vi = 10.0 L, and Vf = 50.0 L

This leads to the same value for both ΔG and ΔA, showing that the only driving force behind the reversible isothermal expansion is entropy.

(b) Isothermal Expansion Against a Constant External Pressure

When expanding against a constant external pressure of 0.750 bar, the pressure-volume work becomes different. However, ΔG can still be calculated using the formula for isothermal processes in the presence of external pressure, but would yield a different context compared to ΔA which is not typically affected by external pressure in the same straightforward manner.

ΔG is defined by the system's ability to do non-PV work, and in isothermal conditions against constant external pressures, its calculation might deviate, illustrating differences in real-world versus ideal conditions. ΔA would not generally change because it pertains to the work done in a closed system at constant temperature and volume.

Answer:

C.) HOCl Ka=3.5x10^-8

Explanation:

In order to a construct a buffer of pH= 7.0 we need to find the pKa values of all the acids given below

we Know that

pKa= -log(Ka)

therefore

A) pKa of  HClO2 = -log(1.2 x 10^-2)

=1.9208

B) similarly PKa of HF= -log(7.2 x 1 0^-4)= 2.7644

C)  pKa of HOCl= -log(3.5 x 1 0^-8)= 7.45

D) pKa of HCN = -log(4 x 1 0^-10)=  9.3979

If we consider the  Henderson- Hasselbalch equation for the calculation of the pH of the buffer solution

The weak acid for making the buffer must have a pKa value near to the desired pH of the weak acid.

So, near to value, pH=7.0. , the only option is HOCl whose pKa value is 7.45.

Hence, HOCl will be chosen for buffer construction.

Answer:

D

Explanation:

Final answer:

The bonding in metals is best described by the delocalization of valence electrons over all of the atoms in the piece of metal. This creates a strong bond between the positive metal ions and the delocalized electrons, allowing metals to conduct electricity and heat well.

Explanation:

The bonding in metals is best described by option D: The valence electrons of each metal atom are delocalized over all of the atoms in the piece of metal.

In metallic bonding, the positive metal ions are surrounded by a sea of delocalized electrons, which are free to move throughout the metal lattice. The attraction between the positive ions and the delocalized electrons creates a strong bond that holds the metal atoms together. This type of bonding allows metals to conduct electricity and heat well.

For example, in a piece of copper metal, the copper atoms form a crystal lattice structure, with the delocalized electrons moving freely between the atoms. This bonding also gives metals their malleability and ductility, as the mobile electrons allow the metal atoms to slide past each other without breaking the bonds.

With the increase in branching solubility in glycogen. The number of sites for enzyme action is increased through α‑1,6 linkages. The reaction that forms α‑1,6 linkages is catalyzed by a branching enzyme. Therefore, option A, C, and E are correct.

Glycogen can be described as a multibranched polysaccharide of glucose that facilitates a form of energy storage in animals, and bacteria.  In humans, glycogen is composed and stored in the cells of the liver and skeletal muscle.

In the liver, glycogen can make up 5 to 6% of the fresh weight, and the liver of an adult can store roughly 100 to 120 grams of glycogen. Glycogen is found in skeletal muscles in a low concentration.

The amount of glycogen that can be stored in the body, particularly within the muscles and liver. Branches in glycogen are linked to the chains from α-1,6 glycosidic bonds between the first glucose of the new branch and glucose on the stem chain.

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