Final answer:
Without specific numeric values for the individual atomic heat capacities of the elements in sodium carbonate, directly answering the heat capacity multiple-choice question using Kopp's rule is not feasible. Additional data is required to estimate the heat capacity of solid sodium carbonate.
Explanation:
The question regarding the heat capacity of solid sodium carbonate (Na2CO3) can be answered by applying Kopp's rule, which is a method used to estimate the heat capacities of solids. According to Kopp's rule, the heat capacity of a compound in the solid state is the sum of the atomic heat capacities of the individual elements that compose the compound. Atomic heat capacities can be estimated using the Law of Dulong and Petit, which suggests that at room temperature, the molar heat capacity of many solid elements roughly equates to 3R, where R is the gas constant with a value approximately equal to 8.314 J/(mol·K).
Applying Kopp's rule would involve calculating the total molar heat capacity for sodium, carbon, and oxygen in sodium carbonate, based on their atomic weights and expected contributions. However, since specific numeric values are not provided in the question or reference information to apply Kopp's rule directly, directly answering the multiple-choice question is not feasible without additional data on the individual atomic heat capacities of the elements involved in sodium carbonate.
Give the structure of the expected organic product in the reaction of 3−phenylpropanal with sodium hydroxide in ethanol at 70°C. If your structure contains an aldehyde do NOT use the condensed formula (−CHO), draw it out.
Answer:The expected product formed is 2-benzyl-5-phenylpent-2-enal .
Kindly refer the attachment for structure.
Explanation:
3-phenylpropanal on reaction with sodium hydroxide undergoes an self- aldol condensation reaction and leads to formation of 2-benzyl-5-phenylpent-2-enal as final product.
In the first step of the reaction the highly basic hydroxide anions abstract a acidic hydrogen available at the carbon next to carbonyl carbon.These hydrogens are acidic because of the electron withdrawing effect of carbonyl group.
The proton abstraction leads to the generation of a carbanion and it further delocalizes forming an enolate anion.
The carbanion can behave as a nucleophile and can attack at the electrophilic carbon centers.
The carbonyl carbon is electrophilic in nature due to the electron withdrawl from oxygen which generates a partial positive charge on the carbonyl carbon.
The carbanion further reacts with another molecule of 3-phenylpropanal at its electrophilic carbonyl center and forms 2-benzyl-3-hydroxy-5-phenylpentanal. This reaction is known as self- aldol condensation reaction
2-benzyl-3-hydroxy-5-phenylpentanal has a OH group and this OH group is protonated through the available acidic protons from the solvent .
As OH is protonated it easily leaves leading to the formation of C=C double bond.
This reaction is one of the examples of Carbon-Carbon bond forming reaction.
Kindly refer attachments for reaction, structure and mechanism.
Which of the following aqueous solutions are good buffer systems?
0.25 M ammonium nitrate + 0.40 M ammonia
0.23 M nitric acid + 0.20 M sodium nitrate
0.16 M potassium hydroxide + 0.22 M potassium chloride
0.39 M hypochlorous acid + 0.25 M potassium hypochlorite
0.12 M hydrofluoric acid + 0.14 M sodium fluoride
Answer:
0.25 M ammonium nitrate + 0.40 M ammonia
0.39 M hypochlorous acid + 0.25 M potassium hypochlorite
0.12 M hydrofluoric acid + 0.14 M sodium fluoride
Explanation:
A buffer consists of a weak acid and its salt or a weak base and its salt.
NH₃ is a weak base, and HClO and HF are weak acids.
B is wrong. HNO₃ is a strong acid.
C is wrong. KOH is a strong base.
How many grams of Br are required to react completely with 29.5 g of NO? 2NO+Br2- 2NOBr
Answer: 78.54 grams
Explanation:
According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
[tex]\text{Number of molesof} NO=\frac{29.5g}{30.01g/mol}=0.98moles[/tex]
[tex]2NO+Br_2\rightarrow 2NOBr[/tex]
According to stoichiometry :
2 moles of [tex]NO[/tex] react with 1 mole of [tex]Br_2[/tex]
0.98 moles of [tex]NO[/tex] react with =[tex]\frac{1}{2}\times 0.98=0.49[/tex] moles of [tex]Br_2[/tex]
Mass of [tex]Br_2=moles\times {\text{Molar mass}=0.49\times 159.8=78.54g[/tex]
Thus 78.54 g of [tex]Br_2[/tex] are required to react completely with 29.5 g of NO.
1) For this reaction,
SiCl4 (l)+ 2H2O(g) <-> SiO2 (s) + 4HCl (g) DeltaH = -127 KJ
Predict the effect increase (i), decrease(d) or no change(nc) that each change has on the QUANTITY for the equilibrium mixture. Assume that each change is applied separately to the system.
Change Quantity Effect
increase in pressure amount of H2O
i, d or nc
decrease temp amount of SiCl4 i d or nc
increase temp Kc i d or nc
increase temp amount of H2O i d or nc
add H2O amount of SiCl4 i d or nc
add SiO2 amount of HCl i d or nc
add HCl amount of SiO2 i d or nc
Answer:
a) Increase
b) the amount of SiCl₄ will decrease.
c) K will increases.
d) H₂O will increase.
e) SiCl₄ will decrease
f) HCl will decrease.
g) SiO₂ will decrease
Explanation:
The equilibrium changes are explained by Le-Chatelier's Principle.
If we apply a change to an equilibrium system then it shifts to the side where the effect of stress can be released.
a) Change : Increase in pressure
Quantity : amount of water
Change: We are increasing the pressure it will decrease the volume and hence will increase the moles per unit volume. so the system will move in the direction where the number of gaseous moles are less. It will move towards reactant side. thus the amount of water will increase.
b) Change : decrease in temperature
Quantity : amount of SiCl₄
As this is an exothermic reaction, the decrease in temperature will shift the reaction towards product side and hence the amount of SiCl₄ will decrease.
c)
Change : Increase in Temperature
Quantity: Kc
The equilibrium constant will increase as it increases with temperature.
d) Change : increase in temperature
Quantity : amount of H₂O
As this is an exothermic reaction, the increase in temperature will shift the reaction towards reactant side and hence the amount of H₂O will increase.
e) Change : Add H₂O
Quantity: Amount of SiCl₄
As we are adding reactant to the equilibrium, the equilibrium will shift towards product side thus the amount of SiCl₄ will decrease
f) Change : Add SiO₂
Quantity: Amount of HCl
As we are adding product, the equilibrium will shift in reactant side and thus there will be decrease in amount of HCl.
g) Change : Add HCl
Quantity: Amount of SiO₂
As we are adding product to the equilibrium the equilibrium will shift in reactant side and thus there will be decrease in amount of SiO₂
For all gas law calculations, the temperature must be in kelvins. (T/F)
Answer:
True
Explanation:
For, all the gas law calculations, Temperature must be used in Kelvins. This statement is true.
The reason is:
Ideal gas equation is made up of gas laws like P/T (Amonton's Law), V/T (Charle's Law) and combined gas laws like PV/T ( Boyle's Laws, Amonton's Law and Charle's Law).
Charle's Law and Amonton's Law are only valid when temperature values are put in Kelvin and only then V/T or P/T will be constant.
Also, in all the cases, Temperature occurs in the denominator. If we measure temperature values in Celcius, then it will lead to wrong calculations. Also, if we put [tex]0^0C[/tex] in the equation, then the equation will have zero in the denominator which will solve as no solution.
But, If we will put 0K in the equation, then it will achieve absolute state where all the things stop and there will be zero entropy (Third law of thermodynamics). In that case, we dont have to think about any of the parameters to be calculated.
The depletion of ozone (O3) in the stratosphere has been a matter of great concern among scientists in recent years. It is believed that ozone can react with nitric oxide (NO) that is discharged from the high-altitude jet plane, the SST. The reaction is
O3 + NO ---> O2 + NO2
If 0.827 g of O3 reacts with 0.635 g of NO, how many grams of NO2 will be produced? g NO2 Which compound is the limiting reagent? ozone (O3) nitric oxide (NO) Calculate the number of moles of the excess reagent remaining at the end of the reaction.
Answer: Nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.
Explanation:
To calculate the number of moles, we use the equation
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ....(1)
For ozone:Given mass of ozone = 0.827 g
Molar mass of ozone = 48 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of ozone}=\frac{0.827g}{48g/mol}=0.0172mol[/tex]
For nitric oxide:Given mass of nitric oxide = 0.635 g
Molar mass of nitric oxide = 30.01 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of nitric oxide}=\frac{0.635g}{30.01g/mol}=0.0211mol[/tex]
For the given chemical equation:
[tex]O_3+NO\rightarrow O_2+NO_2[/tex]
By Stoichiometry of the reaction:
1 mole of ozone reacts with 1 mole of nitric oxide.
So, 0.0172 moles of ozone will react with = [tex]\frac{1}{1}\times 0.0172=0.0172moles[/tex] of nitric oxide
As, given amount of nitric oxide is more than the required amount. So, it is considered as an excess reagent.
Thus, ozone is considered as a limiting reagent because it limits the formation of product.
Amount of excess reagent (nitric oxide) left = 0.0211 - 0.0172 = 0.0039 molesBy Stoichiometry of the reaction:
1 mole of ozone produces 1 mole of nitrogen dioxide.
So, 0.0172 moles of ozone will react with = [tex]\frac{1}{1}\times 0.0172=0.0172moles[/tex] of nitrogen dioxide
Now, calculating the mass of nitrogen dioxide from equation 1, we get:
Molar mass of nitrogen dioxide = 46 g/mol
Moles of nitrogen dioxide = 0.0172 moles
Putting values in equation 1, we get:
[tex]0.0172mol=\frac{\text{Mass of nitrogen dioxide}}{46g/mol}\\\\\text{Mass of nitrogen dioxide}=0.7912g[/tex]
Hence, nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.
Final answer:
To find the mass of NO₂ produced in the reaction between O₃ and NO, and to identify the limiting reagent, one must convert the mass of each reactant to moles, compare these to the reaction stoichiometry, and perform subsequent calculations.
Explanation:
The concern over the depletion of ozone (O3) in the stratosphere due to reactions with nitric oxide (NO) from high-altitude jet planes is a crucial environmental issue. The reaction of interest is O3 + NO → O2 + NO2. To calculate the amount of NO2 produced from 0.827 grams of O₃ and 0.635 grams of NO, we must first determine the limiting reagent, which will dictate the maximum amount of product formed. This involves converting the masses of the reactants to moles, using their molar masses (O₃ = 48.00 g/mol and NO = 30.01 g/mol), and comparing the mole ratios to the stoichiometry of the reaction.
Calculations show that NO is the limiting reagent. From the stoichiometry of the reaction, it's clear that 1 mole of NO produces 1 mole of NO2. Thus, the mass of NO₂ produced can be directly calculated from the moles of NO reacted, taking into account the molar mass of NO2 (46.01 g/mol). The remaining amount of the excess reagent, O3, can also be determined by subtracting the moles of O₃ that reacted (equal to the moles of NO reacted) from the initial moles of O₃.
Why can't methanol, CH3OH, be used as a solvent for sodium amide, NaNH2? Sodium amide is nonpolar and methanol is polar. Sodium amide is polar and methanol is nonpolar. Sodium amide undergoes an acid-base reaction with methanol. There would be no ion-dipole attractive forces between the two compounds.
Answer: sodium amide undergoes an acid -base reaction
Explanation:
sodium amide is a ionic compound and basically exists as sodium cation and amide anion. Amide anion is highly basic in nature and hence as soon as there is amide anion generated in the solution , Due to its very pronounced acidity it very quickly abstracts the slightly acidic proton available on methanol.
This leads to formation of ammonia and sodium methoxide.
Hence sodium amide reacts with methanol and abstracts its only acidic proton and form ammonia and sodium Methoxide.
Hence the 3rd statement is a corrects statement.
So we cannot use methanol for sodium amide because sodium amide itself would react with methanol and the inherent molecular natur of sodium amide would then change.
The 1st and 2nd statements both are incorrect because both the compounds methanol as well as sodium amide have dipole moments and hence are polar molecules.
The 4th statement is also incorrect as both the molecules have dipole moment and hence there would be ion-dipole forces operating between them.
The following reaction occurs:
NaNH₂+CH₃OH→NH₃+CH₃ONa
Sodium amide undergoes an acid -base reaction
Why methanol is not used as a solvent for sodium amide?Methanol is acidic in nature whereas sodium amide is a strong base so when we add or combine these two chemicals, they undergoes an acid -base reaction so that's why we can't use methanol as a solvent for sodium amide.
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When 26.0 mL of 0.500 M H2SO4 is added to 26.0 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate ΔH of this reaction. (Assume that the total volume is the sum of the individual volumes and that the density and specific heat capacity of the solution are the same as for pure water.) (d for water = 1.00 g/mL; c for water = 4.184 J/g·°C.)
The enthalpy of reaction is -55.8 KJ/mol.
From the equation of the reaction;
2KOH(aq) + H2SO4(aq) ------> K2SO4(aq) + 2H2O(l)
Number of moles of H2SO4 = 26.0/1000 × 0.500 M = 0.013 moles
Number of moles of KOH = 26.0/1000 × 1.00 M = 0.026 moles
2 moles of KOH produces 2 moles of water
Hence 0.0026 moles of KOH produces 0.026 moles of water.
Total volume of solution = 26.0 mL + 26.0 mL = 52 mL
Mass of water = density × volume = 1.00 g/mL × 52 mL = 52 g
Using the formula;
ΔH = mcθ
Mass of solution (m) = 52 g
Specific heat capacity of solution (c) = 4.184 J/g·°C
Temperature difference(θ) = 30.17°C - 23.50°C = 6.67°C
Substituting values;
ΔH = -() 52 g × 4.184 J/g·°C × 6.67°C/ 0.026 moles
ΔH = -(1.45 KJ/0.026 moles)
ΔH = -55.8 KJ/mol
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Select all the true statements. When an atom gains an electron, it becomes a cation. Anions carry a positive charge. The Cl− and Br− ions have the same number of electrons. The K+ ion is formed when a potassium atom loses one electron. The Fe2+ and Fe3+ ions have the same number of protons. The Cu+ and Cu2+ ions have the same number of electrons.
Answer:
The statements 4 and 5 are true.
Explanation:
1. When an atom gains an electron it becomes negatively charged. This negatively charged species is called anion.
A + e⁻ → A⁻ (anion)
Therefore, the statement 1 is false.
2. An anion is formed when an atom gains an electron and becomes negatively charged. Therefore, an anion is a negatively charged species.
A + e⁻ → A⁻ (anion)
Therefore, the statement 2 is false.
3. The atomic number of chlorine atom Cl is 17 and atomic number of bromine atom Br is 35.
Since, for neutral atom, the atomic number of an atom is equal to the number of electrons present in that atom.
Therefore, the number of electrons in Cl atom is 17 and the number of electrons in Br atom is 35.
When the Cl atom gains one electron it forms Cl⁻ ion and when the Br atom gains one electron it forms Br⁻ ion.
Therefore, the number of electrons in Cl⁻ ion is 17 + 1 = 18 electrons
and the number of electrons in Br⁻ ion is 35 + 1 = 36 electrons
Therefore, Cl⁻ and Br⁻ ions do not have the same number of electrons.
Therefore, the statement 3 is false.
4. When potassium atom (K) loses one electron it forms a positively charged species called potassium cation (K⁺).
K → K⁺ + e⁻
Therefore, the statement 4 is true.
5. The atomic number of Fe atom is 26.
Since, the atomic number of an atom is equal to the number of protons present in that atom.
When the Fe atom loses two electrons to form Fe²⁺ and when the Fe atom loses three electrons to form Fe³⁺ ion, the number of protons remains the same.
Therefore, the ions Fe²⁺ and Fe³⁺ have the same number of protons.
Therefore, the statement 5 is true.
6. The atomic number of copper atom Cu is 29.
Since, for neutral atom, the atomic number of an atom is equal to the number of electrons present in that atom.
Therefore, the number of electrons in Cu atom is 29
When the Cu atom loses one electron it forms Cu⁺ ion and when the Cu atom loses two electrons it forms Cu²⁺ ion.
Cu → Cu⁺ + e⁻ and Cu → Cu²⁺ + 2e⁻
Therefore, the number of electrons in Cu⁺ ion is 29 - 1 = 28 electrons
and the number of electrons in Cu²⁺ ion is 29 - 2 = 27 electrons
Therefore, Cu⁺ ion and Cu²⁺ ion do not have the same number of electrons.
Therefore, the statement 6 is false.
Atom is the smallest constituent of any chemical species and contains protons and electrons. The true statements are, [tex]\rm K^{+}[/tex] ion is formed when a potassium atom loses one electron. The [tex]\rm Fe^{2+}[/tex] and [tex]\rm Fe^{3+}[/tex] ions have the same number of protons.
What are cations and anions?When an atom acquires electrons from some other atom then they become negatively charged and are called anions.
Cations and anions are formed when the atom loses and gains an electron from another species. When an atom acquires an electron they are called an anion and when loose electrons are called a cation.
When potassium atom (K) relinquishes an electron then a positively charged species formed is called a cation. It can be shown as,
[tex]\rm K \rightarrow K ^{+} + e^{-}[/tex]
The atomic number of an iron atom is 26 and is equal to the number of protons. The number of protons remains the same when the iron atom relinquishes two electrons yields ferrous ions and when loses three electrons yields ferric ion. Hence, ferrous and ferric have the same number of protons.
Therefore, option 4. K+ ion is formed when a potassium atom loses one electron and option 5. ferrous and ferric ions have the same number of protons are correct.
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During the winter months, many locations experience snow and ice storms. It is a common practice to treat roadways and sidewalks with salt, such as CaCO3 . If a 11.3 kg bag of CaCO3 is used to treat the sidewalk, how many moles of CaCO3 does this bag contain?
Answer:
The number of moles of CaCO3 on the bag is 112.90 moles
Explanation:
number mole (n) = mass (m) divided by molecular mass (Mm)
Mm of CaCO3 = 100.0869 g/mole
mass in grams = 11.3 Kg x (10^3 g/1 Kg) = 11300 grams
number of moles (n) = 11300 grams divided by 100.0869 grams per mole = 112.90 moles of CaCO3 in the bag.
The number of moles of CaCO3 in an 11.3 kg bag is calculated by dividing the mass of the CaCO3 by its molar mass.
First, we need to determine the molar mass of CaCO3. The molar mass of calcium (Ca) is approximately 40.08 g/mol, carbon (C) is approximately 12.01 g/mol, and oxygen (O) is approximately 16.00 g/mol. Since there is one atom of calcium, one atom of carbon, and three atoms of oxygen in CaCO3, the molar mass of CaCO3 is calculated as follows:
Molar mass of CaCO3 = [tex](40.08 g/mol) + (12.01 g/mol) + (3 -16.00 g/mol)[/tex]
Molar mass of CaCO3 = [tex]40.08 g/mol + 12.01 g/mol + 48.00 g/mol[/tex]
Molar mass of CaCO3 = [tex]100.09 g/mol[/tex]
Now, we convert the mass of the bag from kilograms to grams because the molar mass is given in grams per mole:
[tex]11.3 kg - 1000 g/kg = 11300 g[/tex]
Finally, we calculate the number of moles of CaCO3 in the bag:
Number of moles = mass of CaCO3 / molar mass of CaCO3
[tex]Number of moles[/tex] =[tex]11300 g / 100.09 g/mol[/tex]
[tex]Number of moles = 113 moles[/tex]
Therefore, the bag contains approximately 113 moles of CaCO3.
The correct answer is [tex]\boxed{113 \text{ moles}}.[/tex]
The answer is: [tex]113 \text{ moles}.[/tex]
The lowest pressure attainable using the best available vacuum techniques is about 10−12 N/m2. Part A At such a pressure, how many molecules are there per cm3 at 2 ∘C?
Hey there!:
We are given P = 10⁻²² N/m2 ,
Volume, V = 1 cm3= 10⁻⁶ m³ and
T= 13ºC= 13+273= 286 K
Using gas law, P*V = n R T, where n is number of moles and R=8.314JK⁻¹.
n = PV/(RT) => n = 10-12*10⁻⁶/ (8.314*286) = 4.205*10₋²² moles
Hence number of molecules = number of moles * Avogadro's number
= 4.205*10-22 moles * 6.023*1023 molecules/mole
= 253
Number of molecules = 250 (upto 2 significant figures)
Number of molecules: N = 263.31
Further explanationSome of the laws regarding gas can apply to ideal gas (volume expansion does not occur when the gas is heated),:
Boyle's law at constant T,
[tex] \displaystyle P = \dfrac {1} {V} [/tex]
Charles's law, at constant P,[tex] \displaystyle V = T [/tex]
Avogadro's law, at constant P and T,[tex] \displaystyle V = n [/tex]
So that the three laws can be combined into a single gas equation, the ideal gas equation
In general, the gas equation can be written
[tex] \large {\boxed {\bold {PV = nRT}}} [/tex]
where
P = pressure, atm , N/m²
V = volume, liter
n = number of moles
R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m2, v= m³)
T = temperature, Kelvin
n = N / No
n = mole
No = Avogadro number (6.02.10²³)
n = m / m
m = mass
M = relative molecular mass
Known
P = 10−12 N / m2
V = 1 cm3 = 10-6 m3
T = 2 ºC = 2 + 273 = 275 K
R = 8,314 J / mol. K
[tex]\rm n=\dfrac{PV}{RT}\\\\n=\dfrac{10^{-12}\times 10^{-6}}{8.314\times 275}\\\\n=\boxed{\bold{4.374\times 10^{-22}}}[/tex]
then the number of molecules (N):
N = n x No
N = 4,374.10⁻²² x 6.02.10²³
N = 263.31
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Consider the reaction of 2-methyl-1,3-cyclohexadiene with HCl. Assuming that the reaction takes place at a very low temperature: (a) draw the intermediate and product structures, including any formal charges
Answer:.The product formed in this reaction would be a kinetic product formed from 1,2-addition.
Kindly refer attachments for mechanism and structures of product.
Explanation:
In organic chemistry once the intermediate is generated the reaction can go in two ways. In one way product formed would be rate dependent and would be known as kinetic product and in the other way product formed would be stable and would be know as thermodynamic product.
A kinetic product is formed faster but it is generally not that stable as a thermodynamic product so a kinetic product is formed at lower temperatures where the molecular energy is very less.
A thermodynamic product formation takes time to form and the reaction is carried out at higher temperatures where the molecules have energy. The thermodynamic product is relatively stable.
In this case since we are doing our reaction at very low temperature so the major product formed in the reaction would be under kinetic control and hence product formed would be rate dependent.
The reaction of 2-methyl-1,3-cyclohexadiene with HCl would be a electrophilic addition reaction in which the pi bond would attack HCl to generate a carbocation intermediate. After the formation of carbocation chloride anion can attack the carbocation and can form the product.
Once the carbocation is formed it can be stabilised by rearrangement or other stabilizing mechanisms.
In this case initially 1-methylcyclohex-2-en-1-ylium carbocation is generated which is at tertiary center as well as allylic position. This carbocation formed initially can stabilize itself through resonance as the charge can be delocalised with the allyl group.
The reaction can happen in two ways :
In the first way the initially formed 1-methylcyclohex-2-en-1-ylium carbocation is been attacked by the chloride anion and this leads to the 1,2 addition product.
In the other way the initially formed 1-methylcyclohex-2-en-1-ylium carbocation delocalizes its positive charge with the allyl group as it is in conjugation with the allyl group thereby generating a positive charge at 4 postion and form 3-methylcyclohex-2-en-1-ylium. Now this carbocation is attacked by the chloride anion and this leads to the 1,4 addition product.
The 1,2 addition product is a Kinetic product as it can quickly lead to products wheras the 1,4 addition product is a thermodynamic product.
The product formed in this reaction would be a kinetic product formed form 1,2-addition.
Kindly refer the attachments for reaction mechanism.
What material would be the best for a hip implant?
Fiber glass
Titanium
ABS
Steel
Rubber
Answer:
titanium
Explanation:
An exothermic reaction has a positive enthalpy (heat) of reaction.(T/F)
Answer:
True.
Explanation:
An exothermic reaction has a positive enthalpy (heat) of reaction. However, it can be negative in some circumstances.
Enter your answer in the provided box. One of the half-reactions for the electrolysis of water is 2H2O(l) → O2(g) + 4H+(aq) + 4e− If 3.696 L of O2 is collected at 25°C and 755 mmHg, how many faradays of electricity had to pass through the solution?
Answer : The number of faradays of electricity had to pass through the solution will be, 0.596 F
Explanation :
First we have to calculate the moles of oxygen gas.
using ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = pressure of gas = 755 mm Hg = 0.99 atm
conversion used : (1 atm = 760 mmHg)
V = volume of gas = 3.696 L
T = temperature of gas = [tex]25^oC=273+25=298K[/tex]
n = number of moles of gas = ?
R = gas constant = 0.0821 L.atm/mole.K
Now put all the given values in the ideal gas equation, we get the number of moles of oxygen gas.
[tex](0.99atm)\times (3.696L)=n\times (0.0821L.atm/mole.K)\times (298K)[/tex]
[tex]n=0.149mole[/tex]
Now we have to calculate the number of faradays of electricity had to pass through the solution.
The balanced half-reactions for the electrolysis of water is,
[tex]2H_2O(l)\rightarrow O_2(g)+4H^+(aq)+4e^-[/tex]
From this we conclude that,
As, 1 mole of oxygen gas require 4 mole of electrons that means 4 F (faraday) of electricity.
So, 0.149 mole of oxygen gas require [tex]0.149\times 4=0.596F[/tex] of electricity.
Therefore, the number of faradays of electricity had to pass through the solution will be, 0.596 F
Calculate the solubility of nitrogen in water at an atmospheric pressure of 0.480 atm (a typical value at high altitude). Atmospheric Gas Mole Fraction kH mol/(L*atm) N2 7.81 x 10-1 6.70 x 10-4 O2 2.10 x 10-1 1.30 x 10-3 Ar 9.34 x 10-3 1.40 x 10-3 CO2 3.33 x 10-4 3.50 x 10-2 CH4 2.00 x 10-6 1.40 x 10-3 H2 5.00 x 10-7 7.80 x 10-4
Answer : The solubility of nitrogen in water at an atmospheric pressure will be, [tex]2.5125\times 10^{-4}mole/L[/tex]
Explanation :
First we have to calculate the partial pressure of nitrogen.
Formula used :
[tex]p_{N_2}=X_{N_2}\times P_{atm}[/tex]
where,
[tex]p_{N_2}[/tex] = partial pressure of nitrogen = ?
[tex]X_{N_2}[/tex] = mole fraction of nitrogen = [tex]7.81\times 10^{-1}[/tex]
[tex]p_{atm}[/tex] = atmospheric pressure = 0.480 atm
Now put all the given values in the above formula, we get :
[tex]p_{N_2}=7.81\times 10^{-1}\times 0.480 atm[/tex]
[tex]p_{N_2}=0.375atm[/tex]
Now we have to calculate the solubility of nitrogen in water.
Formula used :
[tex]s_{N_2}=p_{N_2}\times K_H[/tex]
where,
[tex]p_{N_2}[/tex] = partial pressure of nitrogen = 0.375 atm
[tex]s_{N_2}[/tex] = solubility of nitrogen in water = ?
[tex]K_H[/tex] = Henry's constant = [tex]6.70\times 10^{-4}mole/L.atm[/tex]
Now put all the given values in the above formula, we get :
[tex]s_{N_2}=0.375atm\times 6.70\times 10^{-4}mole/L.atm[/tex]
[tex]s_{N_2}=2.5125\times 10^{-4}mole/L[/tex]
Therefore, the solubility of nitrogen in water at an atmospheric pressure will be, [tex]2.5125\times 10^{-4}mole/L[/tex]
To calculate the solubility of nitrogen in water at an atmospheric pressure of 0.480 atm, we can use Henry's Law. Multiply the mole fraction of nitrogen by the atmospheric pressure to calculate the partial pressure. Then, multiply the Henry's Law constant for nitrogen by the partial pressure to find the solubility of nitrogen in water.
Explanation:To calculate the solubility of nitrogen in water at a given atmospheric pressure, we can use Henry's Law. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
The equation for Henry's Law is:
C = kH × P
where C is the concentration of the gas in the liquid, kH is the Henry's Law constant, and P is the partial pressure of the gas above the liquid.
In this case, we need to use the mole fraction of nitrogen and the Henry's Law constant for nitrogen to calculate the solubility. The mole fraction of nitrogen is 7.81 x 10^-1 and the Henry's Law constant for nitrogen is 6.70 x 10^-4 mol/(L×atm).
Convert the mole fraction of nitrogen to partial pressure using the given atmospheric pressure of 0.480 atm. Multiply the mole fraction by the atmospheric pressure: 7.81 x 10^-1 ×0.480 atm = 3.7572 x 10^-1 atm.Use the Henry's Law constant for nitrogen and the partial pressure of nitrogen calculated in step 1 to find the solubility of nitrogen in water. Multiply the Henry's Law constant by the partial pressure: 6.70 x 10^-4 mol/(L×atm) ×3.7572 x 10^-1 atm = 2.5108 x 10^-4 mol/L.Therefore, the solubility of nitrogen in water at an atmospheric pressure of 0.480 atm is 2.5108 x 10^-4 mol/L.
Suppose 23.8 g of oxygen (O2) is heated at constant atmospheric pressure from 27.4°C to 149°C. (a) How many moles of oxygen are present? (Take the molar mass of oxygen to be 32.0 g/mol) (b) How much energy is transferred to the oxygen as heat? (The molecules rotate but do not oscillate.) (c) What fraction of the heat is used to raise the internal energy of the oxygen?
Answer:
For a: The number of moles of oxygen gas is 0.74375 moles.
For b: The energy transferred to oxygen as heat is [tex]2.631\times 10^3[/tex]
For c: The fraction of heat used to raise the internal energy of oxygen is 0.714.
Explanation:
For a:To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of oxygen gas = 23.8 g
Molar mass of oxygen gas = 32 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of oxygen gas}=\frac{23.8g}{32g/mol}=0.74375mol[/tex]
Hence, the number of moles of oxygen gas is 0.74375 moles.
For b:Oxygen is a diatomic gas.
To calculate the amount of heat transferred, we use the equation:
[tex]Q=nC_p\Delta T[/tex]
where,
Q = heat absorbed or released
n = number of moles of oxygen gas = 0.74375 moles
[tex]C_p[/tex] = specific heat capacity at constant pressure = [tex]\frac{7}{2}R[/tex] (For diatomic gas)
R = gas constant = 8.314 J/mol K
[tex]\Delta T[/tex] = change in temperature = [tex](149-27.4)^oC=121.6^oC=121.6K[/tex]
Putting values in above equation, we get:
[tex]Q=0.74375mol\times (\frac{7}{2})\times 8.314J/mol.K\times 121.6K\\\\Q=2631.71J=2.631\times 10^3J[/tex]
Hence, the energy transferred to oxygen as heat is [tex]2.631\times 10^3[/tex]
For c:To calculate the fraction of heat, we use the equation:
[tex]f=\frac{U}{Q}[/tex]
where,
U = internal energy = [tex]nC_v\Delta T[/tex]
Calculating the value of U:
n = number of moles of oxygen gas = 0.74375 moles
[tex]C_v[/tex] = specific heat capacity at constant pressure = [tex]\frac{5}{2}R[/tex] (For diatomic gas)
R = gas constant = 8.314 J/mol K
[tex]\Delta T[/tex] = change in temperature = [tex](149-27.4)^oC=121.6^oC=121.6K[/tex]
Putting values in above equation, we get:
[tex]U=0.74375mol\times (\frac{5}{2})\times 8.314J/mol.K\times 121.6K\\\\U=1879.79J=1.879\times 10^3J[/tex]
Taking the ratio of 'U' and 'Q', we get:
[tex]f=\frac{1.879\times 10^3}{2.631\times 10^3}\\\\f=0.714[/tex]
Hence, the fraction of heat used to raise the internal energy of oxygen is 0.714.
The compound AX2 decomposes according to the equation, 2 AX2(g) => 2 AX(g) + X2(g). In one experiment, AX2 was measured at various times, and these data recorded: Time (s) {AX2} mol/L 0 0.0500 2.0 0.0448 6.0 0.0300 8.0 0.0249 10.0 0.0209 20.0 0.0088 What is the average rate from 8.0 to 20.0 seconds in mol / L s? Use 4 decimal places.
Answer:
0.001341 mol/L s is the average rate from 8.0 to 20.0 seconds.
Explanation:
[tex]2 AX_2(g)\rightarrow 2 AX(g) + X_2(g)[/tex]
Average rate of the reaction =[tex]R_a[/tex]
[tex]R_a=-\frac{\Delta [x]}{\Delta T}=-\frac{x_2-x_1}{t_2-t_1}[/tex]
[tex]R_a[/tex] =Average rate of the reaction during the given time interval.
[tex]\Delta [x][/tex] = Change in concentration of reactant with respect to time.
[tex]\Delta T[/tex] = Change in time.
[tex]x_1[/tex]=Concentration of reactant at time[tex]t_1[/tex]
[tex]x_2[/tex]=Concentration of reactant at time[tex]t_2[/tex]
So, at [tex]t_1=8.0 sec[/tex] the concentration of [tex]AX_2[/tex] :
[tex]x_1=0.0249 mol/L[/tex]
And at [tex]t_2=2.0 sec[/tex] the concentration of [tex]AX_2[/tex] :
[tex]x_2=0.0088 mol/L[/tex]
The average rate of the reaction at given interval will be given as:
[tex]R_a=-\frac{x_2-x_1}{t_2-t_1}=-\frac{0.0088 mol/L-0.0249mol/L}{20.0s-8.0 s}=0.001341 mol/L s[/tex]
0.001341 mol/L s is the average rate from 8.0 to 20.0 seconds.
Given the recipe: 2 cups flour+ 1 egg+3 oz blueberries 4 muffins. Assuming you have sufficient flour and blueberries, you can make 16 muffins from 3 eggs (T/F)
Answer:
False, we cannot made 16 muffins from 3 eggs.
Explanation:
2 cups flour + 1 egg + 3 oz blueberries → 4 muffins
According top recipe, 2 cups of flour, 1 egg and 3 oz of blueberries are made into batter to give 4 muffins.
Then 16 muffins will made from:
For each muffin we need =[tex]\frac{1}{4} egg[/tex]
Then for 16 muffins we will need:
[tex]\frac{1}{4}\times 14 eggs=4 eggs[/tex]
4 eggs will be required to make 16 muffins
From 1 egg we can made 4 muffins
Then from 3 eggs we can made = 3 × 4 muffins = 12 muffins
12 muffins will made from 3 eggs.
We cannot make 16 muffins from 3 eggs.
A hydrogen molecule (diameter 1.38 × 10-8 cm), traveling at the rms speed, escapes from a 4200 K furnace into a chamber containing cold argon atoms (diameter 3.43 × 10-8 cm) at a density of 4.43 × 1019 atoms/cm3(a) What is the speed of the hydrogen molecule
Answer: The speed of hydrogen molecule is [tex]7.199\times 10^3m/s[/tex]
Explanation:
The equation used to calculate the root mean square speed of a molecule is:
[tex]V_{rms}=\sqrt{\frac{3RT}{M}}[/tex]
where,
[tex]V_{rms}[/tex] = root mean square speed of the molecule = ?
R = Gas constant = 8.314 J/mol.K
T = temperature = 4200 K
M = molar mass of hydrogen molecule = [tex]2.02g=2.02\times 10^{-3}kg[/tex] (Conversion factor: 1 kg = 1000 g)
Putting values in above equation, we get:
[tex]V_{rms}=\sqrt{\frac{3\times 8.314\times 4200}{2.02\times 10^{-3}}}\\\\V_{rms}=7.199\times 10^3m/s[/tex]
Hence, the speed of hydrogen molecule is [tex]7.199\times 10^3m/s[/tex]
Question 5 A solution is prepared at 25°C that is initially 0.35M in chlorous acid HClO2, a weak acid with =Ka×1.110−2, and 0.29M in sodium chlorite NaClO2. Calculate the pH of the solution. Round your answer to 2 decimal places.
Answer : The pH of the solution is, 1.88
Explanation : Given,
[tex]K_a=1.1\times 10^{-2}[/tex]
Concentration of [tex]HClO_2[/tex] = 0.35 M
Concentration of [tex]NaClO_2[/tex] = 0.29 M
First we have to calculate the value of [tex]pK_a[/tex].
The expression used for the calculation of [tex]pK_a[/tex] is,
[tex]pK_a=-\log [K_a][/tex]
Now put the value of [tex]K_a[/tex] in this expression, we get:
[tex]pK_a=-\log (1.1\times 1-^{-2})[/tex]
[tex]pK_a=2-\log (1.1)[/tex]
[tex]pK_a=1.96[/tex]
Now we have to calculate the pH of the solution.
Using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]
[tex]pH=pK_a+\log \frac{[NaClO_2]}{[HClO_2]}[/tex]
Now put all the given values in this expression, we get:
[tex]pH=1.96+\log (\frac{0.29}{0.35})[/tex]
[tex]pH=1.88[/tex]
Therefore, the pH of the solution is, 1.88
The pH of a buffer solution consisting of a weak acid and its conjugate base can be calculated using the Henderson–Hasselbalch equation, with the concentrations of the weak acid and base, and the Ka or pKa of the weak acid.
Explanation:In this scenario, we are dealing with a weak acid, chlorous acid (HClO2), and the salt of its conjugate base, sodium chlorite (NaClO2). When a weak acid is in solution with the salt of its conjugate base, a buffer solution is formed. Buffer solutions resist significant changes in pH upon the addition of small quantities of acid or base. The pH of the buffer solution can be calculated using the Henderson–Hasselbalch equation:
pH = pKa + log([A-]/[HA]).
Where,
pKa is the negative log of the acid dissociation constant Ka,[A-] is the concentration of the base form, here it is sodium chlorite NaClO2,[HA] is the concentration of the weak acid, here it is chlorous acid HClO2.Using the given concentrations and Ka or pKa, substitute them into the formula to calculate the pH of the solution.
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Joan’s initial nickel (II) chloride sample was green and weighed 4.3872 g. After the dehydration reaction and removal of excess thionyl chloride, the residue was yellow, and had a mass of 2.3921 g.
(a) Was the nickel chloride sample a hydrate? ___________________
(b) If the data indicates a hydrate, calculate x, the number of waters of hydration.
Show your work.
Answer:
a) yes, it was an hydrate
b) the number of waters of hydration, x = 6
Explanation:
a) yes it was an hydrate because the mass decreased after the process of dehydration which means removal of water thus some water molecules were present in the sample.
b) NiCl2. xH2O
mass if dehydrated NiCl2 = 2.3921 grams
mass of water in the hydrated sample = mass of hydrated - mass of dehydrated = 4.3872 - 2.3921 = 1.9951 g which represent the mass of water that was present in the hydrated sample.
NiCl2.xH2O
mole of dehydrated NiCl2 = m/Mm = 2.3921/129.5994 = 0.01846 mole
mole of water = m/Mm = 1.9951/18.02 = 0.11072 mole
Divide both by the smallest number of mole (which is for NiCl2) to find the coefficient of each
for NiCl2 = 0.01846/0.01846 = 1
for H2O = 0.11072/0.01846 = 5.9976 = 6
thus the hydrated sample was NiCl2. 6H2O
A 3.4 g sample of sodium hydrogen carbonate is added to a solution of acetic acid weighing 10.9 g. The two substances react, releasing carbon dioxide gas to the atmosphere. After the reaction, the contents of the reaction vessel weigh 11.6 g. What is the mass of carbon dioxide released during the reaction?
Answer: 2.7 grams
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
[tex]NaHCO_3(aq)+CH_3COOH(aq)\rightarrow CH_3COONa(aq)+H_2O(l)+CO_2(g)[/tex]
Given: mass of sodium hydrogen carbonate = 3.4 g
mass of acetic acid = 10.9 g
Mass of reactants = mass of sodium hydrogen carbonate+ mass of acetic acid = 3.4 + 10.9= 14.3 g
Mass of reactants = Mass of products in reaction vessel + mass of carbon dioxide (as it escapes)
Mass of carbon dioxide = 14.3 - 11.6 =2.7 g
Thus the mass of carbon dioxide released during the reaction is 2.7 grams.
The feed to an ammonia synthesis reactor contains 25 mole% nitrogen and the balance hydrogen. The fow rate of the stream is 3000 kg/h. Calculate the rate of flow of nitrogen into the reactor in kg/h. (Suggestion: First calculate the average molecular weight of the mixture.)
Answer:
The rate of flow of nitrogen into the reactor is 2,470.588 kg/h.
Explanation:
Mole percentage of nitrogen gas = 25 mole%
Mole percentage of hydrogen gas = 75 mole%
Average molecular weight of the mixture:
[tex](0.25\times 28 g/mol)+(0.75\times 2 g/mol)=8.5 g/mol[/tex]
Rate of flow of the stream = 3000 kg/h
Mass of stream in 1 hour = 3000 kg = 3,000,000 g
Moles of stream :
[tex]\frac{3000 g}{8.5 g/mol}=352,941.17 mol[/tex]
Moles of nitrogen gas in 352,941.17 moles of stream be x
[tex]25\%=\frac{\text{Moles of nitrogen gas}}{\text{Moles of stream}}\times 100[/tex]
[tex]25=\frac{x}{352,941.17mol}\times 100[/tex]
x = 88,235.294 mol
Mass of 8,823,529.4 mole of nitrogen gas:
[tex]88235.294 mol\times 28 g/mol=2.470588.2 g=2,470.588 kg[/tex]
The rate of flow of nitrogen into the reactor is 2,470.588 kg/h.
To calculate the rate of flow of nitrogen into the reactor, you need to calculate the average molecular weight of the mixture and use the ideal gas law.
Explanation:To calculate the rate of flow of nitrogen into the reactor, we first need to calculate the average molecular weight of the mixture. The average molecular weight (MW) is calculated by summing the product of the mole fraction of each component and its molecular weight:
MW = (mole fraction of N2 * molecular weight of N2) + (mole fraction of H2 * molecular weight of H2)
Now, we can calculate the rate of flow of nitrogen using the ideal gas law:
Rate of flow of nitrogen = (mole fraction of N2 * flow rate of the stream * MW of N2) / molecular weight of N2
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For the reaction H2(g) + CO2(g) ⇌ H2O(g) + CO(g) at 700ºC, Kc = 0.534. Calculate the number of moles of H2 that are present at equilibrium if a mixture of 0.300 mole of CO and 0.300 mole of H2O is heated to 700ºC in a 10.0-L container.
Answer:
[tex][H_2]_{eq}=0.0173M[/tex]
Explanation:
Hello,
In this case, for the given reaction, the law of mass action turns out:
[tex]Kc=\frac{[H_2O][CO]}{[H_2][CO_2]}[/tex]
In such a way, the reactants initial concentrations are:
[tex][CO_2]_0=[H_2]_0=\frac{0.300mol}{10.0L} =0.030M[/tex]
And considering the change [tex]x[/tex] due to the equilibrium, the law of mass action takes the following form:
[tex]Kc=\frac{(x)(x)}{(0.030M-x)(0.030M-x)} =\frac{x^2}{0.0009-0.06x+x^2} \\Kc(0.0009-0.06x+x^2)=x^2\\0.0004806-0.03204x+0.534x^2-x^2=0\\0.466x^2+0.03204x-0.0004806=0\\x_1=-0.0814M\\x_2=0.0127M[/tex]
The feasible answer is [tex]x=0.0127M[/tex]
Therefore, the hydrogen moles at equilibrium are:
[tex][H_2]_{eq}=0.030M-0.0127M=0.0173M[/tex]
Best regards.
The number of moles of H2 present at equilibrium is 0.04moles
Data;
Kc = 0.534CO = 0.3molH2O = 0.30molV = 10LT = 700ºCEquilibrium Concentrationlet's find the concentration of the species in the reaction.
The concentration of CO is
[tex][CO] = \frac{0.3}{10} = 0.03M[/tex]
The concentration of H2O is
[tex][H_2O] = \frac{0.3}{10} 0.03M[/tex]
The equation of this reaction is
H2(g) + CO2(g) ⇌ H2O(g) + CO(g)
final -x -x 0.03 + x 0.03 + x
The Kc i.e the equilibrium constant of this reaction is
[tex]K_c = \frac{[product]}{[reactant]}\\[/tex]
let's substitute the values and solve.
[tex]K_c = \frac{product}{reactant} \\0.534 = \frac{(0.03+x)(0.03+x)}{x * x} \\x= 0.04M[/tex]
The number of moles of H2 present at equilibrium can be calculated as
[tex]0.04 = \frac{0.04}{1}[/tex]
The number of moles of H2 present at equilibrium is 0.04moles
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On another planet, the isotopes of titanium have the given natural abundances. Isotope Abundance Mass (amu) 46Ti 72.000% 45.95263 48Ti 14.300% 47.94795 50Ti 13.700% 49.94479 What is the average atomic mass of titanium on that planet?
The average atomic mass of titanium on the given planet is 46.800 amu.
Explanation:In order to calculate the average atomic mass of titanium on the given planet, we multiply the mass of each isotope by its natural abundance and sum them up.
For 46Ti: 72.000% x 45.95263 amu = 33.105 amuFor 48Ti: 14.300% x 47.94795 amu = 6.839 amuFor 50Ti: 13.700% x 49.94479 amu = 6.856 amuAdd the results together: 33.105 amu + 6.839 amu + 6.856 amu = 46.800 amuTherefore, the average atomic mass of titanium on the planet is 46.800 amu.
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A solution is prepared by diluting 43.5 mL of 6.5×10?2M Ba(OH)2 to a volume of 270.5 mL .
Express the pH to two decimal places.
Answer:
The pH of the solution is 12.31 .
Explanation:
Initial molarity of barium hydroxide =[tex]M_1=6.5\times 10^{-2}M[/tex]
Initial volume of barium hydroxide =[tex]V_1=43.5 mL[/tex]
Final molarity of barium hydroxide =[tex]M_2[/tex]
Final volume of barium hydroxide =[tex]V_2=270.5 mL[/tex]
[tex]M_1V_1=M_2V_2[/tex]
[tex]M_2=\frac{6.5\times 10^{-2}M\times 43.5 mL}{270.5 mL}[/tex]
[tex]M_2=0.0104 M[/tex]
[tex]Ba(OH)_2\rightarrow Ba^{2+}+2OH^-[/tex]
1 mol of barium gives 2 mol of hydroxide ions.
Then 0.0104 M of barium hydroxide will give:
[tex]2\times 0.0104 M=0.0208 M[/tex] of hydroxide ions
[tex][OH^-]=0.0208 M[/tex]
[tex]pH=14-pOH=14-(-\log[OH^-])[/tex]
[tex]pH=14-(-\log[0.0208 M])=12.31[/tex]
The pH of the solution is 12.31 .
3-methyl-1-hexene is reacted with HCl. Select ALL statements below that are true for this reaction. a) Reaction is stereoselective b) Reaction involves carbocation intermediate c) A 1,2-shift is expected in reaction mechanism d) Reaction is regioselective
Answer:The options b, c and d are True.
Explanation:
The above reaction is an example of electrophilic addition to alkenes and is a typical reaction known as Markownikoffs addition reaction.
In the above reaction we are using Hydrogen chloride as an acid and as HCl is a strong acid so it will provide acidic protons.
The acidic protons (H⁺) acts as electrophile and the the pi-bond in 3-methyl-1-hexene has sufficient electron density to attack the electrophilic protons. As a result of the attack hydrogen is added on one of the carbon atoms across the pi bond and a generation of carbocation takes place.
Once the carboation is formed then it rearranges into a more stable carbocation if it can and hence a stable carbocation is generated. The chloride anion can attack this stable carbocation leading to a regio-selective product.
The carbocation formed is the reaction intermediate in this case.
The statement (a) is incorrect as the above reaction involves carbocation as intermediate and as carbocation is a planar molecule so there are two faces available for the chloride ion to attack hence a racemic product would be formed resulting from the attack on both the sides. So the re action is not stereoselective as not a specific isomer is formed.
The statement (b) is correct as the intermediate formed in the reaction is a carbocation.
The statement (c) is correct as carbocation is formed and we know that carbocations undergo rearrangements to form more stable carbocations so 1,2 shifts can occur in the reaction. As 1,2 shift is a way through rearrangements can occur in carbocation.
The statement (d) is correct as the reaction would lead to formation of a stable carbocation and hence the chloride ion will only attack the stable carbocation so the reaction would be regioselective in nature.
Predict the sign for AS for each of the following systems: Water freezing Water evaporating Crystalline urea dissolving Assembly of the plasma membrane from individual lipids. Assembly of a protein from individual amino acids.
Answer: 1. Water freezing : [tex]\Delta S[/tex] is -ve.
2. Water evaporating : [tex]\Delta S[/tex] is +ve.
3. Crystalline urea dissolving : [tex]\Delta S[/tex] is +ve.
4. Assembly of the plasma membrane from individual lipids: [tex]\Delta S[/tex] is -ve.
5. Assembly of a protein from individual amino acids: [tex]\Delta S[/tex] is -ve.
Explanation:
Entropy is defined as the measurement of degree of randomness in a system.
It is represented by symbol S and we can only measure a change in entropy which is given by [tex]\Delta S[/tex].
If there is decrease in randomness , the sign for [tex]\Delta S[/tex] is -ve and If there is increase in randomness , the sign for [tex]\Delta S[/tex] is +ve.
1. Water freezing: Entropy decreases as we move from liquid state to to solid state and thus [tex]\Delta S[/tex] is -ve.
2. Water evaporating : Entropy increases as we move from liquid state to gaseous state and thus [tex]\Delta S[/tex] is +ve.
3. Crystalline urea dissolving : The molecules convert from solid and ordered state to aqueous phase and random state. Thus the entropy increases and thus [tex]\Delta S[/tex] is +ve.
4. Assembly of the plasma membrane from individual lipids: random lipids are associating to form a single large polymer and thus entropy decreases and thus [tex]\Delta S[/tex] is -ve.
5. Assembly of a protein from individual amino acids: random amino acids are associating to form a single large polymer and thus entropy decreases and thus [tex]\Delta S[/tex] is -ve.
Magnesium and nitrogen react in a combination reaction to produce magnesium nitride: 3 Mg N2 → Mg3N2 In a particular experiment, a 8.33-g sample of N2 reacts completely. The mass of Mg consumed is ________ g.
Answer:
The mass of Mg consumed is 21.42g
Explanation:
The reaction is
[tex]3Mg+N_{2}-->Mg_{3}N_{2}[/tex]
As per balanced equation, three moles of Mg will react with one mole of nitrogen to give one mole of magnesium nitride.
as given that mass of nitrogen reacted = 8.33g
So moles of nitrogen reacted = [tex]\frac{mass}{molarmass}=\frac{8.33}{28}=0.2975mol[/tex]
moles of Mg required = 3 X moles of nitrogen taken = 3X0.2975 = 0.8925mol
Mass of Mg required = moles X molar mass = 0.8925 X 24 = 21.42 g
The stoichiometric reaction between magnesium and nitrogen to produce magnesium nitride requires 3 moles of magnesium for each mole of nitrogen. Therefore, for an 8.33g sample of nitrogen, approximately 21.66g of magnesium will be consumed.
Explanation:The subject question involves a stoichiometric calculation regarding a reaction between magnesium (Mg) and nitrogen (N2) to produce magnesium nitride (Mg3N2). The balanced equation for the reaction is: 3 Mg + N2 → Mg3N2. This ratio tells us that for every 1 mole of N2, 3 moles of Mg are needed.
The molar mass of N2 is 28.014 g/mol. In an 8.33-g sample, there are (8.33 g)/(28.014 g/mol) = 0.297 mol of N2.
As per the reaction's stoichiometry, three times this amount in moles of Mg are needed, so 0.297 mol x 3 = 0.891 mol of Mg are required.
The molar mass of Mg is 24.305 g/mol. Thus, the mass of Mg consumed is (0.891 mol) x (24.305 g/mol) = 21.66 g.
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