Answer: The cell potential of the cell is +0.118 V
Explanation:
The half reactions for the cell is:
Oxidation half reaction (anode): [tex]Ag(s)+Cl^-(aq.)\rightarrow AgCl(s)+e^-[/tex]
Reduction half reaction (cathode): [tex]AgCl(s)+e^-\rightarrow Ag(s)+Cl^-(aq.)[/tex]
In this case, the cathode and anode both are same. So, [tex]E^o_{cell}[/tex] will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}]_{diluted}}{[Cl^{-}]_{concentrated}}[/tex]
where,
n = number of electrons in oxidation-reduction reaction = 1
[tex]E_{cell}[/tex] = ?
[tex][Cl^{-}]_{diluted}[/tex] = 0.0222 M
[tex][Cl^{-}]_{concentrated}[/tex] = 2.22 M
Putting values in above equation, we get:
[tex]E_{cell}=0-\frac{0.0592}{1}\log \frac{0.0222M}{2.22M}[/tex]
[tex]E_{cell}=0.118V[/tex]
Hence, the cell potential of the cell is +0.118 V
When radioactive uranium decays to produce thorium, it also emits a particle. As seen in the balanced nuclear equation, this particle can BEST be described as
A) a helium atom.
B) an alpha particle or a helium atom.
C) a beta particle or a hydrogen nucleus.
D) an alpha particle or a helium nucleus.
The radioactive uranium decays to produce thorium and it emits an alpha particle or helium atom. Thus, option A is correct.
What is radioactive decay?Unstable heavy isotopes of elements undergo nuclear decay to produce stable atoms by the emission of charged particle such as alpha or beta particles.
Based on the emitted particle, there are two types of decay process namely alpha decay and beta decay. In alpha decay atoms emits alpha particles which are helium nuclei and the atom losses its mass number by 4 units and atomic number by two units,
In beta decay, electrons are emitted by the atom, where no change occurs in mass number and atomic number increases by one unit. Uranium undergo alpha decay by emitting alpha particle or helium nuclei.
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Use the law of conservation of mass to answer the questions. Consider a hypothetical reaction in which A and B are reactants and C and D are products. If 21 g of A completely reacts with 22 g of B to produce 17 g of C, how many grams of D will be produced?
In a chemical reaction, the total mass of reactants is equals to the total mass of the products. Here, A and B react to produce C and D. Given that 21g of A and 22g of B (43g in total) are converted into 17g of C and x grams of D, the mass of D is therefore 43g - 17g = 26g.
Explanation:The law of conservation of mass states that matter cannot be created or destroyed in a chemical reaction. In your hypothetical reaction, 21g of A and 22g of B react to create C and D. Combined, the reactants have a total mass of 43g. In the product, we have 17g of C and the mass of D which we're trying to find. Now, given that mass of reactants = mass of products, we can say that 43g (total mass of A+B) = 17g (mass of C) + mass of D. By rearranging the equation, we find that mass of D = 43g - 17g = 26g. So, 26g of D will be produced according to the law of conservation of mass.
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A voltaic cell utilizes the following reaction: 4Fe2+(aq)+O2(g)+4H+(aq)→4Fe3+(aq)+2H2O(l).
The EMF under standard contions is .46 V.
What is the emf of this cell when [Fe2+]= 2.0M , [Fe3+]= 1.9
Answer:
The Emf of the given cell at [Fe2+] = 2.0 M and [Fe3+] = 1.9 M is 0.48 V
Explanation:
The half cell reaction can be written as :
Anode-Half (oxidation) :
[tex]Fe^{2+}\rightarrow Fe^{3+} + 1e^{-}[/tex] ......E = 0.77 V
(multiply this equation by 4 to balance the electrons)
Cathode-half (reduction)
[tex]4H^{+} +O_{2} + 4e^{-} \rightarrow 2H_{2}O[/tex]....E= 1.23 V
[tex]E^{0}_{cell} = E_{cathode} - E_{anode}[/tex]
[tex]E^{0}_{cell} = 1.23 - 0.77[/tex]
[tex]E^{0}_{cell} = 0.46 V[/tex]
According to Nernst Equation
[tex]E_{cell} = E^{0} - \frac{RT}{nF}lnQ[/tex]
[tex]E_{cell} = E^{0} - \frac{0.059}{n}logQ[/tex]
n = number of electron transferred in the cell reaction = 4
The balanced equation is :
[tex]4Fe^{2+} + 4H^{+} +O_{2} \rightarrow 4Fe^{3+} + 2H_{2}O[/tex]
[tex]E_{cell} = 0.46 - \frac{0.059}{4}logQ[/tex]
[tex]log Q = \frac{[Fe^{3+}]^{4}}{[Fe^{2+}]^{4}}[/tex]
[tex]log Q = \left ( \frac{[Fe^{3+}]}{[Fe^{2+}]} \right )^{4}[/tex]
[tex]log Q = \left ( \frac{1.9}{2.0} \right )^{4}[/tex]
Insert the value of log Q in Nernst Equation:
[tex]E_{cell} = 0.46 - \frac{0.059}{4} log\left ( \frac{1.9}{2.0} \right )^{4}[/tex]
(using :[tex]log^{a}b = a\log b[/tex]
)
[tex]E_{cell} = 0.46 - log\frac{1.9}{2.0}[/tex]
[tex]E_{cell} = 0.46 - log(0.95)[/tex]
[tex]E_{cell} = 0.46 -(-0.0227)[/tex]
[tex]E_{cell} = 0.482 V[/tex]
The cell EMF is obtained from Nernst equation as 0.451 V.
The reaction equation is given by;
4Fe2+ (aq) + O2(g) + 4H+ (aq)-------->4Fe3+(aq)+2H2O(l)
We have the following information;
EMF under standard conditions = 0.46 V
[Fe2+]= 2.0M
[Fe3+]= 1.9 M
From Nernst equation;
Ecell = E°cell - 0.0592/n logQ
Where;
E°cell = 0.46 V
n = 4
Q = [Fe3+]^4/ [Fe2+]^4
Q = [1.9]^4/[2.0]^4
Q = 0.8
Substituting values;
Ecell = 0.46 - 0.0592/4 log (0.8)
Ecell = 0.451 V
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Write the values for the four quantum numbers for a 4f6 electron.2.Write the values for the four quantum numbers for a 3s1 electron.3.Write the values for the four quantum numbers for the last electron to fill 28Ni.4.Write the values for the four quantum numbers for the outermost electron in 22Ti.
The electron configuration shows the arrangement of electrons in atoms.
The electron configuration gives a description of the location of the electron in an atom.
Note that;
n = Principal quantum number
l = orbital quantum number
m = magnetic quantum number
s = spin quantum number
For the 4f electron;n = 4, l = 3, m = 3, -2, -1, 0, 1, 2, 3, s= ±1/2
For the 3s electron;n = 3, l = 0, m = 0, s = ±1/2
For the last electron in Ni of electron configuration; [Ar] 3d⁸ 4s²n = 4, l = 0, m = 0, s = ±1/2
For the outermost electron in Ti of electron configuration; [Ar] 3d² 4s²n = 4, l = 0, m = 0, s = ±1/2
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Final answer:
The quantum numbers for specific electrons in the 4f6, 3s1, 28Ni, and 22Ti configurations have been provided according to their respective principal, azimuthal, magnetic, and spin quantum numbers.
Explanation:
The four quantum numbers are principal (n), azimuthal (l), magnetic (ml), and spin (ms). Here are their specified values for the given electrons:
Final answer:
The quantum numbers for specific electrons in the 4f6, 3s1, 28Ni, and 22Ti configurations have been provided according to their respective principal, azimuthal, magnetic, and spin quantum numbers.
Explanation:
The four quantum numbers are principal (n), azimuthal (l), magnetic (ml), and spin (ms). Here are their specified values for the given electrons:
Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
(a) HNO3+H2O⟶H3O++NO3−
(b) CN− +H2O ⟶ HCN+OH−
(c) H2SO4+Cl− ⟶HCl+HSO4−
(d) HSO4−+OH− ⟶SO42−+H2O
(e) O2− + H2 O ⟶ 2OH−
Explanation:
As per Brønsted-Lowry concept of acids and bases, chemical species which donate proton are called Brønsted-Lowry acids.
The chemical species which accept proton are called Brønsted-Lowry base.
(a) [tex]HNO_3 + H_2O \rightarrow H_3O^+ + NO_3^-[/tex]
[tex]HNO_3[/tex] is Bronsted lowry acid and [tex]NO_3^-[/tex] is its conjugate base.
[tex]H_2O[/tex] is Bronsted lowry base and [tex]H_3O^+[/tex] is its conjugate acid.
(b)
[tex]CN^- + H_2O \rightarrow HCN + OH^-[/tex]
[tex]CN^-[/tex] is Bronsted lowry base and HCN is its conjugate acid.
[tex]H_2O[/tex] is Bronsted lowry acid and [tex]OH^-[/tex] is its conjugate base.
(c)
[tex]H_2SO_4 + Cl^- \rightarrow HCl + HSO_4^-[/tex]
[tex]H_2SO_4[/tex] is Bronsted lowry acid and [tex]HSO_4^-[/tex] is its conjugate base.
Cl^- is Bronsted lowry base and HCl is its conjugate acid.
(d)
[tex]HSO_4^-+OH^- \rightarrow SO_4^{2-}+H_2O[/tex]
[tex]HSO_4^-[/tex] is Bronsted lowry acid and [tex]SO_4^{2-}[/tex] is its conjugate base.
OH^- is Bronsted lowry base and [tex]H_2O[/tex] is its conjugate acid.
(e)
[tex]O_{2-}+H_2O \rightarrow 2OH^-[/tex]
[tex]O_{2-}[/tex] is Bronsted lowry base and OH- is its conjugate acid.
[tex]H_2O[/tex] is Bronsted lowry acid and OH- is its conjugate base.
The identification and labelling of the Brønsted-Lowry acid and base, and their respective conjugate are done below.
What is Brønsted-Lowry acid and base?Brønsted-Lowry acids are chemical species that donate proton (H+) while the chemical species which accept proton (H+) are called Brønsted-Lowry base
Based on this question, the conjugate acid and base of the equations given are as follows:
For equation a:
HNO3 is the Bronsted lowry acid and NO3- is its conjugate base.H2O is Bronsted lowry base while H3O+ is its conjugate acid.For equation b:
Cyanide ion is the Bronsted lowry base while hydrogen cyanide is its conjugate acid. H2O is the Bronsted lowry acid while OH- is its conjugate base.For equation c:
H2SO4 is the Bronsted lowry acid while HSO4- is its conjugate base.Cl- is the Bronsted lowry base and HCl is its conjugate acid.For equation d:
HSO4- is the Bronsted lowry acid while SO4²- its conjugate base.OH- is the Bronsted lowry base and H2O is its conjugate acidFor equation e:
O²- is the Bronsted lowry base and OH- is its conjugate acid.H2O is the Bronsted lowry acid and OH- is its conjugate baseLearn more about Brønsted-Lowry base and acid at: https://brainly.com/question/21736327
Predict the products of the decomposition of lithium nitride, Li3N.
Final answer:
Upon decomposition, lithium nitride (Li₃N) breaks down into lithium (Li) and nitrogen (N₂) gas. The balanced chemical equation for this decomposition is 2 Li₃N(s) → 6 Li(s) + N₂(g). This is an example of chemical decomposition in chemistry.
Explanation:
Predicting the products of the decomposition of lithium nitride, Li₃N, involves understanding the reactions of ionic compounds. Lithium nitride is comprised of lithium (Li) ions and nitride (N³⁻) ions. Upon decomposition, lithium nitride would likely break down into its constituent elements, lithium (Li) and nitrogen (N₂). The balanced equation for this decomposition would be: 2 Li₃N(s) → 6 Li(s) + N₂(g)
Here, solid lithium nitride (Li₃N) decomposes into solid lithium (Li) and nitrogen gas (N₂) when subjected to suitable conditions such as heating. This type of reaction demonstrates basic principles of chemical decomposition and stoichiometry in chemistry.
Which of the following correctly represents the third ionization of aluminum?
A) Al^2+ (g) + e^- -----> Al^+ (g)
B) Al^2+ (g) --------> Al^3+(g) + e^-
C) Al^2- (g) + e^- ---------> A1^3- (g)
D) Al (g) --------> Al^+ (g) + e^-
E) Al^2+ (g) + e^- --------> Al^3+ (g)
Answer:
Option B
Explanation:
Energy required to remove outermost electron from an neutral atom is called ionization energy or first ionization energy ([tex]IE_1[/tex])
Energy required to remove an electron from M+ ion is called second ionization energy ([tex]IE_2[/tex]).
Energy required to remove an electron from M2+ ion is called third ionization energy [tex]IE_2[/tex]).
Therefore, among the given options:
[tex]Al+e^- \rightarrow Al^{+}(g),\ IE_1[/tex]
[tex]Al^++e^- \rightarrow Al^{2+}(g),\ IE_2[/tex]
[tex]Al^{2+}+e^- \rightarrow Al^{3+}(g),\ IE_3[/tex]
Therefore, the correct option is b
The third ionization of aluminum refers to the removal of the third electron from an aluminum atom. The correct representation is: B) Al2+ (g) --------> Al3+(g) + e-. This represents the transition from a double positive ion to a triple positive ion of aluminum.
Explanation:In Chemistry, the ionization energy of an element refers to the energy required to remove an electron from an atom or an ion. The process of ionization occurs in successive stages for each electron removed. When considering the third ionization of aluminum, it implies that two electrons have already been removed, and we're focusing on the removal of a third electron from the resulting ion.
The correct option that represents the third ionization of aluminum is:
B) Al
2+
(g) --------> Al
3+
(g) + e
-
This is because option B represents the transition of aluminum from a double positive ion (Al2+) to a triple positive ion (Al3+) with the removal of an electron.
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________ is an example of an element. Question 4 options: A) Water B) Nitrogen C) Glucose D) Salt
Answer:
Nitrogen is an example of an element (option B)
Explanation:
Water, glucose and salt are sort of compound.
They are compounds that have certain elements linked by specific bonds.
H₂O - Water
C₆H₁₂O₆ - Glucose
NaCl - Salt
Nitrogen is an example of an element. Unlike water, glucose, and salt, which are compounds, nitrogen exists as a diatomic molecule with the chemical formula N₂.
An example of an element is B) Nitrogen. Nitrogen is a chemical substance that is represented by the symbol N and is found in the air we breathe as a diatomic molecule (N₂).
In the context of other options presented in such questions, water (H₂O) is a compound consisting of hydrogen and oxygen, glucose (C₆HO₆) is a carbohydrate compound, and salt commonly refers to sodium chloride (NaCl), which is also a compound.
A sample weighing 3.110 g is a mixture of Fe 2 O 3 (molar mass = 159.69 g/mol) and Al 2 O 3 (molar mass = 101.96 g/mol). When heat and a stream of H 2 gas is applied to the sample, the Fe 2 O 3 reacts to form metallic Fe and H 2 O ( g ) . The Al 2 O 3 does not react. If the sample residue (the solid species remaining after the reaction) weighs 2.387 g, what is the mass fraction of Fe 2 O 3 in the original sample?
Answer:
The mass fraction of ferric oxide in the original sample :[tex]\frac{723}{3110}[/tex]
Explanation:
Mass of the mixture = 3.110 g
Mass of [tex]Fe_2O_3=x[/tex]
Mass of [tex]Al_2O_3=y[/tex]
After heating the mixture it allowed to react with hydrogen gas in which all the ferric oxide reacted to form metallic iron and water vapors where as aluminum oxide did not react.
[tex]Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)[/tex]
Mass of mixture left after all the ferric oxide has reacted = 2.387 g
Mass of mixture left after all the ferric oxide has reacted = y
[tex]x=3.110 g- y=3.110 g - 2.387 g = 0.723 g[/tex]
The mass fraction of ferric oxide in the original sample :
[tex]\frac{0.723 g}{3.110 g}=\frac{723}{3110}[/tex]
Furan and maleimide undergo the Diels-Alder reaction at 25 °C to give the endo product. When the reaction takes place at 90 °C, however, the major product is the exo isomer. Further study shows that the endo product isomerizes to the exo product at 90 °C. Draw the exo product of the Diels-Alder reaction of furan with maleimide.
Answer:see attached image
Explanation:
In organic chemistry, a product may be kinetically or thermodynamically favoured. A kinetically favours product forms faster, it may not necessarily be the more stable product. The thermodynamically favoured product forms at a slower rate but is more stable. Often times, the kinetically favoured product rearranges itself to form the thermodynamically favoured product at equilibrium. The endo product of the Diels Alder reaction mentioned in the question is first formed (kinetically favoured) but rearranges to form the exo product (thermodynamically favoured) at equilibrium. This is clear shown in the reaction mechanism attached below.
Refrigerant-134a enters the coils of the evaporator of a refrigeration system as a saturated liquid-vapor mixture at a pressure of 160 kPa. The refrigerant absorbs 180 kJ/kg of heat from the cooled space, which is maintained at -5°C, and leaves as a saturated vapor at the same pressure. Determine
(a) the entropy change of the refrigerant,
(b) the entropy change of the cooled space, and
(c) the total entropy change for this process.
Answer:
(a) 0.699 kJ/K
(b) -0.671 kJ/K
(c) 0.028 kJ/K
Explanation:
The Refrigerant-134a flows into the evaporator as a saturated liquid-vapor mixture and flows out as a saturated vapor at a saturation pressure of 160 kPa and temperature of -15.64°C (estimated from the Saturated Refrigerant-134a Temperature Table).
(a) The entropy change of the refrigerant (ΔS[tex]_{R-134a}[/tex]) = Q/T[tex]_{1}[/tex]
Q = 180 kJ
T[tex]_{1}[/tex] = -15.64 + 273.15 = 257.51 K
ΔS[tex]_{R-134a}[/tex] = Q/T[tex]_{1}[/tex] = 180/257.51 = 0.699 kJ/K
(b) The entropy change (ΔS[tex]_{c}[/tex]) of the cooled space (ΔS[tex]_{c}[/tex]) = -Q/T[tex]_{2}[/tex]
Q = -180 kJ
T[tex]_{2}[/tex] = -5 + 273.15 = 268.15 K
ΔS[tex]_{c}[/tex] = Q/T[tex]_{2}[/tex] = -180/268.15 = -0.671 kJ/K
(c) The total entropy change for this process (ΔS[tex]_{t}[/tex]) = ΔS[tex]_{R-134a}[/tex] + ΔS[tex]_{c}[/tex] = 0.699 - 0.671 = 0.028 kJ/K
One of the components of natural crude oil and coal deposits is benzo[a]pyrene, a compound with a molecular mass of about 252.3 amu, containing only carbon and hydrogen. A 3.649 mg sample of benzo[a]pyrene burns to give 12.73 mg of CO2. Determine its empirical and molecular formulas. (Omit states-of-matter from your answer.) Empirical formula
Answer:
Empirical C5H3
Molecular
C20H12
Explanation:
Like all other hydrocarbons, benzopyrene would burn in oxygen or air to form carbon iv oxide and water only.
From the mass of carbon iv oxide produced, we can get the mass of carbon and hence the mass of hydrogen.
It can be seen from the question that 12.73mg of carbon iv oxide was produced. Let us convert this to grammes, we simply divide by 1000 = 0.01273g
Now we need to calculate the number of moles of carbon iv oxide produced. To do this, we simply divide the mass by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol
The number of moles is thus 0.01273/44 = 0.00028392moles
As we can see that there is only one atom of carbon in a molecule of carbon iv oxide. Hence, the number of moles of carbon iv oxide present is equal to 0.00028392moles
From here we can get the mass of carbon which is equal to the number of moles of carbon multiplied by the atomic mass of carbon. The atomic mass of carbon is 12 a.m.u
The mass of carbon in the benzopyrene is thus 0.00028392 * 12 = 0.003472g
The mass of hydrogen in the compound is 0.003649g - 0.003472g = 0.000177g
We can now deduce the empirical formula by dividing the masses by the atomic masses.
H = 0.000177/1 = 0.000177
C = 0.003572/12 = 0.000289333333
We now divide by the smallest which is that of hydrogen.
H = 0.000177/0.000177= 1
C = 0.000289333333/0.000177 = apprx 1.63
Multiply both by 3 to yield C5H3
The molecular formula is as follows:
[(12* 5) + ( 3 * 1)]n = 252.3
63n = 252.3
n = 4
The molecular formula is thus C20H12
Commercial silver-plating operations frequently use a solution containing the complex Ag+ ion. Because the formation constant (Kf) is quite large, this procedure ensures that the free Ag+ concentration in solution is low for uniform electrodeposition. In one process, a chemist added 9.0 L of 5.0 M NaCN to 90.0 L of 0.21 M AgNO3.
Calculate the concentration of free Ag+ ions at equilibrium.
See your textbook for Kf values.
The concentration of free Ag+ ions at equilibrium is calculated by assuming full complexation of Ag+ with CN- due to a large formation constant, and then considering any small dissociation of the complex back to Ag+ and CN-. The exact concentration can be determined by applying the formation constant for the complex, which is not provided here.
Explanation:To calculate the concentration of free Ag+ ions at equilibrium in a silver plating operation, we need to consider the following reaction:
Ag+(aq) + 2 CN-(aq) \<--> Ag(CN)2-(aq)
First, we determine the initial concentrations of Ag+ and CN-. AgNO3 completely dissociates in water, so the initial concentration of Ag+ is 0.21 M in 90.0 L. NaCN also dissociates completely, providing 5.0 M of CN- ion in 9.0 L.
Mixing the solutions, we have a total volume of 99.0 L (90.0 L + 9.0 L). The concentration of CN- is calculated as follows:
(5.0 M)(9.0 L) / 99.0 L = 0.4545 M
Since the formation constant (Kf) of the Ag(CN)2- complex is large, we can assume that all the Ag+ will react to form the complex, leaving a negligible concentration of free Ag+ ions. Any remaining CN- ions will affect the equilibrium concentration of free Ag+ ions due to the shift in reaction to the left, as described by Le Chatelier's principle.
We assume that essentially all 0.21 M of the Ag+ ion forms the complex. To find the equilibrium concentration of Ag+, we use the dissociation of the complex, simplified by assuming that the dissociation is so small that it's negligible compared to the remaining CN- concentration.
Since we do not have the exact Kf value provided here, we would normally express the equilibrium concentration of Ag+ in terms of Kf and the concentration of remaining CN-, but to give an exact numerical answer, the specific Kf value for the Ag(CN)2- complex is required from a reliable source like a textbook or scientific database.
Compared to the average diameter of a hydrogen atom, the average diameter of a helium atom is:
Answer:
The average diameter of a helium atom is smaller.
Explanation:
Step 1: Data given
An atom gets larger as the number of electronic shells increase; therefore the radius of atoms increases as you go down a certain group in the periodic table of elements.
In general, the size of an atom will decrease as you move from left to the right of a certain period.
Step 2:
Since hydrogen is more left on the periodic table than helium, the diameter of the atom will decrease moving from hydrogen to helium.
Hydrogen has a bigger diameter than helium.
The average diameter of a helium atom is smaller.
Watch the animation and select the interactions that can be explained by hydrogen bonding. Check all that apply.
a. CH4 molecules interact more closely in the liquid than in the gas phase.
b. HF is a weak acid neutralized by NaOH.
c. Ice, H2O, has a solid structure with alternating H−O interactions.
d. H2Te has a higher boiling point than H2S.
e. HF has a higher boiling point than HCl.
Answer:
c. Ice, H₂O, has a solid structure with alternating H−O interactions.
e. HF has a higher boiling point than HCl.
Explanation:
For molecules to interact through hydrogen bonding, it is required that there is an H atom bonded to a more electronegative atom, such as N, O or F.
Select the interactions that can be explained by hydrogen bonding. Check all that apply.
a. CH₄ molecules interact more closely in the liquid than in the gas phase. NO. The electronegativity of C is not high enough to form hydrogen bondings.
b. HF is a weak acid neutralized by NaOH. NO. This reaction occurs in water and it is better explained by ion-ion forces.
c. Ice, H₂O, has a solid structure with alternating H−O interactions. YES. This structure is a consequence of the hydrogen bonding.
d. H₂Te has a higher boiling point than H₂S. NO. The electronegativities of Te and S are not high enough to form hydrogen bondings.
e. HF has a higher boiling point than HCl. YES. The stronger hydrogen bonding interactions in HF explain the higher boiling point.
Final answer:
Hydrogen bonding explains interactions such as HF being a weak acid, the solid structure of ice (H2O), and the higher boiling point of HF compared to HCl.
Explanation:
In the context of the provided information, the interactions that can be explained by hydrogen bonding are:
b. HF is a weak acid neutralized by NaOH.c. Ice, H2O, has a solid structure with alternating H-O interactions.e. HF has a higher boiling point than HCl.Hydrogen bonding occurs when a hydrogen atom is bonded to a highly electronegative atom such as oxygen, nitrogen, or fluorine, creating a strong dipole-dipole interaction.
These interactions are responsible for the higher boiling points and particular properties of substances like water (H2O) and hydrogen fluoride (HF).
Addition of 50. J to a 10.0-g sample of a metal will cause the temperature of a metal to rise from 25ºC to 35ºC. The specific heat of the metal is closest to0.0005 J/(g·ºC)0.50 J/(g·ºC)2.5 J/(g·ºC)4.2 J/(g·ºC)
Answer:
b) C = 0.50 J/(g°C)
Explanation:
Q = mCΔT∴ Q = 50 J
∴ m = 10.0 g
∴ ΔT = 35 - 25 = 10 °C
specific heat (C) :
⇒ C = Q / mΔT
⇒ C = 50 J / (10.0 g)(10 °C)
⇒ C = 0.50 J/(g°C)
Most metals are very reactive, as are the elements in the halogen group. Aluminum, for instance, For example, reacts with elemental chlorine to form aluminum chloride. If you have a 13.5 g sample of Al, which choice below is true?
A. you will need 23.6 g Cl2 for complete reaction and will produce 66.7 g of AlCl3.
B. you will need 53.2 g Cl2 for complete reaction and will produce 66.7 g of AlCl3.
C. you will need 11.8 g Cl2 for complete reaction and will produce 49.0 g of AlCl3.
D. you will need 26.6 g Cl2 for complete reaction and will produce 49.0 g of AlCl3. Reset Selection
Answer: B. you will need 53.2 g Cl2 for complete reaction and will produce 66.7 g of AlCl3.
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the moles, we use the equation:
[tex]\text{Number of moles of aluminium}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{13.5g}{27g/mol}=0.5moles[/tex]
The balanced reaction is:
[tex]2Al+3Cl_2(g)\rightarrow 2AlCl_3[/tex]
2 moles of aluminium react with= 3 moles of chlorine
Thus 0.5 moles of aluminium react with=[tex]\frac{3}{2}\times 0.5=0.75[/tex] moles of chlorine
Mass of chlorine=[tex]moles\times {\text{Molar Mass}}=0.75\times 71=53.2g[/tex]
2 moles of aluminium produce = 2 moles of aluminium chloride
Thus 0.5 moles of aluminium react with=[tex]\frac{2}{2}\times 0.5=0.5[/tex] moles of aluminium chloride
Mass of aluminium chloride=[tex]moles\times {\text{Molar Mass}}=0.5\times 133.34=66.7g[/tex]
Thus 53.2 g of chlorine is used and 66.7 g of aluminium chloride is produced.
Challenge question: This question is worth 6 points. As you saw in problem 9 we can have species bound to a central metal ion. These species are called ligands. In the past we have assumed all the d orbitals in some species are degenerate; however, they often are not. Sometimes the ligands bound to a central metal cation can split the d orbitals. That is, some of the d orbitals will be at a lower energy state than others. Ligands that have the ability to cause this splitting are called strong field ligands, CN- is an example of these. If this splitting in the d orbitals is great enough electrons will fill low lying orbitals, pairing with other electrons in a given orbital, before filling higher energy orbitals. In question 7 we had Fe2+, furthermore we found that there were a certain number (non-zero) of unpaired electrons. Consider now Fe(CN)64-: here we also have Fe2+, but in this case all the electrons are paired, yielding a diamagnetic species. How can you explain this?
A. There are 3 low lying d orbitals, which will be filled with 6 electrons before filling the 2, assumed to be degenerate, higher energy orbitals.
B. There are 4 low lying d orbitals, which will be filled with 8 electrons before filling the 1 higher energy orbital.
C. There is 1 low lying d orbital, which will be filled with two electrons before filling the 4, assumed to be degenerate, higher energy orbitals.
D. All the d orbitals are degenerate.
E. There are 2 low lying d orbitals, which will be filled with 4 electrons before filling the 3, assumed to be degenerate, higher energy orbitals.
Answer:
A
Explanation:
Iron has the ground state electronic configuration [Ar]3d64s2
Fe2+ has the electronic configuration [Ar]3d6.
In an octahedral crystal field, there are two sets of degenerate orbitals; the lower lying three t2g orbitals, and the higher level two degenerate eg orbitals. Strong field ligands cause high octahedral crystal field splitting, there by separating the two sets of degenerate orbitals by a tremendous amount of energy. This energy is much greater than the pairing energy required to pair the six electrons in three degenerate orbitals. Since CN- is a strong field ligand, it leads to pairing of six electrons in three degenerate orbitals
The Fe(CN)64- ion is diamagnetic due to the significant energy difference caused by strong field ligands like CN-, resulting in pairing of electrons in the 3 low energy d orbitals before filling the higher energy ones.
Explanation:The answer is A. There are 3 low lying d orbitals, which will be filled with 6 electrons before filling the 2, assumed to be degenerate, higher energy orbitals. When a metal ion is coordinated to ligands, such as in Fe(CN)64-, the degeneracy of the 3d orbitals is broken (they have different energies), due to the electrostatic interactions between the ligands and the orbitals. In the case of strong-field ligands like CN-, the energy difference is significant enough to cause pairing of electrons in the 3 low energy d orbitals before the 2 high energy orbitals are populated, resulting in a diamagnetic species.
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A mixture contains 25 g of cyclohexane (C6H12) and 44 g of 2-methylpentane (C6H14). The mixture of liquids is at 35 oC . At this temperature, the vapor pressure of pure cyclohexane is 150 torr, and that of pure 2-methylpentane is 313 torr. Assume this is an ideal solution. What is the mole fraction of cyclohexane in the liquid phase?
Answer:
The mol fraction of cyclohexane in the liquid phase is 0.368
Explanation:
Step 1: Data given
Mass of cyclohexane = 25.0 grams
Mass of 2-methylpentane = 44.0 grams
Temperature = 35.0 °C
The pressure of cyclohexane = 150 torr
The pressure of 2-methylpentane = 313 torr
The pressure we only need for the mole fraction in gas phase.
Step 2: Calculate moles of cyclohexane
Moles cyclohexane = mass cyclohexane / molar mass
Moles cyclohexane = 25.0 g / 84 g/mol = 0.298 mol of cyclohexane
Step 3: Calculate moles of 2-methylpentane
Moles = 44.0 grams / 86 g/mol = 0.512 mol of 2-methylpentane
Step 4: Calculate mole fraction of cyclohexane in the liquid phase
Mole fraction of C6H12:
0.298 / (0.298 + 0.512) = 0.368
The mol fraction of cyclohexane in the liquid phase is 0.368
The mole fraction of cyclohexane in the liquid phase is 36.8%.
Explanation:To find the mole fraction of cyclohexane in the liquid phase, we need to calculate the total moles of cyclohexane and 2-methylpentane in the mixture. First, we calculate the moles of each component using their molar masses:
moles of cyclohexane = 25 g / 84.18 g/mol = 0.297 mol
moles of 2-methylpentane = 44 g / 86.18 g/mol = 0.509 mol
Next, we calculate the mole fraction of cyclohexane:
mole fraction of cyclohexane = moles of cyclohexane / (moles of cyclohexane + moles of 2-methylpentane)
mole fraction of cyclohexane = 0.297 mol / (0.297 mol + 0.509 mol) = 0.368 or 36.8%
Based on the reduction potentials listed in the textbook appendix, which of the following redox reactions do you expect to occur spontaneously?
W. 2Al(s)+3Pb2+ (aq) → 2Al3+ (aq)+3Pb(s)
X. Fe(s)+Cr3+ (aq) → Fe3+ (aq)+Cr(s)
Y. Ca2+ (aq)+Zn(s) → Ca(s)+Zn2+(aq)
Z. 2Cu+(aq)+Co(s) → 2Cu(s)+Co2+ (s)
a. W only
b. X, Y and Z
c. Y only
d. X and Z
e. Z only
f. X and Y
g. W, X and Z
h. X and Y
i.W and Z
Answer: The redox reactions that occur spontaneously are Reaction W and Reaction Z.
Explanation:
For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.
Relationship between standard Gibbs free energy and standard electrode potential follows:
[tex]\Delta G^o=-nFE^o_{cell}[/tex]
For a reaction to be spontaneous, the standard electrode potential must be positive.
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex] .......(1)
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
For reaction W:The chemical reaction follows:
[tex]2Al(s)+3Pb^{2+}(aq.)\rightarrow 2Al^{3+}(aq.)+3Pb(s)[/tex]
We know that:
[tex]E^o_{Al^{3+}/Al}=-1.66V\\E^o_{Pb^{2+}/Pb}=-0.13V[/tex]
Calculating the [tex]E^o_{cell}[/tex] using equation 1, we get:
[tex]E^o_{cell}=-0.13-(-1.66)=1.53V[/tex]
As, the standard electrode potential is coming out to be positive. So, the reaction is spontaneous.
For reaction X:The chemical reaction follows:
[tex]Fe(s)+Cr^{3+}(aq.)\rightarrow Fe^{3+}(aq.)+Cr(s)[/tex]
We know that:
[tex]E^o_{Fe^{3+}/Fe}=0.77V\\E^o_{Cr^{3+}/Cr}=-0.74V[/tex]
Calculating the [tex]E^o_{cell}[/tex] using equation 1, we get:
[tex]E^o_{cell}=-0.74-(0.77)=-1.51V[/tex]
As, the standard electrode potential is coming out to be negative. So, the reaction is not spontaneous.
For reaction Y:The chemical reaction follows:
[tex]Zn(s)+Ca^{2+}(aq.)\rightarrow Zn^{2+}(aq.)+Ca(s)[/tex]
We know that:
[tex]E^o_{Ca^{2+}/Ca}=-2.87V\\E^o_{Zn^{2+}/Zn}=-0.76V[/tex]
Calculating the [tex]E^o_{cell}[/tex] using equation 1, we get:
[tex]E^o_{cell}=-2.87-(-0.76)=-2.11V[/tex]
As, the standard electrode potential is coming out to be negative. So, the reaction is not spontaneous.
For reaction Z:The chemical reaction follows:
[tex]Co(s)+2Cu^{+}(aq.)\rightarrow Co^{2+}(aq.)+2Cu(s)[/tex]
We know that:
[tex]E^o_{Cu^{+}/Cu}=0.34V\\E^o_{Co^{2+}/Co}=-0.28V[/tex]
Calculating the [tex]E^o_{cell}[/tex] using equation 1, we get:
[tex]E^o_{cell}=0.34-(-0.28)=0.62V[/tex]
As, the standard electrode potential is coming out to be positive. So, the reaction is spontaneous.
Hence, the redox reactions that occur spontaneously are Reaction W and Reaction Z.
The drug molecules bind the protein in a 1:1 ratio to form a drug-protein complex. The protein concentration in aqueous solution at 25 ∘C is 1.60×10−6 M . Drug A is introduced into the protein solution at an initial concentration of 2.00×10−6M. Drug B is introduced into a separate, identical protein solution at an initial concentration of 2.00×10−6M. At equilibrium, the drug A-protein solution has an A-protein complex concentration of 1.00×10−6M, and the drug B solution has a B-protein complex concentration of 1.40×10−6M.a. Calculate the Kc value for the A-protein binding reaction.b. Calculate the Kc value for the B-protein binding reaction.c. Assuming that the drug that binds more strongly will be more effective, which drug is the better choice for further research?
Answer:
a. kc = 1,67x10⁶
b. kc = 1,17x10⁷
c. Drug B is the better choice.
Explanation:
The bind of drug-protein is described as:
Protein + Drug ⇄ Drug-protein
Where kc is:
kc = [Drug-protein] / [Protein] [Drug] (1)
a. For A, the equilibrium concentration of each specie is:
[Protein]: 1,60x10⁻⁶M - x
[Drug]: 2,00x10⁻⁶M - x
[Drug-protein]: x = 1,00x10⁻⁶M
-Where x is reaction coordinate-
Thus:
[Protein]: 1,60x10⁻⁶M - 1,00x10⁻⁶M = 0,60x10⁻⁶M
[Drug]: 2,00x10⁻⁶M - 1,00x10⁻⁶M = 1,00x10⁻⁶M
Replacing in (1):
kc = [1,00x10⁻⁶] / [1,00x10⁻⁶] [0,60x10⁻⁶]
kc = 1,67x10⁶
b. For B:
[Protein]: 1,60x10⁻⁶M - x
[Drug]: 2,00x10⁻⁶M - x
[Drug-protein]: x = 1,40x10⁻⁶M
-Where x is reaction coordinate-
Thus:
[Protein]: 1,60x10⁻⁶M - 1,40x10⁻⁶M = 0,20x10⁻⁶M
[Drug]: 2,00x10⁻⁶M - 1,40x10⁻⁶M = 0,60x10⁻⁶M
Replacing in (1):
kc = [1,40x10⁻⁶] / [0,20x10⁻⁶] [0,60x10⁻⁶]
kc = 1,17x10⁷
c. The drug with the bigger kc will be the more effective because will be the drug that binds more strongly with the protein. Thus, drug B is the better choice.
I hope it helps!
The Kc values for the A-protein and B-protein binding reactions are 3.125 and 4.375, respectively. Drug B, which has the higher Kc value, is the better choice for further research as it binds more strongly to the protein.
Explanation:a. To calculate the Kc value for the A-protein binding reaction, we need to use the equilibrium concentrations of drug A and the A-protein complex. The Kc value is given by [A-protein complex] / ([A] * [protein]). Plugging in the values, we get Kc = (1.00×10^6) / ((2.00×10^6) * (1.60×10^−6)) = 3.125.
b. Similarly, to calculate the Kc value for the B-protein binding reaction, we use the equilibrium concentrations. Kc = (1.40×10^6) / ((2.00×10^6) * (1.60×10^−6)) = 4.375.
c. The drug that binds more strongly will form a higher concentration of drug-protein complex at equilibrium, indicating greater efficacy. In this case, drug B has a higher Kc value, suggesting that it binds more strongly and would be a better choice for further research.
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Example: Serum containing Na gave a signal of 4.27 mV in an atomic emission analysis. Then, 5 mL of 2.08 M NaCl were added to 95.0 mL of serum. The spiked serum gave a signal of 7.98 mV. Find the original concentration of Nat in serum.
Explanation:
It is known that,
No. of moles = Concentration × volume
As 5 ml of 2.08 M NaCl NaS added to 95 ml of serum then,
Total volume = (95 + 5) ml = 100 ml
Therefore, total moles of [tex]Na^{+}[/tex] = (5 × 2.08) + (95 × x)
where, x = concentration of [tex]Na^{+}[/tex] in original serum
Hence, signal obtained for this NaS = 7.8 mV
Let us assume same volume, that is, 100 ml Serum then [tex]Na^{+}[/tex] will be (100 × x) mol.
Signal obtained for this is 4.27 mV.
Now, the ratio of signal is equal to the ratio of mole.
So, [tex]\frac{4.27}{7.98} = \frac{100x}{(5 \times 2.08) + 95x}[/tex]
x = 0.113
Thus, we can conclude that original concentration of [tex]Na^{+}[/tex] in Serum is 0.113.
The original concentration of Na+ in serum is found by using the provided signals before and after spiking with NaCl, identifying the change in signal due to added Na+, and back-calculating to find the initial Na+ level in serum. The calculation reveals that the original concentration of Na+ is approximately 0.126 M.
Explanation:To find the original concentration of Na+ in serum, we use the information provided by the atomic emission analysis before and after spiking the serum with a known amount of NaCl. We are given that the unspiked serum produced a signal of 4.27 mV and the spiked serum produced a signal of 7.98 mV after adding 5 mL of 2.08 M NaCl to 95.0 mL of serum.
Step 1: Calculate the moles of Na+ added from the NaCl solution.
Moles of NaCl = 2.08 mol/L × 0.005 L = 0.0104 mol
Step 2: Since the reaction of NaCl in water leads to one mole of Na+ for every mole of NaCl, moles of Na+ added = 0.0104 mol.
Step 3: Calculate the increase in signal due to the added Na+.
Signal increase = 7.98 mV - 4.27 mV = 3.71 mV
Step 4: Determine the signal per mole of Na+ added.
Signal per mole = 3.71 mV / 0.0104 mol = 356.73 mV/mol
Step 5: Calculate the original moles of Na+ in serum from its signal.
Moles of Na+ in serum = 4.27 mV / 356.73 mV/mol = 0.01196 mol
Step 6: Calculate the concentration of Na+ using the serum volume.
Concentration of Na+ = 0.01196 mol / 0.095 L = 0.126 mol/L
The original concentration of Na+ in serum is approximately 0.126 M.
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Stan is a member of the Association for Consumer Research, but when he attends the conferences, he doesn't identify with the group. Even though his research is received favorably, he often feels detached and perceives himself as different from the other members. Stan can best be described as having a(n) _____ self-schema with respect to this group.idealizedseparatedactualizedconnectedambiguous
Answer:
Separated
Explanation:
Stan does not feel he identifies closely with the group. He feels the other members are close knit with each other but he is not part of the group.
For Stan, it would be an illusion to say he has a close relationship with the members of the group
This infers that Stan has a separated self-schema with respect to the group.
Radon is a radioactive noble gas that can sometimes be found in unventilated basements. A sample of 1.35×10−4mol of radon gas is held in a container with a volume of 3.03mL. A quantity of radon gas is added to the container, which is then found to have a volume of 7.79mL at the same temperature and pressure. How many moles of radon were added to the container?
Answer:
2.12*10^-4 moles
Explanation:
If 3.03ml contains 1.35*10^-4mol then 7.79 will contain 7.79ml*1.35*10^-4/3.03= 3.47*10^-4 moles
Amount added= (3.47-1.35)*10^-4=2.12*10^-4moles
Answer: 2.12 X 10^-4
Explanation:
First, find the final number of moles of radon in the container after the addition by rearranging Avogadro's law to solve for n2.
n2= V2 × n1 / V1
Substitute the known values ofn1, V1, and V2.
n2 = 7.79 mL × 1.35 × 10^−4 mol / 3.03 mL = 3.47 × 10^−4 mol
Find the difference between the final number of moles (n2) and the initial number of moles (n1).
n2−n1 = 3.47 × 10^-4 mol −1 .35 × 10^−4 mol= 2.12 × 10^−4mol
Lithium aluminum hydride reduces carboxylic acids to primary alcohols via what intermediate?A) a ketone. B) a methyl ester. C) an aldehyde. D) a secondary alcohol. E) an acid chloride.
Final answer:
Lithium aluminum hydride reduces carboxylic acids to primary alcohols via an aldehyde intermediate, although the aldehyde is not isolated and is reduced further to a primary alcohol in one step. So the correct option is C.
Explanation:
Lithium aluminum hydride reduces carboxylic acids to primary alcohols by first forming an aldehyde as the intermediate. The reduction process does not stop at the aldehyde stage but proceeds to reduce the aldehyde further to a primary alcohol. When a carboxylic acid is reduced, the process typically bypasses the isolation of the intermediate aldehyde due to the strong reducing power of lithium aluminum hydride (LiAlH4), which is capable of fully reducing the carboxylic acid to the alcohol in one step.
A mixture of gaseous reactants is put into a cylinder, where a chemical reaction turns them into gaseous products. The cylinder has a piston that moves in or out, as necessary, to keep a constant pressure on the mixture of . The cylinder is also submerged in a large insulated water bath. (See sketch at right.) From previous experiments, this chemical reaction is known to release of energy. The temperature of the water bath is monitored, and it is determined from this data that of heat flows out of the system during the reaction.
a)Exothermic
b) it goes up
c) The piston must move out
d)The energy is released by the system
e)ΔE = -616 kJ
Explanation:
sign convention for heat flow:
if the heat flows into the system it is marked with positive sign
when the heat flows out the system it is marked with negative sign
sign convention for work
if work is done on the system it is marked positive
if work is done by the system it is marked negative
sign convention for internal energy
if energy absorbed by system mark positive
if energy released by system mark negative
a)
the heat flow flowing out of the system (q= -300 kJ)
thus the reaction is Exothermic
b)
In exothermic reaction, heat flows from the system to surroundings.
surrounding here is water bath.
As the system is exothermic, hence the temperature of water bath goes up
c)
It is given that 316 kJ of work is done by the system, applying expansion.
To maintain constant pressure of 1 atm the decreased pressure can be compensated with an increase on volume . To increase volume the piston must move out
d)
According to 1st law of thermodynamics,
ΔE =q +w ⇒1
here,
ΔE = q +w
ΔE = -300 kJ -316 kJ
ΔE = -616 kJ ⇒2
the negative sign denotes the energy is released by the system
e)
from ΔE = -616 kJ the amount of energy released by the reaction is ΔE = -616kJ
A sample of methane at a pressure of 1.00 atm and a temperature of 93.1 K is heated at constant pressure to a temperature of 158 K. Which of the following are true? Choose all that apply
a. The liquid initially present will solidify.
b. The final state of the substance is a gas.
c. The sample is initially a solid.
d. The sample is initially a liquid. One or more phase changes will occur.
Answer:
b. The final state of the substance is a gas.
d. The sample is initially a liquid. One or more phase changes will occur.
Explanation:
Methane has the following properties:
Normal melting point: 90.7 KNormal boiling point: 111.65 K*"Normal" refers to normal pressure (1 atm).
According to this, we can affirm:
Below 90.7 K, methane is solid.Between 90.7 K and 111.65 K, methane is liquid.Above 111.65 K, methane is gas.A sample of methane at a pressure of 1.00 atm and a temperature of 93.1 K is heated at constant pressure to a temperature of 158 K. Which of the following are true? Choose all that apply.
a. The liquid initially present will solidify. FALSE. The liquid will vaporize.
b. The final state of the substance is a gas. TRUE.
c. The sample is initially a solid. FALSE. The sample is initially a liquid.
d. The sample is initially a liquid. One or more phase changes will occur. TRUE.
The substance would start as a gas, then transition to a solid without becoming a liquid due to temperature conditions.
The substance would begin as a gas and as the pressure increases, it would compress and eventually solidify without liquefying as the temperature is below the triple point temperature.
Describe the phase changes from -80°C to 500°C at 2 atm.
The sample is initially a gas, condenses to a solid as the pressure increases, and then melts when the pressure is increased further to give a liquid.
Show the determination of the charge of the ion formed by the Mg12 atom.
Explanation:
It is known that the atomic number of magnesium is 12 and its electronic distribution is 2, 8, 2. And, in order to attain stability a magnesium atom will tend to lose its valence electrons.
That is, being a metal magnesium atom will lose its 2 valence electrons and hence it forms a [tex]Mg^{2+}[/tex] ion.
Equation for this type of loss is as follows.
[tex]Mg \rightarrow Mg^{2+} + 2e^{-}[/tex]
Therefore, we can conclude that the charge of the ion formed by the [tex]_{12}Mg[/tex] atom is +2.
The equilibrium constant is given for one of the reactions below.
Determine the value of the missing equilibrium constant.
2 HD(g) ⇌ H2(g) + D2(g) Kc = 0.28
2 H2(g) + 2 D2(g) ⇌ 4 HD(g)
To find the missing equilibrium constant, we can use stoichiometry and the first given equilibrium constant. The missing equilibrium constant for the second reaction is the square of the given Kc value.
Explanation:To determine the missing equilibrium constant, you can use the concept of equilibrium constant expressions.
In the first reaction, 2 HD(g) ⇌ H2(g) + D2(g), the equilibrium constant is given as Kc = 0.28.
To find the value of the missing equilibrium constant for the second reaction, 2 H2(g) + 2 D2(g) ⇌ 4 HD(g), we can use the stoichiometry.
Since the equilibrium constant is a ratio of product concentrations to reactant concentrations, and the stoichiometric coefficients for the second reaction are all multiplied by 2 compared to the first reaction, the missing equilibrium constant would be the square of the given Kc value.
Therefore, the missing equilibrium constant is 0.28 squared, which is approximately 0.0784.
The equilibrium constant for the reaction 2 H₂(g) + 2 D₂(g) ⇌ 4 HD(g) is D) 13
To determine the equilibrium constant for the reaction 2 H₂(g) + 2 D₂(g) ⇌ 4 HD(g) given the equilibrium constant for the reverse reaction, 2 HD(g) ⇌ H₂(g) + D₂(g), which is Kc = 0.28, we can use the concept that the equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction.
The given reaction is:
2 HD(g) ⇌ H₂(g) + D₂(g), Kc = 0.28
The reverse reaction is:
H₂(g) + D₂(g) ⇌ 2 HD(g), K'c = 1 / 0.28
K'c = 3.57
The target reaction is twice the reverse reaction:
2 H₂(g) + 2 H₂(g) ⇌ 4 HD(g)
When the reaction coefficients are doubled, the equilibrium constant is squared:
Kc = K'c²
Kc = 3.57² = 12.75
Therefore, the equilibrium constant for the reaction 2 H₂(g) + 2 D₂(g) ⇌ 4 HD(g) is 12.798
Rounding of to two significant figures, the Equilibrium constant is D) 13
Complete question is - The equilibrium constant is given for one of the reactions below. Determine the value of the missing equilibrium constant.
2 HD (g) = H2 (g) + D2 (g) Kc = 0.28
2 H2 (g) + 2 D2 (g) = 4 HD (g) Kc = ?
A) 7.8 x 10⁻²
B) 3.6
C) 0.53
D) 13
E) 1.9
A saturated solution contains more solute than solvent. contains more solvent than solute. contains equal moles of solute and solvent. contains the maximum amount of solute that will dissolve in that solvent at that temperature. contains a solvent with only sigma bonds and no pi bonds (i.e. only single bonds, with no double or triple bonds).
Answer:
contains the maximum amount of solute that will dissolve in that solvent at that temperature.
Explanation:
Solution= solute+ solvent
The solubility of a solute depends on temperature. A solution containing just the right (maximum) amount of solute that can normally dissolve at a given temperature is said to be saturated.
A saturated solution contains the maximum amount of solute that dissolves in a solvent at a given temperature. Exceeding the solubility leads to a supersaturated solution and solute precipitation. Solubility varies with temperature.
Explanation:A saturated solution is one that contains the maximum amount of solute that will dissolve in that solvent at that particular temperature. The solubility of a solute in a particular solvent is the maximum concentration that may be achieved under given conditions when the dissolution process is at equilibrium. In a saturated solution, the solute concentration equals its solubility. If a solute's concentration exceeds its solubility, it forms a supersaturated solution which is an unstable condition and results in solute precipitation when perturbed. The extent to which a substance may be dissolved in water, or any solvent, is quantitatively expressed as its solubility. Solubilities for gaseous solutes decrease with increasing temperature, while those for most, but not all, solid solutes increase with temperature.
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