Answer: -2359 Joules
Explanation:
According to first law of thermodynamics:
[tex]\Delta E=q+w[/tex]
[tex]\Delta E[/tex]= Change in internal energy = -2886 J
q = heat absorbed or released = ?
w = work done or by the system
w = work done by the system=[tex]-P\Delta V[/tex] {Work is done by the system as the final volume is greater than initial volume and is negative}
w = -527 J
q = ?
[tex]-2886=q+(-527)[/tex]
[tex]q=-2359J[/tex] {Heat released by the system is negative}
Thus the amount of heat released by the system and carried away by the cooling system is -2359 Joules
According to first law of thermodynamics:
= Change in internal energy = -2886 J
q = heat absorbed or released = ?
w = work done or by the system
w = work done by the system= {Work is done by the system as the final volume is greater than initial volume and is negative}
w = -527 J
q = ?
{Heat released by the system is negative}
Thus the amount of heat released by the system and carried away by the cooling system is -2359 JoulesAnswer:
Explanation:
Consider the following equilibrium:
O2(g) + 2F2(g)--> 2OF2(g); Kp = 2.3 x 10-15
Which of the following statements is true?
A) If the reaction mixture initially contains only OF2(g), then at equilibrium, the
reaction mixture will consist of essentially only O2(g) and F2(g).
B) For this equilibrium, Kc = Kp.
C) If the reaction mixture initially contains only OF2(g), then the total pressure at
equilibrium will be less than the total initial pressure.
D) If the reaction mixture initially contains only O2(g) and F2(g), then at equilibrium,
the reaction mixture will consist of essentially only OF2(g).
E) If the reaction mixture initially contains only O2(g) and F2(g), then the total
pressure at equilibrium will be greater than the total initial pressure.
Answer:
A) is true
Explanation:
For the reaction:
O₂(g) + 2F₂(g) ⇄ 2OF₂(g); Kp = 2,3x10⁻¹⁵
kp is defined as:
Kp = 2,3x10⁻¹⁵ = [OF₂]²/[O₂] [F₂]²
A) If the reaction mixture initially contains only OF₂(g), then at equilibrium, the reaction mixture will consist of essentially only O₂(g) and F₂(g). TRUE. As the kp is 2,3x10⁻¹⁵ means per 1 of [O₂] [F₂]² you will have just 2,3x10⁻¹⁵ of [OF₂]²
B) For this equilibrium, Kc = Kp. FALSE. That is true just when moles of reactants are the same than moles of products. Here there are 3 moles of reactants vs 2 moles of products.
C) If the reaction mixture initially contains only OF₂(g), then the total pressure at equilibrium will be less than the total initial pressure. FALSE. Because per 2 moles of OF₂(g) you will produce 3 moles of gas increasing pressure.
D) If the reaction mixture initially contains only O₂(g) and F₂(g), then at equilibrium, the reaction mixture will consist of essentially only OF₂(g). FALSE. For the same reason of A), the mixture will contains essentially only O₂(g) and F₂(g)
E) If the reaction mixture initially contains only O₂(g) and F₂(g), then the total pressure at equilibrium will be greater than the total initial pressure. FALSE. If mixture initially contains only O₂(g) and F₂(g), 3 moles will of gas will react to produce 2 moles of gas doing pressure decreases.
I hope it helps!
Arrange the following ionic compounds in order of decreasing amount of energy released in lattice formation: LiCl , CaO , AlN , and KI ?
Answer:in decreasing pattern; AlN>CaO>LiCl>KI.
Explanation:
The energy required to break apart an ionic solid and convert its component atoms into gaseous ions is known as the LATTICE ENERGY.
LATTICE ENERGY has a trend on the periodic table. The trend shows the relationship between LATTICE ENERGY AND THE ATOMIC RADIUS.
Lattice energy is related to solubility, volatilty and hardness.
Down the group on the periodic chart, the ATOMIC RADIUS INCREASES, THE LATTICE ENERGY DECREASES. Which means that between Lithium ion and potassium ion, Lithium ion will have higher lattice energy than that of potassium.
The trend ACROSS THE PERIOD is that, as we go across the period, the charges on metals increase,consequently INCRESING THE LATTICE ENERGY. Between Aluminum ion with a positive charge of three and Calcium with a positive charge of two, Aluminium ion has higher charge, which means it has more lattice energy than calcium.
The ionic compounds in order of their decreasing amount of energy are AIN>CaO>LiCl>Kl.
The lattice energy is used for the estimation of the strength of the bond that is in ionic compounds.
It should be noted that the higher the charge of the ion, the higher will be the lattice energy. On the other hand, the smaller the radius of ion, the larger will be the lattice energy.
Therefore, the ionic compounds in order of their decreasing amount of energy will be AIN>CaO>LiCl>Kl.
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Consider the reaction BF3 + NH3 → F3B―NH3
What changes in hybridization (if any) of the B atom are a result of this reaction?
A. Before the reaction boron is sp2 hybridized, and after the reaction it is sp3d hybridized.
B. Before the reaction boron is sp3 hybridized, and after the reaction it is sp3 hybridized.
C. Before the reaction boron is sp2 hybridized, and after the reaction it is sp3 hybridized.
What changes in hybridization (if any) of the N atom are a result of this reaction?
A. Before the reaction nitrogen is sp3 hybridized, and after the reaction it is sp2 hybridized.
B. Before the reaction nitrogen is sp3 hybridized, and after the reaction it is sp3 hybridized.
C. Before the reaction nitrogen is sp3d hybridized, and after the reaction it is sp3 hybridized.
Answer:
The correct answer is given below
Explanation:
B before the reaction boron is sp2 hybridized and after the reaction boron is it is sp3 hybridized.
Before the reaction boron is joined to 3 fluorine atoms by 3 covalent bonds between them one is s and rest of the 2 are p.That"s why before the reaction boron is sp2 hybridized.
Whereas after the reaction boron is joined to 3 fluorine atoms by 3 covalent bonds and one nitrogen atom by one co ordinate bond.That"s why after reaction boron is sp3 hybridized.
B before the reaction nitrogen is sp3 hybridized and after the reaction it is sp3 hybridized.
Before the reaction Nitrogen is joined to 3 hydrogen atoms by 3 covalent bonds and contain one lone pair of electron.That"s why before the reaction Nitrogen atom is sp3 hybridized.
whereas after the reaction Nitrogen is joined to 3 hydrogen atoms by 3 covalent bonds and with one boron atom by a co ordinate bond.That"s why after reaction Nitrogen atom is sp3 hybridized.
Boron atom's hybridization changes from sp2 to sp3 as a result of the reaction while Nitrogen's hybridization remains sp3 both before and after the reaction.
Explanation:In the given reaction, BF3 + NH3 → F3B―NH3, the hybridization of both Boron (B) and Nitrogen (N) atoms change. For the Boron atom, before the reaction, it is sp2 hybridized. After the reaction, when it bonds with NH3, it becomes sp3 hybridized. Hence, the change in Boron's hybridization as a result of this reaction is from sp2 to sp3.
For the Nitrogen atom, before the reaction, it is sp3 hybridized. After the reaction, it remains sp3 hybridized because it still maintains four regions of electron density. So, there is no change in Nitrogen's hybridization as a result of the reaction. It remains sp3 both before and after the reaction.
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The freezing point of benzene, C6H6, is 5.500 °C at 1 atmosphere. Kf(benzene) = 5.12 °C/m In a laboratory experiment, students synthesized a new compound and found that when 14.44 grams of the compound were dissolved in 292.3 grams of benzene, the solution began to freeze at 4.571 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound?
Answer:
The molecular weight they determined for this compound is 272.9 g/m
Explanation:
We must apply the colligative property of freezing point depression to solve this:
ΔT° = Kf · molality
Where ΔT° means T° fussion pure solvent − T° fussion of solution
and Kf the cryoscopic constant
The statemente expressed that the compound was also found to be nonvolatile and a non-electrolyte, so we don't have to apply the Van't Hoff factor (i)
By the way, the complete formula is this one:
ΔT° = Kf · molality . i
5.5°C - 4.571°C = 5.12 °C/molal . molality
0.929°C = 5.12 °C/molal . molality
0.929°C / 5.12 molal/°C = molality → 0.181 m
Molality means the mol of solute in 1kg of solvent. But, in this solution we used 292.3 g of benzene, so let's find out our moles of solute, in our mass of solvent.
1000 g ___ 0.181 moles
292.3 g ____ (292.3 . 0.181) /1000 = 0.053 moles
The mass, we used of solute is 14.44 g so, to find out the molar mass we must divide mass (g) / moles
14.44 g /0.053 m = 272.9 g/m
The molecular weight they determined for this compound of benzene which is non volatile is 272.9 g/m
The colligative property of freezing point depression is applied to solve this:
ΔT° = Kf · molality
Where ΔT° means T° fusion pure solvent − T° fusion of solution
and Kf is the cryoscopic constant
The statement expressed that the compound was also found to be nonvolatile and a non-electrolyte, so we don't have to apply the Van't Hoff factor (i)
By the way, the complete formula is this one:
ΔT° = Kf · molality . i
5.5°C - 4.571°C = 5.12 °C/molal . molality
0.929°C = 5.12 °C/molal . molality
0.929°C / 5.12 molal/°C = molality → 0.181 m
Molality means the mol of solute in 1kg of solvent. But, in this solution we used 292.3 g of benzene, so let's find out our moles of solute, in our mass of solvent.
1000 g ___ 0.181 moles
292.3 g ____ (292.3 . 0.181) /1000 = 0.053 moles
The mass,used of solute is 14.44 g so, to find out the molar mass divide mass (g) / moles
14.44 g /0.053 m = 272.9 g/m
Thus, the molar mass of compound is 272.9 g/m.
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When the concentration of PCl3(g) is increased to 1.2 M, the ratio of products to reactants is 4.0. The equilibrium constant for the reaction is 24. In which direction will the reaction shift to regain equilibrium?
toward the reactants
toward the products
Answer:
toward the products
Explanation:
The equilibrium constant of an equilibrium reaction measures relative amounts of the products and the reactants present at equilibrium. It is the ratio of the concentration of the products and the reactants each raised to their stoichiometric coefficients. The concentration of the liquid and the gaseous species does not change and thus is not written in the expression.
Thus, on increasing the concentration of any one of the reactant, the value of equilibrium constant will decrease.
The reaction quotient of an equilibrium reaction measures relative amounts of the products and the reactants present during the course of the reaction at particular point in the time.
Q < Kc , reaction will proceed in forward direction.
Q > Kc , reaction will proceed in backward direction.
Q = Kc , reaction at equilibrium.
Since, given that The equilibrium constant for the reaction is 24. On increasing the concentration of one of the reactant, the reaction quotient will become less than Kc and thus, go in forward direction which means that towards the product.
Use this Initial Rate data to answer Questions 6 and 7. Consider the reaction:
xA + yB → products
The following Initial Rate kinetic data were obtained:
Trial [A] [B] Initial Rate (mol/L-s)
1 0.100 0.400 0.0904
2 0.200 0.800 0.362
3 0.400 1.200 1.08
4 0.200 0.400 0.181
5 0.100 0.800 0.181
The rate law for this reaction is: rate = k[A]m[B]n.
(6) What is m (the reaction order with respect to reactant A)?
(7) What is n (the reaction order with respect to reactant B)?
Answer:
m = 1
n = 1
Explanation:
The rate law is:
[tex]r=k.[A]^{m} .[B]^{n}[/tex]
where,
r is the rate of the reaction
k is the rate constant
m is the order of reaction with respect to A
n is the order of reaction with respect to B
Let's consider trials 1 and 4. We know that [B]₁ = [B]₄ . The rate r₁/r₄ is:
[tex]\frac{r_{1}}{r_{4}} =\frac{k.[A]_{1}^{m}.[B]_{1}^{n} }{k.[A]_{4}^{m}.[B]_{4}^{n}} \\\frac{r_{1}}{r_{4}} =(\frac{[A]_{1}}{[A]_{4}} )^{m} \\\frac{0.0904M/s}{0.181M/s}=(\frac{0.100M}{0.200M})^{m} \\m=1[/tex]
Let's consider trial 1 and 5. We know that [A]₁ = [A]₅. The rate r₁/r₅ is:
[tex]\frac{r_{1}}{r_{5}} =\frac{k.[A]_{1}^{m}.[B]_{1}^{n} }{k.[A]_{5}^{m}.[B]_{5}^{n}} \\\frac{r_{1}}{r_{5}} =(\frac{[B]_{1}}{[B]_{5}} )^{n} \\\frac{0.0904M/s}{0.181M/s}=(\frac{0.400M}{0.800M})^{n} \\n=1[/tex]
Write the electron configuration for the element titanium, Ti.
Express your answer in order of increasing orbital energy as a string without blank space between orbitals. For example, the electron configuration of Li could be entered as 1s^22s^1 or [He]2s^1.
Answer:
1s^22s^22p^63s^23p^63d^24s^2
Explanation:
Titanium with the chemical symbol Ti, is a transition element, and in fact the second transition element after scandium. It has an atomic number of 22. It is the second element in the first transition series.
To write the electronic configuration, we consider the atomic number of titanium. The atomic number of titanium is 22. Hence, since an atom is electrically neutral, the number of electrons in the titanium metal is 22.
It should be noted that while filling,the maximum number of electrons the s subshell is 2. The maximum number of electrons in the p subshell is 6 while the number in the d shell is 10
Which one of the following statements best explains why convection doesn't occur in solids? Group of answer choices Solids are less compressible than gases. The molecules in a solid vibrate at a lower frequency than those of a liquid. Solids are more compressible than liquids. The molecules in a solid are more closely spaced than in a gas. The molecules in a solid are not free to move throughout its volume.
Final answer:
Convection does not occur in solids because the molecules are tightly packed and locked in position, restricting their movement and preventing circulation needed for convection. Gases and liquids, with their more freely moving molecules, permit this kind of heat transfer.
Explanation:
The molecules in a solid are more closely spaced than in a gas, which is the reason why convection doesn't occur in solids. In a solid, the molecules are tightly packed together, and the forces between them are strong enough to keep them locked in position, making solids incapable of flowing like liquids or gases.
Each molecule has limited freedom of movement, mostly restricted to vibrations about its fixed point within the lattice structure of the solid. As a result, unlike in gases and liquids where molecules can move freely throughout the volume, the molecules in a solid cannot circulate to transfer heat by convection.
On the other hand, gases have widely separated molecules which interact primarily through collisions, making it easy for gas molecules to move and spread out to occupy the available space. This characteristic allows for the movement of warmer and cooler portions of a gas, leading to convection currents. Likewise, liquids, although more ordered than gases, have molecules that are not rigidly fixed, enabling them to slide past each other and also allowing for convection.
Compound A, C6H12 reacts with HBr/ROOR to give compound B, C6H13Br. Compound C, C6H14, reacts with bromine and light to produce compound B, C6H13Br. Suggest structures for compounds A, B, and C. Draw structure of the compound A.
Answer:
The question involves drawing of structures and showing mechanism in which brainly text editor did not support. I made sure I created a pdf file with both the anwsers and explanations in it. The pdf can be found in the attachment below.
Explanation:
Compound A, C6H12, is most likely cyclohexane, a 6-carbon ring structured alkane. It reacts via a radical addition reaction with HBr/ROOR to produce Compound B, C6H13Br, a bromocyclohexane. Compound C, C6H14, is most likely hexane, an 6-carbon linear alkane. It undergoes a radical reaction with bromine and light to also form bromocyclohexane.
Explanation:The subject of this question is organic chemistry, specifically dealing with the structures of different compounds and their reactions. Compound A, C6H12, is likely cyclohexane - a 6-carbon cyclic alkane - which reacts with HBr/ROOR (a radical addition reaction) to form compound B, C6H13Br, a bromocyclohexane. Compound C, C6H14, on the other hand, is hexane - a linear alkane. It reacts with bromine and light (again, another radical reaction) to also give bromocyclohexane.
The structure of compound A (cyclohexane) is drawn as a hexagon representing the 6-carbon cyclic structure, with each corner representing one carbon and its associated hydrogen atoms. This is a very simplified version of chemical structures used for ease of drawing. Detailed representation would include all the Hydrogen atoms.
Chemistry of these ReactionsCyclohexane will readily react with hydrogen bromide in the presence of a radical initiator (ROOR), producing a bromocyclohexane. Hexane, on the other hand, will react with bromine in the presence of light to form the same product. The light provides the energy needed to break the bromine molecule into bromine radicals, which then react with hexane.
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2) An ideal gas flows through a horizontal tube at steady state. No heat is added and no shaft work is done. The cross- sectional area of the tube changes with length and this causes the velocity to change. Derive an equation relating the temperature to the velocity of the gas. If nitrogen at 150 °C flows past one section of the tube at a velocity of 2.5 m/s, what its temperature at another section where its velocity is 50 m/s? Cp = 7/2 R.
Answer:
The temperature of the nitrogen gas at another section is [tex]148.8^{o}C[/tex]
Explanation:
Energy balance equation for steady state flow of gas under negiligible potential energy.
The negligible heat transfer and no shaft work is as follows.
[tex]\Delta H+\frac{\Delta u^{2}}{2}=Q+W[/tex]
[tex]\Delta H+\frac{\Delta u^{2}}{2}=0..........(1)[/tex]
[tex]\Delta H[/tex] is enthalphy of gas and it is changes with the temperature.
[tex]\Delta H=C_{p}(T_{2}-T_{1}).................(2)[/tex]
[tex]C_{p}[/tex]= Molar heat capacity of the gas at constant pressure.[tex]T_{1}[/tex]= Initial temperature at section 1
[tex]T_{2}[/tex] = Final temperature at section 2
Substitute the equation (2) in equation (1)
[tex]C_{p}(T_{2}-T_{1})+\frac{u_{2}^2-u_{1}^2}{2}=0[/tex]
Solve the above equation is as follows.
[tex]T_{2}=T_{1}-\frac{u_{2}^{2}-u_{1}^{2}}{2}=0...............(3)[/tex]
From the given,
[tex]T_{1}=150+273=423K[/tex]
[tex]C_{p}=\frac{7R}{2}[/tex]
[tex]u_{1}=2.5\,m/s[/tex]
[tex]u_{2}=50\,m/s[/tex]
Molar mass of nitrogen gas = 0.02802 kg/mol
Substitute the all values in the equation (3)
[tex]T_{2}=423K-\frac{(50m/s)^{2}-(2.5m/s)^{2}}{2\times \frac{7}{2}\times8.314\,J\,mol^{-1}K^{-1}}\times \frac{J/kg}{m^{2}/s^{2}}\times \frac{0.02802\,kg}{mol}[/tex]
=421.8K=148.8^{o}C
Therefore,The temperature of the nitrogen gas at another section is [tex]148.8^{o}C[/tex].
Final answer:
The equation that relates temperature to velocity of an ideal gas flowing through a horizontal tube with changing cross-sectional area is T1/V1 = T2/V2. Using the given values, we can solve for the temperature at the other section where the velocity is given as 50 m/s, which is found to be 300 °C.
Explanation:
Based on the principle of continuity of flow, the equation that relates temperature to velocity of an ideal gas flowing through a horizontal tube with changing cross-sectional area is:
T1/V1 = T2/V2
To solve for the temperature at another section where the velocity is 50 m/s, we can use the given temperature and velocity at the initial section (T1 = 150 °C, V1 = 2.5 m/s) and plug them into the equation:
150 / 2.5 = T2 / 50
By solving for T2, we find that the temperature at the other section is 300 °C.
The standard free-energy changes for the reactions below are given.Phosphocreatine → creatine + Pi ∆ G'° = –43.0 kJ/molATP → ADP + Pi ∆ G'° = –30.5 kJ/mol
What is the overall ∆ G'° for the following reaction?
Phosphocreatine + ADP → creatine + ATP
Answer:
Gibbs free-energy of the reaction = (–12.5 kJ/mol)
Explanation:
The Gibbs free-energy of a reaction predicts the spontaneity or feasibility of a given chemical reaction.
Given the standard Gibbs free energy changes:
Phosphocreatine → creatine + Pi, ∆G° = –43.0 kJ/mol ...(1)
ATP → ADP + Pi , ∆G° = –30.5 kJ/mol ....(2)
Now to calculate the Gibbs free-energy of the given chemical reaction: Phosphocreatine + ADP → creatine + ATP; the equation (2) is reversed to give:
ADP + Pi → ATP, ∆G° = + 30.5 kJ/mol ...(3)
Now the equation (3) and (1) are added, to give:
Phosphocreatine + ADP + Pi→ creatine + ATP + Pi
⇒ Phosphocreatine + ADP → creatine + ATP
Therefore, to calculate the Gibbs free-energy of the reaction, the standard Gibbs free energy changes of the equations (1) and (3) are added similarly:
Gibbs free-energy of the reaction: ∆G° = (–43.0 kJ/mol) + ( + 30.5 kJ/mol) = (–12.5 kJ/mol)
Therefore, the Gibbs free-energy of the reaction = (–12.5 kJ/mol)
The overall standard free-energy change for the reaction Phosphocreatine + ADP to form creatine and ATP is the sum of the changes for the individual reactions, resulting in −12.5 kJ/mol.
Explanation:To determine the overall standard free-energy change (Δ G'°) for the reaction where phosphocreatine and ADP react to form creatine and ATP (“Phosphocreatine + ADP → creatine + ATP”), we need to consider the given free-energy changes for the individual reactions:
Phosphocreatine → creatine + Pi (Δ G'° = −43.0 kJ/mol)ATP → ADP + Pi (Δ G'° = −30.5 kJ/mol, which should actually be considered as ADP + Pi → ATP since it occurs in the reverse direction)By reversing the second reaction, we invert the sign of its standard free-energy change. Therefore, for the reaction ADP + Pi → ATP, the Δ G'° becomes +30.5 kJ/mol.
Now, to find the overall Δ G'° for the reaction Phosphocreatine + ADP → creatine + ATP, we sum up the standard free-energy changes of the two reactions:
Δ G'° (overall) = Δ G'° (Phosphocreatine → creatine + Pi) + Δ G'° (ADP + Pi → ATP)
Δ G'° (overall) = (−43.0 kJ/mol) + (+30.5 kJ/mol) = −12.5 kJ/mol
The result is −12.5 kJ/mol, which indicates that the coupled reaction is also exergonic, releasing energy.
Which of the following statements concerning galvanic cells is/are true?
A.) The two half-cells are connected by a salt bridge
B.) Electrons flow from the anode to the cathode
C.) Reduction occurs at the cathode
D.) All of the above are true
Answer: Correct answer is D.) Every option is true and correct for a galvanic cell.
All the given statements are true: a galvanic cell consists of two half-cells connected by a salt bridge, electrons flow from the anode to the cathode, and reduction occurs at the cathode.
Explanation:The statement 'D.) All of the above are true' is the correct selection regarding galvanic cells. A.) In a galvanic cell, the two half-cells are indeed connected by a salt bridge. The salt bridge serves to balance the charge, allowing ions to migrate and maintain a neutral charge in the cell. B.) Electrons flow from the anode to the cathode. This is the basic concept of electricity – the flow of electrons. The anode is the electrode where oxidation (loss of electrons) happens, and the electrons released from the oxidation reaction at the anode flow through a wire to the cathode. C.) Reduction occurs at the cathode, meaning that it gains electrons. The cathode is the electrode where reduction (gain of electrons) happens in redox (oxidation-reduction) reactions.
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Which of the statements applies to oxidation?
(A) Ketones cannot be oxidized further.
(B) Carboxylic acids are oxidized to ketones.
(C) Alcohols are oxidized to alkanes.
(D) Ketones are oxidized to alcohols.
Answer:
The correct option is: (A) Ketones cannot be oxidized further
Explanation:
Oxidation refers to the gain of oxygen or the formation of carbon-oxygen bond (C-O bond).
Alcohols are organic compounds containing at least one hydroxyl group. These compounds can be further classified into primary (R-CH₂-OH), secondary (R¹R²CH-OH) and tertiary alcohols (R¹R²R³C-OH).
The partial oxidation of primary alcohol gives aldehyde (R-CHO). Further oxidation of aldehyds gives carboxylic acid (R-COOH). Primary alcohol can also be oxidized directly to carboxylic acid. However, the carboxylic acid can not be further oxidized.[tex]R-CH_{2}-OH \overset{[O]}{\rightarrow} R-CHO \overset{[O]}{\rightarrow} R-COOH \\R-CH_{2}-OH \overset{[O]}{\rightarrow} R-COOH[/tex]
Oxidation of secondary alcohols gives ketones (R¹R²C=O), which can not be oxidized further.[tex]R^{1}R^{2}CH-OH \overset{[O]}{\rightarrow} R^{1}R^{2}C=O[/tex]
Whereas, tertiary alcohols are resistant to oxidation.Therefore, the correct statement regarding oxidation is (A) Ketones cannot be oxidized further.
On a lab exam, you have to find the concentrations of the monoprotic (one proton per molecule) acids HA and HB. You are given 43.5 mL of HA solution in one flask.You titrate this flask of HA solution with 87.3 mL of 0.0906 M NaOH. Calculate the molarity of the HA solution
To find the molarity of the HA solution, calculate the number of moles of NaOH used in the titration and use the ratio of moles of HA to moles of NaOH to determine the number of moles of HA in the solution. Then, calculate the molarity of the HA solution using the moles of HA and the volume of the solution.
Explanation:To find the molarity of the HA solution, we can first calculate the number of moles of NaOH used in the titration. The balanced chemical equation for the reaction between HA and NaOH is:
HA + NaOH → NaA + H2O
From the equation, we can see that the ratio of moles of HA to moles of NaOH is 1:1. We can use the molarity and volume of NaOH used in the titration to calculate the number of moles of NaOH, and since the ratio is 1:1, this will also be the number of moles of HA in the solution.
First, we calculate the number of moles of NaOH:
Moles of NaOH = Molarity of NaOH * Volume of NaOH
= 0.0906 M * 0.0873 L
= 0.0078958 mol NaOH
Since the ratio of moles of HA to moles of NaOH is 1:1, the number of moles of HA in the solution is also 0.0078958 mol.
Now, we can calculate the molarity of the HA solution:
Molarity of HA = Moles of HA / Volume of HA solution
= 0.0078958 mol / 0.0435 L
= 0.181 M
The molarity of the HA solution is [tex]\boxed{0.1817 \text{ M}}[/tex]
To calculate the molarity of the HA solution, we need to use the concept of a neutralization reaction. The reaction between the monoprotic acid HA and the strong base NaOH is as follows:
[tex]\[ \text{HA} + \text{NaOH} \rightarrow \text{NaA} + \text{H}_2\text{O} \][/tex]
In this reaction, one mole of HA reacts with one mole of NaOH to produce one mole of NaA (the salt) and one mole of water. The moles of NaOH used in the titration can be calculated using the formula
[tex]\[ \text{moles of NaOH} = \text{volume of NaOH} \times \text{molarity of NaOH} \][/tex]
Given that the volume of NaOH is 87.3 mL (which we convert to liters by dividing by 1000) and the molarity of NaOH is 0.0906 M, we can calculate the moles of NaOH:
[tex]\[ \text{moles of NaOH} = 0.0873 \text{ L} \times 0.0906 \text{ M} \][/tex]
[tex]\[ \text{moles of NaOH} = 0.00790358 \text{ mol} \][/tex]
Since the stoichiometry of the reaction is 1:1, the moles of HA that reacted with NaOH are equal to the moles of NaOH used. Now we can calculate the molarity of the HA solution using the formula:
[tex]\[ \text{molarity of HA} = \frac{\text{moles of HA}}{\text{volume of HA solution}} \][/tex]
The volume of the HA solution is given as 43.5 mL, which we convert to liters:
[tex]\[ \text{volume of HA solution} = 0.0435 \text{ L} \][/tex]
Now we can calculate the molarity of HA:
[tex]\[ \text{molarity of HA} = \frac{0.00790358 \text{ mol}}{0.0435 \text{ L}} \][/tex]
[tex]\[ \text{molarity of HA} = 0.18167287 \text{ M} \][/tex]
Rounding to a reasonable number of significant figures, the molarity of the HA solution is approximately 0.1817 M.
Therefore, the molarity of the HA solution is [tex]\boxed{0.1817 \text{ M}}[/tex]
The answer is: [tex]0.1817 \text{ M}.[/tex]
Determine whether each redox reaction occurs spontaneously in the forward direction.
a. Ni(s) + Zn2 + (aq) ¡ Ni2 + (aq) + Zn(s)
b. Ni(s) + Pb2 + (aq) ¡ Ni2 + (aq) + Pb(s)
c. Al(s) + 3 Ag+ (aq) ¡ Al3 + (aq) + 3 Ag(s)
d. Pb(s) + Mn2 + (aq) ¡ Pb2 + (aq) + Mn(s)
Answer:
a) non spontaneous
b) spontaneous
c) spontaneous
d) non spontaneous
Answer:
Explanation:
a. Ni(s) + Zn2 + (aq) ¡ Ni2 + (aq) + Zn(s) non spontaneous
b. Ni(s) + Pb2 + (aq) ¡ Ni2 + (aq) + Pb(s) spontaneous
c. Al(s) + 3 Ag+ (aq) ¡ Al3 + (aq) + 3 Ag(s) Reduction spontaneous
d. Pb(s) + Mn2 + (aq) ¡ Pb2 + (aq) + Mn(s) non spontaneous
Alcohols can be oxidized by chromic acid derivatives. One such reagent is pyridinium chlorochromate, (C5H5NH )(ClCrO3–), commonly known as PCC. Draw the proposed (neutral) intermediate and the organic product in the oxidation of 1-butanol by PCC when carried out in an anhydrous solvent such as CH2Cl2."
Answer:
See figure attached.
Explanation:
Pyridinium chlorochromate (PCC) is a weaker oxidizing agent when compared with oth chromic acid derivatives as dichromate, for example. This results in the fact that PCC wil oxidize a primary alcohol transforming it to a aldehyde.
The entire mechanism of the reaction of 1-butanol with PCC in an anhydrous solvent is shown in the figure attached.
The neutral intermediate and the organic product (1-butanal) are both represented in the figure.
You have 41.6 g of O2 gas in a container with twice the volume as one with CO2 gas in another container. The pressure and temperature of both containers are the same. Calculate the mass of carbon dioxide gas you have in the container.
Answer:
The mass of carbon dioxide is 28.6 grams
Explanation:
Step 1: Data given
Mass of O2 = 41.6 grams
Volume of the O2 container = 2V
Volume of the CO2 container = V
Pressure and temperature are the same
Step 2: Ideal gas law
The ideal gas law = p*V=nRT
For O2: p*2V = n(O2)*R*T
For CO2: p*V = n(CO2)*R*T
n(O2)*R*T / P*2V = n(CO2)*R*T / P*V
Since P, R and T are contstant, we can simplify this to:
n(O2)/2 = n(CO2)
Step 3: Calculate moles of O2
Moles O2 = mass O2/ Molar mass O2
Moles O2 = 41.6 grams / 32 g/mol
Moles O2 = 1.3 mol O2
The number of moles CO2 = 0.65 mol
Mass of CO2 = Moles CO2 * Molar Mass CO2
Mass of CO2 = 0.65 mol * 44.01 g/mol
Mass of CO2 = 28.6 grams
The mass of carbon dioxide is 28.6 grams
The electrons formed from the aerobic oxidation of glucose are
I. ultimately transferred to O2 after several other transfer reactions.
II. transferred to the coenzymes NAD and FAD.
III. directly transferred to O2 during the citric acid cycle.
IV. transferred to succinate and arachidonic acid.
Answer:
i and ii
Explanation:
In the aerobic oxidation of glucose, the electrons formed are transferred to O2 after several others transfer reactions like passing through coenzymes NAD+ and FAD
If 45.0 mL of ethanol (density=0.789 g/mL) initially at 9.0 C is mixed with 45.0 mL of water (density=1.0 g/mL) initially at 28.6 C in an insulated beaker, and assuming that no heat is lost, what is the final temperature of the mixture?
I tried many times in trying to get the answer, but I keep getting it wrong. I appreciate who answers this writes it step by step. Thank you.
Applying the principle of conservation of energy and the heat equation, the final temperature of the ethanol-water mixture is calculated to be 22.7°C
The specific example used in the image is for a mixture of ethanol and water.
Step 1: Understand the principle
The underlying principle is that heat lost by the hot object is equal to the heat gained by the cold object. This principle is known as the law of conservation of energy.
Step 2: Form an equation
We can express this principle mathematically using the following equation:
q₁ = -q₂
where:
q₁ is the heat lost by the hot object
q₂ is the heat gained by the cold object
We can further expand this equation using the following relationship between heat, mass, specific heat capacity, and temperature change:
q = mcΔT
where:
m is the mass of the object
c is the specific heat capacity of the object
ΔT is the change in temperature
Substituting this relationship into the first equation, we get:
m₁c₁(T₁ - T_f) = -m₂c₂(T_f - T₂)
where:
m₁ and c₁ are the mass and specific heat capacity of the hot object, respectively
m₂ and c₂ are the mass and specific heat capacity of the cold object, respectively
T₁ and T₂ are the initial temperatures of the hot and cold objects, respectively
T_f is the final temperature of the mixture
Step 3: Apply the formula to the specific case
In the example given, the hot object is ethanol and the cold object is water. The following values are provided:
m₁ = 45.0 g (mass of ethanol)
c₁ = 2.3 J/g°C (specific heat capacity of ethanol)
T₁ = 9.0°C (initial temperature of ethanol)
m₂ = 45.0 g (mass of water)
c₂ = 4.18 J/g°C (specific heat capacity of water)
T₂ = 28.6°C (initial temperature of water)
T_f = ? (final temperature of the mixture)
Step 4: Solve for the unknown
Substituting the known values into the equation and solving for T_f, we get:
(0.789 g/mL x 45.0 mL) x (2.3 J/g°C) x (T_f - 9.0°C) = -(1.0 g/mL x 45.0 mL) x (4.18 J/g°C) x (T_f - 28.6°C)
Solving this equation, we get:
T_f = 22.7°C
Therefore, the final temperature of the mixture is 22.7°C.
Consider the following reaction: 2NO(g)+O2(g)→2NO2(g) Estimate ΔG∘ for this reaction at each of the following temperatures and predict whether or not the reaction will be spontaneous. (Assume that ΔH∘ and ΔS∘ do not change too much within the give temperature range.) 718k
The Gibbs free energy change (ΔG∘) for the reaction can be estimated using the equation ΔG∘ = ΔH∘ - TΔS∘. At a temperature of 718K, if ΔG∘ is negative, the reaction will be spontaneous.
Explanation:The Gibbs free energy change (ΔG∘) for a reaction can be estimated using the equation ΔG∘ = ΔH∘ - TΔS∘, where ΔH∘ is the change in enthalpy and ΔS∘ is the change in entropy. At a temperature of 718K, you can estimate ΔG∘ by substituting the given values into the equation. If the value of ΔG∘ is negative, the reaction will be spontaneous. If it is positive or zero, the reaction will not be spontaneous.
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The reaction is spontaneous at 298 K and 723 K but non-spontaneous at 842 K as determined by calculating ΔG∘ at each temperature using the Gibbs free energy equation. Negative ΔG∘ values indicate spontaneity while positive values indicate non-spontaneity.
To determine the spontaneity of the reaction 2NO(g) + O₂(g) → 2NO₂(g) at different temperatures (298 K, 723 K, and 842 K), we use the Gibbs free energy equation:
ΔG∘ = ΔH∘ - TΔS∘
Given that ΔH∘ (standard enthalpy change) and ΔS∘ (standard entropy change) are assumed to be constant over the temperature range, we need to calculate ΔG∘ for each temperature:
At 298 K:If ΔG∘ is negative, the reaction is spontaneous at that temperature. Conversely, if ΔG∘ is positive, the reaction is non-spontaneous.
Determining Spontaneity:
Using standard thermodynamic tables or given data for ΔH∘ and ΔS∘:
Suppose: ΔH∘ = -114.1 kJ/mol and ΔS∘ = -146.5 J/(mol·K)
At 298 K:In summary, the reaction is spontaneous at 298 K and 723 K, but non-spontaneous at 842 K
complete question.
Consider the reaction:
2NO(g)+O₂(g)→2NO₂(g)
Estimate ΔG∘ for this reaction at each temperature and predict whether or not the reaction will be spontaneous. (Assume that ΔH∘ and ΔS∘ do not change too much within the given temperature range.)
A) 298 K
B) 723 K
C) 842 K
Elimination reactions are favored over nucleophilic substitution reactions:
O at high temperatures.
O when tert-butoxide ion is used.
O when 3° alkyl halides are used as substrates.
O when nucleophiles are used which are strong bases and the substrate is a 2° alkyl halide.
O in all of these cases.
In all of these cases.
Elimination reaction compete with nucleophilic substitution reaction and is favored under conditions of
The heat help to reach the energy required for the dehydrohalogenation reaction with a strong base.
terc-butoxide favor the Hoffman elimination product, terc-butoxide tendency is to break out Zaitseff rule, so in this case the eliminitation in alkyl halides less substituted is favored.
If the substrate of the reaction is an 3rd alkyl halides, the carbocation is more stable so according to Zaitsev's rule in an elimination reaction, the most substituted product will be the most stable, so will be favored.
The nucleophile might to be a strong base because should be strong enough to form a bond with a hydrogen from the alkyl halide and let reaction continues. Regarding to 2° alkyl halide, 3° alkyl halide is favored towards 2° alkyl halide. Being the relative rate of reaction with 3° alkyl halide higher than 2° alkyl halide. That is because provides a greater number of hydrogens for attack by base and hence more favorable for elimination. If the alkyl halide are more branched more stable the alkene.
Elimination reactions are favored over substitution reactions at high temperatures, with tert-butoxide ion, with tertiary alkyl halides, and with strong bases and secondary alkyl halides. Therefore, correct option is: elimination is favored in all these situations.
Elimination reactions are often favored over nucleophilic substitution reactions under certain conditions. The key factors that favor elimination reactions include:
High temperatures: At higher temperatures, elimination (E₂) is favored over substitution (SN₂).Use of tert-butoxide ion: This strong, bulky base favors elimination due to steric hindrance preventing substitution.Use of tertiary (3°) alkyl halides: These substrates are less reactive in SN₂ but highly reactive in E₂.Strong bases with secondary (2°) alkyl halides: Strong bases can favor E₂ over SN₂.Based on these points, elimination reactions are favored in all of these cases.
Which is not true of the standard conditions for electrochemical cell measurements?
Gases are at 1 bar pressure.
Liquids are in the pure state.
Solids are in the pure state.
Solutes are at 1.0 M concentration.
The cell voltage is always positive.
Answer:
Gases are at 1 bar pressure.
Explanation:
Option a is incorrect . Standard Pressure in chemistry is always taken in atm and not in bar. The standard condition for electro chemical cell measurements are
Gases must be at 1 atm pressure.
Liquids must be in pure state
Solids must also be in pure state
Solutes are at 1.0 M concentration
And the cell voltages are positive.
In the context of electrochemical cells, the statement 'The cell voltage is always positive' is incorrect as cell voltage can be negative in an electrolytic cell.
Explanation:The statement that is not true of the standard conditions for electrochemical cell measurements is that 'The cell voltage is always positive.' While the cell voltage can be positive in a galvanic/voltaic cell as it produces an electrical current, it can be negative in an electrolytic cell, where electrical energy is being consumed to bring about a chemical change.
The other statements about gases at 1 bar pressure, liquids and solids being in a pure state, and solutes at a 1.0 M concentration are indeed part of the standard conditions for measuring the properties of an electrochemical cell.
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What can be said about an Endothermic reaction with a negative entropy change?The reaction isa. spontaneous at all temperatures.b. spontaneous at high temperatures.c. spontaneous at low temperatures.d. spontaneous in the reverse direction at all temperatures.e. nonspontaneous in either direction at all temperatures.What can be said about an Exothermic reaction with a negative entropy change?The reaction isa. spontaneous at all temperatures.b. spontaneous at high temperatures.c. spontaneous at low temperatures.d. spontaneous in the reverse direction at all temperatures.e. nonspontaneous in either direction at all temperatures.
Answer:
1. d. The reaction is spontaneous in the reverse direction at all temperatures.
2. c. The reaction is spontaneous at low temperatures.
Explanation:
The spontaneity of a reaction is associated with the Gibbs free energy (ΔG). When ΔG < 0, the reaction is spontaneous. When ΔG > 0, the reaction is non-spontaneous. ΔG is related to the enthalpy (ΔH) and the entropy (ΔS) through the following expression:
ΔG = ΔH - T. ΔS [1]
where,
T is the absolute temperature (T is always positive)
1. What can be said about an Endothermic reaction with a negative entropy change?
If the reaction is endothermic, ΔH > 0. Let's consider ΔS < 0. According to eq. [1], ΔG is always positive. The reaction is not spontaneous in the forward direction at any temperature. This means that the reaction is spontaneous in the reverse direction at all temperatures.
2. What can be said about an Exothermic reaction with a negative entropy change?
If the reaction is exothermic, ΔH < 0. Let's consider ΔS < 0. According to eq. [1], ΔG will be negative when |ΔH| > |T.ΔS|, that is, at low temperatures.
Final answer:
An endothermic reaction with a negative entropy change is nonspontaneous at all temperatures. An exothermic reaction with a negative entropy change can be spontaneous at low temperatures but not necessarily at high temperatures.
Explanation:
An endothermic reaction with a negative entropy change is one that absorbs energy from its surroundings and results in a decrease in the randomness or disorder of the system. According to thermodynamics, the spontaneity of a reaction depends on the change in free energy (ΔG), which is influenced by enthalpic (ΔH) and entropic (-TΔS) factors.
An endothermic reaction with a negative entropy change would have a positive ΔH and a negative TΔS, making the ΔG positive at all temperatures. Therefore, such a reaction is nonspontaneous at either direction at all temperatures (e).
In contrast, an exothermic reaction with a negative entropy change releases energy and also results in a decrease in the system's disorder. This reaction would have a negative ΔH and a negative TΔS. For this reaction, ΔG could be negative at low temperatures, resulting in spontaneity (spontaneous at low temperatures (c)). At high temperatures, the positive TΔS may outweigh the negative ΔH, resulting in a positive ΔG and nonspontaneity.
What is the new solution concentration when 150. mL of water is added to 200. mL of a 3.55 M HBr solutions.
Answer:
The new concentration is 2.03M
Explanation:
Step 1: Data given
A 200 mL 3.55 M HBr is diluted with 150 mL
Step 2: The dilution
In a dilution, the ratio that exists between the concentration of the stock solution and the concentration of the diluted solution equals the ratio that exists between the volume of the diluted solution and the volume of the stock solution.
Dilution factor = [stock sample]/[diluted sample] = diluted volume / stock volume
In this case, the volume of the stock solution is 200 mL
Adding 150 mL of water to the stock solution will dilute it to a final volume of 200 + 150 = 350 mL
The dilution factor wll be 350/200 = 1.75
This makes the diluted concentration:
3.55/1.75 = 2.03M
The new concentration is 2.03M
What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.15 M buffer solution at pH 5.0? Note that the concentration and/or pH value may differ from that in the first question. STRATEGY 1. Rearrange the Henderson-Hasselbalch equation to solve for the ratio of base (acetate) to acid (acetic acid), [A–]/[HA]. 2. Use the mole fraction of acetate to calculate the concentration of acetate. 3. Calculate the concentration of acetic acid. Step 1: Rearrange the Henderson Hasselbalch equation to solve for (A-)/(HA) if the solution is at ph 5.0
Answer:
Acetic acid 0,055M and acetate 0,095M.
Explanation:
It is possible to prepare a 0,15M buffer of acetic acid/acetate at pH 5,0 using Henderson-Hasselblach formula, thus:
pH = pka + log₁₀ [A⁻]/[HA] -Where A⁻ is acetate ion and HA is acetic acid-
Replacing:
5,0 = 4,76 + log₁₀ [A⁻]/[HA]
1,7378 = [A⁻]/[HA] (1)
As concentration of buffer is 0,15M, it is possible to write:
[A⁻] + [HA] = 0,15M (2)
Replacing (1) in (2):
1,7378[HA] + [HA] = 0,15M
2,7378[HA] = 0,15M
[HA] = 0,055M
Thus, [A⁻] = 0,095M
That means you need acetic acid 0,055M and acetate 0,095M to obtain the buffer you need.
i hope it helps!
To make a 0.15 M buffer solution at a pH of 5.0 using acetic acid (pKa=4.76) and acetate, the concentrations of acetic acid and acetate need to be about 0.064 M and 0.086 M respectively, determined through rearranging the Henderson-Hasselbalch equation.
Explanation:
To prepare a 0.15 M buffer solution at pH 5.0 using acetic acid (pKa = 4.76) and acetate, we first utilize the Henderson-Hasselbalch equation to determine the ratio of acetate (A-) to acetic acid (HA). This rearranged equation is: [A-]/[HA] = 10^(pH - pKa) = 10^(5.0 - 4.76).
From this, the ratio of [A-]/[HA] comes out to be approximately 1.74. Since an ideal buffer has roughly equal concentrations of acid and base, we know that the sum of the acetic acid and acetate concentrations should be 0.15 M. Therefore, to achieve a buffer at pH 5.0, the acetic acid concentration is approximately 0.064 M (0.037 M * (1/(1+1.74))) and the acetate concentration is approximately 0.086 M (0.15 M - 0.064 M).
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For the following species, draw a Lewis structure for one important resonance form. Include all lone pair electrons and assign formal charge to any atom with a non-zero formal charge
NO2+
NO2F (N is central)
NO2+ has a Lewis structure with a single bond between the Nitrogen and one Oxygen, and a double bond with the other Oxygen. Nitrogen has a +1 formal charge. NO2F's Lewis structure has Nitrogen double bonded with one Oxygen and single bonded with the other Oxygen and Fluorine; all atoms have three lone electron pairs, and there is no non-zero formal charge.
Explanation:In Chemistry, the process of drawing Lewis structures involves representing the molecules' bonding electrons and any lone pairs. The Lewis structure for NO2+ is drawn with a single bond between the Nitrogen (N) and one of the Oxygen (O), as well as a double bond between Nitrogen and the other Oxygen. Nitrogen has a +1 formal charge due to the extra positive charge on the molecule.
For NO2F, Nitrogen (N) is the central atom which is double bonded with one Oxygen (O) atom and single bonded with the other Oxygen and Fluorine (F) atoms. Every Oxygen atom has three lone pairs of electrons and Fluorine has three lone pairs as well. The Nitrogen has one lone pair of electrons. There is no atom with a non-zero formal charge in this molecule.
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To draw the Lewis structure for NO2+ and NO2F, count the valence electrons and distribute them to satisfy the octet rule.
Explanation:To draw the Lewis structure for NO2+, we start by counting the valence electrons. N has 5 valence electrons, and each O contributes 6 valence electrons, giving us a total of 24 valence electrons. We place N in the center and connect each O to it using a single bond. The remaining electrons are distributed as lone pairs and double bonds to satisfy the octet rule. In one important resonance form, the double bond can be between N and any of the O atoms.
For NO2F, we again count the valence electrons, which are 5 for N, 6 for each O, and 7 for F. So we have a total of 24+7=31 valence electrons. We place N in the center and connect each O and F to it using a single bond. The remaining electrons are distributed as lone pairs and double bonds. In one important resonance form, we have a double bond between N and one of the O atoms, and a single bond between N and the other O atom, with F having a single bond to one of the O atoms.
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When 1 mole of Fe2O3(s) reacts with Al(s) to form Al2O3(s) and Fe(s) according to the following equation, 852 kJ of energy are evolved. Fe2O3(s) + 2Al(s)Al2O3(s) + 2Fe(s) Is this reaction endothermic or exothermic? What is the value of q? kJ g
Answer: The reaction is exothermic in nature and the value of q is 852 kJ
Explanation:
Endothermic reactions are defined as the reactions in which energy of products is more than the energy of the reactants. In these reactions, energy is absorbed by the system. The energy term is written on the reactant side of the reaction.
Exothermic reactions are defined as the reactions in which energy of reactants is more than the energy of the products. In these reactions, energy is released by the system. The energy term is written on the product side of the reaction.
We are given:
Heat released in the reaction = 852 kJ/mol
The chemical equation follows:
[tex]Fe_2O_3(s)+2Al(s)\rightarrow Al_2O_3(s)+2Fe(s)+852kJ[/tex]
As, heat is released during the reaction, it is considered as an exothermic reaction.
To calculate the enthalpy change of the reaction, we use the equation:
[tex]\Delta H_{rxn}=\frac{q}{n}[/tex]
where,
[tex]q[/tex] = amount of heat released
n = number of moles = 1 mole
[tex]\Delta H_{rxn}[/tex] = enthalpy change of the reaction = 852 kJ/mol
Putting values in above equation, we get:
[tex]852kJ/mol=\frac{q}{1mol}\\\\q=(852kJ/mol\times 1mole)=852kJ[/tex]
Hence, the reaction is exothermic in nature and the value of q is 852 kJ
What is the concentration in molarity of a solution made using 30.0 grams of CH3OH in 300.0 mL of water?
Please help me right now!!! :(
Toluene (C6H5CH3 ), an organic compound often used as a solvent in paints, is mixed with a similar organic compound, benzene (C6H6 ). Calculate the molarity, molality, mass percent, and mole fraction of toluene in 200. mL of solution that contains 75.8 g of toluene and 95.6 g of benzene. The density of the solution is 0.857 g/cm3.
Explanation:
The given data is as follows.
Weight of solute = 75.8 g, Molecular weight of solute (toulene) = 92.13 g/mol, volume = 200 ml
Therefore, molarity of toulene is calculated as follows.Molarity = [tex]\frac{\text{weight of solute}}{\text{molecular weight of solute}} \times \frac{1000}{\text{volume of solution in ml}}[/tex]
= [tex]\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{200 ml}[/tex]
= 4.11 M
Hence, molarity of toulene is 4.11 M.
As molality is the number of moles of solute present in kg of solvent.So, we will calculate the molality of toulene as follows.
Molality = [tex]\frac{\text{given weight of solute}}{\text{given molecular weight of solute}} \times \frac{1000}{\text{weight of solvent in grams}}[/tex]
= [tex]\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{95.6 g}[/tex]
= 8.6 m
Hence, molality of given toulene solution is 8.6 m.
Now, calculate the number of moles of toulene as follows.No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{75.8 g}{92.13 g/mol}[/tex]
= 0.8227 mol
Now, no. of moles of benzene will be as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{95.6 g}{78.11 g/mol}[/tex]
= 1.2239 mol
Hence, the mole fraction of toulene is as follows.
Mole fraction = [tex]\frac{\text{moles of toulene}}{\text{total moles}}[/tex]
= [tex]\frac{0.8227 mol}{(0.8227 + 1.2239) mol}[/tex]
= 0.402
Hence, mole fraction of toulene is 0.402.
As density of given solution is 0.857 [tex]g/cm^{3}[/tex] so, we will calculate the mass of solution as follows.Density = [tex]\frac{mass}{volume}[/tex]
0.857 [tex]g/cm^{3}[/tex] = [tex]\frac{mass}{200 ml}[/tex] (As 1 [tex]cm^{3}[/tex] = 1 g)
mass = 171.4 g
Therefore, calculate the mass percent of toulene as follows.
Mass % = [tex]\frac{\text{mass of solute}}{\text{mass of solution}} \times 100[/tex]
= [tex]\frac{75.8 g}{171.4 g} \times 100[/tex]
= 44.22%
Therefore, mass percent of toulene is 44.22%.
The molarity, molality, mass percent, and mole fraction of toluene in a solution can be calculated using the given information.
Explanation:To calculate the molarity of toluene in the solution, we first need to calculate the moles of toluene and benzene. The molarity is then the moles of toluene divided by the volume of the solution in liters. The molality is calculated by dividing the moles of toluene by the mass of the solvent in kilograms. The mass percent is calculated by dividing the mass of toluene by the total mass of the solution and multiplying by 100. Finally, the mole fraction of toluene is calculated by dividing the moles of toluene by the total moles of all components in the solution.
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When a double replacement reaction produced one compound that is soluble and one compounds that is insoluble,a. only the solution of dissolved ion resultsb. only a solid is formedc. the soluble compound exists as dissolved ions in a solution while insoluble compound forms as a precipitated. none
Answer:
c. the soluble compound exists as dissolved ions in a solution while insoluble compound forms as a precipitated.
Explanation:
Let's consider an example of a double displacement reaction that produces a soluble compound and an insoluble one.
BaCl₂(aq) + Na₂SO₄(aq) ⇒ NaCl(aq) + BaSO₄(s)↓
NaCl is a soluble compound so it is dissolved in the solution. It is also an electrolyte so it ionizes in water according to the following equation:
NaCl(aq) ⇒ Na⁺(aq) + Cl⁻(aq)
BaSO₄ is insoluble so it remains as a precipitate.