Answer:
The percent recovery from re crystallization for both compounds A and B is 69.745 and 81.44 % respectively.
Explanation:
Mass of compound A in a mixture = 119 mg
Mass of compound A after re-crystallization = 83 mg
Percent recovery from re-crystallization :
[tex]\frac{\text{Mass after re-crystallization}}{\text{Mass before re-crystallization}}\times 100[/tex]
Percent recovery of compound A:
[tex]\frac{83 mg}{119 mg}\times 100=69.74\%[/tex]
Mass of compound B in a mixture = 97 mg
Mass of compound B after re-crystallization = 79 mg
Percent recovery of compound B:
[tex]\frac{79 mg}{97 mg}\times 100=81.44\%[/tex]
Consider the neutralization reaction 2HNO3(aq)+Ba(OH)2(aq)⟶2H2O(l)+Ba(NO3)2(aq) A 0.105 L sample of an unknown HNO3 solution required 46.5 mL of 0.100 M Ba(OH)2 for complete neutralization. What is the concentration of the HNO3 solution?
Answer: The concentration of [tex]HNO_3[/tex] comes out to be 0.088 M.
Explanation:
To calculate the concentration of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HNO_3[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ba(OH)_2[/tex]
We are given:
Conversion factor: 1 L = 1000 mL
[tex]n_1=1\\M_1=?M\\V_1=0.105L=105mL\\n_2=2\\M_2=0.100M\\V_2=46.5mL[/tex]
Putting values in above equation, we get:
[tex]1\times M_1\times 105=2\times 0.100\times 46.5\\\\M_1=0.088M[/tex]
Hence, the concentration of [tex]HNO_3[/tex] comes out to be 0.088 M.
The concentration of the HNO3 solution is 0.443 M.
Explanation:To determine the concentration of the HNO3 solution, we can use the equation, concentration = (volume of solution titrated)/(volume of solution required for complete neutralization). In this case, the volume of solution titrated is 0.105 L and the volume of 0.100 M Ba(OH)2 solution required for complete neutralization is 46.5 mL (or 0.0465 L). Plugging these values into the equation, we get: concentration = 0.0465 L / 0.105 L = 0.443 M.
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Sodium nitrite (NaNO2) reacted with 2−iodooctane to give a mixture of two constitutionally isomeric compounds of molecular formula C8H17NO2 in a combined yield of 88%. Draw reasonable structures for these two isomers.
Hey there!:
the chemical reaction between 2-iodoctane and sodium nitrite is as follows:
Answer:
On the attached picture.
Explanation:
Hello,
In the case, the reaction between 2-iodooctane and sodium nitrite, leads to the formation of an alkyl nitrite and a nitro alkane as shown on the attached picture. Once the reaction began, the salt breaks and the sodium bonds with the iodine from the 2-iodooctane to form sodium iodide, in such a way, a free radical in the second carbon is formed so the NO₂ could bond both as a nitrite and as a nitro radical; therefore, the formed species are octyl 2-nitrite and 2-nitrooctane.
Best regards.
"Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it is used to make the refrigerant carbon tetrafluoride, CF4 via the reaction 2COF2(g)⇌CO2(g)+CF4(g), Kc=9.00 If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?
Answer:
[tex]\boxed{\text{0.50 mol/L}}[/tex]
Explanation:
The balanced equation is
2COF₂ ⇌ CO₂+CF₄; Kc = 9.00
1. Set up an ICE table
[tex]\begin{array}{ccccc}\rm 2COF_{2} & \, \rightleftharpoons \, & \rm CO_{2} & +&\rm CF_{4}\\2.00& & 0& & 0\\-x& & +x & & +x\\2.00 - x& & x & &x \\\end{array}[/tex]
2. Solve for x
[tex]K_{c} = \dfrac{[\rm CO][ \rm CF_{4}]}{[\rm COF_{2}]^{2}} = 9.00\\\\\begin{array}{rcl}\dfrac{x^{2}}{(2.00 - x)^{2}} & = & 9.00\\\dfrac{x}{2.00 - x} & = & 3.00\\x & = &3.00(2.00 - x)\\x & = & 6.00 - 3.00x\\4.00x & = & 6.00\\x & = & \mathbf{1.50}\\\end{array}[/tex]
3. Calculate the equilibrium concentration of COF₂
c = (2.00 - x) mol·L⁻¹ = (2.00 - 1.50) mol·L⁻¹ = 0.50 mol
[tex]\text{The equilibrium concentration of COF$_{2}$ at equilibrium is $\boxed{\textbf{0.50 mol/L}}$}[/tex]
Check:
[tex]\begin{array}{rcl}\dfrac{1.50^{2}}{0.50^{2}} & = & 9.00\\\\\dfrac{2.25}{0.25}& = & 9.00\\\\9.00 & = & 9.00\\\end{array}[/tex]
OK.
The equilibrium concentration of COF2 can be found by solving a quadratic equation derived from the equilibrium constant expression and initial concentration, and then subtracting the amount that reacted from the initial concentration.
Explanation:The student asked about the equilibrium concentration of carbonyl fluoride, COF2, in a specific chemical reaction where it is converted to carbon dioxide and carbon tetrafluoride. To determine the concentration of COF2 that remains at equilibrium starting with an initial concentration of 2.00 M, we must use the equilibrium constant (Kc) value provided, which is 9.00.
The equilibrium expression for the reaction 2COF2(g) ⇌ CO2(g) + CF4(g) is Kc = [CO2][CF4] / [COF2]2. Let the change in concentration of COF2 at equilibrium be 'x'. Thus, at equilibrium, we'll have [COF2] = 2.00 - 2x, [CO2] = x, and [CF4] = x. Plugging these into the equilibrium expression, we have 9.00 = x2 / (2.00 - 2x)2. Solving for x will give us the change in concentration of COF2, from which we can determine the equilibrium concentration of COF2 by subtracting 'x' from the initial concentration.
Therefore, to find the equilibrium concentration of COF2, one would need to solve the quadratic equation resulting from the substitution of equilibrium concentrations in terms of 'x' into the Kc expression, then calculate 2.00 M - 2x.
Now let M react, and at time = 0, [M] = 0.102 M, A(545nm) = 0.491; at time = 30 seconds, A(545nm) = 0.404. What is the [M] at 30 seconds? What is the slope of the Absorbance vs. [M] function? What is the (average) rate of the reaction during this time period.
If a new reaction was run leaving [N] the same, but beginning (time = 0) with [M] = 0.204 M, A(545nm) = 0.981. After 30 seconds, A (545 nm) is 0.807. What is the order of the reaction with respect to [M] ?
Answer:
election of Cell conduct of your daily India Cases, itu is'n give you some reason=✓
Explanation:
[tex] \geqslant { \sqrt[ \geqslant \sqrt[ \geqslant \leqslant \sqrt[ \geqslant ]{0.time} \times \frac{ log_{30.0m}(3.71m) }{204.9m} ]{3528t} ]{3} }^{8} \times \frac{4}{3} [/tex]
Ammonia is a principal nitrogen fertilizer. It is prepared by the reaction between hydrogen and nitrogen. 3H2(g) + N2(g) → 2NH3(g) In a particular reaction, 8.00 moles of NH3 were produced. How many moles of H2 and how many moles of N2 were reacted to produce this amount of NH3?
To produce 8 moles of ammonia, 4 moles of nitrogen and 12 moles of hydrogen would have been reacted, according to the stoichiometry of the reaction.
Explanation:The balanced chemical reaction for the production of ammonia (NH3) is given by: N2(g) + 3H2(g) → 2NH3(g). This implies that for every 2 moles of ammonia produced, 1 mole of nitrogen (N2) and 3 moles of hydrogen (H2) are reacted. Therefore, if 8 moles of NH3 were produced, the amount of nitrogen reacted would be 8/2 moles of N2, which is 4 moles of N2. Using the stoichiometric ratio, the moles of hydrogen reacted would be 3 x 4, which is 12 moles of H2.
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To produce 8.00 moles of NH3, the reaction would consume 4.00 moles of N2 and 12.00 moles of H2, as inferred from the stoichiometric ratios provided by the balanced equation.
Explanation:The chemical equation for the creation of ammonia, a principle nitrogen fertilizer, from nitrogen and hydrogen is shown as: N2(g) + 3H2(g) → 2NH3(g). To find out how many moles of H2 and N2 were reacted to produce 8.00 moles of NH3, we use the stoichiometric ratios provided by the balanced equation. The balanced equation shows that one mole of N2 reacts with three moles of H2 to produce two moles of NH3. Therefore, if 8.00 moles of NH3 were produced, we can infer that the reaction consumed 4.00 moles of N2 and 12.00 moles of H2. This is because the ratio of N2 to NH3 in the reaction is 1:2 and the ratio H2 to NH3 is 3:2, so we divide 8 by 2 to find the moles of N2, and multiply 8 by 3/2 to find moles of H2.
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Arrange the complexes in order of decreasing stability. Rank from most stable to least stable. To rank items as equivalent, overlap them. [Ni(en)(H2O)4] [FeF6] [Fe(NH3)6] [Ni(en)3])
Answer: [Ni(en)3] > [Ni(en)(H2O)4] > [Fe(NH3)6] > [FeF6]
Explanation:
Generally chelating ligands stabilize the complex more than non chelated ligands and more the no of chelated ligands more the stability.
Here en (ethylenediamine) is a chelated ligand and stabilze the complex more by chelation.
And Strong field ligand (NH3) also stabilze the complex more than weak field ligand (F).
Hence F containing complex is least stable.
The rank of the coordination compounds in the increasing order of the stability is [tex]\rm [Ni(en)_3] > [Ni(en)(H_2O)_4] > [Fe(NH_3)_6] > [FeF_6][/tex].
What are coordination compounds?The coordination compound are the complexes in which the central metal atom has been bounded by the nonmetal to the complexes through the chemical bond.
The stability of the coordination complexes is attributed to the number of chelating agents to the central atoms.
The increased number of atoms to the central atoms adds to the stability of the complex. Thus, the increasing order of the stability is [tex]\rm [Ni(en)_3] > [Ni(en)(H_2O)_4] > [Fe(NH_3)_6] > [FeF_6][/tex].
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during a chemical reacion, an iron atom beacme the ion Fe+2. what happened to the iron atom?
Answer:
That iron atom is oxidized. It loses two electrons.
Explanation:
Compare the formula of an iron atom and an iron(II) ion:
Iron atom: [tex]\mathrm{Fe}[/tex];Iron(II) ion: [tex]\mathrm{Fe^{2+}}[/tex].The superscript [tex]+2[/tex] in the iron(II) ion is the only difference between the two formulas. This superscript indicates a charge of [tex]+2[/tex] on each ion. Atoms and ions contain protons. In many cases, they also contain electrons. Each proton carries a positive charge of [tex]+1[/tex] and each electron carries a charge of [tex]-1[/tex]. Atoms are neutral for they contain an equal number of protons and electrons.
Protons are located at the center of atoms inside the nuclei. They cannot be gained or lost in chemical reactions. However, electrons are outside the nuclei and can be gained or lost. When an atom loses one or more electrons, it will carry more positive charge than negative charge. It will becomes a positive ion. Conversely, when an atom gains one or more electrons, it becomes a negative ion.
An iron atom [tex]\mathrm{Fe}[/tex] will need to lose two electrons to become a positive iron(II) ion [tex]\mathrm{Fe^{2+}}[/tex] with a charge of [tex]+2[/tex] on each ion. That is:
[tex]\rm Fe \to Fe^{2+} + 2\;e^{-}[/tex].
Oxidation is Losing one or more electrons;Reduction is Gaining one or more electrons.This definition can be written as the acronym OILRIG. (Khan Academy.)
In this case, each iron atom loses two electrons. Therefore the iron atoms here are oxidized.
Final answer:
When an iron atom becomes an ion Fe+2, it loses two electrons, increasing its oxidation number to +2. This is oxidation and is common for transition metals like iron which can create ions of different charges.
Explanation:
During a chemical reaction, when an iron atom becomes the ion Fe+2, it means that the iron atom has lost two electrons. This process is known as oxidation and results in an increase in the oxidation number of iron from 0 to +2. Iron atoms can lose electrons from their outermost shell when they react, and transition metals like iron can have more than one possible charged state, commonly forming Fe2+ or Fe3+ ions. In the provided reaction, 4 Fe + 3O2 → 2 Fe2O3, iron is in the +3 oxidation state, indicating it has lost three electrons. Therefore, in different chemical environments, iron atoms can lose varying numbers of electrons, resulting in various oxidation states such as Fe2+ or Fe3+.
Which of the following values would you expect for the ratio of half-lives for a reaction with starting concentrations of 0.05M and 0.01M, t1/2(0.05M) / t1/2 (0.01M), if a reaction is known to be zero order?
Answer:
The expected ratio of half-lives for a reaction will be 5:1.
Explanation:
Integrated rate law for zero order kinetics is given as:
[tex]k=\frac{1}{t}([A_o]-[A])[/tex]
[tex][A_o][/tex] = initial concentration
[A]=concentration at time t
k = rate constant
if, [tex][A]=\frac{1}{2}[A_o][/tex]
[tex]t=t_{\frac{1}{2}}[/tex], the equation (1) becomes:
[tex]t_{\frac{1}{2}}=\frac{[A_o]}{2k}[/tex]
Half life when concentration was 0.05 M=[tex]t_{\frac{1}{2}}[/tex]
Half life when concentration was 0.01 M=[tex]t_{\frac{1}{2}}'[/tex]
Ratio of half-lives will be:
[tex]\frac{t_{\frac{1}{2}}}{t_{\frac{1}{2}}'}=\frac{\frac{[0.05 M]}{2k}}{\frac{[0.01 M]}{2k}}=\frac{5}{1}[/tex]
The expected ratio of half-lives for a reaction will be 5:1.
The elimination of 5−bromononane on treatment with potassium ethoxide is stereoselective. Select the correct Newman projections of 5−bromononane showing the conformations that lead to cis−4−nonene and trans−4−nonene, respectively. Identify the proton that is lost in each case by selecting the Newman projections that have this hydrogen colored blue. Suggest a mechanistic explanation for the observed stereoselectivity. Leads to cis−4−nonene:
hey There!:
Compound B is the structure that leads to Cis-nonene as major product because in this case the Leaving group i.e Br and the electrophile i.e H is not anti-periplanar to each other .
Hope this helps!
Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it is used to make the refrigerant carbon tetrafluoride, CF4 via the reaction 2COF2(g)⇌CO2(g)+CF4(g), Kc=4.20 If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?
Answer: The concentration of [tex]COF_2[/tex] remains at equilibrium is 0.40 M.
Explanation:-
Initial concentration of [tex]COF_2[/tex] = 2 M
The given balanced equilibrium reaction is,
[tex]2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g[/tex]
Initial conc. 2 M 0 0
At eqm. conc. (2-2x) M x M x M
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[CO_2]\times [CF_4]}{[COF_2]^2}[/tex]
Now put all the given values in this expression, we get :
[tex]4.20=\frac{(x)\times (x)}{(2-2x)^2}[/tex]
By solving the term 'x', we get :
x = 0.80 M
Thus, the concentrations of [tex]COF_2[/tex] at equilibrium is : (2-2x) = 2-2(0.80)=0.40 M
The concentration of [tex]COF_2[/tex] remains at equilibrium is 0.40 M.
Chemistry is the study of ________. Chemistry is the study of ________. matter transformations of matter only macroscopic phenomena only microscopic phenomena both A and B
Answer: Correct answer is both A and B.
Explanation:
Chemistry is a branch of science that deals with different substances in recognizing the matter present in them.
For example, chemistry deals with states of matter like solid, liquid or gas.
Also, it tells us about how different substances present in nature react or what is their composition and properties.
Basically, chemistry helps us knowing nature of a matter and how it can be transformed.
Thus, we can conclude that chemistry is the study of matter and its transformation.
Final answer:
Chemistry is the study of matter and its properties, including macroscopic and microscopic phenomena.
Explanation:
Chemistry is the study of matter and the changes it undergoes, encompassing both macroscopic and microscopic phenomena. It focuses on substances, their properties, and how they interact and transform during chemical reactions. There is a vast array of specializations within chemistry, including but not limited to physical chemistry, organic chemistry, inorganic chemistry, analytical chemistry, and biochemistry.
Matter consists of anything that has mass and occupies space. This broad definition includes the air we breathe, the food we eat, and even materials like plastics and metals. Understanding the changes matter undergoes, whether through physical or chemical processes, is central to this science. Moreover, chemistry is often known as the central science because it intersects with other scientific disciplines such as physics, biology, and environmental science.
In a hydrogen-oxygen fuel cell, A) no transfer of electrons occurs.B) both oxygen and hydrogen atoms lose electrons.C) both oxygen and hydrogen atoms gain electrons.D) oxygen atoms lose electrons and hydrogen atoms gain electrons.E) oxygen atoms gain electrons and hydrogen atoms lose electron
Answer:
oxygen atoms gain electrons and hydrogen atoms lose electron
Explanation:
Hydrogen oxygen fuel cell involves redox reactions.
It is an electrochemical cell, which is used for many applications like rocket propellant.
The actual reaction is
[tex]H_{2}+\frac{1}{2}O_{2}--->H_{2}O[/tex]
Here hydrogen undergoes oxidation as it loses electrons
Oxygen undergoes reduction as it gains electrons.
The redox reactions are
At anode:
[tex]H_{2}--->2H^{+}+2e[/tex] [loss of electrons by hydrogen]
At cathode:
[tex]O_{2}+4H^{+}+4e--->2H_{2}O[/tex] [gain of electrons by oxygen]
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.
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.
The reaction 2 H:O 2 H3 + O, would be correctly classified as a decomposition reaction (T/F)
Answer:
Explanation:
It's false, but not because it is not a decomposition equation. It does resemble one.
As near as I can tell, it should be
2H2O >>> 2H2 + O2
Calculate the vapor pressure of a solution containing 24.6 g of glycerin (C3H8O3) in 134 mL of water at 30.0 ∘C. The vapor pressure of pure water at this temperature is 31.8 torr. Assume that glycerin is not volatile and dissolves molecularly (i.e., it is not ionic) and use a density of 1.00 g/mL for the water.
Answer : The vapor pressure of a solution is, 30.687 torr
Explanation :
First we have to calculate the moles of glycerin.
[tex]\text{Moles of }C_3H_8O_3=\frac{\text{Mass of }C_3H_8O_3}{\text{Molar mass of }C_3H_8O_3}=\frac{24.6g}{92.09g/mole}=0.267moles[/tex]
Now we have to calculate the mass of water.
[tex]\text{Mass of }H_2O=\text{Density of }H_2O\times \text{Volume of }H_2O[/tex]
[tex]\text{Mass of }H_2O=(1.00g/ml)\times (134ml)=134g[/tex]
Now we have to calculate the moles of water.
[tex]\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{134g}{18g/mole}=7.444moles[/tex]
Now we have to calculate the mole fraction of water.
[tex]\text{Mole fraction of water}=\frac{\text{Moles of water}}{\text{Moles of water}+\text{Moles of glycerin}}[/tex]
[tex]\text{Mole fraction of water}=\frac{7.444mole}{7.444mole+0.267mole}=0.965[/tex]
Now we have to calculate the vapor pressure of the solution.
According to the Raoult's law,
[tex]p_A=X_A\times p^o_A[/tex]
where,
[tex]p_A[/tex] = vapor pressure of solution = ?
[tex]p^o_A[/tex] = vapor pressure of pure water= 31.8 torr
[tex]X_A[/tex] = mole fraction of water = 0.965
Now put all the given values in this formula, we get the vapor pressure of solution.
[tex]p_A=0.965\times 31.8\text{ torr}[/tex]
[tex]p_A=30.687\text{ torr}[/tex]
Therefore, the vapor pressure of a solution is, 30.687 torr
Considering the Raoult's Law, the vapor pressure of a solution that containing 24.6 g of glycerin in 134 mL of water at 30.0 ∘C is 30.687 torr.
Raoult's LawRaoult's Law is a gas law that relates the vapor pressure and mole fraction of each gas in a solution (solution).
If a solute has a measurable vapor pressure, the vapor pressure of its solution is always less than that of the pure solvent. Thus, the relationship between the vapor pressure of the solution and the vapor pressure of the solvent depends on the concentration of the solute in the solution.
Raoult's law allows us to calculate the vapor pressure of a substance when it is part of an ideal solution, knowing its vapor pressure when it is pure (at the same temperature) and the composition of the ideal solution in terms of molar fraction.
Then, this law establishes as a conclusion that: “In an ideal solution, the partial pressures of each component in the vapor are directly proportional to their respective mole fractions in the solution”.
Mathematically, Raoult's law states that the partial pressure of a solvent above a solution P₁ is given by the vapor pressure of the pure solvent P₁⁰ multiplied by the mole fraction of the solvent in the solution X₁:
P₁= P₁⁰ × X₁
Vapor pressure in this caseThe mole fraction is defined as the ratio of moles of solute to total moles of solution.
So, the moles of glycerin (solute) is calculated as follow:
[tex]moles of glycerin=\frac{mass of glycerin}{molar mass of glycerin}=\frac{24.6 g}{92.09 \frac{g}{mole} } =[/tex] 0.267 moles
Being the density the measure of the amount of mass in a certain volume of a substance, the mass of water (solvent) in the solution is calculated as:
mass of water= volume of water× density of water
mass of water= 134 mL× 1 [tex]\frac{g}{mL}[/tex]= 134 g
So, the moles of water (solvent) is calculated as follow:
[tex]moles of water=\frac{mass of water}{molar mass of water}=\frac{134 g}{18 \frac{g}{mole} } =[/tex] 7.44 moles
Then, the mole fraction of water can be calculated as:
[tex]Mole fraction of water=\frac{moles of water}{moles of glycerin + moles of water}[/tex]
[tex]Mole fraction of water=\frac{7.44 moles}{0.267 moles + 7.44 moles}[/tex]
Solving:
mole fraction of water= 0.965
Then, you know that:
[tex]P_{water} =[/tex] ?[tex]P_{water} ^{0}=[/tex] 31.8 torr[tex]x_{water}=[/tex] 0.965Replacing in Raoult's Law:
[tex]P_{water} =[/tex] 31.8 torr × 0965
Solving:
[tex]P_{water} =[/tex] 30.687 torr
In summary, the vapor pressure of a solution that containing 24.6 g of glycerin in 134 mL of water at 30.0 ∘C is 30.687 torr.
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A flexible container at an initial volume of 6.13 L contains 8.51 mol of gas. More gas is then added to the container until it reaches a final volume of 18.5 L. Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.
Explanation:
According to Avogadro's law, equal volumes of all the gases at same temperature and pressure will also, have same number of molecules.
That is, [tex]V \propto n[/tex]
or [tex]\frac{V}{n}[/tex] = k
Since, it is given that [tex]V_{1}[/tex] is 6.13 L, [tex]n_{1}[/tex] is 8.51 mol, and [tex]V_{2}[/tex] is 18.5 L.
Hence, calculate the [tex]n_{2}[/tex] as follows.
[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}[/tex]
[tex]\frac{6.13 L}{8.51 mol} = \frac{18.5 L}{n_{2}}[/tex]
[tex]{n_{2}}[/tex] = 0.35 mol
Thus, we can conclude that the number of moles of gas added to the container is 0.35 mol.
Final answer:
To calculate the moles of gas added, we use the ratio of the initial and final volumes to determine the final moles, and then subtract the initial moles to find that 17.16 mol of gas were added to the container.
Explanation:
The subject of this question is Chemistry, specifically it is dealing with gas laws and the concept of molar volume at constant pressure and temperature (ideal gas law).
To find the number of moles of gas added to the container, we will apply the ideal gas law in its simplest form, which is Boyle's Law (P1V1 = P2V2), assuming constant temperature and pressure. Since P and T are constant, the relationship between volume and moles is direct according to Avogadro's Law. This means we can simply use the ratio of the initial and final volumes to find the moles added:
Initial moles (n1) = 8.51 mol
Initial volume (V1) = 6.13 L
Final volume (V2) = 18.5 L
Since the volumes are proportional to moles at constant temperature and pressure, we have:
V1 / n1 = V2 / n2
Using a simple cross-multiplication:
n2 = (V2 / V1) × n1
n2 = (18.5 L / 6.13 L) × 8.51 mol
n2 = 3.016 × 8.51 mol
n2 = 25.67 mol
The total moles in the container after addition of gas is 25.67 mol. To get the moles of gas added, we subtract the initial moles from the total moles:
Moles of gas added = 25.67 mol - 8.51 mol
Moles of gas added = 17.16 mol
Calculate the standard potential for the following galvanic cell: Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s) which has the overall balanced equation: Ni(s)+2Ag+(aq)→Ni2+(aq)+2Ag(s) Express your answer to three significant figures and include the appropriate units.
Answer:
1.06 V
Explanation:
The standard reduction potentials are:
Ag^+/Ag E° = 0.7996 V
Ni^2+/Ni E° = -0.257 V
The half-cell and cell reactions for Ni | Ni^2+ || Ag^+ | Ag are
Ni → Ni^2+ + 2e- E° = 0.257 V
2Ag^+ 2e- → 2Ag E° = 0.7996 V
Ni + 2Ag^+ → Ni^2+ + 2Ag E° = 1.0566 V
To three significant figures, the standard potential for the cell is 1.06 V .
The standard cell potential for the galvanic cell is 1.05 V.
The overall balanced equation of the reaction is;
Ni(s) + 2Ag+(aq) →Ni2+(aq) + 2Ag(s)
Since it is a galvanic cell, Nickel is the anode and silver is the cathode.
We know that;
E°cell = E°cathode - E°anode
E°cathode = -0.25 V
E°anode = 0.80 V
E°cell = 0.80 V - (-0.25 V)
E°cell = 1.05 V
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In a paper chromatography analysis, three pigments, A, B, and C, were dissolved in a solvent. A is slightly polar, B is highly polar, and C is moderately polar. List in order how these will appear on the surface of the chromatography strip starting from the applied spot at the origin line.
Answer:It would depend on the eluent solvent what we use ,If we would use a polar solvent than B will be appear first followed by c and then A on the chromatography strip.
If we use non-polar eluent solvent than A will be separated first followed by C and thenB on the chromatography strip.
Explanation:
Separation using paper chromatography is dependent upon the polarity of various pigments .
A polar pigment would move ahead in case of polar solvent used and a non-polar pigment would move ahead when we use a non-polar solvent.
So separation would occur in order of polarity of various pigments in a given solvent.
The amount of distance travelled by each component (or pigment or spot) can be calculated using the formula for retention factor:
Rf= Distance travelled by pigment spot or solute/Distance travelled by eluent solvent
Rf= Retention factor
Retention factor is basically the ratio of distance tarvelled by the pigment or spot to the ratio of distance travelled by the solvent.
Final answer:
In a paper chromatography analysis, pigment B (highly polar) will appear closest to the origin line, followed by pigment C (moderately polar), and then pigment A (slightly polar).
Explanation:
In a paper chromatography analysis, the pigments A, B, and C will appear on the surface of the chromatography strip in the following order:
Pigment B (highly polar)
Pigment C (moderately polar)Pigment A (slightly polar)This order is determined by the polarity of the pigments. In chromatography, more polar substances tend to travel more slowly through the stationary phase, while less polar substances travel faster. Therefore, pigment B, being highly polar, will have the slowest migration and appear closest to the origin line, followed by pigment C and then pigment A.
To avoid dehydration during or after a long distance athletic event many athletes suggest sponging down with isopropanol. Why would this procedure help avoid the loss of water?
Final answer:
Sponging down with isopropanol can help lower the body temperature through rapid evaporation, which can assist in preventing dehydration. Nevertheless, athletes should prioritize proper hydration by consuming adequate fluids, particularly sports drinks that contain a balance of water, electrolytes, and sugar, to maintain endurance and performance.
Explanation:
To avoid dehydration during or after a long-distance athletic event, athletes may sponge down with isopropanol because it accelerates the evaporation process when applied to the skin. This rapid evaporation causes a cooling effect which can help to lower the body temperature. However, it is important to note that the best way to prevent dehydration is through proper hydration, which involves drinking enough fluids, such as water or sports drinks, before, during, and after exercise. These fluids should contain the necessary electrolytes like sodium, which is crucial for fluid absorption and replacing what is lost through sweat.
The American College of Sports Medicine recommends that the goal of drinking fluids during exercise is to prevent dehydration without over-hydrating. Replacing lost fluids without overconsumption is key to maintaining optimal endurance and performance. To ensure proper rehydration, it's essential to consume sports drinks that contain the correct proportions of sugar, water, and sodium. Sodium in sports drinks not only enhances fluid absorption but also aids in maintaining blood-glucose levels for muscle fuel and replaces some of the electrolytes lost in sweat.
Consider the following elementary reaction: NO2(g) + F2(g) --> NO2F(g) + F (g) Suppose we let k1 stand for the rate constant of this reaction, and k -1 stand for the rate constant of the reverse reaction. Write an expression that gives the equilibrium concentration of F2 in terms of k1, k -1, and the equilibrium concentrations of NO2, NO2F, and F.
To express the equilibrium concentration of F2, the equation [F2] = (k -1[NO2F][F]) / (k1[NO2]) is used, based on the principle that the forward reaction rate equals the reverse reaction rate at equilibrium.
Explanation:To write an expression for the equilibrium concentration of F2 in terms of the rate constants k1 and k -1, and the equilibrium concentrations of NO2, NO2F, and F, one can use the principle that at equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction. For the given reaction NO2(g) + F2(g) → NO2F(g) + F(g), the forward rate is given by k1[NO2][F2] and the reverse rate by k -1[NO2F][F]. Setting these equal to each other gives us:
k1[NO2][F2] = k -1[NO2F][F]
Solving for the equilibrium concentration of F2, we have:
[F2] = (k -1[NO2F][F]) / (k1[NO2])
This expression relates the equilibrium concentration of F2 to the rate constants and the equilibrium concentrations of NO2, NO2F, and F.
Find the fugacity coefficient of a gaseous species of mole fraction 0.4 and fugacity 25 psia in a mixture at a total pressure of 50 psia
Answer: The fugacity coefficient of a gaseous species is 1.25
Explanation:
Fugacity coefficient is defined as the ratio of fugacity and the partial pressure of the gas. It is expressed as [tex]\bar{\phi}[/tex]
Mathematically,
[tex]\bar{\phi}_i=\frac{\bar{f_i}}{p_i}[/tex]
Partial pressure of the gas is expressed as:
[tex]p_i=\chi_iP[/tex]
Putting this expression is above equation, we get:
[tex]\bar{\phi}_i=\frac{\bar{f_i}}{\chi_iP}[/tex]
where,
[tex]\bar{\phi}_i[/tex] = fugacity coefficient of the gas
[tex]\bar{f_i}[/tex] = fugacity of the gas = 25 psia
[tex]\chi_i[/tex] = mole fraction of the gas = 0.4
P = total pressure = 50 psia
Putting values in above equation, we get:
[tex]\bar{\phi}_i=\frac{25}{0.4\times 50}\\\\\bar{\phi}_i=1.25[/tex]
Hence, the fugacity coefficient of a gaseous species is 1.25
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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Carry out an energy minimization to calculate the relative steric energies or heats of formation of three of the ten possible isomeric dibenzalacetones
Answer:
Calculate the steric energy or heat of formation for one single bond isomel of trans-benzAlacetone using the usual energy minimization procedure. The result should be a planar molecule, Then deliberately hold the dihedral angle defined by atoms 1, 2, 3, and 4 at 0 90, and 180 and again calculate the energies of the molecule.
How many ATP are produced from a fatty acid that is 14 carbons long?
Answer:
92 ATP
Explanation:
Fatty acid oxidation results in the formation of large number of ATP molecules. Three important process of fatty acid are activation of the fatty acid, beta oxidation and entry of acetyl CoA in Krebs cycle.
14 carbon fatty acid is Miristic acid. The complete oxidation of Miristic acid results in the formation of 7 acetyl CoA + 6NADH and [tex]6FADH_2[/tex]
1 Acetyl CoA gives 10 molecules of ATP then 7 acetyl CoA gives 70 molecules of ATP.
1NADH = 2.5 ATP, 6NADH = 15 ATP.
[tex]1FADH_2[/tex] = 1.5 ATP, [tex]6FADH_2[/tex] = 9 ATP.
2 ATP has been consumed in the activation of fatty acid.
Total ATP = 70+15+9-2
=92 ATP.
Thus, the total ATP generated from the oxidation of 14 carbon fatty acid is 92.
Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)→2AlCl3(s) What is the maximum mass of aluminum chloride that can be formed when reacting 15.0 g of aluminum with 20.0 g of chlorine? Express your answer to three significant figures and include the appropriate units.
The problem posed involves stoichiometry, requiring calculations related to reactants and the products in the given chemical reaction. Utilizing molar masses and the balanced chemical equation, we identify chlorine as the limiting reactant. Therefore, the maximum mass of aluminum chloride that can be produced when reacting 15.0g of aluminum with 20.0g of chlorine is approximately 25.0g.
Explanation:The subject of this question relates to an area of chemistry known as stoichiometry, which involves calculating the amounts of reactants and products in chemical reactions. To determine the maximum mass of aluminum chloride that can be produced from 15.0g of aluminum and 20.0g of chlorine, we first need to convert these amounts to moles. We can do this by dividing the mass of each compound by its molar mass. The molar mass of aluminum (Al) is approximately 27 g/mol, and the molar mass of chlorine (Cl2) is approximately 71 g/mol. Therefore, we have approximately 0.56 moles of aluminum and 0.28 moles of chlorine.
Looking at the balanced chemical equation, we see that the reaction ratio between Al and Cl2 is 2:3. We have more than twice the number of moles of Al as Cl2, which means that Cl2 is the limiting reactant – it will be completely used up before all of the aluminum is reacted. Therefore, the amount of product (AlCl3) formed is determined by the amount of Cl2 present.
The balanced chemical equation also shows that the reaction ratio between Cl2 and AlCl3 is 3:2, so for every 3 moles of Cl2 reacted, we form 2 moles of AlCl3. Therefore, we can multiply the number of moles of Cl2 (0.28 moles) by 2/3 to find the number of moles of AlCl3 produced. We then convert this number of moles back to grams of AlCl3 by multiplying by the molar mass of AlCl3 (approximately 133 g/mol). This gives us approximately 25.0g of AlCl3 as the maximum amount that can be formed under these conditions.
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The following data were obtained from experiments to find the molecular formula of procaine, a local anesthetic, which contains only carbon, hydrogen, nitrogen, and oxygen. Complete combustion of a 3.54g sample of procaine with excess O2 formed 8.58 g of CO2 and 2.70 g H2O. Another sample of mass 2.35 g was found to contain 0.279 g of N. The molar mass of procaine was found to be 236.Find the molar formula of procaine
By calculating the number of moles of each element (carbon, hydrogen, nitrogen, and oxygen) and determining the empirical formula as C9H15N1O11, and seeing that the molar mass of this empirical formula is close to the given molar mass of procaine, we can conclude that its molecular formula is indeed C9H15N1O11.
Explanation:Given CO2 mass = 8.58 g, H2O mass = 2.70 g, and N mass = 0.279 g.
To find the amount of C and H, we'll use the molar ratios derived from the formulas of CO2 and H2O respectively. For CO2, it’s 1 mole of Carbon per mole of CO2 (MW of C = 12.01 g/mol, MW of CO2 = 44.01 g/mol). Hence moles of C = (8.58 g CO2)*(1 mole C/44.01 g CO2) = 0.195 mol C. Similar calculation for H from H2O yields 0.300 mol H. Given the mass of nitrogen (0.279 g), we find 0.02 mol N. We then subtract the masses of C, H and N from the total mass (3.54 g procaine) to get the mass of O, and consequently, 0.218 mol of O.
Now, to find the empirical formula, divide the obtained moles of each element by the smallest mole value, which is N = 0.02. Thus, the empirical formula is C9H15N1O11.
Last step for finding the molecular formula involves dividing the given molar mass of procaine (236 g/mol) by the molar mass of the above empirical formula (~207 g/mol). Getting approximately 1 as a result, the molecular formula will be the same as the empirical, that is C9H15N1O11.
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Select the false statement below : Fe2+ is more paramagnetic (has more unpaired electrons) than Fe3+ Fe3+ is isoelectronic with Mn2+ Fe has 2 outer electrons, 8 valence electrons, and 18 core electrons Fe3+ is predicted to be a stronger potential oxidizing agent (can be reduced more) than Fe2+ none of these ions (Fe2+, Fe3+, Mn2+) have any electrons in the 4s subshell
hey there!:
1) none of these ions ( Fe²⁺ , Fe³⁺ , Mn²⁺ 0 have any electrons in the 4s subshell . ( TRUE )
2) Fe²⁺ is more paramagnetic ( has more unpaired electrons ) than Fe³⁺.
( FALSE ) explanation :
* Fe²⁺ has no free electrons and is diamagnetic
* Fe³⁺ has one unpaired electron and is paramagnetic
3) Fe³⁺ is isoeletronic with Mn²⁺ . ( TRUE )
4) Fe has 2 outer electrons , 8 valence electrons , and 18 core eletrons. (TRUE )
electron configuration Fe = 1s², 2s², 2p⁶, 3s² , 3p⁶, 4s² , 3d⁶ or :
[Ar] = 4s² , 3d⁶
5) Fe³⁺ is predicted to be a stronger potencial oxidizing agent ( can be reduced more ) than Fe²⁺ ( TRUE )
Hope this helps!
The false statement is that; "Fe2+ is more paramagnetic (has more unpaired electrons) than Fe3+."
Iron is a transition element that has 26 valence electrons in the ground state. The ground state electron configuration of iron is [Ar] 3d6 4s2. The 4s electrons are the outer electrons hence Fe2+ is a d6 specie while Fe3+ is a d5 specie.
This means that Fe3+ has five unpaired electrons while Fe2+ has only four unpaired electrons. Hence, the false statement is that; "Fe2+ is more paramagnetic (has more unpaired electrons) than Fe3+."
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Which of the following is the correct balanced equation for the neutralization of barium hydroxide with hydrochloric acid? Ba(OH)2 + 2 HCl → BaCl2 + 2 H2O BaOH2 + 2 HCl → BaCl2 + H2O BaOH + HCl → BaCl + H2O Ba(OH)2 + HCl → BaCl2 + H2O
Answer: The correct answer is [tex]Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O[/tex]
Explanation:
Neutralization reaction is defined as the reaction in which an acid reacts with a base to produce a salt and also releases water molecule.
The general equation for this reaction follows:
[tex]BOH+HX\rightarrow BX+H_2O[/tex]
The balanced chemical equation for the reaction of barium hydroxide and hydrochloric acid follows:
[tex]Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O[/tex]
By Stoichiometry of the reaction:
1 mole of barium hydroxide reacts with 2 moles of hydrochloric acid to produce 1 mole of barium chloride and 2 moles of water molecule.
The balanced chemical equation for the neutralization of barium hydroxide with hydrochloric acid is Ba(OH)2 + 2 HCl → BaCl2 + 2 H2O, leading to the formation of water and a salt, BaCl2.
Explanation:The correct balanced equation for the neutralization of barium hydroxide with hydrochloric acid is Ba(OH)2 + 2 HCl → BaCl2 + 2 H2O. In this equation, Ba(OH)2 is the base and HCl is the acid. A neutralization reaction between an acid and a base results in the formation of water and salt. In this case, the salt is BaCl2. Each reactant and product in the equation is balanced, with the same number of each type of atom appearing on both sides of the equation, hence it's a balanced chemical equation.
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which type of heat exchanger shell and tube or plate heat exchanger is suitable for evaporation?
Answer:
shell and tube type heat exchanger
Explanation:
for evaporation the shell and tube type heat exchanger is best suited.
in the plate heat exchanger there is gaskets in between every part so this part become weak part in heat echanger and there is possibilities of leakage through this part, there is no such problem in shell and tube type.the plate type cant be used when there is high temperature and high pressure drop but shell and tube type can be usedin evaporation there the liquids change into vapors due to which there is sudden change in pressure and in which plate type is not used because there is chances of leakageThe suitable type of heat exchanger for evaporation is the shell and tube heat exchanger.
Evaporation processes typically involve the boiling of a liquid to produce vapor, which requires a heat exchanger that can handle phase change and the associated volume changes. Shell and tube heat exchangers are well-suited for this application due to their ability to withstand high pressures and temperatures, as well as their design flexibility to accommodate the two-phase flow of vapor and liquid.
In a shell and tube heat exchanger, the tubes can be designed to allow for the expansion of vapor and the efficient separation of the vapor phase from the liquid phase. The tubes also provide a large surface area for heat transfer, which is necessary for the evaporation process. Additionally, shell and tube heat exchangers can be designed with multiple passes to increase the time of contact between the heating or cooling medium and the product, thereby improving the efficiency of the evaporation process.
On the other hand, plate heat exchangers are generally not suitable for evaporation processes that involve boiling and phase change. This is because the plates in a plate heat exchanger are not designed to handle the high pressure differentials and volume changes associated with the production of vapor. Plate heat exchangers are more suitable for applications involving low to medium pressure and temperature, and where the fluids remain in a single phase.
In summary, for processes involving evaporation, the robust design and flexibility of the shell and tube heat exchanger make it the more suitable choice compared to the plate heat exchanger."