The energy absorbed as the bonds in the reactants are broken is less than the energy released as the bonds in the products are formed. This is indicative of an exothermic reaction.
Explanation:In the given reaction, CO + 2 H2 ---> CH3OH ΔH < 0, this is an exothermic reaction as indicated by the negative delta H value. The negative ΔH value signifies that the total energy released when the new bonds are formed in the product (CH3OH) is more than the total energy absorbed to break the bonds in the reactants (CO and H2).
Therefore, the correct statement is: The energy absorbed as the bonds in the reactants are broken is less than the energy released as the bonds in the products are formed.
This corresponds to the fact that in an exothermic reaction, the products are at a lower energy level than the reactants, therefore, energy is released in the reaction.
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The correct statement is that the energy absorbed as the bonds in the reactants is broken is less than the energy released as the bonds in the products are formed, indicating an exothermic reaction.
The correct statement is: The energy absorbed as the bonds in the reactants is broken is less than the energy released as the bonds in the products are formed.
In this specific reaction, energy is required to break a C-O triple bond and two H-H single bonds in the reactants. Energy is then released as three C-H single bonds, a C-O single bond, and an O-H single bond are formed in the product, CH₃OH.
Since ΔH is negative, it indicates that the total energy released in bond formation is greater than the total energy absorbed in bond breaking, reaffirming that the reaction is exothermic.
The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and 269.2 J/K*mol, respectively. Calculate ΔH° , ΔS° , and ΔG° for the following process at 25.00°C.
C6H6(l) ------> C6H6(g)
Final answer:
The standard enthalpy change (ΔH°) for the process C6H6(l) → C6H6(g) is 82.93 kJ/mol. The standard entropy change (ΔS°) for the process is 269.2 J/K*mol. The standard free energy change (ΔG°) for the process is 7361.033 kJ/mol.
Explanation:
The standard enthalpy change, ΔH°, for the process C6H6(l) → C6H6(g) can be calculated using the standard enthalpy of formation of benzene in the gaseous state and the liquid state. The equation for ΔH° is ΔH° = ΣΔH°(products) - ΣΔH°(reactants). In this case, since benzene is the only product and there are no reactants, the equation simplifies to ΔH° = ΔH°(C6H6(g)). Therefore, ΔH° = 82.93 kJ/mol.
The standard entropy change, ΔS°, for the process can be calculated using the standard entropy of benzene in the gaseous state and the liquid state. The equation for ΔS° is ΔS° = ΣΔS°(products) - ΣΔS°(reactants). Similar to the calculation for ΔH°, since benzene is the only product and there are no reactants, the equation simplifies to ΔS° = ΔS°(C6H6(g)). Therefore, ΔS° = 269.2 J/K*mol.
The standard free energy change, ΔG°, for the process can be calculated using the equation ΔG° = ΔH° - TΔS°, where T is the temperature in Kelvin. Since the temperature is given as 25.00°C, we need to convert it to Kelvin by adding 273.15. Therefore, T = 25.00°C + 273.15 = 298.15 K. Substituting the values into the equation, we get ΔG° = 82.93 kJ/mol - (298.15 K)(269.2 J/K*mol) = 7361.033 kJ/mol.
A sample consisting of n mol of an ideal gas undergoes a reversible isobaric expansion from volume Vi to volume 3Vi. Find the change in entropy of the gas by calculating, ∫dQ / T, where dQ = nCPdT. (Use the following as necessary: Cp and n.)
Answer:
The change in entropy of gas is [tex]\Delta S= nC_{P}ln3[/tex]
Explanation:
n= Number of moles of gas
Change in entropy of gas = [tex]ds= \int \frac{dQ}{T}[/tex]
[tex]dQ= nC_{p}dT[/tex]
From the given,
[tex]V_{i}=V[/tex]
[tex]V_{f}=3V[/tex]
Let "T" be the initial temperature.
[tex]\frac {V_{i}}{T_{i}}=\frac {V_{f}}{T_{f}}[/tex]
[tex]\frac {V}{T}=\frac {3V}{T_{f}}[/tex]
[tex]{T_{f}} = 3T[/tex]
[tex]\int ds = \int ^{T_{f}}_{T_{i}} \frac{nC_{P}dT}{T}[/tex]
[tex]\Delta S = nC_{p}ln(\frac{T_{f}}{T_{i}})[/tex]
[tex]\Delta S = nC_{p}ln3[/tex]
Therefore, The change in entropy of gas is [tex]\Delta S= nC_{P}ln3[/tex]
The shape of the BF3 molecule is best described as
T-shaped
see-saw
tetrahedral
trigonal planar
Answer:
The shape of the BF3 molecule is best described as trigonal planar.
Explanation:
The Lewis Structure for BF3 is like this:
_ _
| F | | F |
\ /
B
|
| F |
---
It forms three angles of 120° each. The bonds are in the same planar that's why it is trigonal planar and they are exactly the same.
Boron and Fluorine have 3 covalent bonds, produced by electronic promotion that enables the 2py and 2pz orbitals, leaving an electron to pair in the 2px. So boron will have 3 possible electrons to pair in 2s1, 2px and 2py, remember that electronic configuration for B is 1s2, 2s2, 2p1
By hybridization between the orbitals 2s2 and 2p1, the electrons of F, can joined to make the covalent bond. The new B configuration is 1s2, 2s1, 2px1, 2py1 (these last three, hybrid orbitals)
The shape of the BF3 molecule is trigonal planar.
Explanation:The shape of the BF3 molecule is trigonal planar.
To determine the shape of a molecule, we need to look at its electron geometry and molecular geometry. The central atom in BF3 is Boron, with three Fluorine atoms surrounding it. Boron has three valence electrons, and each Fluorine atom contributes one electron. When we draw the Lewis structure for BF3, we can see that Boron forms three bonds with the Fluorine atoms, resulting in a trigonal planar electron geometry.
Since there are no lone pairs on the central atom, the molecular geometry of BF3 is also trigonal planar.
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Draw the Lewis structure for SF2. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons. To change the symbol of an atom, double-click on the atom and enter the letter of the new atom.
Sulfur and Fluorine are nonmetals so they will form covalent bonds to gain stability. To do so, they will follow the octet rule: they will share enough electrons so as to have their valence shell complete with 8 electrons.
Sulfur is in the Group 16 in the Periodic Table and has 6 valence electrons. Thus it must share 2 pairs of electrons to reach the octet.
Fluorine is in the Group 17 in the Periodic Table so each F has 7 valence electrons. Thus, each F needs to share 1 pair of electrons to reach the octet.
As a consequence, they will be bonded in the order F - S - F, with a single bond between each pair of atoms.
Consider the neutralization reaction 2HNO3(aq) + Ba(OH)2 ( aq ) ⟶ 2H2O ( l ) + Ba ( NO3)2 ( aq ). A 0.125 L sample of an unknown HNO 3 solution required 32.3 mL of 0.200 M Ba ( OH ) 2 for complete neutralization. What is the concentration of the HNO 3 solution?
Answer:
The concentration of the HNO3 solution is 0.103 M
Explanation:
Step 1: Data given
Volume of the unknow HNO3 sample = 0.125 L
Volume of 0.200 M Ba(OH)2 = 32.3 mL = 0.0323 L
Step 2: The balanced equation
2HNO3(aq) + Ba(OH)2 ( aq ) ⟶ 2H2O ( l ) + Ba( NO3)2 (aq)
Step 3:
n2*C1*V1 = n1*C2*V2
⇒ n2 = the number of moles of Ba(OH)2 = 1
⇒ C1 = the concentration of HNO3 = TO BE DETERMINED
⇒ V1 = the volume of the HNO3 solution = 0.125 L
⇒ n1 = the number of moles of HNO3 = 2
⇒ C2 = the concentration of Ba(OH)2 = 0.200 M
⇒ V2 = the volume of Ba(OH)2 = 0.0323 L
1*C1 * 0.125 L = 2*0.200M * 0.0323 L
C1 = (2*0.200*0.0323)/0.125
C1 = 0.103 M
The concentration of the HNO3 solution is 0.103 M
A two-phase system of liquid water and water vapor in equilibrium at 8000 kPa con sists of equal volumes of liquid and vapox, If the total volume V = 0.15 m, what is the total enthalpy H' and what is the total entropy S?
The total enthalpy H' is 2758.611 kJ/kg and the the total entropy S is 5.745 kJ/ kg . K.
What is entropy?Entropy is defined as the amount of thermal energy per unit temperature in a system that cannot be used for productive work. Entropy, which measures disorder, has an impact on every element of our daily life. In actuality, you could consider it to be nature's tax. Disorder, if unchecked, gets worse over time.
Water vapor will be in its saturated form because the question states that water and water vapor are in equilibrium.
According to the steam table, the total enthalpy Ht = 2758.611 kJ/kg for saturated steam at 8000 kPa (8MPa).
Entropy in its entirety St = 5.745 kJ/ kilogram. K
The saturated steam table at that particular pressure is seen in this figure.
Thus, the total enthalpy H' is 2758.611 kJ/kg and the the total entropy S is 5.745 kJ/ kg . K.
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Given the conditions stated, the water is in a supercritical fluid state where distinct liquid and gas phases don't exist. However, we can't determine the total enthalpy and entropy with the provided information, as we require additional data and the use of thermodynamic equations.
Explanation:The question pertains to a two-phase system of liquid water and water vapor at a specific pressure and volume. The system exists at 8000 kPa, which is beyond the critical point of water (22.064 Mpa or 218 atm). This signifies that under such conditions, the sample of water is in the state of a supercritical fluid, where distinct liquid and gas phases cease to exist and instead a single phase with properties intermediate of both liquid and gas forms.
Unfortunately, based on the available data, determining the total enthalpy (H) and the total entropy (S) isn't feasible. For that, we need extra data such as the specific heat capacity, the exact temperatures of the different phases of water, and/or the number of moles involved. Calculations for H and S would also involve complex thermodynamic equations and experimentally determined constants for water.
Supercritical fluids are noteworthy substances in which the line separating liquid and gas phases disappears at a critical temperature and pressure, leading to the creation of a new phase with intermediate characteristics between a liquid and a gas. The vapor pressure, the liquid-gas equilibrium, and the thermodynamic properties of substances are crucial concepts in this context.
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Carbon dioxide gas at 320 K is mixed with nitrogen at 280 K in a thermally insulated chamber running in steady state. Both flows are coming in at 100 kPa, and the mole ratio of carbon dioxide to nitrogen is 2:1. Find the exit temperature and the total entropy generation per kmole of the exit mixture.
Ansfdfggggggggggggggggggggggggggggghhhhhhhhhhhhhhhhhkkkkkkkkkkkkkkkkkkvvvvvvvvvvvvvvvgggggggggggg
Explanation:
If you mix different amounts of two ideal gases that are originally at different temperatures, what must be true of the final state after the temperature stabilizes? (There may be more than one correct choice.)
a) Both gases will reach the same final temperature.
b) The final rms molecular speed will be the same for both gases.
c) The final average kinetic energy of a molecule will be the same for both gases.
Answer:
a,c are correct
Explanation:
a) On mixing two gases the final temperature of both the gases becomes the same. The heat will flow from high temp. gas to lower temp gas till the temp of both gases become equal (Thermal equilibrium). This is correct.
b) The rms speed of the molecule is inversely proportional to its molar mass so the final rsm will not be the same. This is incorrect.
c) The average kinetic energy of the system will remain the same. Hence this is also correct.
For an aqueous solution of sodium chloride (NaCl), determine the molarity of 4.15 L of a solution that contains 173 g of sodium chloride. Determine the volume of this solution that would contain 3.93 moles of sodium chloride. Determine the number of moles of sodium chloride in 22.45 L of this solution.
Answer:
a) Molarity = 0.713 M
b) volume = 5.51 L
c) Number of moles = 16.01 moles
Explanation:
A) What is the molarity of a 4.15L solution that contains 173 g of sodium chloride?
Step 1: Data given
Volume = 4.15 L
Mass of NaCl = 173 grams
Molar mass of NaCl = 58.44 g/mol
Step 2: Calculate moles of NaCl
Number of moles NaCl = Mass NaCl / molar mass
Moles NaCl = 173 grams / 58.44 g/mol
Moles NaCl = 2.96 moles
Step 3: Calculate molarity of solution
Molarity = moles NacL/ volume
Molarity = 2.96 moles / 4.15L
Molarity = 0.713 M
b) Determine the volume of this solution that would contain 3.93 moles of sodium chloride.
Step 1: Data given
Number of moles = 3.93 moles
Molarity = 0.713 M
Step 2: Calculate volume
Volume = Moles/ Molarity
Volume = 3.93 mol/0.713 M
volume = 5.51 L
c) Determine the number of moles of sodium chloride in 22.45 L of this solution.
Step 1: Data given
Volume = 22.45 L
Molarity = 0.713 M
Step 2: Calculate number of moles
Moles NaCl = Molarity * volume
Moles NaCl = 0.713 * 22.45 L
Moles NaCl = 16.01 moles
The molarity of a solution containing 173 g of sodium chloride (NaCl) in 4.15 L of solution is 0.71 M. The volume of the solution that would contain 3.93 moles of NaCl is 5.52 L. There are 15.95 moles of NaCl in 22.45 L of the solution.
Explanation:The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution in liters. To determine the molarity of a solution containing 173 g of sodium chloride (NaCl) in 4.15 L of solution, we first need to calculate the number of moles of NaCl. The molar mass of NaCl is 58.44 g/mol, so 173 g of NaCl is equal to 2.96 moles. Dividing this by the volume of the solution (4.15 L), we find that the molarity is 0.71 M.
To determine the volume of the solution that would contain 3.93 moles of NaCl, we can rearrange the equation for molarity: Moles = Molarity x Volume. Solving for Volume, we find that Volume = Moles / Molarity. Plugging in the given values, we have Volume = 3.93 moles / 0.71 M = 5.52 L.
Finally, to determine the number of moles of NaCl in 22.45 L of the solution, we can use the molarity formula again: Moles = Molarity x Volume. Plugging in the given values, we have Moles = 0.71 M x 22.45 L = 15.95 moles. Therefore, there are 15.95 moles of NaCl in 22.45 L of the solution.
The lines on Absorption Atomic Spectra correspond to energies needed for electrons to be excited from a lower energy level to a higher energy level. Assume that the energy needed for an electron in 2p orbital in an O atom to jump to 3s orbital is 3.6*10^-19 J, what is its wavelength of the line atomic spectra in nanometer (nm)?
Note: please use whole numbers and 3 sig figs, or no decimal place.
Answer:
553 nm
Explanation:
When an electron from O absorbs radiation with an energy (E) of 3.6 × 10⁻¹⁹ J, it is excited from orbital 2p to orbital 3s. The wavelength (λ) associated with that radiation can be calculated using the Planck-Einstein equation.
E = h. ν = h . c . λ⁻¹
where,
h is the Planck's constant
c is the speed of light
ν is the frequency
[tex]\lambda = \frac{h.c}{E} =\frac{6.63 \times 10^{-34}J.s \times 3.00 \times 10^{8} m/s}{3.6 \times 10^{-19}J } .\frac{10^{9}nm }{1m} =553nm[/tex]
Explain why cis−1−chloro−2−methylcyclohexane undergoes E2 elimination much faster than its trans isomer.
The reacting conformation of the cis isomer has ___.
group(s) axial, making it ___ stable and present in a ____.
concentration than the reacting conformation of the trans isomer.
cis−1−chloro−2−methylcyclohexane undergoes E2 elimination more quickly than its trans isomer due to its axial configuration allowing for an anti-periplanar orientation, which is favorable for E2 elimination.
Explanation:The E2 elimination reaction of cis−1−chloro−2−methylcyclohexane happens faster than its trans isomer due to the orientation of its reacting groups. In the cis isomer, the leaving group (the chlorine) and the beta-proton are both axial. This axial positioning on the same side of the cyclohexane ring allows for an anti-periplanar orientation, which is favorable for an E2 elimination.
However, for the trans isomer, the beta-proton is equatorial while the leaving group (the chlorine) is axial, preventing them from assuming the anti-periplanar configuration. Hence, it becomes less stable and present in a lower concentration. Because the cis isomer can more readily achieve this favored geometry, it reacts more quickly in E2 reactions.
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A student performs an experiment to determine the volume of hydrogen gas produced when a given mass of magnesium reacts with excess HCl(aq) , as represented by the net ionic equation above. The student begins with a 0.0360 g sample of pure magnesium and a solution of 2.0 M HCl(aq).
(a) Calculate the number of moles of magnesium in the 0.0360 g sample
Answer:
1.48x10⁻³ mol
Explanation:
The balanced equation is
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
The aqueous solution will disociate, so:
Mg(s) + 2H⁺(aq) + 2Cl⁻(aq) → Mg²⁺(aq) + 2Cl⁻(aq) + H₂(g)
Simplifying by eliminating the bolded substances that have the same amount in both sides, we have the net ionic equation:
Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g).
a) The number of moles of Magnesium is the mass divided by the molar mass (24.305 g/mol):
n = 0.0360/24.305
n = 1.48x10⁻³ mol
Ag+ (aq) + Cl- (aq) → AgCl (s) Silver chloride is virtually insoluble in water so that the reaction appears to go to completion. How many grams of solid NaCl must be added to 25.0 mL of 0.251 M AgNO3 solution to completely precipitate the silver?
Answer:
0.367 g
Explanation:
Precipitation of silver and Cl requires 1:1 mol ratio of silver and Cl each, so the amount of mol NaCl (provides 1 mol of Cl⁻) and AgNO₃ (privides 1 mol of Ag⁺) must be equal for complete precipitation.
mol of Ag⁺ we have is
[tex]mol_{Ag^+}=[AgNO_3]*V_{solution}\\ =0.251*0.025=0.006275\ mol[/tex]
Thus we need 0.00625 mol of NaCl as well
The molar mass of NaCl is 58.44 g/mol
Therefore the total grams of NaCl needed is
[tex]m_{NaCl}=58.44*0.006275\\ =0.367\ g[/tex]
To completely precipitate the silver, approximately 0.900 grams of solid NaCl must be added to the 25.0 mL of 0.251 M AgNO3 solution.
Explanation:The question asks how many grams of solid NaCl must be added to 25.0 mL of 0.251 M AgNO3 solution to completely precipitate the silver. To find the number of grams of NaCl needed, we need to calculate the number of moles of AgNO3 in the 25.0 mL solution. Then, using the stoichiometry of the reaction, we can determine the number of moles of AgCl that will precipitate. Finally, we can convert the moles of AgCl to grams of NaCl using the molar mass of NaCl.
First, we calculate the number of moles of AgNO3:
Volume of solution = 25.0 mL = 0.0250 LMolarity of solution = 0.251 MNumber of moles of AgNO3 = volume × molarity = 0.0250 L × 0.251 mol/L = 0.00628 molAccording to the balanced equation, 1 mole of AgNO3 produces 1 mole of AgCl. Therefore, 0.00628 mol of AgNO3 will produce 0.00628 mol of AgCl.
Next, we convert moles of AgCl to grams of NaCl:
Molar mass of AgCl = 143.32 g/molGrams of NaCl = moles of AgCl × molar mass of AgCl = 0.00628 mol × 143.32 g/mol = 0.900 g of NaClTherefore, approximately 0.900 grams of solid NaCl must be added to the 25.0 mL of 0.251 M AgNO3 solution to completely precipitate the silver.
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Consider a galvanic cell composed of the SHE and a half-cell using the following reaction: Ag+(aq) + e− → Ag(s)
(a) Calculate the standard cell potential. E o cell = V
(b) What is the spontaneous cell reaction under standard-state conditions?
(c) Calculate the cell potential when [H+] in the hydrogen electrode is changed to the following concentrations, while all other reagents are held at standard-state conditions:
(i) 4.2 × 10−2 M E = V
(ii) 9.6 × 10−5 M E = V
(d) Based on this cell arrangement, suggest a design for a pH meter.
Answer:
a. 0.80V b. 2Ag⁺(aq) + H2(g) ⇄ 2Ag(s) +2H⁺(aq) c. i) 0.88 ii) 1.03 d. Cell is a ph meter with the potential being a function of hydrogen ion concentration
Explanation:
a. The two half cell reactions are
1. 2H⁺(aq) +2e⁻ → H₂(g) Eanode = 0.00V
2. Ag⁺(aq) + e⁻ → Ag(s) Ecathode = 0.80V
The balanced cell reaction is
2Ag(aq)⁺ + H₂(g) ⇄ 2Ag(s) + 2H⁺(aq)
therefore Ecell = Ecathode - Eanode = 0.80 - 0.00 = +0.80V
b. Since the Ecell is positive, the spontaneous cell reaction under standar state conditions is
2Ag(aq)⁺ + H₂(g) ⇄ 2Ag(s) + 2H⁺(aq)
c. Use Nernst Equation
E = Ecell - (0.0592/n)log([H⁺]/[Ag⁺]²[P H₂]), where n is the number of moles of Ag and P H₂= 1.0 atm
i) E = 0.80 - (0.0592/2)log(4.2x10^-2)/(1.0)²(1.0) = 0.88V
ii) E = 0.80 - (0.0592/2)log(9.6x10^-5)/(1.0)²(1.0) = 1.03V
d . From the above calculation we can conclude that the cell acts as a pH meter as a change in hydrogen ion concentration results in a change in the potential of the cell. A change of ph of 2.64 changes the E of cell by 0.15 V.
At constant pressure, which of these systems do work on the surroundings?
a. 2A(g) + 3B (g) ------> 4C (g)
b. 2A(g) + 2B (g) -------> 5C(g)
c. 2A (g) + B (g) -----> C(g)
d. A(s) + B(g) -------> 2C (g)
Answer:
The system does work on the surroundings.
Explanation:
The work (w) exerted in a chemical reaction depends on the change in the number of gaseous moles (Δn(g)), where
Δn(g) = n(gas, products) - n(gas, reactants)
These variables are linked through the following expression.
w = -R.T.Δn(g)
where,
R is the ideal gas constant
T is the absolute temperature
If Δn(g) > 0, w < 0 and the system does work on the surroundings.If Δn(g) < 0, w > 0 and the surroundings do work on the system.If Δn(g) = 0, w = 0 and no work is done.At constant pressure, which of these systems do work on the surroundings?
a. 2A(g) + 3B (g) ------> 4C (g)
Δn(g) = 4 -5 = -1. The surroundings do work on the system.
b. 2A(g) + 2B (g) -------> 5C(g)
Δn(g) = 5 -4 = 1. The system does work on the surroundings.
c. 2A (g) + B (g) -----> C(g)
Δn(g) = 1 - 3 = -2. The surroundings do work on the system.
d. A(s) + B(g) -------> 2C (g)
Δn(g) = 2 -2 = 0. No work is done.
As the temperature of a reaction is increased, the rate of the reaction increases because the______________. a.reactant molecules collide less frequently and with greater energy per collision b. reactant molecules collide more frequently and with greater energy per collision c. reactant molecules collide less frequently d. activation energy is lowered e. reactant molecules collide more frequently with less energy per collision
Answer:
b. reactant molecules collide more frequently and with greater energy per collision
Explanation:
As the temperature of a reaction is increased, the rate of the reaction increases because the reactant molecules collide more frequently and with greater energy per collision. When temperature is increased there is an increase in the kinetic energy of the molecules. The more the molecules move the more they collide. According to the collision theory there should be enough energy to allow bonds to be formed during a chemical reaction hence the need for greater energy per collision. Also the rate of reaction is directly proportional to the number of collisions that occur.
A substance is found to be nonconductive, to have a relatively low melting point, and to be insoluble in water. This is most likely
A.) A metallic solid
B.) An ionic solid
C.) A molecular solid
D.) A network covalent solid
Answer:
C.) A molecular solid
Explanation:
Molecular solids have low melting points, are non-conductive and are insoluble in water. The interactions between the molecules or atoms can be hydrogen bonds, dipole dipole or London dispersion forces.Because they have relatively weak bonds there are easily vaporized and therefore have a low melting point. Although a network covalent solid is non-conductive it has a high melting point due to the strong covalent bonds. The metallic and the ionic solids are both conductive, with the metallic solids having a high melting point.
A substance that is nonconductive, with a low melting point, and insoluble in water is most likely a molecular solid, due to its weak intermolecular forces and neutral molecular composition.
Explanation:The substance described as nonconductive, having a relatively low melting point, and insoluble in water is most likely C.) A molecular solid. Molecular solids are typically composed of molecules held together by relatively weak intermolecular forces such as London dispersion forces, dipole-dipole interactions, or hydrogen bonds. These characteristics result in their low melting points and the inability to conduct electricity. Since they are made up of neutral molecules, they usually do not dissolve in water which is a polar solvent. On the other hand, metallic solids conduct electricity and are malleable, ionic solids conduct electricity when molten and are usually water soluble, and network covalent solids are typically hard and have high melting points.
The carbon atoms of acetic acid (CH3COOH) exhibit what type of hybridization?A.) spB.) sp^2C.) sp^3D.) sp^2 and sp^3
Answer:
D.) sp² and sp³
Explanation:
Acetic acid CH₃COOH
H3C-----sp³ C(O)OH----sp²
Consider the following thermochemical equation for the combustionof acetone, C3H6O, the main ingredient innail polish remover.C3H6O(l) + 4O2 (g) --->3CO2 (g) + 3H2O (g), ΔHoof the reaction = -1790 kcalIf a bottle of nail polish remover contains 177 mL of acetone,how much heat would be released by its complete combustion? Thedensity of acetone is 0.788 g/mL.
Answer:
-4.30 × 10³ kcal
Explanation:
Let's consider the following thermochemical equation for the combustion of acetone.
C₃H₆O(l) + 4 O₂ (g) → 3 CO₂(g) + 3 H₂O(g) ΔH°of the reaction = -1790 kcal
When 1 mole of C₃H₆O burns, 1790 kcal of heat are released. We have to find out how many moles of C₃H₆O reacted in 177 mL. Considering the density of acetone is 0.788 g/mL and its molar mass is 58.08 g/mol, the moles of acetone are:
[tex]177mL.\frac{0.788g}{mL} .\frac{1mol}{58.08g} =2.40mol[/tex]
The heat released when 2.40 moles of acetone burn is:
[tex]2.40mol.\frac{(-1790kcal)}{1mol} =-4.30 \times 10^{3} kcal[/tex]
if a chlorine atom were to attract an electron from sodium, the sodium atom would become blank charged.
When a chlorine atom attracts an electron from sodium, the sodium atom becomes positively charged, forming a sodium cation with a +1 charge, while chlorine becomes a chloride ion with a -1 charge, resulting in the formation of NaCl.
Explanation:If a chlorine atom were to attract an electron from sodium, the sodium atom would become positively charged. This occurs because chlorine has a high affinity for electrons due to its seven valence electrons, and it is more energy-efficient for chlorine to gain one electron than to lose seven. When chlorine gains an extra electron, it becomes a chloride ion with a net negative charge. Conversely, when sodium loses its single valence electron, it becomes a sodium ion with a +1 charge, also known as a cation.
The transfer of an electron from sodium to chlorine results in the formation of two oppositely charged ions that are held together by an ionic bond, creating the ionic compound NaCl. This electron transfer satisfies the octet rule for both ions, resulting in complete outermost shells with stable electron configurations.
Lauryl alcohol is a nonelectrolyte obtained from coconut oil and is used to make detergents. A solution of 6.80 g of lauryl alcohol in 0.200 kg of benzene freezes at 4.5 ∘C. What is the approximate molar mass of lauryl alcohol?
What is the approximate molar mass of lauryl alcohol?
Answer:
The approximate molar mass of lauryl alcohol is 174.08 g/m
Explanation:
An excersise to apply the colligative property of Freezing-point depression.
This is the formula: ΔT = Kf . m
First of all, think the T° of fusion of benzene → 5.5°C
ΔT = T° pure solvent - T° fusion solution
Kf for benzene: 5.12 °C/m
5.5°C - 4.5°C = 5.12 °C /m . m
1°C / 5.12 m /°C = m
0.195 m = molality
This moles of lauryl alcohol, solute, are in 1 kg of benzene, solvent.
I have to find out in 0.2 kg.
1 kg sv ____ 0.195 moles solute
0.2 kg sv ____ (0.195 . 0.2)/1 = 0.039 moles solute
The mass for these moles is 6.80 g, so if I want to know the molar mass, I have to divide mass / moles
6.80 g/ 0.039 moles = 174.08 g/m
In the industrial "chlor-alkali" process, pure chlorine and sodium hydroxide are produced by electrolyzing brine, essentially an aqueous solution of sodium chloride.
Suppose a current of 18.0 A is passed through an aqueous solution of NaCl for 67.0 seconds.
Calculate the mass of pure chlorine produced.
Be sure your answer has a unit symbol and the correct number of significant digits.
Answer:
0.443 g
Explanation:
In the electrolysis of an aqueous solution of NaCl, the following half-reactions take place:
Reduction: Na⁺(aq) + 1 e⁻ ⇒ Na(s)
Oxidation: 2 Cl⁻(aq) ⇒ Cl₂(g) + 2 e⁻
Let's consider the following relations:
1 A = 1 c/s1 mole of e⁻ has a charge of 96468 c (Faraday's constant)1 mole of Cl₂(g) is released when 2 moles of e⁻ circulateThe molar mass of Cl₂ is 70.9 g/molSuppose a current of 18.0 A is passed through an aqueous solution of NaCl for 67.0 seconds. The mass of chlorine produced is:
[tex]67.0s.\frac{18.0c}{s} .\frac{1mole^{-} }{96468c} .\frac{1molCl_{2}}{2mole^{-} } .\frac{70.9gCl_{2}}{1molCl_{2}} =0.443gCl_{2}[/tex]
Using Faraday's law of electrolysis, we can calculate the mass of chlorine produced by an 18.0 A current over 67.0 seconds in the chlor-alkali process. The result is approximately 0.44 grams.
Explanation:To determine the mass of chlorine produced in the chlor-alkali process using a current of 18.0 A for a duration of 67.0 seconds, we need to use Faraday's law of electrolysis. This states that the amount of a substance produced at an electrode during electrolysis is directly proportional to the quantity of electricity that passes through the solution.
The amount of a substance produced can be calculated using the formula: It = zF, where I is the current in amperes, t is the time in seconds, z is the ionic charge(1 for Cl-), and F is Faraday’s constant (96500 coulombs per mole of electrons).
First, let’s find the quantity of electric charge (Q) that has passed through the solution using the formula Q = It. In this case, I = 18 A and t = 67.0 seconds, so Q = 18 * 67.0 = 1206 Coulombs.
Now we can use this Q in the formula of m = QM/zF, where M is the molar mass of chlorine, which is approximately 35.5 g/mol. After plugging the numbers in, we find the mass (m) of the chlorine produced is 0.44 g.
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Are the bonds in each of the following substances ionic, nonpolar covalent, or polar covalent? (a)KCl (b)P4 (c)BF3 (d)SO2 (e)Br2 (f)NO2 For those substances with polar covalent bonds, which has the least polar bond? For those substances with polar covalent bonds, which has the most polar bond?
Answer:
(a) Ionic
(b) Nonpolar covalent
(c) Polar covalent
(d) Polar covalent
(e) Nonpolar covalent
(f) Polar covalent
For those substances with polar covalent bonds, which has the least polar bond? NO₂
For those substances with polar covalent bonds, which has the most polar bond? BF₃
Explanation:
Are the bonds in each of the following substances ionic, nonpolar covalent, or polar covalent?
The nature of a bond depends on the modulus of the difference of electronegativity (|ΔEN|) between the atoms that form it.
If |ΔEN| = 0, the bond is nonpolar covalent.If 0 < |ΔEN| ≤ 2, the bond is polar covalent.If |ΔEN| > 2, the bond is ionic.(a) KCl |ΔEN| = |EN(K) - EN(Cl)| = |0.8 - 3.0| = 2.2. The bond is ionic.
(b) P₄ |ΔEN| = |EN(P) - EN(P)| = |2.1 - 2.1| = 0.0. The bond is nonpolar covalent.
(c) BF₃ |ΔEN| = |EN(B) - EN(F)| = |2.0 - 4.0| = 2.0. The bond is polar covalent.
(d) SO₂ |ΔEN| = |EN(S) - EN(O)| = |2.5 - 3.5| = 1.0. The bond is polar covalent.
(e) Br₂ |ΔEN| = |EN(Br) - EN(Br)| = |2.8 - 2.8| = 0.0. The bond is nonpolar covalent.
(f) NO₂ |ΔEN| = |EN(N) - EN(O)| = |3.0 - 3.5| = 0.5. The bond is polar covalent.
The nature of bonding between atoms depends on the electro negativity between the atoms in the bond.
An ionic bond is formed by transfer of electrons from one atom to another. It commonly occurs between a metal and a nonmetal. Covalent bonds are formed when electrons are shared between bonding atoms. If the electronegativity difference between the two bonding atoms is small, electrons are equally shared in the molecule and the bonds are nonpolar. However, when the difference in electro negativity is significant (about 0.5) a significant magnitude of polarity of the bond is observed and the bond us polar because electrons lie closer to the more electronegative atom.
The classification of the compounds according to nature of bonding between atoms is done as follows;
KCl - ionic bond
P4 - nonpolar covalent bond
BF3 - polar covalent bond
SO2 - polar covalent bond
Br2 - nonpolar covalent
NO2 - polar covalent bond.
The most polar bond occurs between bromine and bromine because of a large electro negativity difference between the both atoms.
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Assume 0.18 L of a 5.0 M solution of lead (II) nitrate, Pb(NO3)2, reacts with a 2.6 M solution of sodium phosphate, Na3PO4, to produce lead (II) phosphate, Pb3(PO4)2, and sodium nitrate, NaNO3. The problem requires that you determine the volume of sodium phosphate, Na3PO4, needed for the reaction to occur.
Answer:
0.23 L
Explanation:
Let's consider the following balanced equation.
3 Pb(NO₃)₂(aq) + 2 Na₃PO₄(aq) ⇄ Pb₃(PO₄)₂(s) + 6 NaNO₃(aq)
The moles of Pb(NO₃)₂ are:
[tex]0.18L\times \frac{5.0mol}{L} =0.90mol[/tex]
The molar ratio of Pb(NO₃)₂ to Na₃PO₄ is 3:2. The moles of Na₃PO₄ are:
[tex]0.90molPb(NO_{3})_{2}.\frac{2molNa_{3}PO_{4}}{3molPb(NO_{3})_{2}} =0.60molNa_{3}PO_{4}[/tex]
The volume of Na₃PO₄ required is:
[tex]\frac{0.60mol}{2.6mol/L} =0.23L[/tex]
White phosphorus is composed of tetrahedral molecules of P4 in which every P atom is connected to three other P atoms. In the Lewis structure of P4, there are 1. 6 bonding pairs and 4 lone pairs of electrons. 2. 6 bonding pairs and no lone pairs of electrons. 3. 5 bonding pairs and 4 lone pairs of electrons. 4. 3 bonding pairs and 4 lone pairs of electrons. 5. 6 bonding pairs and 2 lone pairs of electrons.
Answer:
In the Lewis structure of P4 there are 6 bonding pairs and 4 lone pairs of electrons.
Explanation:
The structure of tetrahedral molecule of P4 is provided below.
Each phosphorus atom has 5 valence electrons out of which 3 electrons involve in bonding and the rest 2 electrons exist as a lone pair that does not involve in bonding.Hence each phosphorus atom has one lone pair.In P4 molecule there are phosphorus atoms and hence 4 lone pairs in total.
As you can see in the figure, each phosphorus atom is bonded to the other three atoms.A bond is formed when two atoms share one electron each and the pair is called bonding pair.
From the figure we can see that there are 6 bonds in total.Each bond consist of one bonding pair of electrons and hence in total there are 6 bonding pairs of electrons.
Hence in a P4 molecule there are six bonding pairs and 4 lone pairs of electrons.
A chemical reaction occurring in a cylinder equipped with a moveable piston produces 0.621 mol of a gaseous product.
If the cylinder contained 0.120 mol of gas before the reaction and had an initial volume of 2.18 L, what was its volume after the reaction?
(Assume constant pressure and temperature and that the initial amount of gas completely reacts.)
Answer:
11.3 L
Explanation:
Initially, the cylinder had n₁ = 0.120 mol of gas in an initial volume V₁ = 2.18 L. At the end, it had n₂ = 0.621 mol in an unknown volume V₂. According to the Avogadro's law, in the same conditions of pressure and temperature, the volume is directly proportional to the number of moles.
[tex]\frac{V_{1}}{n_{1}} =\frac{V_{2}}{n_{2}} \\V_{2}=\frac{V_{1}\times n_{2} }{n_{1}} =\frac{2.18L \times 0.621mol}{0.120mol} =11.3L[/tex]
1) A certain amount of chlorine gas was placed inside a cylinder with a movable piston at one end. The initial volume was 3.00 L \rm L and the initial pressure of chlorine was 1.45atm The piston was pushed down to change the volume to 1.00 LCalculate the final pressure of the gas if the temperature and number of moles of chlorine remain constant..
2) In an air-conditioned room at 19.0C a spherical balloon had the diameter of 50.0 cm When taken outside on a hot summer day, the balloon expanded to 51.0cm cm in diameter. What was the temperature outside? Assume that the balloon is a perfect sphere and that the pressure and number of moles of air molecules remains the same.
Answer:
1) 4.35 atm
2) 36.88 °C
Explanation:
1) Because the temperature and number of moles remained constant, we can use the formula P₁V₁=P₂V₂
3.00 L * 1.45 atm = P₂ * 1.00 L
P₂ = 4.35 atm
2) First we use a geometrical formula to calculate the volume of the spherical balloon when it has a diameter of 50.0 and of 51.0 cm.
V₁ = 4/3 * π*(50/2)³ = 65449.85cm³
V₂ = 4/3 * π*(51/2)³ = 69455.90cm³
Then we use T₁V₂=T₂V₁, keeping in mind using Kelvin as the unit for temperatures:
292.16 K * 69455.90cm³ = T₂ * 65449.85cm³
T₂ = 310.04 K = 36.88 °C
The final pressure of the chlorine gas in the cylinder is 4.35 atm as per Boyle's Law. The temp on a hot summer day outside can be found using Charles's Law after a series of steps starting with converting the diameter to radius then to volume, adjusting for Kelvin, and then solving for T2 with Charles's Law.
Explanation:The subjects of these problems pertain to gas laws, specifically, Boyle's Law and Charles's Law.
1) According to Boyle's Law, the pressure and volume of a gas have an inverse relationship when the temperature and the number of moles remain constant. Thus, if the volume is decreased, the pressure should increase. We can calculate the final pressure with the formula P1*V1 = P2*V2. Substituting the given figures: 1.45 atm * 3.00 L = P2 * 1.00 L. Solving for P2, the final pressure of the gas is 4.35 atm.
2) With Charles's Law, the volume and temperature of a gas have a direct relationship when pressure and the number of moles remain constant. Converting the diameters to radii (25.0 cm to 26.0 cm) and using the volume formula of a sphere, you find the volume before and after. The formula for Charles's Law is V1/T1 = V2/T2. However, all temperatures need to be in Kelvin, so 19.0 C converts to 292.15 K. Substituting the calculated volumes and temperatures, solve for T2. This will give you the temperature outside in Kelvin. Convert to Celsius for a meaningful interpretation.
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the value of delta G at 141.0 degrees celsius for the formation of phosphorous trichloride from its constituent elements,
P2(g) + 3Cl2(g) ------> 2PCl3(g)
is __________ kj/mol. At 25.0 degrees Celsius for this reaction, delta H is -720.5 kj/mol, delta G is -642.9 kj/mol, and delta S is -263.7 J/K.
A.) -829.7
B.) 1.08 x 10^5
C.) 3.65 x 10^4
D.) -683.3
E.) -611.3
Answer:
The correct answer is option E.
Explanation:
The Gibbs free energy is given by expression:
ΔG = ΔH - TΔS
ΔH = Enthalpy change of the reaction
T = Temperature of the reaction
ΔS = Entropy change
We have :
ΔH = -720.5 kJ/mol = -720500 J/mol (1 kJ = 1000 J)
ΔS = -263.7 J/K
T = 141.0°C = 414.15 K
[tex]\Delta G = -720500 J/mol - (414.15 K\times (-263.7 J/K))[/tex]
[tex]= -611,288.64 J/mol = -611.28 kJ/mol\approx -611.3 kJ/mol[/tex]
The Gibb's free energy of the given reaction at 141.0°C is -611.3 kJ/mol.
Using the temperature in Kelvin and the given values, ΔG is calculated to be -611.3 kJ/mol, corresponding to option E.
To determine the Gibbs free energy (ΔG) at 141.0 degrees Celsius for the reaction involving the formation of phosphorous trichloride (PCl3) from its elements, we can use the following formula:
ΔG = ΔH - TΔS
Given data at 25.0 degrees Celsius (298.15 K):
ΔH = -720.5 kJ/mol
ΔG = -642.9 kJ/mol
ΔS = -263.7 J/K = -0.2637 kJ/K
Convert temperature from Celsius to Kelvin for 141.0 degrees Celsius: T = 141.0 + 273.15 = 414.15 K
Now calculate ΔG at 414.15 K:
ΔG = ΔH - TΔS = -720.5 kJ/mol - (414.15 K * -0.2637 kJ/K) = -720.5 kJ/mol + 109.19 kJ/mol = -611.31 kJ/mol
The closest answer choice is -611.3 kJ/mol which aligns with option E.
Rank the following in order of decreasing acid strength in aqueous solution: HCl, HOCl, HOBr,HOI.
A) HCl > HClO > HBrO > HIO
B) HIO > HBrO > HClO > HCl
C) HCl > HIO > HBrO > HClO
D) HClO > HCl > HBrO > HIO
E) HClO > HBrO > HCl > HIO
The correct order of decreasing acid strength in an aqueous solution among HCl, HOCl, HOBr, and HOI is HCl > HIO > HBrO > HClO. This order is based on the strength of the H-X bond and the stability of the X- ion, where HCl is the strongest acid.
Explanation:The correct order of decreasing acid strength in aqueous solution: HCl, HOCl, HOBr, and HOI, is option C: HCl > HIO > HBrO > HClO. This order is determined by two main factors: the strength of the H-X bond and the stability of the X-ion. Hydrochloric acid (HCl) is the strongest because it can easily donate protons in solution. The other compounds are oxyacids, and their strength increases with increasing electronegativity and size of the halogen attached to oxygen. So, when comparing HOCl, HOBr, and HOI, the more polarizable (or larger in size) the halogen, the stronger the acid. Iodine is the largest here and thus HOI is more acidic than HOBr and HOCl.
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An atom of silver has a radius of and the average orbital speed of the electrons in it is about . Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of silver. Write your answer as a percentage of the average speed, and round it to significant digits.
Answer:
Percentage of uncertainty in average speed of an electron is 0.1756%.
Explanation:
Using Heisenberg uncertainty principle:
[tex]\Deltax\times \Delta p\geq \frac{h}{4\pi }[/tex]
[tex]\Delta p=m\times \Delta v[/tex]
[tex]\Deltax\times m\times \Delta v\geq \frac{h}{4\pi }[/tex]
Δx = Uncertainty in position
Δp = Uncertainty in momentum
Δv = Uncertainty in average speed
h = Planck's constant = [tex]6.626\times 10^{-34} kg m^2/s[/tex]
m = mass of electron = [tex]9.1\times 10^{-31} kg[/tex]
We have
Δx = 2 × 165 pm = 330 pm = [tex]3.3\times 10^{-10} m [/tex]
[tex]1 pm = 10^{-12} m[/tex]
Average orbital speed of electron = v = [tex]=1.0\times 10^8 m/s[/tex]
[tex]3.3\times 10^{-10} 9.1\times 10^{-31} kg \times \Delta v\geq \frac{6.626\times 10^{-34} kg m^2/s}{4\pi }[/tex]
[tex]\Delta v\geq \frac{6.626\times 10^{-34} kg m^2/s}{4\pi \times 3.3\times 10^{-10}\times 9.1\times 10^{-31} kg}[/tex]
[tex]\Delta v\geq 1.756\times 10^5 m/s[/tex]
Percentage of uncertainty in average speed:
[tex]=\frac{\Delta v}{v}\times 100[/tex]
[tex]=\frac{1.756\times 10^5 m/s}{1.0\times 10^8 m/s}\times 100=0.1756\%[/tex]