A 24.201 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 72.053 g of water. A 13.774 g aliquot of this solution is then titrated with 0.1045 M HCl. It required 28.02 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.

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:

The answer to this question has been described in details on the screenshots attached to this question.

Thanks. Hope it helps

The final temperature of the mixed solution is approximately 19.5°C, assuming complete reaction and constant heat capacity of the solution.

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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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The question involves balancing acid-base reactions in chemistry. These are neutralization reactions. The balanced equations are: 1) H2SO4 (aq) + Ca(OH)2 (aq) --> 2H2O (l) + CaSO4(s), 2) HClO4 (aq) + KOH (aq) --> H2O (l) + KClO4(aq), 3) H2SO4 (aq) + 2NaOH (aq) --> 2H2O (l) + Na2SO4(aq).

The original equations are examples of acid-base reactions, specifically neutralization reactions. Here are the balanced equations:

In each of these reactions, the acid (H2SO4, HClO4) reacts with the base (Ca(OH)2, KOH, NaOH) to form water and a salt. The resulting phases of the products are indicated with each product (l for liquid, aq for aqueous, or s for solid).

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The given problems are examples of acid-base neutralization reactions where an acid reacts with a base to form water and a salt. The balanced equations for the reactions are: H2SO4 (aq) + Ca(OH)2 (aq) --> CaSO4 (s) + 2H2O (l), HClO4 (aq) + KOH (aq) --> KClO4 (aq) + H2O (l), and H2SO4 (aq) + 2NaOH (aq) --> Na2SO4 (aq) + 2H2O (l).

The reactions you are asking about are examples of acid-base neutralization reactions. In these reactions, an acid reacts with a base to produce water and a salt.

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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:

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 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.

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:

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.

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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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:

We need 42.4 mL of benzene to produce 1.5 *10³ kJ of heat

Explanation:

Step 1: Data given

Density of benzene = 0.88 g/mL

Molar mass of benzene = 78.11 g/mol

Heat produced = 1.5 * 10³ kJ

Step 2: The balanced equation

2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g)         ΔH°rxn = -6278 kJ

Step 3: Calculate moles of benzen

1.5 * 10³ kJ * (2 mol C6H6 / 6278 kJ) = 0.478 mol C6H6

Step 4: Calculate mass of benzene

Mass benzene : moles benzene * molar mass benzene

Mass benzene= 0.478 * 78.11 g

Mass of benzene = 37.34 grams

Step 5: Calculate volume of benzene

Volume benzene = mass / density

Volume benzene = 37.34 grams / 0.88g/mL

Volume benzene = 42.4 mL

We need 42.4 mL of benzene to produce 1.5 *10³ kJ of heat

The volume of benzene required is 42.5 mL of benzene.

The equation of the reaction is;  2C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) ΔH°rxn = -6278 kJ

If 2 mole of benzene produces -6278 kJ of heat

x moles of benzene produces 1.5 x 103 kJ

x =  2 mole ×  -1.5 x 10^3 kJ/ -6278 kJ

x = 0.48 moles of benzene

The mass of benzene required = number of moles of benzene x molar mass of benzene

Molar mass of benzene = 78 g/mol

Mass of benzene required = 0.48 moles of benzene × 78 g/mol

= 37.44 g

Density of benzene = 0.88 g/mL

But; density = mass/volume

Volume = mass/density

volume = 37.44 g/0.88 g/mL

volume = 42.5 mL of benzene

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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:

1) When 6.97 grams of sodium(s) react with excess water(l), 56.0 kJ of energy are evolved.

2) When 10.4 grams of carbon monoxide(g) react with excess water(l), 1.04 kJ of energy are absorbed.

Explanation:

1) The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g).

2 Na(s) + 2H₂O(l) ⇒ 2NaOH(aq) + H₂(g) ΔH = -369 kJ

The enthalpy of the reaction is negative, which means that 369 kJ of energy are evolved per 2 moles of sodium. The energy evolved for 6.97 g of Na (molar mass 22.98 g/mol) is:

[tex]6.97g.\frac{1mol}{22.98g} .\frac{-369kJ}{2mol} =-56.0kJ[/tex]

2) The following thermochemical equation is for the reaction of carbon monoxide(g) with water(l) to form carbon dioxide(g) and hydrogen(g).

CO(g) + H₂O(l) ⇒ CO₂(g) + H₂(g)  ΔH = 2.80 kJ

The enthalpy of the reaction is positive, which means that 2.80 kJ of energy are absorbed per mole of carbon monoxide. The energy evolved for 10.4 g of CO (molar mass 28.01 g/mol) is:

[tex]10.4g.\frac{1mol}{28.01g} .\frac{2.80kJ}{mol} =1.04kJ[/tex]

To find the energy evolved in each reaction, use stoichiometry and the molar mass of the reactant to calculate the moles reacted and then use the stoichiometric ratio between the reactant and energy.

For the first reaction, we are given the thermochemical equation 2Na(s) + 2H2O(l)2NaOH(aq) + H2(g) with ΔH = -369 kJ. To find the energy evolved when 6.97 grams of sodium react with excess water, we need to use stoichiometry and the molar mass of sodium to calculate the moles of sodium reacted. Then, we can use the stoichiometric ratio between sodium and energy to find the energy evolved.

For the second reaction, we are given the thermochemical equation CO(g) + H2O(l)CO2(g) + H2(g) with ΔH = 2.80 kJ. To find the energy evolved when 10.4 grams of carbon monoxide react with excess water, we need to use stoichiometry and the molar mass of carbon monoxide to calculate the moles of carbon monoxide reacted. Then, we can use the stoichiometric ratio between carbon monoxide and energy to find the energy evolved.

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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:

A mole is 6.02 X 1023 of any particles and the SI unit for measuring the quantity of a substance

Explanation:

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:

69.8 kilo Pasacl is the pressure of the hydrogen gas.

Explanation:

[tex]Mg+2HCl\rightarrow MgCl_2+H_2[/tex]

Pressure at which hydrogen gas collected = p = 101.2 kilo Pascals

Vapor pressure water = [tex]p^o[/tex] = 31.4 kilo Pascals

The pressure of hydrogen gas = P

The pressure at which gas was collected was sum of vapor pressure of water and hydrogen gas.

[tex]p=P+p^o[/tex]

[tex]P =p-p^o=101.2 kPa-31.4 kPa=69.8 kPa[/tex]

69.8 kilo Pasacl is the pressure of the hydrogen gas.

The pressure of the hydrogen gas is 69.8 kPa.

The pressure of the hydrogen gas can be calculated by subtracting the vapor pressure of water from the total pressure. In this case, the total pressure is 101.2 kPa and the vapor pressure of water is 31.4 kPa. So, the pressure of the hydrogen gas is 101.2 kPa - 31.4 kPa = 69.8 kPa.

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Answer:

[tex]Energy=1\times 10^{15}h\ J[/tex]

Explanation:

Considering:-

[tex]E=\frac {h\times c}{\lambda}[/tex]

Where,  

h is Planck's constant having value [tex]6.626\times 10^{-34}\ Js[/tex]

c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]

[tex]\lambda[/tex] is the wavelength of the light

Given, Wavelength = 300 nm

Also, 1 m = [tex]10^{-9}[/tex] nm

So,  

Wavelength = [tex]300\times 10^{-9}[/tex] m

Applying in the equation as:-

[tex]E=\frac {h\times c}{\lambda}[/tex]

[tex]E=\frac{h\times 3\times 10^8}{300\times 10^{-9}}\ J[/tex]

[tex]Energy=1\times 10^{15}h\ J[/tex]

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:

The Ka for this acid is 1.4 * 10^-4 (option C)

Explanation:

Step 1: Data given

In a 0.20M aqueous solution, lactic acid is 2.6% dissociated.

Step 2: The equation for the dissociation of HAc is:

HAc ⇌ H+ + Ac¯

The Ka expression is:

Ka = ([H+] [Ac¯]) / [HAc]

1) [H+] using the concentration and the percent dissociation:

(0.026) (0.2) = 0.0052 M

2) Calculate [Ac-] and [H+]

[Ac¯] = [H+] = x = 0.0052 M

3) Calculate [HAc]

[HAc] = 0.20M - x ( since x << 0.20, we can assume [HAc]  = 0.20 M

4) Calculate Ka

Ka= [( 0.0052) ( 0.0052)] / 0.20

Ka = 0.0001352 = 1.4 * 10^-4

The Ka for this acid is 1.4 * 10^-4

Final answer:

The value of Ka for lactic acid, given that a 0.20 M solution is 2.6% dissociated, can be calculated using the concentration of dissociated ions; the closest value to our calculation is 1.4x10^-4.

Explanation:

When you exercise, the burning sensation in your muscles is often due to the buildup of lactic acid. To find the acid dissociation constant (Ka) for lactic acid given a 0.20 M solution that is 2.6% dissociated, we use the dissociation equation of a weak acid:

HC₃H₅O₃ → H⁺ + C₃H₅O₃⁻

The concentration of H⁺ and C₃H₅O₃⁻ will also be 0.0052 M, since it's a one-to-one dissociation. Thus, the expression for Ka becomes:

The closest answer to our calculated Ka value is C.) 1.4x10^-4.

The Brønsted acid-base equation for the reaction between vinegar (acetic acid, HC2H3O2) and bleach (sodium hypochlorite, NaClO) is HC2H3O2 + NaClO → C2H3O2^- + HClO, where HC2H3O2 is the acid and NaClO is the base.

The acid-base reaction between vinegar (which contains acetic acid, HC2H3O2) and bleach (which contains sodium hypochlorite, NaClO) can be represented as a Brønsted acid-base equation:

HC2H3O2 + NaClO → C2H3O2^- + HClO

In this reaction, HC2H3O2 is the acid as it donates a proton (H^+) and NaClO is the base as it accepts a proton (H+). Remember, spectator ions, such as Na^+ from sodium hypochlorite, have been omitted because they don't participate in the reaction.

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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

Forming dimers, like acetic acid's molecules under specific conditions, would double the experimental molar mass as the measurements would account for the mass of the dimer, which is twice that of the monomer.

When molecules form dimers, such as acetic acid (CH3COOH) under certain conditions, it can significantly impact the determination of the molar mass experimentally. Typically, the molar mass is determined by measuring the mass of a certain number of moles of a substance. However, if the molecules are forming dimers, the actual molar mass calculated will be double the expected molar mass of the monomer because the experiments would be measuring the mass of the dimer (A2) rather than the individual monomer (A).

In other words, the experimental molar mass would be higher than it should be if one assumes that no dimers are present. This is because the mass being measured corresponds to the molar mass of the dimer rather than that of the monomer. When dealing with substances that can form dimers, this possibility must be taken into account to ensure accurate molar mass determination.

The formation of dimers in a liquid affects the experimental molar mass by potentially doubling the value if the molecular associations are not accounted for, leading to a measured molar mass that reflects the dimer rather than the monomer.

When molecules in a liquid form dimers due to strong intermolecular attractions, such as hydrogen bonding, the observed experimental molar mass would be affected. If the tendency to form dimers is not accounted for in molar mass determination, the measured molar mass would be roughly double that of the monomeric form because the dimer is made up of two monomer units. For example, in acetic acid, when dimers are formed, the molar mass derived from experimental measurements would reflect the mass of the dimer, rather than the monomer.

In cases where dissociation occurs, like A2 dissociating into two A molecules in solution, monitoring the concentration of both the dimer and monomer can be critical. Incorrect assumptions about the degree of dimerization can lead to inaccurate molar mass calculations.

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: A

Explanation:

Please check the attachment for the answer.

Answer:

a. Isopropyl pentanoate.

Explanation:

Hello,

In this case, such esterification is:

[tex]CH_3CH_2CH_2CH_2COOH+OHCH(CH_3)_2\rightarrow CH_3CH_2CH_2CH_2COOCH(CH_3)_2+H_2O[/tex]

Wherein the first reactant is the pentanoic acid, the second one the isopropanol and the first product is the result, which is a. isopropyl pentanoate. This is because the isopropyl get separated from the hydroxile and bonds with the carboxile group of the pentanoic acid by also releasing water as the byproduct.

Best regards.

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:

(A) endothermic

(A) Yes, absorbed

Explanation:

Let's consider the following thermochemical equation.

2 Fe₂O₃(s) ⇒ 4 FeO(s) + O₂(g)  ΔH = 560 kJ

Since ΔH > 0, the reaction is endothermic.

We can establish the following relations:

Suppose 66.6 g of Fe₂O₃ react. The heat absorbed is:

[tex]66.6g.\frac{1mol}{160g} .\frac{560kJ}{2mol} =117kJ[/tex]

Final answer:

The reaction 2Fe2O3(s) to 4FeO(s) + O2(g) is endothermic with a \\Delta H of +560 kJ, meaning it absorbs heat. For 66.6 g of Fe2O3 react, heat will be absorbed.

Explanation:

When assessing a chemical reaction's energy change, the sign of \\Delta H\ indicates whether the reaction is exothermic or endothermic. For the reaction given, 2Fe2O3(s) \\rightarrow\ 4FeO(s) + O2(g), the \\Delta H\ is 560 kJ. This positive value means that the reaction is endothermic, as it absorbs heat from the surroundings.

For the specific amount of 66.6 g of Fe2O3, since the reaction is endothermic, energy in the form of heat will indeed be absorbed. Therefore, heat will be absorbed when 66.6 g of Fe2O3 react.

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.

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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