Answer:
[tex]\Delta G^{\circ}=-15902 J/mol[/tex]
Explanation:
In this problem we only have information of the equilibrium, so we need to find a expression of the free energy in function of the constant of equilireium (Keq):
[tex]\Delta G^{\circ}=-R*T*ln(K_{eq})[/tex]
Being Keq:
[tex]K_{eq}=\frac{[fructose][Pi]}{[Fructose-1-P]}[/tex]
Initial conditions:
[tex][Fructose-1-P]=0.2M[/tex]
[tex][Fructose]=0M[/tex]
[tex][Pi]=0M[/tex]
Equilibrium conditions:
[tex][Fructose-1-P]=6.52*10^{-5}M[/tex]
[tex][Fructose]=0.2M-6.52*10^{-5}M[/tex]
[tex][Pi]=0.2M-6.52*10^{-5}M[/tex]
[tex]K_{eq}=\frac{(0.2M-6.52*10^{-5}M)*(0.2M-6.52*10^{-5}M)}{6.52*10^{-5}M}[/tex]
[tex]K_{eq}=613.1[/tex]
Free-energy for T=298K (standard):
[tex]\Delta G^{\circ}=-8.314\frac{J}{mol*K}*298K*ln(613.1)[/tex]
[tex]\Delta G^{\circ}=-15902 J/mol[/tex]
A quantity of 200 mL of 0.862 M HCl (aq) is mixed with 200 mL of 0.431
M Ba(OH)2 (aq) in a constant-pressure calorimeter of negligible heat
capacity. The initial temperature of the HCl and Ba(OH)2 solutions is the
same at 20.5
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.
To calculate the final temperature of the mixed solution when 200 mL of 0.862 M HCl (aq) is mixed with 200 mL of 0.431 M Ba(OH)2 (aq), you can use the principle of heat transfer, specifically the equation: q = m * C * ΔTWhere: q is the heat transferred (in joules). m is the mass of the solution (in grams). C is the specific heat capacity of the solution (in J/g°C). ΔT is the change in temperature (in °C).Since the solutions are initially at the same temperature (20.5°C), the heat gained by one solution equals the heat lost by the other: q(HCl) = -q(Ba(OH)2)Now, calculate the heat transfer for each solution: q(HCl) = (200 mL * 0.862 mol/L * 36.46 g/mol) * C * ΔT q(Ba(OH)2) = -(200 mL * 0.431 mol/L * 171.34 g/mol) * C * ΔTSince the specific heat capacity (C) of the mixed solution is assumed to be constant, you can equate the two expressions for heat transfer: (200 mL * 0.862 mol/L * 36.46 g/mol) * ΔT = -(200 mL * 0.431 mol/L * 171.34 g/mol) * ΔTNow, solve for ΔT: ΔT = -(200 mL * 0.431 mol/L * 171.34 g/mol) / (200 mL * 0.862 mol/L * 36.46 g/mol) ΔT ≈ -1.00°CSince the negative sign indicates that the temperature of the solution decreases, the final temperature is 20.5°C - 1.00°C ≈ 19.5°C.For more such questions on temperature
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Suppose that coal of density 1.5 g/cm^3 is pure carbon. (It is, in fact, much more complicated, but this is a reasonable first approximation.) The combustion of carbon is described by the equation:
C(s) + O2(g) -----> CO2(g) ∆H° = −394 kJ
What is the value of q (heat) when a lump of coal of size 5.6 cm x 5.1 cm x 4.6 cm is burned?
Answer in units of kJ.
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:
Calculate volume of the coal: Volume = 5.6 cm × 5.1 cm × 4.6 cm = 131.112 cm³.Calculate mass of the coal: Mass = Volume × Density = 131.112 cm³ × 1.5 g/cm³ = 196.668 g.Convert mass to moles: Moles of carbon = Mass / Molar mass of carbon = 196.668 g / 12.01 g/mol ≈ 16.375 moles of C.Calculate the energy released: q = Moles × Energy per mole = 16.375 moles × (-394 kJ/mol) ≈ -6451.75 kJ.Therefore, the value of q when the lump of coal is burned is approximately -6451.75 kJ.
Some liquids have enough attractions between molecules to form dimers. (Dimers are molecules formed from the combination of identical molecules, A+ A → A2.) What effect would this have on the experimental molar mass?
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.
In a particular electroplating process, the metal being plated has a +4 charge. If 861.8 C of charge pass through the cell, how many moles of metal should be plated? Useful information: F = 96,500 C/mol e- Provide your response to four digits after the decimal.
Answer: [tex]2.2326\times 10^{-3}[/tex] moles
Explanation:
We are given:
Moles of electron = 1 mole
According to mole concept:
1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of particles.
We know that:
Charge on 1 electron = [tex]1.6\times 10^{-19}C[/tex]
Charge on 1 mole of electrons = [tex]1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C[/tex]
The metal being plated has a +4 charge, thus the equation will be:
[tex]M^{4+}+4e^-\rightarrow M[/tex]
[tex]4\times 96500C[/tex] of electricity deposits = 1 mole of metal
Thus 861.8 C of electricity deposits =[tex]\frac{1}{4\times 96500}\times 861.8=2.2326\times 10^{-3}[/tex] moles of metal
Thus [tex]2.2326\times 10^{-3}[/tex] moles of metal should be plated
The number of moles of the metal deposited is 0.0022 moles of metal.
What is electroplating?The term electroplating refers to the use of one metal to cover the surface of another metal. Let the metal in question be M, the equation of the reaction is; M^4+(aq) + 4e -----> M(s).
1 mole of the metal is deposited by 4( 96,500) C
x moles will be deposited by 861.8 C
x = 1 mole * 861.8 C/ 4( 96,500) C
x = 0.0022 moles of metal
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Complete and balance each of the following equations for acid-base reactions. Express your answer as a chemical equation. Identify all of the phases in your answer.
1) H2SO4 (aq) + Ca(OH)2 (aq) -->
2) HClO4 (aq) + KOH (aq) -->
3) H2SO4 (aq) + NaOH (aq) -->
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).
Explanation:The original equations are examples of acid-base reactions, specifically neutralization reactions. Here are the balanced equations:
H2SO4 (aq) + Ca(OH)2 (aq) -->> 2H2O (l) + CaSO4(s)HClO4 (aq) + KOH (aq) -->> H2O (l) + KClO4(aq)H2SO4 (aq) + 2NaOH (aq) -->> 2H2O (l) + Na2SO4(aq)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).
Explanation: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.
The reaction H2SO4 (aq) + Ca(OH)2 (aq) produces CaSO4 (s) + 2H2O (l), balanced as H2SO4 (aq) + Ca(OH)2 (aq) --> CaSO4 (s) + 2H2O (l).The reaction HClO4 (aq) + KOH (aq) yields KClO4 (aq) + H2O (l), expressed as: HClO4 (aq) + KOH (aq) --> KClO4 (aq) + H2O (l).For H2SO4 (aq) + NaOH (aq), it also produces a salt and water: Na2SO4 (aq) + 2H2O (l), which you balance as: H2SO4 (aq) + 2NaOH (aq) --> Na2SO4 (aq) + 2H2O (l).Learn more about Acid-Base Reactions here:https://brainly.com/question/15209937
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A mole is 6.02 X 1023 ____________ particles and the SI unit for measuring the ________ of a substance.
Answer:
A mole is 6.02 X 1023 of any particles and the SI unit for measuring the quantity of a substance
Explanation:
The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g). 2Na(s) + 2H2O(l)2NaOH(aq) + H2(g) H = -369 kJ When 6.97 grams of sodium(s) react with excess water(l), kJ of energy are .
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)Arrow.gif CO2(g) + H2(g) delta16-1.GIFH = 2.80 kJ
When 10.4 grams of carbon monoxide(g) react with excess water(l), kJ of energy are _________evolvedabsorbed.
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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The density of lead is 1.13 ✕ 104 kg/m3 at 20.0°C. Find its density (in kg/m3) at 143°C. (Use α = 29 ✕ 10−6 (°C)−1 for the coefficient of linear expansion. Give your answer to at least four significant figures.)
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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Fill in the blank.
Fatty acids _______.
A)can be branched or unbranched.
B) are referred to as "saturated" if they have one or more C–C double bonds.
C) are highly oxidized storage forms of hydrocarbons.
D) by definition, never exceed 20 carbons in length.
Triacylglycerols _______.
A) always have at least one fatty acid with a trans double bond.
B)consist of glycerol and three different fatty acids.
C) are a major energy storage form.
D) consist of glycerol and three identical fatty acids.
Mass spectrometry _______.
A)can be used to identify individual lipids in complex mixtures.
B)cannot be used with lipids other than fatty acids.
C)can determine the mass but not the identity of a lipid.
D) cannot determine the locations of double bonds in a fatty acid.
Which statement is true?
A) Prostaglandins are found only in the prostate gland.
B) Vitamin E is a hormone precursor.
C) Vitamin D regulates milk production in mammary glands.
D) Membrane sphingolipids can be used to produce intracellular messengers.
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
For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 bar for all species. For the reaction
2 NO ( g ) + O 2 ( g ) -------> 2 NO 2 ( g )
the standard change in Gibbs free energy is Δ G ° = − 32.8 kJ / mol . What is Δ G for this reaction at 298 K when the partial pressures are:
PNO = 0.500 bar , PO2 = 0.250 bar , and PNO 2 = 0.800 bar
DeltaG = ?
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
Identify the true statements regarding α‑1,6 linkages in glycogen.
(A) Branching increases glycogen solubility.
(B) New α‑1,6 linkages can only form if the branch has a free reducing end.
(C) The number of sites for enzyme action on a glycogen molecule is increased through α‑1,6 linkages.
(D) Exactly seven residues extend from these linkages.
(E) The reaction that forms α‑1,6 linkages is catalyzed by a branching enzym
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.
What is glycogen?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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Name the ester that is formed when pentanoic acid reacts with isopropanol.a.isopropyl pentanoateb.isopropyl pentanec.pentanoate isopropyld.isopropyl pentanoic acid
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.
What is the energy of a photon (in units of h) that has a wavelength of 300 nm?
(c = 3.00 x 108 m/s)
1 x 10^17 h
1 x 10^15 h
1 x 10^6 h
1 x 10^-2 h
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]
Match the chromatography term with its definition. The volume of solvent traveling through the column per unit time. The elapsed time between sample injection and detection. The time required by a retained solute to travel through the column beyond the time required by the unretained solvent. The distance traveled by the solvent per unit time. Describes the amount of time that a sample spends in the stationary phase relative to the mobile phase. It is sometimes also called the capacity factor or capacity ratio. Volume of the mobile phase required to elute a solute from the column. Ratio of the adjusted retention times or retention factors of two solutes. It is sometimes also called the separation factor. The ratio of the solute concentrations in the mobile and stationary phases.
Answer:
The statements are definitions to chromatography terms which have been highlighted below.
Explanation:
Match the chromatography term with its definition.
Volumetric Flow Rate = The volume of solvent traveling through the column per unit time.
Retention time = The elapsed time between sample injection and detection.
Adjusted Retention Time = The time required by a retained solute to travel through the column beyond the time required by the un -retained solvent.
Linear Flow Rate = The distance traveled by the solvent per unit time.
Retention factor = Describes the amount of time that a sample spends in the stationary phase relative to the mobile phase. It is sometimes also called the capacity factor or capacity ratio.
Relative Volume = Volume of the mobile phase required to elute a solute from the column.
Relative Retention = Ratio of the adjusted retention times or retention factors of two solutes. It is sometimes also called the separation factor.
Partition coefficient = The ratio of the solute concentrations in the mobile and stationary phases.
Chromatography is a laboratory technique used to separate and analyze the components of a mixture based on their different affinities for a stationary phase and a mobile phase. The different terms in chromatography can be defined as given below-
The volume of solvent traveling through the column per unit time - Flow rate
The elapsed time between sample injection and detection - Retention time
The time required by a retained solute to travel through the column beyond the time required by the unretained solvent - Adjusted retention time
The distance traveled by the solvent per unit time - Velocity
Describes the amount of time that a sample spends in the stationary phase relative to the mobile phase. It is sometimes also called the capacity factor or capacity ratio - Retention factor
Volume of the mobile phase required to elute a solute from the column - Elution volume
Ratio of the adjusted retention times or retention factors of two solutes. It is sometimes also called the separation factor - Selectivity factor
The ratio of the solute concentrations in the mobile and stationary phases - Distribution coefficient
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What volume of benzene (C6H6, d= 0.88 g/mL, molar mass = 78.11 g/mol) is required to produce 1.5 x 103 kJ of heat according to the following reaction?2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) ΔH°rxn = -6278 kJ
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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When you exercise, the burning sensation that sometimes occurs in your muscles represents the buildup of lactic acid (HC3H5O3). In a .20M aqueous solution, lactic acid is 2.6% dissociated. What is the value of Ka for this acid?
A.) 4.3x10^-6
B.) 8.3x10^-5
C.) 1.4x10^-4
D.) 5.2x10^-3
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⁻
Initial concentration of lactic acid = 0.20 MPercent dissociation = 2.6%Dissociation of lactic acid (α) = 2.6% of 0.20 M = 0.0026 × 0.20 = 0.0052 MThe 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:
Ka = [H⁺][C₃H₅O₃⁻] / [HC₃H₅O₃]The closest answer to our calculated Ka value is C.) 1.4x10^-4.
Calculate the standard molar enthalpy of formation, in kJ/mol, of NO(g) from the following data:
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]
Calculate the enthalpy of the reaction
4B(s)+3O2(g)→2B2O3(s)
given the following pertinent information:
B2O3(s)+3H2O(g)→3O2(g)+B2H6(g), ΔH∘A=+2035 kJ
2B(s)+3H2(g)→B2H6(g), ΔH∘B=+36 kJ
H2(g)+12O2(g)→H2O(l), ΔH∘C=−285 kJ
H2O(l)→H2O(g), ΔH∘D=+44 kJ
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
During the lab, you measured the pH of mixtures containing strong acid and strong base. Write net Brønsted equations that show the acid-base reactions of common household items. For polyprotic species (such as vitamin C, lemon juice, and washing soda), please show only one proton transfer. Remember that spectator ions are not included. (Use the lowest possible coefficients. Omit states-of-matter in your answer.)
(a) bleach and vinegar (sodium hypochlorite and HC2H3O2)
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.
Explanation: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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If you were to overshoot the endpoint by 1 drop while you were standardizing the NaOH solution, what would be your % error? Assume the actual volume is 30.00 mL and there are exactly 20 drops in 1.00 ml for the sake of this calculation.
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]
Last week you reacted magnesium with a hydrochloric acid aqueous solution and hydrogen gas was produced. Let's say that you collected the gas given off by the reaction and measured it's pressure as 101.2 kPa. If the vapor pressure of water is 31.4 kPa at this temperature, then what is the pressure of the hydrogen gas?
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.
Explanation: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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Be sure to answer all parts. Write an unbalanced equation to represent each of the following reactions: Do not include phase abbreviations. (a) Nitrogen and oxygen react to form nitrogen dioxide. (b) Dinitrogen pentoxide reacts to form dinitrogen tetroxide and oxygen. (c) Ozone reacts to form oxygen. (d) Chlorine and sodium iodide react to form iodine and sodium chloride. (e) Magnesium and oxygen react to form magnesium oxide.
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.
For a: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.
For b: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.
For c: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.
For d: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
For e: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.
Which of the following best describes the bonding in metals?
A. Electrons are transferred from one atom to another, producing charges which then interact with each other.
B. Adjacent atoms share pairs of electrons.
C. The atomic nuclei fuse to produce a continuous malleable array.
D. The valence electrons of a each metal atom are delocalized over all of the atoms in the piece of metal.
E. The protons, neutrons and electrons are pooled to form a homogeneous mass.
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.
The Lineweaver–Burk plot, which illustrates the reciprocal of the reaction rate ( 1 / v ) versus the reciprocal of the substrate concentration ( 1 / [ S ] ) , is a graphical representation of enzyme kinetics. This plot is typically used to determine the maximum rate, V max , and the Michaelis constant, K m , which can be gleaned from the intercepts and slope. Identify each intercept and the slope in terms of the constants V max and K m .
Answer:
x-intercept = [tex]\frac{-1}{Km}[/tex]y-intercept = [tex]\frac{1}{Vmax}[/tex]Slope = [tex]\frac{Km}{Vm}[/tex]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.
Explanation: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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You wish to construct a buffer of pH=7.0. Which of the following weak acids (w/ corresponding conjugate base) would you select?
A.) HClO2 Ka=1.2x10^-2
B.) HF Ka=7.2x10^-4
C.) HOCl Ka=3.5x10-8
D.) HCN Ka=4.0x10^-10
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.
Americium-241 is used in smoke detectors. It has a first-order rate constant for radioactive decay of k = 1.6x10^−3yr−1. By contrast, iodine-125, which is used to test for thyroid functioning, has a rate constant for radioactive decay of k = 0.011 day−1.
(a) What are the half-lives of these two isotopes?
(b) Which one decays at a faster rate?
(c) How much of a 1.00-mg sample of each isotope remains after three half-lives?
(d) How much of a 1.00-mg sample of each isotope remains after 4 days?
Answer:
A. 433 years and 63 days
B. Iodine
C. 0.125mg
D. 1.00mg
Explanation:
A. Americium
The formula of a radioactive decay constant half life is t = 0.693/k
Where k is the decay constant.
For americium, k = 0.0016
t = 0.693/0.0016 = 433.125 apprx 433 years
For iodine, k = 0.011
Half life = 0.693/0.011 = 63 days.
B. Iodine decays at a faster rate.
C. After three half lives
For both, first half life yields a mass of 0.5mg, next yields 0.25mg, next yield 0.125mg
D. 1.00mg still remains.
Half life is high for both so the decay after one day is insignificant.
a) The half-lives of Americium-241 and Iodine-125 are 433 years and 63 days, respectively.
b) Iodine-125 decays faster.
c) After three half-lives, 0.125 mg remains.
d) After 4 days, nearly 1.00 mg of Americium-241 and 0.957 mg of Iodine-125 remain.
To answer the given questions regarding the radioactive decay of Americium-241 and Iodine-125:
(a) Half-lives
For a first-order reaction, the half-life (t1/2) can be calculated using the equation:
t1/2 = 0.693 / k
For Americium-241:
k = 1.6 × 10⁻³ yr⁻¹
t1/2 = 0.693 / (1.6 × 10⁻³) = 433 years
For Iodine-125:
k = 0.011 day⁻¹
t1/2 = 0.693 / 0.011 = 63 days
(b) Decay Rate
Since the half-life of Iodine-125 is shorter (63 days) compared to Americium-241 (433 years), Iodine-125 decays at a faster rate.
(c) Remaining Sample After Three Half-lives
After three half-lives, the remaining amount of a radioactive isotope can be calculated using:
Remaining Amount = Initial Amount / (23)
Remaining Amount = 1.00 mg / 8 = 0.125 mg
Therefore, for both isotopes, 0.125 mg remains after three half-lives.
(d) Remaining Sample After 4 Days
For Americium-241 (with t1/2 about 433 years), 4 days is negligible, so nearly the entire 1.00 mg remains.
For Iodine-125, using the first-order decay formula:
N = N₀ e-kt
t = 4 days, k = 0.011 day⁻¹
N = 1.00 mg × e-(0.011 × 4) ≈ 1.00 mg × e-0.044 ≈ 1.00 mg × 0.957 = 0.957 mg
Thus, after 4 days, 0.957 mg of Iodine-125 remains.
A single bond is almost always a sigma bond, and a double bond is almost always made up of a sigma bond and a pi bond. There are very few exceptions to this rule. Which of the following species have violated this generalization?
A. N2
B. O2
C. F2
D. C2
E. B2.
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.
Explanation: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.
Consider the formation of [Ni(en)3]2+ from [Ni(H2O)6]2+. The stepwise ΔG∘ values at 298 K are ΔG∘1 for first step=−42.9 kJ⋅mol−1 ΔG∘2 for second step=−35.8 kJ⋅mol−1 ΔG∘3 for third step=−24.3 kJ⋅mol−1 Calculate the overall formation constant (Kf) for the complex [Ni(en)3]2+.
Answer:
kf = 1.16 x 10¹⁸
Explanation:
Step 1: [Ni(H₂O)₆]²⁺ + 1en → [Ni(H₂O)₄(en)]²⁺ ΔG°1 = -42.9 kJmol⁻¹
Step 2: [Ni(H₂O)₄(en)]²⁺ + 1en → [Ni(H₂O)₂(en)₂]²⁺ ΔG°2 = -35.8 kJmol⁻¹
Step 3: [Ni(H₂O)₂(en)₂]²⁺ + 1en → [Ni(en)₃]²⁺ ΔG°3 = -24.3 kJmol⁻¹
________________________________________________________
Overall reaction: [Ni(H₂O)₆]²⁺ + 3en → [Ni(en)₃]²⁺ ΔG°r
ΔG°r = ΔG°1 + ΔG°2 + ΔG°3
ΔG°r = -42.9 - 35.8 - 24.3
ΔG°r = -103.0 kJmol⁻¹
ΔG°r = -RTlnKf
-103,000 Jmol⁻¹ = - 8.31 J.K⁻¹mol⁻¹ x 298 K x lnKf
kf = e ^(-103,000/-8.31x298)
kf = e ^41.59
kf = 1.16 x 10¹⁸
A chemist measures the energy change ?H during the following reaction:
2Fe2O3(s) ? 4FeO(s) + O2(g) =?H560.kJ
This reaction is.
(A) endothermic
(B) exothermic
Suppose 66.6 g of Fe2O3 react. Will any heat be released or absorbed?|
(A) Yes,absorbed
(B) Yes released.
(C) No
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:
560 kJ are absorbed when 2 moles of Fe₂O₃ react.The molar mass of Fe₂O₃ is 160 g/mol.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.
Write net Brønsted equations that show the acidic or basic nature of the following solutions in water. For polyprotic species (vitamin C, lemon juice, and washing soda), show only one proton transfer. Remember that spectator ions are not included. (Use the lowest possible coefficients. Omit states-of-matter in your answer.)
(a) vinegar (acetic acid, HC2H3O2)
(b) bleach (sodium hypochlorite, NaOCl)
(c) ammonia (NH3)
(d) Vitamin C (ascorbic acid, H2C6H6O6)
(e) lemon juice (citric acid, H3C6H5O7)
(f) washing soda (sodium carbonate, Na2CO3)
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:
An acid is a species that donates H⁺.A base is a species that accepts H⁺.Let's consider the net Brønsted equations for the following species.
(a) vinegar (acetic acid, HC₂H₃O₂)HC₂H₃O₂ is an acid according to the following equation.
HC₂H₃O₂ + H₂O ⇒ C₂H₃O₂⁻ + H₃O⁺
(b) bleach (sodium hypochlorite, NaOCl)
NaOCl is a base according to the following equation.
OCl⁻ + H₂O ⇒ HClO + OH⁻
(c) ammonia (NH₃)
NH₃ is a base according to the following equation.
NH₃ + H₂O ⇒ NH₄⁺ + OH⁻
(d) Vitamin C (ascorbic acid, H₂C₆H₆O₆)
H₂C₆H₆O₆ is an acid according to the following equation.
H₂C₆H₆O₆ + H₂O ⇒ HC₆H₆O₆⁻ + H₃O⁺
(e) lemon juice (citric acid, H₃C₆H₅O₇)
H₃C₆H₅O₇ is an acid according to the following equation.
H₃C₆H₅O₇ + H₂O ⇒ H₂C₆H₅O₇⁻ + H₃O⁺
(f) washing soda (sodium carbonate, Na₂CO₃)
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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