Answer:
Oxidation half equation;
3Cd(s) -------> 3Cd^2+(aq) + 6e
Reduction half equation;
2In^3+(aq) + 6e -----> 2In(s)
Explanation:
Since the reduction potentials of Indium and Cadmium are -0.34 V and - 0.40 V respectively, we can see that cadmium will be oxidized while indium will the reduced.
We arrived at this conclusion by examining the reduction potential of both species. The specie with more negative reduction potential is oxidized in the process.
Oxidation half equation;
3Cd(s) -------> 3Cd^2+(aq) + 6e
Reduction half equation;
2In^3+(aq) + 6e -----> 2In(s)
From the graph of Density vs. Concentration, created in Graph 1, what was the relationship between the concentration of the sugar solution and the density of the sugar solution?
The graph is not given in the question, so, the required graph is attached below:
Answer:
According to the graph, the relationship between the density of the sugar solution and the concentration of the sugar solution is directly proportional to each other as they both are increasing exponentially.
The graph shows that, the density of sugar solution will increase with the increase in concentration of sugar in the solution.
what is ammonium nitrate
Answer:
Ammonium nitrate is a chemical compound with the chemical formula NH₄NO₃. It is a white crystalline solid consisting of ions of ammonium and nitrate.
Which of the following goes through the largest volumetric change? Question 4 options: A) Water when it's heated from 1oC to 99oC B) Water when it freezes into ice C) Ice when it melts into water D) Water when it boils into steam
Answer:
Water when it freezes into ice
Explanation:
Most liquids expand when heated and contract when cooled, water behaves in an anomalous fashion. Water rather expands when cooled and contracts when heated.
Water usually contracts on cooling from any temperature until 4°C, after 4°C, the water begins to expand rapidly. Hence water has its least volume at 4°C and increases rapidly afterwards.
Thus the largest volume change for water occurs during freezing since it expands when cooled.
The Lewis structure of N2H2 shows ________. Group of answer choices a nitrogen-nitrogen single bond each hydrogen has one nonbonding electron pair each nitrogen has one nonbonding electron pair each nitrogen has two nonbonding electron pairs a nitrogen-nitrogen triple bond
Answer:
one bond between nitrogen and hydrogen and a double bond between the nitrogen atoms.
Explanation:
H-N=N-H
A balloon is filled to a volume of 1.50 L with 3.00 moles of gas at 25 °C. With pressure and temperature held constant, what will be the volume (in L) of the balloon if 0.50 moles of gas are released?
Answer:
Volume : 1.25 L
Explanation:
We are given here that the volume ( V[tex]_1[/tex] ) = 1.50 Liters, the initial moles ( held at 25 °C ) = 3.00 mol, and the final moles ( n[tex]_2[/tex] ) = 3.00 - 0.5 = 2.5 mol. The final mol is calculated given that 0.50 mol of gas are released from the prior 3.00 moles of gas.
Volume ( V[tex]_1[/tex] ) = 1.50 L,
Initial moles ( n[tex]_1[/tex] ) = 3.00 mol,
Final Volume ( n[tex]_2[/tex] ) = 3.00 - 0.5 = 2.5 mol
Applying the combined gas law, we can calculate the final volume ( V[tex]_2[/tex] ).
P[tex]_1[/tex]V[tex]_1[/tex] / n[tex]_1[/tex]T[tex]_1[/tex] = P[tex]_2[/tex]V[tex]_2[/tex] / n[tex]_2[/tex]T[tex]_2[/tex] - we know that the pressure and temperature are constant, and therefore we can apply the following formula,
V[tex]_1[/tex] / n[tex]_1[/tex] = V[tex]_2[/tex] / n[tex]_2[/tex] - isolate V[tex]_2[/tex],
V[tex]_2[/tex] = V[tex]_1[/tex] n[tex]_2[/tex] / n[tex]_1[/tex] = 1.50 L [tex]*[/tex] 2.5 mol / 3.00 mol = ( 1.5 [tex]*[/tex] 2.5 / 3 ) L = 1.25 L
The volume of the balloon will be 1.25 L.
While balancing a chemical equation, we change the _____ to balance the number of atoms on each side of the equation.
Answer:
While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation
Explanation:
While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
What is chemical equation?To summarize in chemistry terms, a chemical equation depicts the initial chemicals, or reactants, on the left-hand side and the final compounds, or products, just on right-hand side, divided by an arrow. In the chemical equation, the number of atoms in each element as well as the total charge are the same on opposite of the equation's sides.
Chemical equations are used in chemistry to depict chemical processes by writing the reactants and products in terms of their corresponding chemical formulas. While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
Therefore, while balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
To know more about chemical equation, here:
https://brainly.com/question/29028257
#SPJ6
Question 8 of 30
When is a redox reaction spontaneous?
A. When a metal electrode is in contact with an electrolyte
B. When a power source supplies an electrical current
C. When the cell potential is positive
D. When the cell potential is negative
According to the concept of redox reactions, the answer to this question is option C.
When the cell potential is positive. A redox reaction is spontaneous when the cell potential is positive.The cell potential is the measure of the driving force of the chemical reaction occurring in the electrochemical cell. In an electrochemical cell, a redox reaction occurs, which leads to the production of an electric potential.
If this potential is positive, then the redox reaction is considered spontaneous. However, if the potential is negative, then the reaction is non-spontaneous.In general, a redox reaction is spontaneous if the potential difference between the two electrodes of the cell is positive. This means that the reaction will occur spontaneously without any external energy input.
Thus, the correct option is C.
Learn more about redox reaction,here:
https://brainly.com/question/28300253
#SPJ4
Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K sp (MgF 2) = 6.9 × 10 –9]
Answer:
No precipitate is formed.
Explanation:
Hello,
In this case, given the dissociation reaction of magnesium fluoride:
[tex]MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-[/tex]
And the undergoing chemical reaction:
[tex]MgCl_2+2NaF\rightarrow MgF_2+2NaCl[/tex]
We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:
[tex]n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2[/tex]
Next, the moles of magnesium chloride consumed by the sodium fluoride:
[tex]n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2[/tex]
Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:
[tex]n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2[/tex]
Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:
[tex][Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M[/tex]
[tex][F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M[/tex]
Thereby, the reaction quotient is:
[tex]Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}[/tex]
In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.
Regards.
Using these metal ion/metal standard reduction potentials Cd2+(aq)|Cd(s) Zn2+(aq)|Zn(s) Ni2+(aq)|Ni(s) Cu2+(aq)/Cu(s) -0.40 V -0.76 V ‑0.25 V +0.34 V Calculate the standard cell potential for the cell whose reaction is Ni2+(aq) + Zn(s) →Zn2+(aq)+ Ni(s)
Answer: The standard cell potential for the cell is +0.51 V
Explanation:
Given : [tex]E^0_{Ni^{2+}/Ni}=-0.25V[/tex]
[tex]E^0_{Zn^{2+}/Zn}=-0.76V[/tex]
The given reaction is:
[tex]Ni^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Ni(s)[/tex]
As nickel is undergoing reduction, it acts as cathode and Zinc is undergoing oxidation, so it acts as anode.
[tex]E^0_{cell}=E^0_{cathode}-E^0_{anode}[/tex]
where both [tex]E^0[/tex] are standard reduction potentials.
Thus putting the values we get:
[tex]E^0_{cell}=-0.25-(-0.76)[/tex]
[tex]E^0_{cell}=0.51V[/tex]
Thus the standard cell potential for the cell is +0.51 V
Calculate the pH of a 0.20 M NH3/0.20 M NH4Cl buffer after the addition of 25.0 mL of 0.10 M HCl to 65.0 mL of the buffer.
Answer:
Explanation:
For pH of a buffer solution , the formula is
pH = pKa + log [ Base ] / [ conjugate acid ]
= pKa + log [ NH₃ ] / [ NH₄⁺ ]
Ka = Kw / Kb
Kb for NH₄OH = 1.8 x 10⁻⁵
Ka = 10⁻¹⁴ / 1.8 x 10⁻⁵
= 5.6 x 10⁻¹⁰
pH = - log ( 5.6 x 10⁻¹⁰ ) + log 0.2 / 0.2
= 10 - log 5.6
= 9.25
Effect of addition of HCl
H⁺ of HCl will react with NH₃ to produce NH₄⁺
25 mL of .1 HCl = 2.5 mM of HCl
25 mL of .1 NH₄⁺ = 2.5 mM of NH₄⁺
65 mL of .2 M NH₃ = 13 mM of NH₃
65 mL of .2 M NH₄⁺ = 13 mM of NH₄⁺
NH₃ + H⁺ = NH₄⁺
NH₄⁺ formed = 2.5 + 13 mM
15.5 mM of NH₄⁺
NH₃ = 13 mM
Concentration of NH₃ = 13 / 90
Concentration of NH₄⁺ = 15.5 / 90
pH of final buffer mixture
= 9.6 + log 13 / 15.5
= 9.25 - .076
= 9.174
The pH value is mathematically given as
pH= -6.332.
What is the pH of a 0.20 M NH3/0.20 M NH4Cl buffer?Question Parameters:
the pH of a 0.20 M NH3/0.20 M NH4Cl buffer
the addition of 25.0 mL of 0.10 M HCl to 65.0 mL of the buffer.
Generally, the equation for the Chemical Reaction is mathematically given as
HCl + NH3 --> NH4^+ + Cl^-
Therefore
pH= pka + log(13/14).
pH= -6.3 + log 0.93.
pH= -6.3+ (-0.032).
pH= -6.332.
Read more about Chemical Reaction
https://brainly.com/question/11231920
candium forms the ion Sc3+. How many bromite ions could bond with Sc3+, and what would be the chemical formula?
a.3 bromite ions, Sc(Broa)2
b.2 bromite ions, Sc(BrO4)3
c.3 bromite ions, Sc(Broz)
d.2 bromite ions, Sc (BrO2)2
Answer: 3 bromite ions and [tex]Sc(BrO_2)_3[/tex]
Explanation:
For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
Here scandium is having an oxidation state of +3 called as [tex]Sc^{3+}[/tex] cation and bromite is an anion with oxidation state of -1 called as [tex]BrO_2^-[/tex]. Thus 1 Scandium ion combines with three bromite ions and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]Sc(BrO_2)_3[/tex]
Answer:
3 bromite ions and
Explanation:
Which of the reactions are spontaneous (favorable)? DHAP↽−−⇀glyceraldehyde-3-phosphateΔG=3.8 kJ/mol C4H4O5⟶C4H2O4+H2OΔG=3.1 kJ/mol C2H4+H2−→−−Rh(I)C2H6ΔG=−150.97 kJ/mol glutamate+NAD++H2O⟶
Answer:
C₂H₄ + H₂ →−−Rh(I) C₂H₆ Δ
what is a chemical that is safe to use in food but in small amounts?
Answer:
Toxins
Explanation:
at 89 ∘C∘C , where [Fe2+]=[Fe2+]= 3.60 MM and [Mg2+]=[Mg2+]= 0.310 MM . Part A What is the value for the reaction quotient, QQQ, for the cell?
Answer:
8.6×10^-2
Explanation:
The reaction is;
Mg(s) + Fe^2+(aq) -----> Mg^2+(aq) + Fe(s)
This implies that;
Q = [Mg^2+]/[Fe^2+]
But;
[Fe2+]= 3.60 M
[Mg2+]= 0.310 M
Q= [0.310 M]/[3.60 M]
Q= 0.086
Q= 8.6×10^-2
Nitric oxide (NO) can be formed from nitrogen, hydrogen and oxygen in two steps. In the first step, nitrogen and hydrogen react to form ammonia: (g) (g) (g) In the second step, ammonia and oxygen react to form nitric oxide and water: (g) (g) (g) (g) Calculate the net change in enthalpy for the formation of one mole of nitric oxide from nitrogen, hydrogen and oxygen from these reactions. Round your answer to the nearest .
Answer: [tex]\Delta H = -272.25kJ[/tex] for 1 mole of NO.
Explanation: Hess' Law of Constant Summation or Hess' Law states that the total enthalpy change of a reaction with multiple stages is the sum of the enthalpies of all the changes.
For this question:
1) [tex]N_{2}_{(g)} + 3H_{2}_{(g)}[/tex] => [tex]2NH_{3}_{(g)}[/tex] [tex]\Delta H=-92kJ[/tex]
2) [tex]4NH_{3}_{(g)}+5O_{2}_{(g)}[/tex] => [tex]4NO_{(g)}+6H_{2}O_{(g)}[/tex] [tex]\Delta H=-905kJ[/tex]
Amonia ([tex]NH_{3}_{(g)}[/tex]) appeares as product in the first equation and as reagent in the 2 reaction, so when adding both, there is no need to inverse reactions. However, in the 2nd, there are 4 moles of that molecule, so to cancel it, you have to multiply by 2 the first chemical equation and enthalpy:
[tex]2N_{2}_{(g)} + 6H_{2}_{(g)}[/tex] => [tex]4NH_{3}_{(g)}[/tex] [tex]\Delta H=-184kJ[/tex]
Now, adding them:
[tex]2N_{2}_{(g)} + 6H_{2}_{(g)}[/tex] => [tex]4NH_{3}_{(g)}[/tex] [tex]\Delta H=-184kJ[/tex]
[tex]4NH_{3}_{(g)}+5O_{2}_{(g)}[/tex] => [tex]4NO_{(g)}+6H_{2}O_{(g)}[/tex] [tex]\Delta H=-905kJ[/tex]
[tex]2N_{2}_{(g)}+6H_{2}_{(g)}+5O_{2}_{(g)}=>4NO_{(g)}+6H_{2}O_{(g)}[/tex] [tex]\Delta H = -185-905[/tex]
[tex]2N_{2}_{(g)}+6H_{2}_{(g)}+5O_{2}_{(g)}=>4NO_{(g)}+6H_{2}O_{(g)}[/tex] [tex]\Delta H = -1089kJ[/tex]
Note net enthalpy is for the formation of 4 moles of nitric oxide.
For 1 mole:
[tex]\Delta H = \frac{-1089}{4}[/tex]
[tex]\Delta H=-272.25kJ[/tex]
To form 1 mol of nitric oxide from nitrogen, oxygen and hydrogen, net change in enthalpy is [tex]\Delta H=-272.25kJ[/tex].
A monoprotic weak acid, HA , dissociates in water according to the reaction HA(aq)+H2O(l)↽−−⇀H3O+(aq)+A−(aq) The equilibrium concentrations of the reactants and products are [HA]=0.260 M , [H3O+]=4.00×10−4 M , and [A−]=4.00×10−4 M . Calculate the Ka value for the acid HA.
Answer:
Ka = 6.15x10⁻⁷
Explanation:
Ka is defined as dissociation constant in the equilibrium of a weak acid with water. The general reaction is:
HA(aq) + H₂O(l) ⇆ H₃O⁺(aq) + A⁻(aq)
And Ka is defined as the ratio between molar concentrations in equilibrium of products over reactants as follows:
Ka = [H₃O⁺] [A⁻] / [HA]
You don't take water in the equilibrium beacuse is a pure liquid
Replacing with the concentrations of the problem:
Ka = [H₃O⁺] [A⁻] / [HA]
Ka = [4.00x10⁻⁴] [4.00x10⁻⁴] / [0.260]
Ka = 6.15x10⁻⁷
Oxygen condenses into a liquid at approximately 90 K. What temperature, in degrees Fahrenheit, does this correspond to?
Answer:
-297.67 °F
Explanation:
Oxygen condenses into a liquid at approximately 90 K. We can convert any temperature in the Kelvin scale (absolute scale) to the Fahrenheit scale using the following expression.
°F = (K − 273.15) × 9/5 + 32
°F = (90 − 273.15) × 9/5 + 32
°F = (-183.15) × 9/5 + 32
°F = -329.67 + 32
°F = -297.67 °F
A strand of messenger RNA coding the synthesis of a protein with 146 amino acids must have a string of___________ bases along its backbone. a. 146 b. 292 c. 438 d. 20 e. 3
Answer:
Since the relationship between amino acid and codon bases would be the values of 3 nitrogenous bases per 1 amino acid.
knowing this relationship what you would do is simply multiply 146 x 3 to find the number of codon bases which would be C. 438.
A strand of messenger RNA coding the synthesis of a protein with 146 amino acids must have a string of 438 bases along its backbone and the correct option is option C.
What is messenger RNA?mRNA or messenger RNA is a single stranded RNA molecule. It is complementary to the DNA and carries genetic information present in the DNA. It is translated to form proteins. The genetic codes (triplet) present on mRNA get translated to amino acids, giving rise to the functional product of a gene.
So mRNA really is a form of nucleic acid, which helps the human genome which is coded in DNA to be read by the cellular machinery. mRNA is actually the translated form of DNA that the machinery can recognize and use to assemble amino acids into proteins.
Each strand has 3 bases, so 146 × 3 = 438 bases
Therefore, A strand of messenger RNA coding the synthesis of a protein with 146 amino acids must have a string of 438 bases along its backbone and the correct option is option C.
Learn more about messenger RNA, here:
https://brainly.com/question/15837874
#SPJ3
When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coefficients are:
________hydrochloric acid (aq) + ___________oxygen (g) → _________water (l) + ________chlorine (g)
Answer:
The coefficients are; 4, 0, 2, 2
Explanation:
The equation is given as;
HCl + O2 --> H2O + Cl2
Upon balancing the equation, we have;
4HCl + O2 --> 2H2O + 2Cl2
conversion of 35 mL to ML
Answer:
1000ml=1l
35ml. = ?
Explanation:
35×1/1000
0.035litres
what is the lewis structure for OP(N3)3
Explanation:
this is the ans
hope this helps
An HCl solution has a concentration of 0.09714 M. Then 10.00 mL of this solution was then diluted to 250.00 mL in a volumetric flask. The diluted solution was then used to titrate 250.0 mL of a saturated AgOH solution using methyl orange indicator to reach the endpoint.
Required:
a. What is the concentration of the diluted HCI solution?
b. If 7.93 mL of the diluted HCI solution was required to reach the endpoint, what is the concentration of OH- in solution?
c. What is the concentration of Ag+ in solution?
d. What is the Ksp expression for the dissolution of AgOH?
Answer:
a. 3.8856x10⁻³M HCl
b. 1.23x10⁻⁴M OH⁻
c. 1.23x10⁻⁴M Ag⁺
d. Ksp = [Ag⁺] [OH⁻]
Explanation:
a. The reaction that you are studying is:
HCl(aq) + AgOH(aq) → H₂O(l) + AgCl(s)
The HCl solution is diluted from 10.00mL to 250.00mL, that is:
250.00mL / 10.00mL = 25 -The solution is diluted 25 times-
As original concentration of HCl is 0.09714M, the concentration of the diluted solution is:
0.09714M / 25 =
3.8856x10⁻³M HClb. 1 mole of HCl reacts per mole of AgOH, moles of HCl that reacts are:
7.93mL = 7.93x10⁻³L × (3.8856x10⁻³mol HCl / L) = 3.0813x10⁻⁵ moles of HCl.
Based on the reaction, you have in solution
3.0813x10⁻⁵ moles of AgOH = Ag⁺ = OH⁻
The AgOH solution was 250.0mL = 0.2500L, its concentration is:
3.0813x10⁻⁵ moles OH⁻ / 0.2500L =
1.23x10⁻⁴M OH⁻c. In solution, AgOH produce Ag⁺ and OH⁻ in equals proportions, that means:
1.23x10⁻⁴M OH⁻ =
1.23x10⁻⁴M Ag⁺d. The solubility product reaction of AgOH(s) is:
AgOH(s) ⇄ Ag⁺(aq) + OH⁻(aq)
Where Ksp for this reaction is defined as:
Ksp = [Ag⁺] [OH⁻]f the Ksp for HgBr2 is 2.8×10−14, and the mercury ion concentration in solution is 0.085 M, what does the bromide concentration need to be for a precipitate to occur?
Answer:
0.057 M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴
Concentration of mercury (II) ion: 0.085 M
Step 2: Write the reaction for the solution of HgBr₂
HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻
Step 3: Calculate the bromide concentration needed for a precipitate to occur
The Ksp is:
Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²
[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M
An SN2 reaction is a type of _____________ in which the nucleophile attacks the electrophile.
Answer:
A bimolecular nucleophilic substitution (SN2) reaction is a type of nucleophilic substitution whereby a lone pair of electrons on a nucleophile attacks an electron deficient electrophilic center and bonds to it, resulting in the expulsion of a leaving group.
Advantages of using a resource person in handling the first aid lesson
The advantage of a resource person would be that it will provide a hands-on activity that will allow the students to experience spacing between organs and on the body of the person.
It will also allow them to identify challenges when doing this and will engage them more in the activity and lesson.
Answer:A resource person add knowledge to the course
Explanation:
An object has a mass of 4.9g and a volume of 14.ml.what is the density of the object?
Answer:
0.35 g/mL
Explanation:
Use the formula D = [tex]\frac{m}{v}[/tex], where D is density, m is mass, and v is volume.
D = 4.9/14
D = 0.35
D = 0.35 g/mL
Two elements represents by the letter Q and R atomic number 9 and 12 respectively. Write the electronic configuration of R
Answer:
Atomic no = 12 = Mg
Explanation:
It is given that,
The atomic number of two elements that are represented by letter Q and R are 9 and 12.
We need to write the electronic configuration of R. Atomic number shows the number of protons in atom.
For R, atomic number = 12
Its electronic configuration is : 2,8,2
It has two valance electrons in its outermost shell. The element is Magnesium (Mg).
The equation represents the decomposition of a generic diatomic element in its standard state. 12X2(g)⟶X(g) Assume that the standard molar Gibbs energy of formation of X(g) is 4.25 kJ·mol−1 at 2000. K and −63.12 kJ·mol−1 at 3000. K. Determine the value of K (the thermodynamic equilibrium constant) at each temperature.
Answer:
[tex]K^{2000K}=0.774\\\\K^{3000K}=12.56[/tex]
Explanation:
Hello,
In this case, considering the reaction, we can compute the Gibbs free energy of reaction at each temperature, taking into account that the Gibbs free energy for the diatomic element is 0 kJ/mol:
[tex]\Delta _rG=\Delta _fG_{X}-\frac{1}{2} \Delta _fG_{X_2}=\Delta _fG_{X}[/tex]
Thus, at 2000 K:
[tex]\Delta _rG=\Delta _fG_{X}^{2000K}=4.25kJ/mol[/tex]
And at 3000 K:
[tex]\Delta _rG=\Delta _fG_{X}^{3000K}=-63.12kJ/mol[/tex]
Next, since the relationship between the equilibrium constant and the Gibbs free energy of reaction is:
[tex]K=exp(-\frac{\Delta _rG}{RT} )[/tex]
Thus, at each temperature we obtain:
[tex]K^{2000K}=exp(-\frac{4250J/mol}{8.314\frac{J}{mol\times K}*2000K} )=0.774\\\\K^{3000K}=exp(-\frac{-63120J/mol}{8.314\frac{J}{mol\times K}*3000K} )=12.56[/tex]
In such a way, we can also conclude that at 2000 K reaction is unfavorable (K<1) and at 3000 K reaction is favorable (K>1).
Best regards.
In the reaction of Zn(s) + 2HCl (aq) Imported Asset ZnCl2 (aq) + H2 (g), if [HCl] increases from 2.6 M to 8.2 M:
The rate at which Zn disappears decreases.
The rate at which H2 appears decreases.
The rate at which ZnCl2 appears increases.
The concentration of Zn (s) also increases.
Answer:
The rate at which ZnCl2 appears increases.
Explanation:
Hello,
In this case, the reaction is:
[tex]Zn(s) + 2HCl (aq) \rightarrow ZnCl_2 (aq) + H_2 (g)[/tex]
Therefore, the law of rate proportions is:
[tex]\frac{1}{-1}r_{Zn}= \frac{1}{-2}r_{HCl}= \frac{1}{1}r_{ZnCl_2}= \frac{1}{1}r_{H_2}}[/tex]
In such a way, since the concentration of hydrochloric acid is increasing The rate at which ZnCl2 appears increases, because the addition of a reactant is directly related with the products formation due to the fact that more reactant will yield more product.
Best regards.
which law states that the pressure and absolue tempeture of a fixed quantity of gas are directly proportional under constant volume conditions?
Answer:
Gay lussacs law
Explanation:
