Answers
Answer:
1. oxidation
2. reduction
3. oxidation
4. oxidation
Explanation:
Oxidation and Reduction in terms of hydrogen
Oxidation and Reduction with respect to Hydrogen Transfer. Oxidation is the loss of hydrogen. Reduction is the gain of hydrogen.Oxidation and Reduction in terms of Oxygen
Oxidation and Reduction with respect to Oxygen Transfer. Oxidation is the gain of Oxygen. Reduction is the loss of Oxygen.Related Questions
Complete the sentences describing the cell.
a. In the nickel-aluminum galvanic cell, the cathode is ____ .
b. Therefore electrons flow from___ to ____.
c. The ____ electrode loses mass, while the ____ electrode gains mass.
Answers
Answer:
a. In the nickel-aluminum galvanic cell, the cathode is nickel electrode.
b. Therefore electrons flow from the aluminium electrode to the nickel electrode.
c. The aluminium electrode loses mass, while the nickel electrode electrode gains mass.
Explanation:
Voltaic or galvanic cells are electrochemical cells in which spontaneous oxidation-reduction reactions. The two halves of the redox reaction are separate and electron transfer is required to occur through an external circuit for the redox reaction to take place. That is, one of the metals in one of the half cells is oxidized while the metal of the other half cell is reduced, producing an exchange of electrons through an external circuit. This makes it possible to take advantage of the electric current.
Given:
E ⁰N i ⁺² = − 0.23 V is the standard reduction potential for the nickel ion
E ⁰ A l ⁺³ = − 1.66 V is the standard reduction potential for the aluminum ion
The most negative potentials correspond to more reducing substances. In this case, the aluminum ion is the reducing agent, where oxidation takes place. In the anodic half cell oxidations occur, while in the cathode half cell reductions occur. So the aluminum cell acts as the anode while the nickel cell acts as the cathode.
So a. In the nickel-aluminum galvanic cell, the cathode is nickel electrode.
The metal that is oxidized gives electrons to the metal that is reduced through the outer conductor. Then the electrons flow spontaneously from the anode to the cathode.
Then b. Therefore electrons flow from the aluminium electrode to the nickel electrode.
Ni⁺², being the cathode, accepts electrons, becoming Ni (s) and depositing on the Ni electrodes.
So, c. The aluminium electrode loses mass, while the nickel electrode electrode gains mass.
A 0.100 M solution of NaOH is used to titrate an HCl solution of unknown concentration. To neutralize the solution, an average volume of the titrant was 38.2 mL. The starting volume of the HCl solution was 20 mL. What's the concentration of the HCl? answer options: A) 0.788 M B) 0.284 M C) 3.34 M D) 0.191 M
Answers
Answer: it is A
Explanation: I am sure
Answer:
0.191 M
Explanation:
i took the test.
Which of the following is not the same as 1,400 mL? a. 1.4 cm³ b 1.4 L c. 1,400 cm³ d. 140 cL
Answers
answer should be 1.4 cm³
1 L = 10 and so
dL = 100 and then
cL = 1,000
mL = 0.001 m³
1 m³ = 1,000
dm³ = 1,000,000
cm³ = 1,000,000,000
mm³ = 1,000 L
So, 1 mL = 1 cm³ = 0.001 L = 0.1 cL
1,400 mL = 1,400 cm³ = 1.4 L = 140 cL
Answer:
1.4 cm^3
Explanation:
If the rate of formation (also called rate of production) of compound C is 2M/s in the reaction A --->2C, what is the rate of consumption of A
Answers
Answer:
[tex]r_A=-1\frac{M}{s}[/tex]
Explanation:
Hello,
In this case, given the rate of production of C, we can compute the rate of consumption of A by using the rate relationships which include the stoichiometric coefficients at the denominators (-1 for A and 2 for C) as follows:
[tex]\frac{1}{-1} r_A=\frac{1}{2}r_C[/tex]
In such a way, solving the rate of consumption of A, we obtain:
[tex]r_A=-\frac{1}{2} r_C=-\frac{1}{2}*2\frac{M}{s}\\ \\r_A=-1\frac{M}{s}[/tex]
Clearly, such rate is negative which account for consumption process.
Regards.
Select True or False: Pi bonds are covalent bonds in which the electron density is concentrated above and below a plane containing the nuclei of the bonding atoms and occurs by sideways overlap of p orbitals.
Answers
Answer:
True
Explanation:
In pi bonds, the electron density concentrates itself between the atoms of the compound but are present on either side of the line joining the atoms. Electron density is found above and below the plane of the line joining the internuclear axis of the two atoms involved in the bond.
Pi bonds usually occur by sideways overlap of atomic orbitals and this leads to both double and triple bonds.
By heating a 93% pure kclo3 sample, what percentage of its mass is reduced?
2KCLO3---->2KCL+3O2
Answers
Explanation:
free your mind drink water and go outside take fresh air you will get answers
What is the pH of 10.0 mL solution of 0.75 M acetate after adding 5.0 mL of 0.10 M HCl (assume a Ka of acetic acid of 1.78x10-5)
Answers
Answer:
5.90
Explanation:
Initial moles of CH3COO- = 10.0/1000 x 0.75 = 0.0075 mol
Moles of HCl added = 5.0/1000 x 0.10 = 0.0005 mol
CH3COO- + HCl => CH3COOH + Cl-
Moles of CH3COO- left = 0.0075 - 0.0005 = 0.007 mol
Moles of CH3COOH formed = moles of HCl added = 0.0005 mol
pH = pKa + log([CH3COO-]/[CH3COOH])
= -log Ka + log(moles of CH3COO-/moles of CH3COOH)
= -log(1.78 x 10^(-5)) + log(0.007/0.0005)
= 5.90
Answer:
The correct answer is 5.895.
Explanation:
The reaction will be,
CHCOO⁻ + H+ ⇔ CH₃COOH
Both the HCl and the acetate are having one n factor.
The millimoles of CH₃COO⁻ is,
= Volume in ml × molarity = 10 × 0.75 = 7.5
The millimoles of HCl = Volume in ml × molarity = 5 × 0.1 = 0.5
Therefore, 0.5 will be the millimoles of CH₃COOH formed, now the millimoles of the CH₃COO⁻ left will be, 7.5-0.5 = 7.0
The volume of the solution is, 10+5 = 15 ml
The molarity of CH₃COO⁻ is, millimoles / volume in ml = 7/15
The molarity of CH₃COOH is 0.5/15
pH = pKa + log[CH₃COO⁻]/[CH₃COOH]
= 4.74957 + 1.146
= 5.895
When the equation MnO₄⁻ + I⁻ + H₂O → MnO₂ + IO₃⁻ is balanced in basic solution, what is the smallest whole-number coefficient for OH⁻?
Answers
Answer:
The smallest whole-number coefficient for OH⁻ is 2
Explanation:
Step 1: The equation redox reaction is divided into two half equations
Reduction half equation: MnO₄⁻ ----> MnO₂
Oxidation half-equation: I⁻ ---> IO₃⁻
Step 2: Next the atoms are balanced by adding OH⁻ ions and H₂O molecules to the appropriate side of each half equation;
MnO₄⁻ + 2H₂O ----> MnO₂ + 4OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O
Step 3 : The charges are then balanced by adding electrons to the appropriate sides of each half equation
MnO₄⁻ + 2H₂O + 3e⁻ ----> MnO₂ + 4OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻
Step 4: Oxidation half equation is multiplied by 2 while reduction half equation is multiplied by 1 to balance the number of electrons gained and lost for the reaction
2MnO₄⁻ + 4H₂O + 6e⁻ ----> 2MnO₂ + 8OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻
Step 5 : addition of the two half equations to yield a net ionic equation
2MnO₄⁻ + I⁻ + H₂O ----> 2MnO₂ + IO₃⁻ + 2OH⁻
The smallest whole number coefficient for OH⁻ is 2
A redox reaction is divided into two half equations which are shown below:
Reduction half equation: MnO₄⁻ ----> MnO₂
Oxidation half-equation: I⁻ ---> IO₃⁻
Atoms are balanced by adding OH⁻ ions and H₂O molecules to the appropriate side of each half equation to make the equation complete ;
MnO₄⁻ + 2H₂O ----> MnO₂ + 4OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O
The charges needs to be balanced and this is done by adding electrons to the appropriate sides of each half equation
MnO₄⁻ + 2H₂O + 3e⁻ ----> MnO₂ + 4OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻
The equation needs to be balanced by multiplying the oxidation half equation by 2 while reduction half equation is multiplied by 1 to balance the number of electrons on both sides of the equations.
2MnO₄⁻ + 4H₂O + 6e⁻ ----> 2MnO₂ + 8OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻
The two half equations are then added and written together to form a net ionic equation
2MnO₄⁻ + I⁻ + H₂O ----> 2MnO₂ + IO₃⁻ + 2OH⁻
The smallest whole-number coefficient for OH⁻ is therefore 2.
Read more on https://brainly.com/question/17156617
A galvanic cell is powered by the following redox reaction:
2Zn2+(aq) + N2H4(aq) 4OH-zn2+ right arrow(aq) 2Zn(s) + N2(g) + 4H2O(I)
1. Write a balanced equation for the half-reaction that takes place at the cathode.
2. Write a balanced equation for the half-reaction that takes place at the anode.
3. Calculate the cell voltage under standard conditions.
Answers
Write the IUPAC and common names, if any, of the carboxylate salts produced in the reaction of each of the following carboxylic acids with NaOH: 2-methylhexanoic acid
Part A
2-bromopropanoic acid Spell out the full name of the compound. If there is a common name separate your answers by a comma.
Part B
2-methylhexanoic acid Spell out the full name of the compound. If there is a common name separate your answers by a comma.
Answers
Answer:
Following are the explanation to this question:
Explanation:
The salts of carboxylate are named by the writing, which is also named as the creation of the first, which is followed by the name of the carboxylic acid were '-ic' of the acid end and replaced by the 'ate'.
Following are the description of the given reaction:
In reaction A:
2-Bromopropanoic acid= [tex]C_3H_5BrO_2[/tex]
[tex]C_3H_5BrO_2+NaOH[/tex]⇄ [tex]C_3H_4BrNaO_2 +H_2O[/tex]
The IUPAC name is Sodium-2-Bromopropanate
In reaction B:
2-Methylhexanoic acid= [tex]C_7H_{14}O_2[/tex]
[tex]C_7H_{14}O_2+NaOH[/tex]⇄ [tex]C_7H_{13}NaO_{2}+H_2O[/tex]
The IUPAC name is Sodium-2-Methyl hexanoate
The IUPAC name of compound is Sodium-2-Bromopropanate in reaction 1 and in reaction 2 it is 2-Methylhexanoic acid.
The salts of carboxylate are named by the writing, which is also named as the creation of the first, which is followed by the name of the compound carboxylic acid were '-ic' of the acid end and replaced by the 'ate'.
Compound is defined as a chemical substance made up of identical molecules containing atoms from more than one type of chemical element.
Molecule consisting atoms of only one element is not called compound.It is transformed into new substances during chemical reactions. There are four major types of compounds depending on chemical bonding present in them.
Thus, the IUPAC name of compound is Sodium-2-Bromopropanate in reaction 1 and in reaction 2 it is 2-Methylhexanoic acid.
Learn more about compound,here:
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If there are a 1000 ml per 1 L and a 1000g per kilogram
a. How many ml are there in 5.0 L?
b. How many kg are there in 230g?
Answers
Answer:
hbchbjH j jas a aa a s ds d as das
Explanation:
An atom of 120In has a mass of 119.907890 amu. Calculate the mass defect (deficit) in amu/atom. Use the masses: mass of 1H atom
Answers
Answer:
a
Explanation:
answer is a on edg
Consider the following reaction at 298.15 K: Co(s)+Fe2+(aq,1.47 M)⟶Co2+(aq,0.33 M)+Fe(s) If the standard reduction potential for cobalt(II) is −0.28 V and the standard reduction potential for iron(II) is −0.447 V, what is the cell potential in volts for this cell? Report your answer with two significant figures.
Answers
Answer:
The correct answer is 0.186 V
Explanation:
The two hemirreactions are:
Reduction: Fe²⁺ + 2 e- → Fe(s)
Oxidation : Co(s) → Co²⁺ + 2 e-
Thus, we calculate the standard cell potential (Eº) from the difference between the reduction potentials of cobalt and iron, respectively, as follows:
Eº = Eº(Fe²⁺/Fe(s)) - Eº(Co²⁺/Co(s)) = -0.28 V - (-0.447 V) = 0.167 V
Then, we use the Nernst equation to calculate the cell potential (E) at 298.15 K:
E= Eº - (0.0592 V/n) x log Q
Where:
n: number of electrons that are transferred in the reaction. In this case, n= 2.
Q: ratio between the concentrations of products over reactants, calculated as follows:
[tex]Q = \frac{ [Co^{2+} ]}{[Fe^{2+} ]} = \frac{0.33 M}{1.47 M} = 0.2244[/tex]
Finally, we introduce Eº= 0.167 V, n= 2, Q=0.2244, to obtain E:
E= 0.167 V - (0.0592 V/2) x log (0.2244) = 0.186 V
what is the valency of element sulphur
Answers
sorry not sure wish I can help u
Experiment:
Part I: Voltaic Cell
Assume that you are provided with the following materials:
Strips of metallic zinc, metallic copper, metallic iron
1M aqueous solutions of ZnSO4, CuSO4, FeSO4, and aqueous iodine (I2)
Other required materials to create Voltaic cells such as beakers, porous containers, graphite rods, a voltmeter, and a few wires with alligator clips.
In this modified version of the lab, after thoroughly studying the lab hand out and watching the videos, identify 4 different combinations of Voltaic cells that are possible to be created with the above materials.
For each cell created, include the following details.
Which electrode was the anode and which was the Cathode?
The anode and cathode half reactions.
Balanced equation for each cell you propose to construct.
Calculated Eocell
Short hand notation (line notation) for each cell (be sure to include the inactive electrode if needed)
Answers
Answer:
Here are four possible voltaic cells.
Explanation:
1. Standard reduction potentials
E°/V
I₂(s) + 2e⁻ ⟶ 2I⁻(aq); 0.54
Cu²⁺(aq) + 2e⁻ ⟶ Cu(s); 0.34
Fe²⁺(aq) + 2e⁻ ⟶ Fe(s); -0.41
Zn²⁺(aq) + 2e⁻ ⟶ Zn(s); -0.76
2. Possible Voltaic cells
(a) Zn/I₂
E°/V
Anode: Zn(s) ⟶ Zn²⁺(aq) + 2e⁻; 0.76
Cathode: I₂(s) + 2e⁻ ⟶ 2I⁻(aq); 0.54
Cell: Zn(s) + I₂(s) ⟶ Zn²⁺(aq) + 2I⁻(aq); 1.30
Zn(s)|Zn²⁺(aq)∥I⁻(aq)|I₂(s)|C(s, graphite)
Zn is the anode; graphite is the cathode.
(b) Zn/Cu²⁺
E°/V
Anode: Zn(s) ⟶ Zn²⁺(aq) + 2e⁻; 0.76
Cathode: Cu²⁺(aq) + 2e⁻ ⟶ Cu(s); 0.34
Cell: Zn(s) + Cu²⁺(s) ⟶ Zn²⁺(aq) + Cu(s); 1.10
Zn(s)|Zn²⁺(aq)∥Cu²⁺(aq)|Cu(s)
Zn is the anode; Cu is the cathode.
(c) Zn/Fe²⁺
E°/V
Anode: Zn(s) ⟶ Zn²⁺(aq) + 2e⁻; 0.76
Cathode: Fe²⁺(aq) + 2e⁻ ⟶ Fe(s); -0.41
Cell: Zn(s) + Fe²⁺(s) ⟶ Zn²⁺(aq) + Fe(s); 0.35
Zn(s)|Zn²⁺(aq)∥Fe²⁺(aq)|Fe(s)
Zn is the anode; Fe is the cathode.
(d) Fe/I₂
E°/V
Anode: Fe(s) ⟶ Fe²⁺(aq) + 2e⁻; 0.41
Cathode: I₂(s) + 2e⁻ ⟶ 2I⁻(aq); 0.54
Cell: Zn(s) + I₂(s) ⟶ Zn²⁺(aq) + 2I⁻(aq); 0.95
Fe(s)|Fe²⁺(aq)∥I⁻(aq)|I₂(s)|C(s, graphite)
Fe is the anode; graphite is the cathode.
A scientist observes that the electrical resistance of a superconducting material drops to zero when the material is cooled to very low temperatures. Which of the following statements best describes what the scientist is observing?
The scientist is observing the electrical power of a superconductor.
The scientist is observing the temperature of a superconductor.
The scientist is observing an intensive property of a superconductor.
The scientist is observing an extensive property of a superconductor
Answers
Answer:
The scientist is observing an intensive property of a superconductor.
Explanation:
An intensive property is a bulk property of matter. This means that an intensive property does not depend on the amount of substance present in the material under study. Typical examples of intensive properties include; conductivity, resistivity, density, hardness, etc.
An extensive property is a property that depends on the amount of substance present in a sample. Extensive properties depend on the quantity of matter present in the sample under study. Examples of extensive properties include, mass and volume.
Resistance of a superconducting material has nothing to do with the amount of the material present hence it is an intensive property of the superconductor.
Answer:
The scientist is observing an intensive property of a superconductor.
Explanation:
its the only one that makes logical sense
If one pound is the same as 454 grams, then convert the mass of 78 grams to pounds.
Answers
Answer:
0.17 lb
Explanation:
78 g * (1 lb/454 g)=0.17 lb
If you want additional help with chemistry or another subject for FREE, check out growthinyouth.org.
Pentanone was treated with excess sodium cyanide in HCl (aq) followed by hydrogen gas has over Pd. This produced:________
A. 2-amino-1-hexanol
B. 1-amino-2-methylpentan-2-ol
C. 1-cyano-1-pentanol
D. 2-aminomethylpentan-1-ol
Answers
Answer:
B. 1-amino-2-methylpentan-2-ol
Explanation:
In this case, the first step, we have the attack of the nucleophile cyanide ([tex] CN^-[/tex] produced by sodium cyanide to the carbon on the carbonyl group (C=O) producing a negative charge in the oxygen.
Then HCl protonates the molecule to produce a cyanohydrin. This cyanohydrin can be reduced by the action of hydrogen gas ([tex]H_2[/tex]) in the presence of a catalyst ([tex]Pd[/tex]), producing an amino group. With this in mind, the final molecule is: 1-amino-2-methylpentan-2-ol.
See figure 1 to further explanations
I hope it helps!
i) Briefly discuss the strengths and weaknesses of the four spectroscopy techniques listed below. Include in your answer the specific structural information you get from each method.
IR
UV-VIS
NMR
Mass Spec
Answers
delete please .....................................
For the reaction 3H 2(g) + N 2(g) 2NH 3(g), K c = 9.0 at 350°C. What is the value of ΔG at this temperature when 1.0 mol NH 3, 5.0 mol N 2, and 5.0 mol H 2 are mixed in a 2.5 L reactor?
Answers
Answer:
ΔG = - 31.7kJ/mol
Explanation:
It is possible to find ΔG of a reaction at certain temperature knowing Kc following the equation:
ΔG = ΔG° + RT ln Q
ΔG° = -RT lnKc
ΔG = -RT lnKc + RT ln Q (1)
Where R is gas constant (8.314J/molK), T absolute temperature (350°C + 273.15 = 623.15K) and Q reaction quotient
For the reaction,
3H₂(g) + N₂(g) ⇄ 2NH₃(g)
Q = [NH₃]² / [H₂]³[N₂]
Where the concentrations of each chemical are:
[NH₃] = 1.0mol / 2.5L = 0.4M
[H₂] = 5.0mol / 2.5L = 2M
[N₂] = 2.5mol / 2.5L = 1M}
Q = [0.4M]² / [2M]³[1M]
Q = 0.02
And replacing in (1):
ΔG = -RT lnKc + RT ln Q
ΔG = -8.314J/molK*623.15K ln 9 + 8.314J/molK*623.15K ln 0.02
ΔG = - 31651J/mol
ΔG = - 31.7kJ/molSelect the correct answer.
Which state of matter is highly compressible, is made of particles moving independently of each other, and is present in large quantities near Earth’s surface?
A.
solid
B.
liquid
C.
gas
D.
plasma
Answers
Answer:
C. Gas
Explanation:
In the laboratory, a general chemistry student measured the pH of a 0.425 M aqueous solution of benzoic acid, C6H5COOH to be 2.270.
Use the information she obtained to determine the Ka for this acid.
Ka(experiment) = _____
Answers
Answer:
Ka = 6.87x10⁻⁵
Explanation:
The equilibrium of benzoic acid in water is:
C₆H₅COOH(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)
The equilibrium constant, Ka, is:
Ka = [C₆H₅COO⁻] [H₃O⁺] / [C₆H₅COOH]
The initial concentration of benzoic acid is 0.425M. In equilibrium its concentration is 0.425M - X and [C₆H₅COO⁻] [H₃O⁺] = X.
X is the reaction coordinate. How many acid produce C₆H₅COO⁻ and H₃O⁺ until reach equilibrium.
Concentrations in equilibrium are:
[C₆H₅COOH] = 0.425M - X[C₆H₅COO⁻] = X [H₃O⁺] = XpH is defined as -log [H₃O⁺]. As pH = 2.270
2.270 = -log [H₃O⁺]
10^-2.270 = [H₃O⁺]
5.37x10⁻³M = [H₃O⁺] = X.
Replacing, concentrations in equilibrium are:
[C₆H₅COOH] = 0.425M - 5.37x10⁻³M = 0.4196M
[C₆H₅COO⁻] = 5.37x10⁻³M
[H₃O⁺] = 5.37x10⁻³M
Ka = [5.37x10⁻³M] [5.37x10⁻³M] / [0.4196M]
Ka = 6.87x10⁻⁵I think I'm typing it into my calculator wrong. I will give brainliest to whoever gets it right.
Answers
Answer:
36.7 mg
Explanation:
The following data were obtained from the question.
Original amount (A₀) = 65.1 mg
Rate constant (K) = 2.47×10¯² years¯¹
Time (t) = 23.2 years
Amount of substance remaining (A) =?
Thus, we can obtain the amount of substance remaining after 23.2 years as follow
ln A = lnA₀ – Kt
lnA = ln(65.1) – (2.47×10¯² × 23.2)
lnA = 4.1759 – 0.57304
lnA = 3.60286
Take the inverse of ln
A = e^3.60286
A = 36.7 mg
Therefore, the amount remaining after 23.2 years is 36.7 mg.
Draw the product that valine forms when it reacts with excess CH3CH2OH and HCl followed by a wash with aqueous base.
Answers
Answer:
Product: ethyl L-valinate
Explanation:
If we want to understand what it is the molecule produced we have to analyze the reagents. We have valine an amino acid, in this kind of compounds we have an amine group ([tex]NH_2[/tex]) and a carboxylic acid group ([tex]COOH[/tex]). Additionally, we have an alcohol ([tex]CH_3CH_2OH[/tex]) in the presence of HCl (a strong acid) in the first step, and a base ([tex]OH^-[/tex]).
When we have an acid and an alcohol in a vessel we will have an esterification reaction. In other words, an ester is produced. As the first step, the oxygen in the C=O (in the carboxylic acid group) would be protonated. In the second step, the ethanol attacks the carbon in the C=O of the carboxylic acid group producing a new bond between the oxygen in the ethanol and the carbon in the carboxylic acid. In step 3, a proton is transferred to produce a better leaving group ([tex]H_2O[/tex]). In step 4, a water molecule leaves the main structure to produce again the double bond C=O. Finally, a base ([tex]OH^-[/tex]) removes the hydrogen from the C=O bond to produce ethyl L-valinate
See figure 1
I hope it helps!
Assume you dissolve 0.235 g of the weak benzoic acid, C6H5CO2H in enough water to make 100.0 mL of the solution and then titrate the solution with 0.108 M NaOH. Benzoic acid is a monoprotic acid.
1. What is the pH of the original benzoic acid solution before the titration is started?
2. What is the pH when 7.00 mL of the base is added? (Hint: This is in the buffer region.)
3. What is the pH at the equivalence point?
Answers
Answer:
1. pH = 2.98
2. pH = 4.02
3. pH = 8.12
Explanation:
1. Initial molarity of benzoic acid (Molar mass: 122.12g/mol; Ka = 6.14x10⁻⁵) is:
0.235 ₓ (1mol / 122.12g) = 1.92x10⁻³ moles / 0.100L = 0.01924M
The equilibrium of benzoic acid with water is:
C6H5CO2H(aq) + H2O(l) → C6H5O-(aq) + H3O+(aq)
And Ka is defined as the ratio between equilibrium concentrations of products over reactants, thus:
Ka = 6.14x10⁻⁵ = [C6H5O⁻] [H3O⁺] / [C6H5CO2H]
The benzoic acid will react with water until reach equilibrium. And equilibrium concentrations will be:
[C6H5CO2H] = 0.01924 - X
[C6H5O⁻] = X
[H3O⁺] = X
Replacing in Ka:
6.14x10⁻⁵ = [X] [X] / [0.01924 - X]
1.1815x10⁻⁶ - 6.14x10⁻⁵X = X²
1.1815x10⁻⁶ - 6.14x10⁻⁵X - X² = 0
Solving for X:
X = -0.0010→ False solution. There is no negative concentrations
X = 0.0010567M → Right solution.
pH = - log [H3O⁺] and as [H3O⁺] = X:
pH = - log [0.0010567M]
pH = 2.982.
pH of a buffer is determined using H-H equation (For benzoic acid:
pH = pka + log [C6H5O⁻] / [C6H5OH]
pKa = -log Ka = 4.21 and [] could be understood as moles of each chemical
The benzoic acid reacts with NaOH as follows:
C6H5OH + NaOH → C6H5O⁻ + Na⁺ + H₂O
That means NaOH added = Moles C6H5O⁻ And C6H5OH = Initial moles (1.92x10⁻³ moles - Moles NaOH added)
7.00mL of NaOH 0.108M are:
7x10⁻³L ₓ (0.108 mol / L) = 7.56x10⁻⁴ moles NaOH = Moles C₆H₅O⁻
And moles C6H5OH = 1.92x10⁻³ moles - 7.56x10⁻⁴ moles = 1.164x10⁻³ moles C₆H₅OH
Replacing in H-H equation:
pH = 4.21 + log [7.56x10⁻⁴ moles] / [ 1.164x10⁻³ moles]
pH = 4.023. At equivalence point, all C6H5OH reacts producing C6H5O⁻. The moles are 1.164x10⁻³ moles
Volume of NaOH to reach equivalence point:
1.164x10⁻³ moles ₓ (1L / 0.108mol) = 0.011L. As initial volume was 0.100L, In equivalence point volume is 0.111L and concentration of C₆H₅O⁻ is:
1.164x10⁻³ moles / 0.111L = 0.01049M
Equilibrium of C₆H₅O⁻ with water is:
C₆H₅O⁻(aq) + H₂O(l) ⇄ C₆H₅OH(aq) + OH⁻(aq)
Kb = [C₆H₅OH] [OH⁻]/ [C₆H₅O⁻]
Kb = kw / Ka = 1x10⁻¹⁴ / 6.14x10⁻⁵ = 1.63x10⁻¹⁰
Equilibrium concentrations of the species are:
C₆H₅O⁻ = 0.01049M - X
C₆H₅OH = X
OH⁻ = X
Replacing in Kb expression:
1.63x10⁻¹⁰ = X² / 0.01049- X
1.71x10⁻¹² - 1.63x10⁻¹⁰X - X² = 0
Solving for X:
X = -1.3x10⁻⁶ → False solution
X = 1.3076x10⁻⁶ → Right solution
[OH⁻] = 1.3076x10⁻⁶
as pOH = -log [OH⁻]
pOH = 5.88
And pH = 14 - pOH
pH = 8.12Does a reaction occur when aqueous solutions of potassium hydroxide and chromium(III) bromide are combined
Answers
Explanation:
Potassium hydroxide = KOH
Chromium(iii)bromide = CrBr3
Yes! A reaction occurs. This is given by the balanced equation;
3 KOH + CrBr3 → 3 KBr + Cr(OH)3
"Aqueous solutions of lead nitrate and ammonium chloride are mixed" together. Which statement is correct
Answers
Answer:
PbCl₂ will precipitate from solution.
Explanation:
Statements are:
Insufficient information is given.
Both NH4NO3 and PbCl2 precipitate from solution.
No precipitate forms.
PbCl2 will precipitate from solution.
NH4NO3 will precipitate from solution.
The reaction of ammonium chloride (NH₄Cl) with lead nitrate (Pb(NO₃)₂) is:
Pb(NO₃)₂ + 2NH₄Cl → PbCl₂ + 2 NH₄NO₃
Talking of rules of solubility, all nitrates are soluble in water, that means NH₄NO₃ is soluble and no precipitate is formed.
In the same way, all chlorides are soluble except silver chloride and lead chloride. That means:
PbCl₂ (Lead chloride) will precipitate from solution.In which of the following compounds does the carbonyl stretch in the IR spectrum occur at the lowest wavenumber?
a. Cyclohexanone
b. Ethyl Acetate
c. λ- butyrolactone
d. Pentanamide
e. Propanoyl Chloride
Answers
Answer:
a. Cyclohexanone
Explanation:
The principle of IR technique is based on the vibration of the bonds by using the energy that is in this region of the electromagnetic spectrum. For each bond, there is a specific energy that generates a specific vibration. In this case, you want to study the vibration that is given in the carbonyl group C=O. Which is located around 1700 cm-1.
Now, we must remember that the lower the wavenumber we will have less energy. So, what we should look for in these molecules, is a carbonyl group in which less energy is needed to vibrate since we look for the molecule with a smaller wavenumber.
If we look at the structure of all the molecules we will find that in the last three we have heteroatoms (atoms different to carbon I hydrogen) on the right side of the carbonyl group. These atoms allow the production of resonance structures which makes the molecule more stable. If the molecule is more stable we will need more energy to make it vibrate and therefore greater wavenumbers.
The molecule that fulfills this condition is the cyclohexanone.
See figure 1
I hope it helps!
10. For the following isotopes that have missing information, fill in the missing informatic
complete the notation: 36P
Answers
Answer:
Krypton.
Explanation:
Krypton is an atom which has 36 protons in its nucleus. There are 31 isotopes of Krypton which have same number of protons i. e. 36, same number of electrons i. e. 36 but different number of neutrons. Isotope refers to those atoms having same atomic number i. e. number of proton but different mass number i. e. number of neutron. For example, in Krypton-78, there 36 protons and 42 neutrons.
What is the primary source of energy in most living communities?
Answers
Answer:
The sun
Explanation:
The sun is the primary source of energy in most living communities. The producers or the green plants that prepare their own food by the use of sunlight and other natural resources. Carbon dioxide, water, and other minerals are used by the plants to make their food in the presence of chlorophyll. Plants are then consumed by the consumers. This chain helps in forming the food chain and the food web.
