Answer:
2nd order reaction
Explanation:
Let us assume the reaction to be:
R → P
Where R is the reactant and P is the product.
So here, say initially we have "a" amount of reactant.
R → P
At t=0: a 0 (initial condition)
At t=t: a-x x
Say x be the amount of reactant which forms the product in time t.
So from the rate law, we have
rate of decomposition = k (R)ⁿ
Where k is rate constant , R is amount of reactant at time t and n is the order of the reaction
From the question, at the instant when 5% and 20% have reacted, we will be left with 95% and 80% of the reactant respectively. So writing the rate law equation:
1.07 = k ( 95a / 100)ⁿ
0.76 = k ( 80a/100)ⁿ
Dividing these two equations, we get:
(1.07 / 0.76 ) = ( 95/80 )ⁿ
Taking logarithm on both sides we get
n = ( ㏒ (1.07 / 0.76) ) ÷ ㏒(95/80) = 2.0067 ≈ 2
Therefore the reaction is of order 2
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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The solubility of gold (III) chloride is 1.00x10^-4 g/L. What is the Ksp for AuCl3 (Molar Mass=303.3 g/mol)
A.) 1.00x10^-16
B.) 2.7x10^-15
C.) 1.2x10^-26
D.) 3.2x10^-25
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.
Explanation: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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(b) Once the ionic solid has dissolved, the anion that is formed is able to react as a base, with water as the acid. Write the net acid-base reaction that occurs when dissolved NaC2H3O2 reacts with water. (Use the lowest possible coefficients. Omit states-of-matter in your answer.)
(e) Write the net acid-base reaction that occurs when HC2H3O2 is added to water.
chemPadHelp
HC2H3O2 + H2O → H3O+ + C2H3O2-
Correct.
(f) What is the relative Ka value for this reaction?
Ka << 1
Ka ≈ 1
Ka >> 1Correct: Your answer is correct.
(g) Which species are present in the greatest concentration? (Select all that apply.) --I need b and g
weak acid molecules
hydronium ions
water molecules
conjugate base ions
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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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]
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.
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.
What is the pH of a 0.0042 M hydrochloric acid solution?
A) 5.47
B) 11.62
C) 7.00
D) 8.53
E) 2.38
Answer:
E) 2.38
Explanation:
The pH of any solution , helps to determine the acidic strength of the solution ,
i.e. ,
Lower the value of pH , higher is its acidic strengthand ,
Higher the value of pH , lower is its acidic strength .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.
Saponification is a chemical reaction _____. A. due to the presence of a strong organic acid. B. that results in an alcohol and a carboxylic acid. C. that explains why soap does not wash off easily in hard water. D. that always results directly in a soap.
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:
The manufacturer specs for a particular balloon indicate the maximum inflated volume is 3.00 L. The balloon is filled with 2.50 L of helium at sea level (assume P= 1.00 atm) and released. When the balloon rises to a higher altitude where the pressure is 450.0 mm Hg, will the balloon burst? Show a calculation to support your answer. Assume constant temperature.
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.
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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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
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.
For the reaction shown here
8H2S(g) ---> <--- 8H2(g) + S8(g)
the equlibrium concentrations were found to be
[H2S] = 0.250 M
[H2] = 0.400 M
[S8] = 0.750 M
What is the equilibrium constant for this reaction?
Kc = ???
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
Equilibrium constant (K꜀) =?[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]
K꜀ = 32.2Therefore, 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.
Explanation: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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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
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
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.
Combustion of hydrocarbons such as butane () produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's atmosphere can trap the Sun's heat, raising the average temperature of the Earth. For this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide. Write a balanced chemical equation, including physical state symbols, for the combustion of gaseous butane into gaseous carbon dioxide and gaseous water. 2. Suppose 0.360 kg of butane are burned in air at a pressure of exactly 1 atm and a temperature of 20.0 °C. Calculate the volume of carbon dioxide gas that is produced Be sure your answer has the correct number of significant digits.
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.
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.
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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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.
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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A sample containing 2.50 mol of an ideal gas at 298 K is expanded from an initial volume of 10.0 L to a final volume of 50.0 L. Calculate ΔG and ΔA for this process for:
(a) an isothermal reversible path and (b) an isothermal expansion against a constant external pressure of 0.750 bar. Explain why ΔG and ΔA do or do not differ from one another.
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.
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]
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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The synthesis of methanol from carbon monoxide and hydrogen gas is described by the following chemical equation:
CO(g)+2H2(g)⇌CH3OH(g)
The equilibrium constant for this reaction at 25 ∘C is Kc=2.3×104. In this tutorial, you will use the equilibrium-constant expression to find the concentration of methanol at equilibrium, given the concentration of the reactants.
Determine the expression for the equilibrium constant, Kc, for the reaction by identifying which terms will be in the numerator and denominator:
The equilibrium-constant expression is a mathematical equation that can be rearranged to solve for any of the variables in it. Rearrange the equilibrium-constant expression to solve for [CH3OH].
[CH3OH]=[CH3OH]=
1Kc[CO][H2]2
Kc[CO][H2]2
Kc[CO][H2]2
[CO][H2]2Kc
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]
From these two reactions at 298 K, V2O3(s) + 3CO(g) → 2V(s) + 3CO2(g); ΔH° = 369.8 kJ; ΔS° = 8.3 J/K V2O5(s) + 2CO(g) → V2O3(s) + 2CO2(g); ΔH° = –234.2 kJ; ΔS° = 0.2 J/K calculate ΔG° for the following at 298 K: 2V(s) + 5CO2(g) → V2O5(s) + 5CO(g
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.
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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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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Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in this list of thermodynamic properties. Mg ( OH ) 2 ( s ) + 2 HCl ( g ) ⟶ MgCl 2 ( s ) + 2 H 2 O ( g )
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.
Explanation:Standard Enthalpy of Formation: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 rate law of a certain reaction is rate=k[X][Y]. The units of k, with measured in seconds, is
A.) s^-1
B.) M^-1 s^-1
C.) M^-2 s^-1
D.) M^-1
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).
Explanation: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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