Answer:
See explanation below
Explanation:
This problem is a little long so I'm gonna be as clear as possible.
a) In this case we have two acids, HBr and HCHO2. Between these two acids, the HBr is the strongest, and does not have a Ka value to dissociate, while HCHO2 do.
In order to calculate pH we need the [H₃O⁺], and in this case, as HBr is stronger, the contribution of the weaker acid can be negligible, therefore, the pH of this mixture will be:
pH = -log[H₃O⁺]
pH = -log(0.115)
pH = 0.93
b) In this case it happens the same thing as part a) HNO₃ is the strongest acid, so the contribution of the HNO₂ which is a weak acid is negligible too, therefore the pH of this mixture will be:
pH = -log(0.085)
pH = 1.07
c) Now in this case, HCHO2 and HC2H3O2 are both weak acids, so to determine which is stronger, we need to see their Ka values. In the case of HCHO2 the Ka is 1.8x10⁻⁴ and for the HC2H3O2 the Ka is 1.8x10⁻⁵. Note that the difference between the two values of Ka is just 10¹ order, so, we can neglect the concentration of either the first or the second acid. We need to see the contribution of each acid, let's begin with the stronger acid first, which is the HCHO2, we will write an ICE chart to determine the value of the [H₃O⁺] and then, use this value to determine the same concentration for the second acid and finally the pH:
HCHO₂ + H₂O <-------> CHO₂⁻ + H₃O⁺ Ka = 1.8*10⁻⁴
i) 0.185 0 0
c) -x +x +x
e) 0.185-x x x
1.8*10⁻⁴ = x² / 0.185-x
As Ka is small, we can assume that "x is small" too, therefore the (0.185-x) can be rounded to just 0.185 so:
1.8*10⁻⁴ = x²/0.185
1.8*10⁻⁴ * 0.185 = x²
x² = 3.33*10⁻⁵
x = 5.77*10⁻³ M = [H₃O⁺]
Now that we have this concentration, let's write an ICE chart for the other acid, but taking account this concentration of [H₃O⁺] as innitial in the chart, and solve for the new concentration of [H₃O⁺] (In this case i will use "y" instead of "x" to make a difference from the above):
HC₂H₃O₂ + H₂O <--------> C₂H₃O₂⁻ + H₃O⁺ Ka = 1.8x10⁻⁵
i) 0.225 0 5.77x10⁻⁶
c) -y +y +y
e) 0.225-y y 5-77x10⁻³+y
1.8x10⁻⁵ = y(5.77x10⁻³+y) / 0.225-y ---> once again, y is small so:
1.8x10⁻⁵ = 5.77x10⁻³y + y² / 0.225
1.8x10⁻⁵ * 0.225 = 5.77x10⁻³y + y²
y² + 5.77x10⁻³y - 4.05x10⁻⁶ = 0
Solving for y:
y = -5.77x10⁻³ ±√(5.77x10⁻³)² + 4*4.05x10⁻⁶ / 2
y = -5.77x10⁻³ ±√4.95x10⁻⁵ / 2
y = -5.77x10⁻³ ± 7.04x10⁻³ / 2
y₁ = 6.35x10⁻⁴ M
y₂ = -6.41x10⁻³ M
We will take y₁ as the value, so the concentration of hydronium will be:
[H₃O⁺] = 5.77x10⁻³ + 6.35x10⁻⁴ = 6.41x10⁻³ M
Finally the pH for this mixture is:
pH = -log(6.41x10⁻³)
pH = 2.19
d) In this case, we have the same as part c, however the Ka values differ this time. The Ka for acetic acid is 1.8x10⁻⁵ while for HCN is 4.9x10⁻¹⁰. In this ocassion, we the difference in their ka is 10⁵ order, so we can neglect the HCN concentration and focus in the acetic acid. Let's do an ICE chart and then, with the hydronium concentration we will calculate pH:
HC₂H₃O₂ + H₂O <--------> C₂H₃O₂⁻ + H₃O⁺ Ka = 1.8x10⁻⁵
i) 0.050 0 0
c) -y +y +y
e) 0.050-y y y
1.8*10⁻⁵ = y² / 0.050-y
As Ka is small, we can assume that "y is small" too
1.8*10⁻⁵ = y²/0.050
1.8*10⁻⁵ * 0.050 = y²
y² = 9*10⁻⁷
y = 9.45*10⁻⁵ M = [H₃O⁺]
Finally the pH:
pH = -log(9.45x10⁻⁵)
pH = 3.02
Three different students determined the density of a metal object. Here are their results: 15.12 g/mL, 15.09 g/mL, and 15.12 g/mL. The actual density of the object was 14.41 g/mL. Calculate the percent error. Make sure to include units with your answer, units are %.
Answer:
The correct answers are 4.93 %, 4.72 % and 4.93 %.
Explanation:
Based on the given question, 14.41 g per ml is the actual density of the object. However, the density determined by three different students of the object is 15.12 g per ml, 15.09 g per ml, and 15.12 g per ml. The percent error can be calculated by using the formula,
% error = (actual value - calculated value) / actual value * 100
By 1st student, the calculated value is 15.12 g per ml, the percent error will be,
% error = (14.41 - 15.12) / 14.41 * 100
= 0.71/14.41 * 100
= 4.93 %
By 2nd student, the calculated value is 15.09 g per ml, the percent error will be,
% error = (14.41-15.09)/14.41 * 100
= 0.68/14.41 * 100
= 4.72 %
By 3rd student, the calculated value is 15.12 g per ml, the percent error will be,
% error = (14.41-15.12)/14.41 * 100
= 0.71/14.41 * 100
= 4.93 %
Suppose that a wave has a period of 0.03 seconds what is its frequency be sure to show the steps for your work
Answer:
f = 33.34 Hz
Explanation:
A wave has a period of 0.03 seconds. It is required to find the frequency of a wave. The relation between time period and frequency is inverse. The time period of a wave is given by :
T = 1/f, f = frequency of wave
[tex]f=\dfrac{1}{T}\\\\f=\dfrac{1}{0.03}\\\\f=33.34\ Hz[/tex]
So, the frequency of the wave is 33.34 Hz.
A salt solution was found to contain 1.50 g of salt dissolved in 50 mL of water. On evaporation, the recovered salt weighed 1.47 g. What percent of salt was recovered?
A) 20.4%
B) 107%
C) 98%
D) 20.0%
Answer:
C = 98%
Explanation:
Hello,
To determine the percentage of salt recovered, we'll divide the mass of the salt recovered over by the original mass of the salt.
Mass of salt recovered = 1.47g
Initial mass of salt = 1.50g
Percentage of salt recovered = (mass recovered/ initial mass of salt) × 100
Percentage of salt recovered = (1.47 / 1.50) × 100
Percentage of salt recovered = 0.98 × 100
Percentage of salt recovered = 98%
The percentage of salt recovered is equal to 98%
When 8.1 g of an unknown non-electrolyte is dissolved in 50.0 g of carbon tetrachloride, the boiling point increased by 3.67 degrees C. If the Kbp of the solvent is 4.95 K/m, calculate the molar mass of the unknown solute.
Answer:
218.3 g/mol
Explanation:
Boiling point elevation occurs when a solute is added to a solvent increasing the boiling point of the solution with regard to the pure solvent.
The law is:
ΔT = Kb×m×i
Where ΔT is change in temperature (3.67°C), Kb is the boiling point constant of the solvent (4.95°C/m), m is molality of the solution and i is Van't Hoff factor (1 for a non-electrolyte).
3.67°C = 4.95°C/m×m×i
0.7414m = molality of the solution (Moles solute / kg solvent).
As the mass of the solvent is 50.0g = 0.0500kg:
0.7414m = Moles solute / 0.0500kg
0.0371 = moles of solute
As the mass of the solute is 8.1g, molar mass of the solute (Ratio between mass in g and moles) is:
8.1g / 0.0371mol =
218.3 g/molA stock solution will be prepared by mixing the following chemicals together:
3.0 mL of 0.00200 M KSCN
10.0 mL of 0.200 M Fe(NO3)3
17.0 mL of 0.5 M HNO3
Determine the molar concentration of Fe(NO3)3 in the stock solution.
Answer:
0.067M Fe(NO3)3
Explanation:
A stock solution is a concentrated solution that is diluted to prepare the solutions that you will use.
The volume of the stock solution is 3.0mL + 10.0mL + 17.0mL= 30.0mL.
The ratio between volume of the aliquot (10.0mL) and total volume (30.0mL) is called dilution factor, that is: 30.0mL / 10.0mL = 3
That means the Fe(NO3)3 is diluted 3 times. That means the molar concentration of the stock solution is:
0.200M / 3 =
0.067M Fe(NO3)3Which statement describes a chemical property of an object? A:The object is white in color.B:The object has a powdery texture.C:The object’s density is 2.11 g/cm3.D:The object reacts with acid to form water.
Answer:
D
Explanation:
Color, texture, and density are all physical properties but reactivity is a chemical property so the answer is D.
Trans-4-hexen-3-ol can be synthesized starting from acetaldehyde. One of the key reagents is ethyl grignard.
1. Synthesize ethyl grignard from acetaldehyde in the steps below using the reagents provided.
2. Synthesize (trans)-4-hexen-3-ol from acetaldehyde.
find the given attachment
What are the relations between Electrochemistry and Cancer?
Answer: if im not wrong the relations are that the electrochemistry can detect the cancer and any other sickness
just like it does with chemical phenomena
=)
What type of bond will be formed for atoms that have a +1 or -1 charge?
carbon dioxide is a non-polar molecule true or false
Answer:
True
Explanation:
Due to the arrangement of the molecule, a carbon dioxide molecule is non-polar.
A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO 3 in a coffee cup calorimeter. If both solutions were initially at 35.00°C and the temperature of the resulting solution was recorded as 37.00°C, determine the ΔH° rxn (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HCl. Assume 1) that no heat is lost to the calorimeter or the surroundings, and 2) that the density and the heat capacity of the resulting solution are the same as water.
Answer:
THE STANDARD HEAT OF NEUTRALIZATION OF THE BASE SODIUM HYDROXIDE BY THE ACID HYDROGEN TRIOXONITRATE V ACID IS -56 kJ / mol.
Explanation:
Volume of 0.3 M NaOh = 100 mL
Volume of 0.3 M HNO3 = 100 mL
Initail temp of NaOH and HNO3 = 35 °C = 35 + 273 K = 308 K
Final temp. of mixture = 37 °C = 37 + 273 K = 310 K
We can make the following assumptions form the question given:
1. specific heat of the reaction mixture is the same as the specific heat of water = 4.2 J/g K
2. the toal mass of the reaction mixture is 200 mL = 200 g since no heat is lost to the calorimeter or surrounding.
3. initail temperature of the reaction mixture is equal to the average temperature of the two reactant solutions
= ( 308 + 308 /2) = 308 K
4. Rise in temeperature for the reaction = 310 -308 K = 2 K
Then the total heat evolved during the reaction = mass * specifc heat capacity * temperature change
Heat = 200 g * 4.2 J/g K * 2 K
Heat = 1680 J
EQUATION FOR THE REACTION
HNO3 + NaOH -------> NaNO3 + H20
From the equation, 1 mole of HNO3 reacts with 1 mole of NaOH to prouce mole of water.
100 mL of 0.5 M HNO3 contains 100 * 0.3 /1000 = 0.03 mole of acid
This result is same for the base NaOH = 0.03 mole of base
So therefore,
0.03 mole of acid will react with 0.03 mole of base to produce 0.03 mole of water to evolved 1680 J of heat energy.
The production of 1 mole of water will evolve 1680 / 0.03 J of heat
= 56 000 J or 56 kJ of heat energy per mole of water.
So therefore, 1the standard heat of neutralization of sodium hydroxide by trioxoxnitrate V acid is -56 kJ/mol.
g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.904 g of precipitate, what is the molarity of lead(II) ion in the original solution
Answer:
[tex]M=0.0637M[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)[/tex]
Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:
[tex]n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2} *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}[/tex]
Finally, the resulting molarity in 30.8 mL (0.0308 L):
[tex]M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M[/tex]
Regards.
The cryosphere is part of which sphere of the Earth system?
atmosphere
biosphere
geosphere
hydrosphere
Answer:
Ice (frozen water) is part of the hydrosphere, but it's given its own name, the cryosphere.
The cryosphere is part of the hydrosphere of the Earth system. The correct option is D.
What is the cryosphere?The cryosphere contains all the frozen parts of the earth. The term is made up of the Greek word “krios” which means cold. All the frozen water of the oceans and snow comes under the cryosphere.
The atmosphere contains all spheres, it is an envelope of gases. The geosphere is the land part of the earth, and the biosphere is the part where the living part is present.
The cryosphere is h habitat of many living creatures, and the climate of the earth is highly dependent on this sphere. The warmth of the earth is increasing and the cryosphere part is decreasing day by day, which is having problems for many animals.
Thus, the correct option is D, hydrosphere.
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Which statement best describes information that would be associated with science rather than with pseudoscience?
Answer:
The information has been changed to support a claim
Explanation:
Which best describes thermal energy? It is the difference between internal energies of two or more substances. It is the sum of internal energies of two or more substances. It is the portion of internal energy that can be transferred from one substance to another. It is the portion of potential energy that can be transferred from one substance to another.
Answer:
It is the portion of internal energy that can be transferred from one substance to another.
Thermal energy is the portion of internal energy that can be transferred from one substance to another.
What is thermal energy?Thermal energy is the energy an object posses which is as a result of particles movement within it.
It is also the internal energy system in a state of thermodynamic equilibrium which is as a result of its temperature. Thermal energy cannot be concert to useful work easily.
Therefore, thermal energy is the portion of internal energy that can be transferred from one substance to another.
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A 25.0 mLsample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 37.5 mL of the base is added. The concentration of acetic acid in the sample was:_______,
A) 0.263
B) 0.365
C) 0.175
D) 1.83×10−4
E) 0.119
Answer:
0.263M of CH₃COOH is the concentration of the solution.
Explanation:
The reaction of acetic acid (CH₃COOH) with NaOH is:
CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O
1 mole of acetic acid reacts per mole of NaOH to produce sodium acetate and water.
In the equivalence point, moles of acetic acid are equal to moles of NaOH and moles of NaOH are:
0.0375L × (0.175 moles / L) = 6.56x10⁻³ moles of NaOH = moles of CH₃COOH.
As the sample of acetic acid had a volume of 25.0mL = 0.025L:
6.56x10⁻³ moles of CH₃COOH / 0.0250L =
0.263M of CH₃COOH is the concentration of the solution
If the Moon rises at 7 P.M. on a particular day, then approximately what time will it rise six days later?
Answer:
below
Explanation:
28th 10;24 am
If the Moon rises at 7 P.M. on a particular day, then approximately what time will it rise six days later at 12A.M.
How much time changes between Moon rises from one day to the next?This movement is from the Moon's orbit, which takes 27 days, 7 hours and 43 minutes to go full circle. It causes the Moon to move 12–13 degrees east every day. This shift means Earth has to rotate a little longer to bring the Moon into view, which is why moonrise is about 50 minutes later each day.
So knowing that moonrise is about 50 minutes later each day, we have:
[tex]7+50 minutes = 7:50\\8:40\\9:30\\10:20\\11:10\\12:00 A.M[/tex]
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How many grams of 02 are needed to react with 7.50g of ethanol
Reactive oxygen species (ROS) are unstable, or reactive, compounds that result from the partial reduction of oxygen. ROS can cause damage to molecules, including membrane lipids and nucleic acids, and may be associated with some diseases. Which of these compounds are reactive oxygen species? Choose all that apply.
a. OH
b. OH-
c. O2-
d. H2O
e. H2O2
f. H-
What did John Dalton publish?
Answer:
An early theory describing properties of atoms.
Explanation:
Apex
Heat is added to a 1.0-kg block of ice at OC. Determine if the process is
endothermic or exothermic. Explain your answer. *
Answer:
endothermic
Explanation:
Heat is added to make the process possible.
what is the sign of Mercury
Answer:
The answer is Hg.
Explanation:
Symbol for Mercury is Hg.
Hydrocarbon X has the formula C6H12. X reacts with one molar equivalent of hydrogen in the presence of a palladium catalyst to form a product having 12 primary hydrogens. Treatment of X with ozone followed by zinc in aqueous acid gives a mixture two aldehydes. What is the structure of X
Answer:
Explanation:
X( C₆H₁₂ )= (CH₃)₃-C-CH=CH₂
(CH₃)₃-C-CH=CH₂ + H₂ = (CH₃)₃-C-CH₂-CH₃ ( 12 primary hydrogen bonds )
(CH₃)₃-C-CH=CH₂ + O₃ = (CH₃)₃-C-CH= O + HCHO
how many grams of H2 are needed to produce 14.34 g of NH3?
please help, this hw is due in a few hours
Answer:
im pretty sure its 2.54g H2
Explanation:
14.34gNH3 / 17.03gNH3 <-- molar mass
.842g x 3 mol <-- mols of H2
2.52 / 2 mol <-- mols of NH3
1.26 x 2.016gH2= 2.54gH2
The substances nitrogen monoxide and hydrogen gas react to form nitrogen gas and water. Unbalanced equation: NO (g) + H2 (g) N2 (g) + H2O (l) In one reaction, 76.2 g of H2O is produced. What amount (in mol) of H2 was consumed? What mass (in grams) of N2 is produced?
Answer:
H2 consumed 4.22 mol
N2 produced 59.107 g
Explanation:
Balanced equation:
2NO (g) + 2H2 (g) N2 (g) + 2H2O (l)
To perform the calculations, the molecular weights of the following compounds must be known:H2O MW = 18.02 g/mol
N2 MW = 28.01 g/mol
To determine the moles of H2O produced, the following formula should be used:
[tex]MW=\frac{mass}{mol}[/tex]
The value of moles is cleared:
[tex]mol=\frac{mass}{MW} =\frac{76.2g}{18.02\frac{g}{mol} } =4.22 mol[/tex]
Now, to calculate the grams of N2 consumed, we look at the balanced equation and note that 2 moles of H2 produce 1 mole of N2. Therefore, through said observation, the amount of moles of H2 consumed can be determined.2 mol H2 ⇒ 1 mol N2
4.22 mol H2 ⇒ X
[tex]X=\frac{4.22mol*1 mol}{2 mol} =2.11 mol[/tex]
To calculate the mass of H2 consumed, the molecular weight equation is used again:
[tex]mass=MW*mol=28.013\frac{g}{mol}*2.11mol=59.107g[/tex]
The acetate ion is the conjugate base of the weak acid acetic acid. The value of Kb for CH3COO-, is 5.56×10-10. Write the equation for the reaction that goes with this equilibrium constant.
Answer: The equation for the reaction that goes with this equilibrium constant is [tex]5.56\times 10^{-10}=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]
Explanation:
[tex]CH_3COOH\rightarrow CH_3COO^-+H^+[/tex]
Here [tex]CH_3COOH[/tex] donates a proton and thus behaves as an acid and forms [tex]CH_3COO^-[/tex] which is called as the conjugate base of [tex]CH_3COOH[/tex]
The dissociation constant of acids is given by the term [tex]K_a[/tex] and the dissociation constant of bases is given by the term [tex]K_b[/tex] and is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.
[tex]K_a[/tex] for [tex]CH_3COOH[/tex] :
[tex]K_a=\frac{[CH_3COO^-]\times [H^+]}{[CH_3COOH]}[/tex]
[tex]CH_3COO^-+H^+\rightarrow CH_3COOH[/tex]
[tex]K_b=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]
[tex]5.56\times 10^{-10}=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]
The equation for the reaction that goes with this equilibrium constant is [tex]K_b=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]
Each of the insoluble salts below are put into 0.10 M hydrobromic acid solution. Do you expect their solubility to be more, less, or about the same as in a pure water solution ?
a. Calcium sulfite
b. Calcium fluoride
c. Silver bromide
Answer:
A. Solubility of calcium sulfite increases
B. Solubility of calcium fluoride increases
C. Solubility of Silver bromide decreases
Explanation:
The solubility factor is proportional to ions' concentration. The solubility of a solution can be predicted from Le Chatelier's principle which states that if an external constraint is imposed on a system in equilibrium, the equilibrium position will shift in order to annul the effect of the external constraint. So, If the reactant's concentration increases, the equilibrium shifts to the right indicating a higher solubility of the solution and if the product's concentration increases, the equilibrium shifts to the left indicating a lesser solubility of the solution.
Case 1. Calcium sulfite
The dissociation reaction of CaSO3 is given below:
CaSO3 ----> Ca²+ + SO3²-
SO3²- is the conjugate base of the weak acid, H2SO3. Therefore, on the addition of hydrobromic acid, some of the sulfite ion is removed from the solution by the following reaction;
H+ + SO3²- ---> HSO3-
This shifts the equilibrium to the right, more dissociation, thereby resulting in more solubility of the solute.
Case 2. Calcium fluoride
The dissociation reaction of calcium fluoride (CaF2) is shown below.
CaF2 ----> Ca²+ + 2F-
Fluoride ion (F-) is a strong conjugate base of the weak acid. Therefore, some of fluoride ions is removed by the addition of hydrobromic acid as shown below:
H+ + F- ---->. HF
Hence, the concentration of fluoride ions reduces, shifting equilibrium in the forward direction. Therefore, the solubility will be more than in pure water solution.
Case 3: Silver bromide
The dissociation reaction of AgBr is as follows:
AgBr ----> Ag+ + Br-
The addition of HBr will increase the concentration of bromide ions. Hence, equilibrium will shift in backward direction resulting in a lesser solubility than in water.
The solubility of calcium sulfite and calcium fluoride is greater in 0.10 M hydrobromic acid solution than in pure water while the solubility of silver bromide is lesser in 0.10 M hydrobromic acid solution than in pure water.
Common ion effect refers to the decrease in solubility of a substance in a solution that contains another solute with which it has a common ion. If a substance is dissolved in a solution that contains a solute with which it has a common ion, the solubility of the substance in that solution is less than its solubility in pure water.
Considering the substances given, the solubility of calcium sulfite and calcium fluoride in 0.10 M hydrobromic acid solution is more than their solubility in pure water the equilibrium position is shifted in the forward direction.
However, solubility of silver bromide in 0.10 M hydrobromic acid solution is less than its solubility in pure water due to common ion effect.
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How do the particles in plasmas compare with the particles in solids?
Answer:
Plasmas and solids are both made up of cation-anion pairs. Solids and plasmas are both made up of electrons and cations. Solids are made up of cation-anion pairs, but plasmas are not.
Answer:
Solids are made up of cation-anion pairs, but plasmas are not.
Explanation:
When vinylcyclohexane is treated with in dichloromethane, the major product is (2-bromo ethylidene)cyclohexane . Account for the formation of this product by drawing the structure of the most stable radical intermediate. Include all valence lone pairs in your answer. Include all valence radical electrons in your answer.
Answer:
Explanation:
Vinylcyclohexane is an example of a cyclic hydrocarbon where the vinyl group (-CH=CH₂ ) attaches itself to an end of a cyclohexane in ring form thereby giving rise to a vinylcyclohexane. The vinyl group are ethylene with a reduction in one hydrogen atom given them the name vinyl.
SOo, when vinylcyclohexane is treated with NBS ( i.e N-Bromosuccinimide a chemical reagent used in organic reactions) ; the bromine in the NBS reacts with the cyclohexane thereby giving rise to a allyl radical first. The allyl radical is resonance stabilized radical with an unpaired electron on the allylic carbon . As a result of stabilization ; a more stable substituted cycloalkene is formed as an intermediate .
This stable substituted cycloalkene intermediate then finally react with a bromine ion to give a major product known as ; (2-bromo ethylidene)cyclohexane.
The diagram emphasizing more on the above explanation can be seen in the attached image below
Interpret the following equation for a chemical reaction using the coefficients given: CO(g) Cl2(g) COCl2(g) On the particulate level: _________ of CO(g) reacts with _________ of Cl2(g) to form _________ of COCl2(g). On the molar level: _________ of CO(g) reacts with _________ of Cl2(g) to form _________ of COCl2(g).
Answer:
On the particulate level: 6.02 * 10²³ particles of CO(g) reacts with 6.02 * 10²³ particles of Cl₂(g) to form 6.02 * 10²³ particles of COCl2(g).
On the molar level: 1 mole of CO(g) reacts with 1 mole of Cl2(g) to form 1 mole of COCl₂(g).
Explanation:
The particulate level refers to the microscopic or atomic level of substances. It also involves the ions, protons, neutrons and molecules present in substances.
The molar level refers to the quantitative measure of substances in terms of the mole, where a mole represents the amount of substances containing the Avogadro number of particles which is equal to 6.02 * 10³ particles.
Equation of the reaction: CO(g) + Cl₂(g) ----> COCl₂(g)
From the equation above, I mole of CO gas reacts with 1 mole of Cl₂ gas to produce 1 mole of COCl₂ gas.
Since 1 mole of a substance contains 6.02 * 10²³ particles, on a particulate level, 6.02 * 10²³ particles of CO gas reacts with 6.02 * 10²³ particles of Cl₂ gas to produce 6.02 * 10²³ particles of COCl₂ gas.
