Heating carvone with aqueous sulfuric acid converts it into carvacrol. The mechanism involves the following steps:
1. The terminal alkene of carvone reacts with acid to form tertiary carbocation 1;
2. A hydride shift results in the formation of tertiary carbocation 2;
3. Deprotonation of the ring leads to conjugated diene 3;
4. Deprotonation at the α carbon leads to the product carvacrol.
Required:
Draw the mechanism and then draw the structure of tertiary carbocation 2.
Answers
Answer:
See figure 1
Explanation:
In this question, we have to start with the protonation of the double bond. In carvone we have two double bonds, so, we have to decide first which one would be protonated.
The problem states that the terminal alkene is the one that would is protonated. Therefore, we have to do the protonation in the double bond at the bottom to produce the carbocation number 1. Then, a hydride shift takes place to produce the carbocation number 2. A continuation, an elimination reaction takes place to produce the conjugated diene. Then the diene is protonated at the carbonyl group and with an elimination reaction of an hydrogen in the alpha carbon we can obtain carvacol.
Related Questions
Which functional group does the molecule below have?
A. Ether
B. Ester
C. Hydroxyl
D. Amino
Answers
Answer:
Hydroxyl
Explanation:
A hydroxyl group is a functional group that attaches to some molecules containing an oxygen and hydrogen atom, bonded together. Also spelled hydroxy, this functional group provides important functions to both alcohols and carboxylic acids.
The functional groups are the part of the organic chemistry that confers the characteristic feature of a molecule. The molecule has a hydroxyl group in its structure. Thus, option C is correct.
What are hydroxyl functional groups?Hydroxyl functional groups are the atoms or molecules that provide a distinctive property to a compound. It has a chemical formula of -OH that has oxygen covalently bonded to the hydrogen atom.
The hydroxyl group is called the alcohol group that is seen in methanol, ethanol, propanol, etc. The presence of hydrogen allows the compound to form a water bond with other molecules and makes them soluble and polar.
Therefore, option C. the molecule has a hydroxyl or alcoholic functional group attached to its carbon atom.
Learn more about the hydroxyl functional group here:
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At what pressure would 11.1 moles of a gas occupy 44.8 L at 300 K?
Answers
Answer:
[tex]P=6.10atm[/tex]
Explanation:
Hello,
In this case, we can study the ideal gas equation that relates temperature, volume, pressure and moles as shown below:
[tex]PV=nRT[/tex]
Thus, since we are asked to compute the pressure y simply solve for it as follows:
[tex]P=\frac{nRT}{V}=\frac{11.1mol*0.082\frac{atm*L}{mol*K}*300K}{44.8L}\\ \\P=6.10atm[/tex]
Best regards.
Which phase change is an example of an exothermic process?
A.
solid to liquid
B.
solid to gas
C.
liquid to solid
D.
liquid to gas
E.
solid to plasma
Reset
Answers
Answer:
C
Explanation:
Turning liquid to a solid is like freezing water to ice and requires the water to LOSE (release) heat causing an exothermic reaction.
An endothermic process is when heat is absorbed to break bonds between particles (liquid-> gas)
which statement describes the use of a flowchart?
Answers
Answer:
A flowchart is a type of diagram that represents a workflow or process
Explanation:
Answer: orders in which steps in a process happen
Explanation:
Calculate the maximum volume in mL of 0.18 M HCl that a tablet containing 340 mg Al(OH)3 and 516 mg Mg(OH)2 would be expected to neutralize. Assume complete neutralization.
Answers
Answer:
171 mL of HCl
Explanation:
The first thing we want to do is consider the reaction between Al(OH)3 and water - as that is the expected reaction that is taking place,
Al(OH)3 + 3HCl → AlCl3 + 3H2O
Knowing this, let's identify the mass of Al(OH)3. Aluminum = 27 g / mol, Oxygen( 3 ) = 16 [tex]*[/tex] 3 = 48, Hydrogen ( 3 ) = 1 [tex]*[/tex] 3 = 3 - 27 + 48 + 3 = 78 g / mol. This value is approximated however ( 78 g / mol ), as the molar mass of each substance is rounded as well. Another key thing we need to do here is to convert 340 mg → grams, considering that that unit is a necessity with respect to moles, as you might know - 340 mg = 0.340 g.
Now we can calculate how much moles of HCl will be present in solution, provided we have sufficient information for that,
(0.340 g Al(OH)3) / (78.0036 g / mol Al(OH)3) [tex]*[/tex] (3 mol HCl / 1 mol Al(OH)3)
⇒ (.004358773185 g^2 / mol Al(OH)3) [tex]*[/tex] (3 HCl / Al(OH)3 )
⇒ .01307632 mol HCl
We can apply this same concept on the reaction of Mg(OH)2 and water, receiving the number of moles of HCl when that takes place. Then we can add the two ( moles of HCl ) and divide by the value " 0.18 mol / L " given to us.
" Mg(OH)2 + 2HCl → MgCl2 + 2H2O "
Molar mass of Mg(OH)2 = 58.3197 g / mol,
516 mg = 0.516 g
(0.516 g Mg(OH)2) / (58.3197 g / mol Mg(OH)2) [tex]*[/tex] (2 mol HCl / 1 mol Mg(OH)2)
= .017695564 mol HCL
___________
( .01307632 + .017695564 ) / ( 0.18 M HCl )
= 0.170954911 L
= 171 mL of HCl
When a 2.75g sample of liquid octane (C8H18) is burned in a bomb calorimeter, the temperature of the calorimeter rises from 22.0 °C to 41.5 °C. The heat capacity of the calorimeter, measured in a separate experiment, is 6.18 kJ/°C. Determine the ΔE for octane combustion in units of kJ/mol octane.
Answers
Answer:
THE HEAT OF COMBUSTION IS 4995.69 kJ/mol OF OCTANE.
Explanation:
Heat capacity = 6.18 kJ/C
Temperature change = 41.5 C - 22.0 C = 19.5 C
Heat required to raise the temperature by 19.5 °C is:
Heat = heat capacity * temperature change
Heat = 6.18 kJ/ C * 19.5 C
heat = 120.51 kJ of heat
120.51 kJ of heat is required to raise the temperature of 2.75 g sample of a liquid octane.
Molar mass of octane = ( 12* 8 + 1 * 18) = 114 g/mol
So therefore, the heat of the reaction per mole of octane will be:
120.51 kJ of heat is required for 2.75 g of octane
x J of heat will be required for 114 g of octane
x J = 120.51kJ * 114 / 2.75
x = 4995.69 kJ of heat per mole.
In conclusion, the heat of the combustion reaction in kJ / mole of octane is 4995.69 kJ/mol
g If you have three identical containers (same volume) at the same temperature and pressure, each with a different gas. Container A has He, container B has Ne, and container C has O2. Which flask contains the largest number of molecules? Group of answer choices
Answers
Answer:
The three gases, in the three identical containers, will all have the same number of molecules
Explanation:
If these three gases (Helium He, Neon Ne, and Oxygen [tex]O_{2}[/tex]) are all contained in separate identical containers with the same volume. And they are all stored at the same temperature, and pressure. Then, they'll all contain the same number of molecules. This is in line with Avogadro's law which states that "Equal volume of all gases, at the same temperature and pressure, have the same number of molecules."
What is an example of a molecular compound
Answers
Answer:
Molecular compounds are inorganic compounds that take the form of discrete (covalent) molecules. Examples include such familiar substances as water (H2O) and carbon dioxide (CO2).
Eugenol is a molecule that contains the phenolic functional group. Which option properly identifies the phenol in eugenol
Answers
Answer:
Explanation:
Hello,
Among the options given on the attached document, since phenolic functional group is characterized by a benzene ring bonded with a hydroxyl group (C₆H₅OH) we can see that the first option correctly points out such description. Thus, answer is on the second attached picture. Other options are related with other sections found in eugenol that are not phenolic.
Best regards.
The first option identified the phenol in eugenol.
Phenolic functional groupAccording to the attached image, since the phenolic functional group should be characterized by a benzene ring bonded along with a hydroxyl group (C₆H₅OH) so here we can see that the first option correctly points out such description. However, other options are related to other sections found in eugenol that are not phenolic.
learn more about molecule here: https://brainly.com/question/13127022
Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a
75.0 L tank with 3.8 mol of sulfur dioxide gas and 7.0 mol of oxygen gas, and when the mixture has come to equilibrium measures the amount of sulfur trioxide
gas to be 1.5 mol
Calculate the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2
significant digits.
Answers
Answer:
[tex]\large \boxed{5.1}[/tex]
Explanation:
1. Initial concentrations of reactants
[SO₂] = (3.8 mol)/(75 L) = 0.0507 mol·L⁻¹
[O₂] = (7.0 mol)/(75 L) = 0.0933 mol·L⁻¹
2. Equilibrium concentration of SO₃
[SO₃] = (1.5 mol)/(75 L) = 0.0200 mol·L⁻¹
3. Set up an ICE table
2SO₂ + O₂ ⇌ 2SO₃
I/mol·L⁻¹: 0.0507 0.0933 0
C/mol·L⁻¹: -2x -x +2x
E/mol·L⁻¹: 0.0507-2x 0.0933-x 2x
4. Calculate x
We know the final concentration of SO₃ is 0.0200 mol·L⁻¹, so
2x = 0.0200
x = 0.0100
5. Find the final concentrations of the reactants
Insert the numbers into the ICE table.
2SO₂ + O₂ ⇌ 2SO₃
I/mol·L⁻¹: 0.0507 0.0933 0
C/mol·L⁻¹: -0.0200 -0.0100 +0.0200
E/mol·L⁻¹: 0.0307 0.0833 0.0200
6. Calculate K
[tex]K_{\text{eq}} = \dfrac{\text{[SO$_{3}$]}^{2}}{\text{[SO}_{2}]^{2}\text{[O$_{2}$]}} = \dfrac{0.0200^{2}}{0.0307^{2}\times0.0833} =\mathbf{5.1}\\\\\text{The value of the equilibrium constant is $\large \boxed{\mathbf{5.1}}$}[/tex]
The Lucas test has _______ results based on the type of alcohol present because the reaction involves a _________, which is ________ stable for tertiary alcohols compared to primary alcohols. Therefore, tertiary alcohols react ________ primary alcohols.
Answers
Answer:
1) positive
2) carbocation
3) most stable
4) faster
Explanation:
A common test for the presence of alcohols can be achieved using the Lucas reagent. Lucas reagent is a mixture of concentrated hydrochloric acid and zinc chloride.
The reaction of Lucas reagent reacts with alcohols leading to the formation of an alkyl chloride. Since the reaction proceeds via a carbocation mechanism, tertiary alcohols give an immediate reaction. Once a tertiary alcohol is mixed with Lucas reagent, the solution turns cloudy almost immediately indicating an instant positive reaction.
Secondary alcohols may turn cloudy within five minutes of mixing the solutions. Primary alcohols do not significantly react with Lucas reagent obviously because they do not form stable carbocations.
Therefore we can use the Lucas reagent to distinguish between primary, secondary and tertiary alcohols.
For each of the following, classify the substance as a strong acid, strong base, weak acid, or weak base (or perhaps not acidic or basic). Then determine the pH of the solution and calculate the concentrations of all aqueous species present in the solution.a. 2.0 × 10 ^–2 M HBrb. 1.0 × 10^–4 M NaOHc. 0.0015 M Ba(OH)2 d. 0.25 M HCN e. 2.0 × 10 ^–10 M KOH f. 0.050 M NH3 g. 0.100 M NH4Cl h. 0.200 M CaF2 i. 0.0500 M Ba(NO3)2 j. 0.100 M Al(NO3)3
Answers
Answer:
a. Strong acid, pH = 1.69
b. Strong base, pH = 10
c. Strong base, pH = 11
d. Weak acid, pH = 4.90
e. Strong base, pH ≅ 7 (pH should be higher than 7, but the base is so diluted)
f. Weak base, pH = 10.96
g. Acidic salt, pH = 5.12
h. Basic salt, pH = 8.38
i. Neutral salt, pH = 7
j. Acidic salt, pH < 7
Explanation:
a. HBr → H⁺ + Br⁻
Hydrobromic acid is a strong acid.
pH = - log [H⁺]
- log 0.02 = 1.69
b. NaOH → Na⁺ + OH⁻
Sodium hydroxide is a strong base.
pH = 14 - pOH
pOH = - log [OH⁻]
pH = 14 - (-log 0.0001) = 10
c. Ba(OH)₂ → Ba²⁺ + 2OH⁻
Barium hydroxide is a strong base
[OH⁻] = 2 . 0.0015 = 0.003M
pH = 14 - (-log 0.003) = 11
d. HCN + H₂O ⇄ H₃O⁺ + CN⁻
This is a weak acid, it reacts in water to make an equilibrium between the given protons and cyanide anion.
To calculate the [H₃O⁺] we must apply, the Ka
Ka = [H₃O⁺] . [CN⁻] / [HCN]
6.2×10⁻¹⁰ = x² / 0.25-x
As Ka is really small, we can not consider the x in the divisor, so we avoid the quadratic formula.
[H₃O⁺] = √(6.2×10⁻¹⁰ . 0.25) = 1.24×10⁻⁵
-log 1.24×10⁻⁵ = 4.90 → pH
e. KOH → K⁺ + OH⁻
2×10⁻¹⁰ M → It is a very diluted concentration, so we must consider the OH⁻ which are given, by water.
In this case, we propose the mass and charges balances equations.
Analytic concentration of base = 2×10⁻¹⁰ M = K⁺
[OH⁻] = K⁺ + H⁺ → Charges balance
The solution's hydroxides are given by water and the strong base.
Remember that Kw = H⁺ . OH⁻, so H⁺ = Kw/OH⁻
[OH⁻] = K⁺ + Kw/OH⁻. Let's solve the quadratic equation.
[OH⁻] = 2×10⁻¹⁰ + 1×10⁻¹⁴ /OH⁻
OH⁻² = 2×10⁻¹⁰. OH⁻ + 1×10⁻¹⁴
2×10⁻¹⁰. OH⁻ + 1×10⁻¹⁴ - OH⁻²
We finally arrived at the answer [OH⁻] = 1.001ₓ10⁻⁷
pH = 14 - (- log1.001ₓ10⁻⁷) = 7
The strong base is soo diluted, that water makes the pH be a neutral value.
Be careful, if you determine the [OH⁻] as - log 2×10⁻¹⁰, because you will obtain as pOH 9.69, so the pH would be 4.31. It is not possible, KOH is a strong base and 4.30 is an acid pH.
f. Ammonia is a weak base.
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻
Kb = OH⁻ . NH₄⁺ / NH₃
1.74×10⁻⁵ = x² / 0.05 - x
We can avoid the x from the divisor, so:
[OH⁻] = √(1.74×10⁻⁵ . 0.05) = 9.32×10⁻⁴
pH = 14 - (-log 9.32×10⁻⁴ ) = 10.96
g. NH₄Cl, an acid salt. We dissociate the compound:
NH₄Cl → NH₄⁺ + Cl⁻. We analyse the ions:
Cl⁻ does not make hydrolisis to water. In the opposide, the ammonium can react given OH⁻ to medium, that's why the salt is acid, and pH sould be lower than 7
NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺ Ka
Ka = NH₃ . H₃O⁺ / NH₄⁺
5.70×10⁻¹⁰ = x² / 0.1 -x
[H₃O⁺] = √ (5.70×10⁻¹⁰ . 0.1) = 7.55×10⁻⁶
pH = - log 7.55×10⁻⁶ = 5.12
As Ka is so small, we avoid the x from the divisor.
h. CaF₂ → Ca²⁺ + 2F⁻
This is a basic salt.
The Ca²⁺ does not react to water. F⁻ can make hydrolisis because, the anion is the strong conjugate base, of a weak acid.
F⁻ + H₂O ⇄ HF + OH⁻ Kb
Kb = x² / 2 . 0.2 - x
Remember that, in the original salt we have an stoichiometry of 1:2, so 1 mol of calcium flouride may have 2 moles of flourides.
As Kb is small, we avoid the x, so:
[OH⁻] = √(1.47×10⁻¹¹ . 2 . 0.2) = 2.42×10⁻⁵
14 - (-log 2.42×10⁻⁵) = pH → 8.38
i . Neutral salt
BaNO₃₂ → Ba²⁺ + 2NO₃⁻
Ba²⁺ comes from a strong base, so it is the conjugate weak acid and it does not react to water. The same situation to the nitrate anion. (The conjugate weak base, from a strong acid, HNO₃)
pH = 7
j. Al(NO₃)₃, this is an acid salt.
Al(NO₃)₃ → Al³⁺ + 3NO₃⁻
The nitrate anion is the conjugate weak base, from a strong acid, HNO₃ so it does not make hydrolisis. The Al³⁺ comes from the Al(OH)₃ which is an amphoterous compound (it can react as an acid or a base) but the cation has an acidic power.
Al·(H₂O)₆³⁺ + H₂O ⇄ Al·(H₂O)₅(OH)²⁺ + H₃O⁺
Modern atomic theory states that atoms are neutral. How is this neutrality achieved in atoms? (2 points)
Answers
Ag+(aq)+2NH3(aq)⇌Ag(NH3)2+(aq) : A g + ( a q ) + 2 N H 3 ( a q ) ⇌ A g ( N H 3 ) 2 + ( a q ) : blank is the Lewis acid and blank is the Lewis base. is the Lewis acid and A g + ( a q ) + 2 N H 3 ( a q ) ⇌ A g ( N H 3 ) 2 + ( a q ) : blank is the Lewis acid and blank is the Lewis base. is the Lewis base.
Answers
Answer:
Silver ion - Lewis acid, Ammonia - Lewis base
Explanation:
The reaction is given as;
Ag+(aq) + 2NH3(aq) ⇌ [Ag(NH3)]2+(aq)
A lewis acid is an electron pair acceptor. While a lewis base is any substance that that can donate a pair of nonbonding electrons.
This reaction however is a complexation reaction, where ammonia is reacting with the silver ion.
Silver ion accepts electrons in this reaction, hence it is the lewis acid. The ammonia on the other hand donates the electrons used in bonding so it is the lewis base.
A crystal lattice formed by positive and negative ions is called a
Answers
Answer:
Ionic Crystal
Explanation:
In each of the three reactions between NaOH and HCl, the sign of q for the water was positive. This means the the sign of q for the reaction was ______ and the reaction was ______.
Answers
Answer:
This means the the sign of q for the reaction was _NEGATIVE _____ and the reaction was _EXOTHERMIC_____.
Explanation:
In calorimetry, when heat is absorbed by the solution, the q-value of the solution will have a positive value. This means that the reaction will produce heat for the solution to absorb and thus the q-value for the reaction will be negative. This is an exothermic reaction.
Whereas, when heat is absorbed from the solution, the q-value for the solution will have a negative value. This means that the reaction will absorb heat from the solution and so the reaction is endothermic, and q value for the reaction is positive.
So, from the question, since the q-value of water is positive, it means that heat is absorbed by the solution and the reaction will produce a negative value of q and it's an exothermic reaction because the reaction produces heat for the solution.
Write the condensed formula from left to right, starting with (CH3)x where x is a number.
Answers
Complete question:
Write the condensed formula from left to right, starting with (CH3)x where x is a number.
See attached image for the structure formula of the compound
Answer:
(CH₃)₂CHC(CH₃)₃ named as 2,2,3-Trimethylbutane
Explanation:
If we number the longest chain of the carbon starting from the left, we will observe that there are four carbons in the straight chain as shown in the image.
Starting from first carbon from the left of the carbon chain, at carbon number number 2, there two alkyl group, that is two methyl (CH3 is two). Also at carbon number 3, there are three alkyl group, that is three methyl (CH3 is three).
The condensed formula will be written as;
(CH₃)₂CHC(CH₃)₃
This compound is named as 2,2,3-Trimethylbutane, an isomer of Heptane
In a mixture of argon and hydrogen, occupying a volume of 1.66 L at 910.0 mmHg and 54.9oC, it is found that the total mass of the sample is 1.13 g. What is the partial pressure of argon
Answers
Answer:
Partial pressure (Ar) = 316.1mmHg
Explanation:
In the mixture of Ar and H₂ you can find the total moles of both gases using general gas law and with the mass of the sample and molar weight of each gas find the mole fraction of Argon and thus, its partial pressure.
Moles of gases:
PV = nRT
P = 910.0mmHg ₓ (1atm / 760mmHg) = 1.1974atm
V = 1.66L
n = Moles gases
R = 0.082atmL/molK
T = 54.9°C + 273.15K = 328.05K
PV = nRT
1.1974atm*1.66L = n*0.082atmL/molK*328.05K
0.0739 moles = total moles of the sampleKnowing H₂ = 2.016g/mol and Ar = 39.948g/mol you can write:
1.13g = 2.016X + 39.948Y (1)
Where X = moles of hydrogen and Y = moles of Argon.
Also we can write:
0.0739moles = X + Y (2)
Total moles of the sample are moles of hydrogen + moles Argon
Replacing 2 in 1:
1.13g = 2.016(0.0739-Y) + 39.948Y
1.13 = 0.1564 - 2.016Y + 39.948Y
0.9736 = 37.932Y
0.02567 = Y = moles of Argon
As total moles are 0.0739moles, mole fraction of Ar in the sample are:
XAr = 0.02567mol / 0.0739mol
X Ar = 0.347
Last, partial pressure of Ar = X Ar * total pressure.
Partial pressure (Ar) = 0.347*910.0mmHg
Partial pressure (Ar) = 316.1mmHg
Which Group is in the second column of the periodic table?
A. Noble gases
B. Halogens
C. Alkali metals
D. Alkaline earth metals
Answers
Answer:
Hey there!
That would be the alkaline earth metals.
Hope this helps :)
Answer: alkaline earth metals
Explanation:
Convert the following measurement
Answers
Answer:
6.9 Kg/mol•dL
Explanation:
To convert 6.9×10⁴ g/mol•L to kg/mol•dL,
First, we shall convert to kg/mol•L.
This can be achieved by doing the following:
Recall: 1 g = 1×10¯³ Kg
1 g/mol•L = 1×10¯³ Kg/mol•L.
Therefore,
6.9×10⁴ g/mol•L = 6.9×10⁴× 1×10¯³
6.9×10⁴ g/mol•L = 69 Kg/mol•L
Finally, we shall convert 69 Kg/mol•L to Kg/mol•dL.
This is illustrated below:
Recall: 1 L = 10 dL
1 Kg/mol•L = 1×10¯¹ Kg/mol•dL
Therefore,
69 Kg/mol•L = 69 × 1×10¯¹
69 Kg/mol•L = 6.9 Kg/mol•dL
Therefore, 6.9×10⁴ g/mol•L is equivalent to 6.9 Kg/mol•dL.
9. Ibuprofen contain which of the following two functional groups: (1 point)
A) benzene
B) halogen
C) carboxyl
D) hydroxyl
Answers
Answer:
A and C
I hope this helps you:)
A 45.0 mL sample of 0.020 M acetic acid (HC2H3O2) is titrated with 0.020 M NaOH.? Determine the pH of the solution after adding 35.0 mL of any NaOH. (Ka of acetic acid is 1.8 x 10-5) HC2H3O2 (aq) + NaOH (aq) D NaC2H3O2(aq) + H2O (l) (Hint: Calculate new concentration and ICE table)
Answers
Answer:
Explanation:
CH₃COOH + NaOH = CH₃COONa + H₂O .
.02M
CH₃COOH = CH₃COO⁻ + H⁺
C xC xC
Ka = xC . xC / C = x² C
1.8 x 10⁻⁵ = x² . .02
x² = 9 x 10⁻⁴
x = 3 x 10⁻²
= .03
concentration of H⁺ = xC = .03 . .02
= 6 x 10⁻⁴ M , volume = 45 x 10⁻³ L
moles of H⁺ = 6 X 10⁻⁴ x 45 x 10⁻³
= 270 x 10⁻⁷ moles
= 2.7 x 10⁻⁵ moles
concentration of NaOH = .0200 M , volume = 35 x 10⁻³ L
moles of Na OH = 2 X 10⁻² x 35 x 10⁻³
= 70 x 10⁻⁵ moles
=
NaOH is a strong base so it will dissociate fully .
there will be neutralisation reaction between the two .
Net NaOH remaining = (70 - 2.7 ) x 10⁻⁵ moles
= 67.3 x 10⁻⁵ moles of NaOH
Total volume = 45 + 35 = 80 x 10⁻³
concentration of NaOH after neutralisation.= 67.3 x 10⁻⁵ / 80 x 10⁻³ moles / L
= 8.4125 x 10⁻³ moles / L
OH⁻ = 8.4125 x 10⁻³
H⁺ = 10⁻¹⁴ / 8.4125 x 10⁻³
= 1.1887 x 10⁻¹²
pH = - log ( 1.1887 x 10⁻¹² )
= 12 - log 1.1887
= 12 - .075
= 11.925 .
bleaching powder reaction, mechanism, use
Answers
Answer:
Bleaching Powder's chemical formula is CaOCl2 and is called Calcium Oxychloride. It is prepared on dry slaked lime by chlorine gas. 2. ... It gives calcium chloride, chlorine and water when bleaching powder reacts with hydrochloric acid.
Explanation:
Which of the following metals has a low melting point?
2 A. Rubidium
B. Potassium
C. Calcium
D. Sodium
Answers
Answer:
Rubidium
Explanation:
At what temperature is the following reaction feasible: Al2O3(s) + 3C(s) -> 2Al(s) + 3CO(g)? Enthalpy (H) = +1287 kJ mol–1 Entropy (S) = +614 J K–1 mol–1 A. 2096.1 K B. 1273.8 K C. 477.1 K D. 1901.0 K
Answers
Answer:
Option A. 2096.1 K
Explanation:
The following data were obtained from the question:
Al2O3(s) + 3C(s) —> 2Al(s) + 3CO(g)? Enthalpy (H) = +1287 kJ mol¯¹
Entropy (S) = +614 JK¯¹ mol¯¹
Temperature (T) =...?
Entropy, enthalphy and temperature are related by the following equation:
Change in Entropy (ΔS) = Change in Enthalphy (ΔH) /Temperature (T)
ΔS = ΔH/T
With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:
Enthalpy (H) = +1287 kJ mol¯¹ = 1287000 Jmol¯¹
Entropy (S) = +614 JK¯¹mol¯¹
Temperature (T) =...?
ΔS = ΔH/T
614 = 1287000/ T
Cross multiply
614 x T = 1287000
Divide both side by 614
T = 1287000/614
T = 2096.1 K
Therefore, the temperature at which the reaction will be feasible is 2096.1 K.
What was Ernest Rutherford experiment
Answers
In movies about space, there is frequently a space battle scene where ships
explode in big fireballs. Why are these scenes unscientific?
O
A. The explosion would not be a fireball, but would point towards the closest gravity
source, a planet or star. They are only fireballs on Earth because we are already on a
gravity source.
B. None of these
C. There is no oxygen in space, so there can be no combustion.
OD. Space is very cold, there would not be enough heat energy for an explosion to occur.
-
Answers
The correct answer is C. There is no oxygen in space, so there can be no combustion.
Explanation:
Fire and flames are the result of a chemical process known as combustion. Moreover, for combustion to occur there are two essential elements. The first one is a fuel or a substance that releases energy and ignites, and the second one is an oxidant, which accepts electrons. This mix and reaction causes high temperatures and release of heat in the form of fire and flames.
This implies, that for fireballs or any other form of fire to exist there must be oxygen or any substance that replaces it. This does not occur in space because the levels of oxygen are extremely low, this means, at least oxygen is added fireballs are not possible in this context as there is no oxygen, and therefore no combustion (Option C).
Answer:
C.) There is no oxygen in space, so there can be no combustion.
Explanation:
I got it correct on founders edtell
Q1) How much heat is released when 6.38 grams of Ag(s) (m.m = 107.9 g/mol) reacts by the equation shown below at
standard state conditions?
4A9 (s) + 2H,Sq) + O2(g)
2Ag $(s) + 2H200)
Substance
AHof (kJ/mol)
-20.6
H259)
Ag2S (5)
H200
-32.6
-285.8
a)
8.80 KI
b) 69.9 kJ
C) 22.1 kJ
d) 90.8 kJ
e) 40.5 kJ
Answers
Answer:
The correct answer is -8.80 kJ.
Explanation:
The ΔH° can be determined by using the formula,
ΔH°rxn = ΔH°f (products) - ΔH°f(reactants)
Based on the given information, the ΔH°f of H2S(g) is -20.6, for Ag2S (s) is -32.6 and for H2O (l) is -285.8 kJ/mole.
Now putting the values we get,
= [2 molΔH°f (Ag2S) + 2 molΔH°f (H2O)] - [4 molΔH°f(Ag) + 2 molΔH°f(H2S) + 1 molΔH°f(O2)]
=[2 mol (-32.6 kJ/mol) + 2 mol(-285.8 kJ/mol)] - [4 mol(0.00 kJ/mol) + 2 mol (-20.6 kJ/mol) + 1 mol (0.00 kJ/mol)
= [(-65.2 kJ) + (-571.6 kJ)] - [(-41.2 kJ)]
= -595.6 kJ
Thus, the enthalpy change of -595.6 kJ takes place when 4 mol of Ag reacts by the equation mentioned.
The mass of Ag given is 6.38 grams, the molecular mass of Ag is 107.9 g/mol. The formula for calculating moles is,
Moles = mass/molar mass
= 6.38 g / 107.9 g/mol
= 0.0591 mol
Now the change in enthalpy when 0.0591 mol of Ag reacts by the given reaction is (-595.6 kJ/4 mol) × 0.0591 mol = -8.80 kJ
The negative sign indicates that the heat is released in the process. Therefore, the -8.80 kJ of heat is released by 6.38 grams of Ag in the given case.
Use your periodic table and calculator as needed for the following question.
How much stock solution is needed to make 250 mL of a 6.0M solution. The molarity of the stock solution is 18M.
Selections may be rounded so choose the best answer.
56 mL
83 mL
2.3 mL
4.7 ml
Answers
Medical implants and high-quality jewelry items for body piercings are frequently made of a material known as G23Ti or surgical-grade titanium. The percent composition of the material is 64.39% titanium, 24.19% aluminum, and 11.42% vanadium. What is the empirical formula for surgical-grade titanium
Answers
Answer:
The Empirical Formular is given as; Ti₆Al₄V
Explanation:
The percent composition of the material is 64.39% titanium, 24.19% aluminum, and 11.42% vanadium.
Elements Titanium Aluminium Vanadium
Percentage 64.39 24.19 11.42
Divide all through by their molar mass
64.39 / 47.87 24.19 / 27 11.42 / 50.94
= 1.345 = 0.896 = 0.224
Divide all though by the smallest number (0.224)
1.345 / 0.224 0.896 / 0.224 0.224 / 0.224
= 6 = 4 = 1
The Empirical Formular is given as; Ti₆Al₄V
Using the stepwise procedure for obtaining the empirical formula of a compound, the empirical formula is [tex] T_{6}Al_{4}V[/tex]
Titanium :
Percentage composition = 64.39%Molar mass = 47.87Divide by Molar mass : = 64.39/47.87 = 1.345
Aluminum :
Percentage composition = 24.19%Molar mass = 27Divide by Molar mass : = 24.19/27 = 0.896
Vanadium :
Percentage composition = 11.42%Molar mass = 50.94%Divide by Molar mass : = 11.42/50.94 = 0.224
Divide by the smallest :
Titanium = 1.345 / 0.224 = 6.00
Aluminum = 0.896 / 0.224 = 4
Vanadium = 0.224 / 0.224 = 1
Hence, the empirical formula is [tex] T_{6}Al_{4}V[/tex]
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