“Vapor pressure increases with temperature” is the true statement from the given statements.
Option: A
Explanation:
Vapor pressure exerted in “thermodynamic equilibrium” with solid or liquid phase called “condensed phase” at a given temperature in packed or closed system. “Liquid particles” are arranged with more inter molecular space than solid and not as fixed as solid. Therefore when temperature is increased “kinetic energy” of the molecules also increases and thereby molecules transitioning into vapor also increases and in this way whole process is responsible for increase in “vapor pressure”.
Final answer:
The statement that vapor pressure increases with temperature is true because higher temperatures lead to increased kinetic energies, causing more molecules to escape into the vapor phase.
Explanation:
The correct statement among those listed is that vapor pressure increases with temperature. As the temperature rises, the kinetic energy of the molecules increases, leading to a greater number of molecules having enough energy to escape the liquid phase and enter the vapor phase, thereby increasing the vapor pressure. On the other hand, hydrogen bonds are not chemical bonds; they are strong intermolecular forces that occur between molecules, not within them as chemical bonds do.
Dispersion forces are generally weaker than dipole-dipole forces, and they increase with the mass and size of the molecules involved due to increased polarizability. Finally, intermolecular forces are responsible for the attractions between molecules, not for holding the atoms within a molecule together; that role is filled by intramolecular forces or chemical bonds.
Which of the following statements explain why the van der Waals equation must be used to describe real gases? X. interactions between gas molecules reduces the temperature of the gas in the sample Y. the non-zero volumes of gas particles effectively decrease the amount of "empty space" between them Z. the molecular attractions between particles of gas decreases the pressure exerted by the gas
Answer:
Statements Y and Z.
Explanation:
The Van der Waals equation is the next one:
[tex] nRT = (P + \frac{an^{2}}{V^{2}})(V -nb) [/tex] (1)
The ideal gas law is the following:
[tex] nRT = PV [/tex] (2)
where n: is the moles of the gas, R: is the gas constant, T: is the temperature, P: is the measured pressure, V: is the volume of the container, and a and b: are measured constants for a specific gas.
As we can see from equation (1), the Van der Waals equation introduces two terms that correct the P and the V of the ideal gas equation (2), by the incorporation of the intermolecular interaction between the gases and the gases volume. The term an²/V² corrects the P of the ideal gas equation since the measured pressure is decreased by the attraction forces between the gases. The term nb corrects the V of the ideal gas equation, taking into account the volume occuppied by the gas in the total volume, which implies a reduction of the total space available for the gas molecules.
So, the correct statements are the Y and Z: the non-zero volumes of the gas particles effectively decrease the amount of "empty space" between them and the molecular attractions between gas particles decrease the pressure exerted by the gas.
Have a nice day!
24.711 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 79.741 g of water. A 11.169 g aliquot of this solution is then titrated with 0.1045 M HCl . It required 28.42 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.
Final answer:
The weight percent of NH3 in the aqueous waste was calculated from the titration data to be 0.2047%. The moles of NH3 were found to be 0.0029711 mol, corresponding to 0.050585g NH3 in the original waste sample.
Explanation:
The weight percent of NH3 in the aqueous waste is calculated using titration analysis. First, determine the moles of HCl that reacted:
# mol HCl = 0.02842 L × 0.1045 M = 0.0029711 mol HCl.
The reaction between HCl and NH3 is:
NH3(aq) + HCl(aq) → NH4Cl(aq).
Hence, 1 mole of HCl reacts with 1 mole of NH3. Therefore, the moles of NH3 in the aliquot are also 0.0029711 mol. To find the mass of NH3:
Mass of NH3 = moles × molar mass
= 0.0029711 mol × 17.031 g/mol
= 0.050585 g NH3.
Since the aliquot is part of the diluted waste, and assuming the ammonia is evenly distributed, we calculate the weight percent in the original waste sample:
Weight percent = (mass of NH3 in aliquot / mass of waste sample) × 100
= (0.050585 g / 24.711 g) × 100
= 0.2047%.
Draw the product of nucleophilic substitution with each neutral nucleophile. When the initial substitution product can lose a proton to form a neutral product, draw the product after proton transfer
Answer:
Neutral nucleophile are: H2O, CH3OH, NH3, RNH2, R2NH, R3N, RCOOH, RSH and PR3. The products by nucleophilic substitution are diverse depending on the different nucleophiles, obtaining alcohol, eter, amines, ester and tioeter considering only the nucleophiles with a hydrogen available.
Explanation:
Please see the images attached.
Nucleophilic subtitution with water occurs under Sn1 mechanism. That's it because water as nucleophile is so weak. With the other neutral nucleophiles, the reaction occur under Sn2 mechanism.
RSH + CH3I -----> RSCH3 + HI
Classify each statement as true or false. Drag the appropriate items to their respective bins. View Available Hint(s) Reset Help The valence electrons of group 4A elements are in the 5s subshell. Period 3 elements have an inner electron configuration of [Ar). The highest principal quantum number of period 4 elements is 4 Period 5 elements have six 4p electrons Period 5 elements have an inner electron configuration of (Kr Group 8A elements have full outer principal s and p subshells. The valence electrons of group 3A elements are in an s and p subshell. The highest principal quantum number of period 4 elements is 5 False True Submit
Explanation:
1) The valence electrons of group 4A elements are in the 5s subshell.
The given statement is false .
The valence electrons of group 4A elements are in the ns and np subshell .Where n is equal to 1 to 7.
2) Period 3 elements have an inner electron configuration of [Ar].
The given statement is false .
The noble gas which comes before third period is neon.Period 3 elements have an inner electron configuration of [Ne].
3)The highest principal quantum number of period 4 elements is 4.
The given statement is true.
The period number of periodic table indicates the value of highest principal quantum number.
4) Period 5 elements have six 4p electrons.
The given statement is true.
This so because 4p subshell is filled before filling of shells of fifth shell. So, the 4p subshell will obviously have 6 electrons.
5) Period 5 elements have an inner electron configuration of (Kr).
The given statement is true.
The noble gas which comes before third period is krypton .
6) Group 8A elements have full outer principal s and p subshells.
The given statement is true.
This is because the the group mentioned in the statement if noble gases and noble gases are inert in nature due to fully filled electronic configurations.
7) The valence electrons of group 3A elements are in an s and p subshell.
The given statement is true.
The valence electrons of group 3A elements are in the ns and np subshell .Where n is equal to 1 to 7. there general electronic configuration is [tex]ns^2np^1[/tex].
8) The highest principal quantum number of period 4 elements is 5
The given statement is false.
The period number of periodic table indicates the value of highest principal quantum number.
The electron address in an atom can be given with the help of quantum mechanics. It helps to locate electrons with quantum numbers.
(1) The valence electrons of group 4A elements are in the 5s subshell.
The ns in group 4A can vary from 1 to 7. Thus the statement is false.
(2) Period 3 elements have an inner electron configuration of [Ar).
The nearest noble gas has been Neon, thus the inner electron configuration is of [Ne]. The statement is false.
(3) The highest principal quantum number of period 4 elements is 4.
The period number has been equal to the highest principal quantum number. Thus, the statement is true.
(4) Period 5 elements have six 4p electrons.
It has 5 subshells, and 4 subshells have been filled prior to 5s. Thus, the elements have a 4p orbital filled with 6 electrons. The statement is true.
(5) Period 5 elements have an inner electron configuration of (Kr).
The nearest noble gas has been Krypton, thus the inner electron configuration is of [Kr]. The statement is true.
(6) Group 8A elements have full outer principal s and p subshells.
Group 8A belongs to a noble gas. The subshells of noble gas are completely filled. The statement is true.
(7) The valence electrons of group 3A elements are in an s and p subshell.
The electrons of the 3A group have a general configuration [tex]\rm ns^2\;np^1[/tex]. The electrons are filled in s and p subshell. The statement is true.
(8) The highest principal quantum number of period 4 elements is 5.
The period number has been equal to the highest principal quantum number. Thus, the statement is false.
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What is the charge on each of the following complex ions? hexaaquairon(II), [Fe(H2O)6]? tris(carbonato)aluminate(III), [Al(CO3)3]? diaquatetrachlorovanadate(III), [V(H2O)2Cl4]? Express the charges numerically, in the order that the complexes are listed, separated by commas. For a nonzero charge, be sure to include the sign (e.g., +1,-2,+3). View Available Hint(s)
Answer:
The charges on the given complexes are: +2, (-3), (-1)
Explanation:
A coordination complex is composed of a central metal atom or metal ion and ligands, bonded by coordinate covalent bonds.
The total charge of the complex is equal to the sum of the charge on central metal and the total charge of the ligands.
1. hexaaquairon(II), [Fe(H₂O)₆]
The oxidation state or charge on the central metal ion, Fe = +2
Charge on the ligand water molecule H₂O = 0
Therefore, the total charge on the complex = +2 + (0 × 6) = +2
2. tris(carbonato)aluminate(III), [Al(CO₃)₃]
The oxidation state or charge on the central metal ion, Al = +3
Charge on the ligand carbonate CO₃²⁻ = (-2)
Therefore, the total charge on the complex = +3 + (-2 × 3) = (-3)
3. diaquatetrachlorovanadate(III), [V(H₂O)₂Cl₄]
The oxidation state or charge on the central metal ion, V = +3
Charge on the ligand water molecule H₂O = 0
Charge on the ligand chloride Cl = (-1)
Therefore, the total charge on the complex = +3 + [(0 × 2)+(-1 × 4)] = +3 + (-4) = (-1)
Therefore, the charge on 1. hexaaquairon(II), [Fe(H₂O)₆] complex is +2
2. tris(carbonato)aluminate(III), [Al(CO₃)₃] complex is (-3)
3. diaquatetrachlorovanadate(III), [V(H₂O)₂Cl₄] complex is (-1)
When 551. mg of a certain molecular compound X are dissolved in 100 g of benzonitrile (CH,CN), the freezing point of the solution is measured to be 13.4 °C. Calculate the molar mass of X. If you need any additional information on benzonitrile, use only what you find in the ALEKS Data resource. Also, be sure your answer has a unit symbol, and is rounded to the correct number of significant digits.
To find the molar mass of compound X, we use the formula moles of solute = (m)(mass of solvent), and then divide the mass of the solute by the moles of solute to calculate the molar mass. In this case, the molar mass of compound X is 5510 g/mol.
Explanation:To find the molar mass of compound X, we need to use the formula:
Molality (m) = moles of solute / mass of solvent (in kg)
In this case, we know the molality (m) = 1.0 m, and the mass of the solvent (benzonitrile) = 100 g = 0.1 kg. We can rearrange the formula to solve for the moles of solute:
moles of solute = (m)(mass of solvent)
moles of solute = (1.0 m)(0.1 kg) = 0.1 mol
To calculate the molar mass, we divide the mass of the solute by the moles of solute:
Molar mass = mass of solute / moles of solute
Molar mass = 551 mg / 0.1 mol = 5510 g/mol
Therefore, the molar mass of compound X is 5510 g/mol.
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Hydrogen peroxide can be prepared by the reaction of barium peroxide with sulfuric acid according to the reaction BaO 2 ( s ) + H 2 SO 4 ( aq ) ⟶ BaSO 4 ( s ) + H 2 O 2 ( aq ) BaO2(s)+H2SO4(aq)⟶BaSO4(s)+H2O2(aq) How many milliliters of 3.00 M H 2 SO 4 ( aq ) 3.00 M H2SO4(aq) are needed to react completely with 92.5 g BaO 2 ( s ) ? 92.5 g BaO2(s)?
Answer:
There is 182 mL of 3.00 M H2SO4 needed
Explanation:
Step 1: Data given
Mass of BaO2 = 92.5 grams
Molarity of 3.00 M H2SO4
Step 2: The balanced equation
BaO2 ( s ) + H2SO4(aq ) ⟶ BaSO4(s) + H2O2(aq )
For 1 mole of BaO2 we need 1 mole of H2SO4 to produce 1 mole of BaSO4 and 1 mole H2SO2
Step 3: Calculate moles of BaO2
Moles BaO2 = mass BaO2 / Molar mass BaO2
Moles BaO2 = 92.5 / 169.33 g/mol
Moles BaO2 = 0.546 moles
Step 4: Calculate moles of H2SO4
For 1 mole BaO2 we need 1 mole of H2SO4
For 0.546 moles of BaO2 we need 0.546 moles of H2SO4
Step 5: Calculate the volume of H2SO4 needed
Volume = moles / molarity
Volume = 0.546 mol/ 3.00 mol/L
Volume = 0.182 L = 182 mL
There is 182 mL of 3.00 M H2SO4 needed
An automobile engine provides 527 Joules of work to push the pistons. In this process the internal energy changes by -2886 Joules. Calculate q for the engine. This represents the amount of heat that must be carried away by the cooling system. q = Joules
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:
Determine the volume fraction of fibers required to obtain a longitudinal tensile strength of 812 MPa in a unidirectional carbon fiber reinforced composite. You are given that the tensile strength of the carbon fibers are 1240 MPa and that the stress in the matrix at fiber failure will be 300 MPa. Answer Format X.XX Unit: Unitless (fraction, example: 50 vol% fiber would be entered as 0.50)
Answer:
V= 0.544
Explanation:
Given that
σ=812 MPa
σ₁=1240 MPa
σ₂=300 MPa
Lets take volume fraction V
The strength of composite fiber given as
σ = σ₁ V + σ₂(1-V)
By putting the values
812 = 1240 V + 300 (1-V)
812 = 1240 V +300 - 300 V
812 - 300 = 1240 V- 300 V
512 = 940 V
V= 0.544
The percentage volume = 54.4 %
When 4.21 grams of potassium hydroxide are added to 250 mL of water in a coffee cup calorimeter, the temperature rises by 4.14°C.
Assume that the density and specific heat of the dilute aqueous solution are the same as those of H2O and calculate the molar heat of solution of potassium hydroxide in kJ/mol. C= 4.184 J/g*K
Answer:
The molar heat of solution of potassium hydroxide = 57.7 kJ /mol
Explanation:
Step 1: Data given
Mass of potassium hydroxide = 4.21 grams
Volume of water = 250 mL
Temperature rise = 4.14 °C
Density = 1g/mL
Specific heat = 4.184 J/g°C
Step 2: Calculate the heat absorbed by water
q = m*c*ΔT
⇒ with m = the mass of water = 250 grams
⇒ with c= the specific heat of the solution = 4.184 J/g°C
⇒ with ΔT = 4.14 °C
q = 250 * 4.184*4.14 = 4330.4 J
Step 3: Calculate moles of KOH
Moles KOH = Mass KOH / Molar mass KOH
Moles KOH = 4.21 grams / 56.106 g/mol
Moles KOH = 0.075 moles
Step 4: Calculate molar heat of solution
4330.4 J / 0.075 moles = 57738.7 j/mol = 57.7 kJ /mol
(Note that the enthalpy change for the reaction is negative because the reaction is exothermic)
The molar heat of solution of potassium hydroxide = 57.7 kJ /mol
Final answer:
To calculate the molar heat of solution of potassium hydroxide, use the formula q = m × C × ΔT and divide the heat absorbed by the moles of potassium hydroxide.
Explanation:
To calculate the molar heat of solution of potassium hydroxide, we first need to find the amount of heat absorbed by the solution. This can be done using the formula:
q = m × C × ΔT
Where q is the heat absorbed, m is the mass of the solution, C is the specific heat capacity, and ΔT is the change in temperature. Rearranging the formula, we get:
q = molar heat of solution × moles of potassium hydroxide
Now, we can solve for the molar heat of solution by dividing the heat absorbed by the moles of potassium hydroxide.
A buret is filled with 0.1517 M A 25.0 mL portion of an unknown acid and two drops of indicator are added to an Erlenmeyer flask, and the titration experiment is carried out. If the initial buret reading was 0.55 mL, and the buret reading at the end point was 22.50 mL, what is the molarity of the unknown acid, HA?
Answer:
Molarity of Unknown Acid = 0.1332 M
Explanation:
Data for solving problem:
Molarity of base in buret (M₁)= 0.1517 M
volume of the acid in Erlenmeyer flask (V₂)= 25.0 mL
Volume of the base in the buret (V₁) = final volume of buret - initial volume in buret
final volume of buret = 22.5 mL
initial volume in buret = 0.55 mL
So
Volume of the base in the buret (V₁) = 22.5 mL -0.55 mL = 21.95 mL
Volume of the base in the buret (V₁) = 21.95 mL
Molarity of Unknown acid in the Erlenmeyer flask (M₂) = To be find
Explanation:
It is acid base titration and formula for this titration is as follows:
Molarity of base x Volume of base = Molarity of acid x volume of acid
it can be written as
M₁V₁ = M₂V₂ -------------------- equation (1)
we have to find M₂
so by rearrangment the equation (1)
M₁V₁ / V₂ = M₂ ------------------ equation (2)
put the values in equation in equation (2)
M₂ = 0.1517 M x 21.95 mL / 25.0 mL
M₂ = 3.3298 /25.0
M₂ = 0.1332 M
so the Molarity of Unknown acid is 0.1332 M
The molarity of the unknown acid HA is calculated to be 0.1331 M by first determining the moles of NaOH that reacted and then using the volume of acid to find its concentration.
To determine the molarity of the unknown acid HA in a titration experiment, first, we calculate the volume of 0.1517 M NaOH used by subtracting the initial buret reading from the final buret reading: 22.50 mL - 0.55 mL = 21.95 mL. This volume is then converted to liters by dividing by 1000: 21.95 mL/1000 = 0.02195 L. Next, we use the molarity of the NaOH solution to find the moles of NaOH that have reacted: 0.1517 M × 0.02195 L = 0.003328665 moles.
Assuming a 1:1 mole ratio between NaOH and HA in the reaction, the moles of unknown acid HA that reacted are also 0.003328665. We then calculate the molarity of the unknown acid, by dividing moles of acid by the volume of acid in liters (25.0 mL = 0.025 L): 0.003328665 moles / 0.025 L = 0.1331466 M.
The molarity of the unknown acid HA is therefore 0.1331 M, rounded to four significant figures to match the precision of the given data.
A sample of hydrogen was collected by water displacement at 23.0°C and an atmospheric pressure of 735 mmHg. Its volume is 568 mL. After water vapor is removed, what volume would the hydrogen occupy at the same conditions of pressure and temperature? (The vapor pressure of water at 23.0°C is 21 mmHg.)
A) 509 mL
B) 539 mL
C) 552 mL
D) 568 mL
E) 585 mL
Answer:
[tex]V_f=552 mL[/tex]
Explanation:
Initially:
Total pressure: 735 mmHg
Water vapor pressure: 21 mmHg
Hydrogen pressure: 714 mmHg
(This is because the total pressure is divided between both gases)
When water vapor is removed:
Total pressure: 735 mmHg
Hydrogen pressure: 735 mmHg
Assuming ideal gases:
Boyle-Mariotte's Law:
[tex]P_i*V_i = P_f * V_f[/tex]
[tex]V_f=\frac{P_i*V_i}{P_f}[/tex]
[tex]V_f=\frac{714 mmHg*568 mL}{735 mmHg}[/tex]
[tex]V_f=552 mL[/tex]
Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing copper and silver. Cu ( s ) ∣ ∣ Cu 2 + ( aq ) ∥ ∥ Ag + ( aq ) ∣ ∣ Ag ( s ) anode: cathode: net cell reaction:
Answer: The half reactions and net cell reaction of the cell is written below.
Explanation:
The given cell is:
[tex]Cu(s)/Cu^{2+}(aq.)||Ag^{+}(aq.)/Ag(s)[/tex]
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
Half reactions for the given cell follows:
Oxidation half reaction (anode): [tex]Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-[/tex]
Reduction half reaction (cathode): [tex]Ag^{+}(aq.)+e^-\rightarrow Ag(s)[/tex] ( × 2)
Net cell reaction: [tex]Cu(s)+2Ag^+(aq.)\rightarrow Cu^{2+}(aq.)+2Ag(s)[/tex]
Hence, the half reactions and net cell reaction of the cell is written above.
The half-reactions at each electrode and the net cell reaction for the electrochemical cell containing copper and silver are provided.
Explanation:The half-reactions at each electrode in this electrochemical cell are:
Anode: Cu (s) → Cu2+ (aq) + 2e−
Cathode: 2Ag+ (aq) + 2e− → Ag (s)
The net cell reaction for this electrochemical cell is:
Cu (s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag (s)
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Which of the following is correctly balanced redox half reaction? Group of answer choices A. 14H+ + 9e- + Cr2O72- ⟶ Cr3+ + 7H2O B. 14H+ + 6e- + Cr2O72- ⟶ 2Cr3+ + 7H2O C. 14H+ + Cr2O72- ⟶ 2Cr3+ + 7H2O + 6e- D. 14H+ + Cr2O72- ⟶ Cr3+ + 7H2O + 9e-
Answer:
The correct option is: B. 14 H⁺ + 6 e⁻ + Cr₂O₇²⁻ ⟶ 2 Cr³⁺ + 7 H₂O
Explanation:
Redox reactions is an reaction in which the oxidation and reduction reactions occur simultaneously due to the simultaneous movement of electrons from one chemical species to another.
The reduction of a chemical species is represented in a reduction half- reaction and the oxidation of a chemical species is represented in a oxidation half- reaction.
To balance the reduction half-reaction for the reduction of Cr₂O₇²⁻ to Cr³⁺:
Cr₂O₇²⁻ ⟶ Cr³⁺
First the number of Cr atoms on the reactant and product side is balanced
Cr₂O₇²⁻ ⟶ 2 Cr³⁺
Now, Cr is preset in +6 oxidation state in Cr₂O₇²⁻ and +3 oxidation state in Cr³⁺. So each Cr gains 3 electrons to get reduced.
Therefore, 6 electrons are gained by 2 Cr atoms of Cr₂O₇²⁻ to get reduced.
Cr₂O₇²⁻ + 6 e⁻ ⟶ 2 Cr³⁺
Now the total charge on the reactant side is (-8) and the total charge on the product side is (+6).
From the given options it is evident that the reaction must be balanced in acidic conditions.
Therefore, to balance the total charge on the reactant and product side, 14 H⁺ is added on the reactant side.
Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺
Now to balance the number of hydrogen and oxygen atoms, 7 H₂O is added on the product side.
Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺ + 7 H₂O
Therefore, the correct balanced reduction half-reaction is:
Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺ + 7 H₂O
If 500.0 mL of 0.10 M Ca2+ is mixed with 500.0 mL of 0.10 M SO42−, what mass of calcium sulfate will precipitate? Ksp for CaSO4 is 2.40×10−5. Express your answer to three significant figures and include the appropriate units.
Answer:
The mass of calcium sulfate that will precipitate is 6.14 grams
Explanation:
Step 1: Data given
500.0 mL of 0.10 M Ca^2+ is mixed with 500.0 mL of 0.10 M SO4^2−
Ksp for CaSO4 is 2.40*10^−5
Step 2: Calculate moles of Ca^2+
Moles of Ca^2+ = Molarity Ca^2+ * volume
Moles of Ca^2+ = 0.10 * 0.500 L
Moles Ca^2+ = 0.05 moles
Step 3: Calculate moles of SO4^2-
Moles of SO4^2- = 0.10 * 0.500 L
Moles SO4^2- = 0.05 moles
Step 4: Calculate total volume
500.0 mL + 500.0 mL = 1000 mL = 1L
Step 5: Calculate Q
Q = [Ca2+] [SO42-]
[Ca2+]= 0.050 M [O42-]
Qsp = (0.050)(0.050 )=0.0025 >> Ksp
This means precipitation will occur
Step 6: Calculate molar solubility
Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)
2.40 * 10^-5 = x²
x = √(2.40 * 10^-5)
x = 0.0049 M = Molar solubility
Step 7: Calculate total CaSO4 dissolved
total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 mol/L = 0.667 g
Step 8: Calculate initial mass of CaSO4
Since initial moles CaSo4 = 0.050
Initial mass of CaSO4 = 0.050 * 136.14 g/mol
Initial mass of CaSO4 = 6.807 grams
Step 9: Calculate mass precipitate
6.807 - 0.667 = 6.14 grams
The mass of calcium sulfate that will precipitate is 6.14 grams
Propane is often used to heat homes. The combustion of propane follows the following reaction: C3H8(g) + 5O2(g) 3CO2 (g) + 4H2O(g), ΔH° = –2044 kJ. How many grams of propane must be reacted by this reaction to release 7563 kJ of heat?
Answer:
To release 7563 kJ of heat, we need to burn 163.17 grams of propane
Explanation:
Step 1: Data given
C3H8 + 5O2 -----------> 3CO2 + 4H2O ΔH° = –2044 kJ
This means every mole C3H8
Every mole of C3H8 produces 2044 kJ of heat when it burns (ΔH° is negative because it's an exothermic reaction)
Step 2: Calculate the number of moles to produce 7563 kJ of heat
1 mol = 2044 kJ
x mol = 7563 kJ
x = 7563/2044 = 3.70 moles
To produce 7563 kJ of heat we have to burn 3.70 moles of C3H8
Step 3: Calculate mass of propane
Mass propane = moles * Molar mass
Mass propane = 3.70 moles * 44.1 g/mol
Mass propane = 163.17 grams
To release 7563 kJ of heat, we need to burn 163.17 grams of propane
The mass of propane needed by the reaction to release 7563 KJ of heat energy is 162.8 g
Balanced equationC₃H₈ + 5O₂ —> 3CO₂ + 4H₂O ΔH° = –2044 KJ
Molar mass of propane C₃H₈ = (3×12) + (1×8) = 44 g/mol
Mass of propane C₃H₈ from the balanced equation = 1 × 44 = 44 g
Heat released = 2044 KJ
SUMMARY
From the balanced equation above,
2044 KJ of heat was released by the reaction of 44 g of propane
How to determine the mass of propane neededFrom the balanced equation above,
2044 KJ of heat was released by the reaction of 44 g of propane
Therefore,
7563 KJ of heat energy will be release by = (7563 × 44) / 2044 = 162.8 g of propane
Thus, 162.8 g of propane is needed for the reaction
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A student is doing a titration using potassium permanganate solution, KMnO4, to determine the amount of H2O2 in a sample. The balanced equation for the reaction in the titration is given below:2 MnO4-(aq) + 6 H+(aq) + 5 H2O2(aq) -->ž 2 Mn2+(aq) + 8 H2O(l) + 5 O2(g)A student calculates an amount of moles of H2O2 that is larger than the actual value. Which of the following errors could correctly explain the larger value?The student failed to wear goggles.The student did not swirl the flask appropriately and therefore stopped short of the endpoint.The student failed to rinse the buret with KMnO4¬ solution after rinsing it with distilled water.The student added an extra 15 mL of distilled water to the H2O2 solution.
Answer:
The student failed to rinse the buret with KMnO₄ solution after rinsing it with distilled water.
Explanation:
In a titration, the equivalent moles of titrant must be the same than analyte. If the student calculates an amount of moles of H₂O₂ that is larger than the actual value:
The student failed to wear goggles FALSE. The use of goggles doesn't change the results in the lab but you must use it!
The student did not swirl the flask appropriately and therefore stopped short of the endpoint. FALSE. The stopped short of the endpoint would explain a lower concentration of the real value.
The student failed to rinse the buret with KMnO₄ solution after rinsing it with distilled water. TRUE. If you don't rinse the buret with KMnO₄ solution the concentration of this solution in the buret will be lower than real concentration doing you spend more titrant volume overestimating the amount of H₂O₂ moles.
The student added an extra 15 mL of distilled water to the H₂O₂ solution. FALSE. The addition of water doesn't change the amount of H₂O₂ moles in the solution.
I hope it helps!
Calculate how many times more soluble Mg(OH)2 is in pure water Based on the given value of the Ksp, 5.61×10−11, calculate the ratio of solubility of Mg(OH)2 dissolved in pure H2O to Mg(OH)2 dissolved in a 0.130 M NaOH solution.
molar solubility in water = 2.41 x 10^-4
molar solubility of NaOH in 0.130M = 3.32 x 10^-9
Final answer:
The molar solubility of Mg(OH)2 in pure water is 2.41 x 10^-4 M. To calculate the molar solubility in a 0.130 M NaOH solution, we can use the common ion effect.
Explanation:
The molar solubility of Mg(OH)2 in pure water is 2.41 x 10^-4 M. The molar solubility of Mg(OH)2 in a 0.130 M NaOH solution can be determined using the common ion effect. Since NaOH dissociates into Na+ and OH-, the concentration of OH- in the NaOH solution is 0.130 M. To calculate the molar solubility, we can set up an ICE table:
Mg(OH)2 ⇌ Mg2+ + 2OH-
Initial: 0.130 M (from NaOH concentration)
Change: -x (due to dissolution of Mg(OH)2) and +2x (due to formation of OH- ions)
Equilibrium: 0.130 - x M
Ksp = (Mg2+)(OH-)²
Using the given value of the Ksp, 5.61x10^-11, we can substitute the equilibrium concentrations into the Ksp expression and solve for x. Then we can calculate the ratio of the solubility of Mg(OH)2 in pure water to the solubility in the NaOH solution by dividing the molar solubility in water by the molar solubility in the NaOH solution.
Mg(OH)₂ is approximately [tex]\( 0.472} \)[/tex] times more soluble in pure water compared to a 0.130 M NaOH solution. The solubility of Mg(OH)₂ in the presence of 0.130 M NaOH, is approximately [tex]\( 5.1 \times 10^{-4} \)[/tex] M.
To determine how many times more soluble Mg(OH)₂ is in pure water compared to a 0.130 M NaOH solution, we need to consider the effect of the common ion (OH-) from NaOH on the solubility of Mg(OH)₂.
Given:
- [tex]\( K_{sp} \)[/tex] for Mg(OH)₂ in pure water = [tex]\( 5.61 \times 10^{-11} \)[/tex]
- Molar solubility of Mg(OH)₂ in pure water = [tex]\( 2.41 \times 10^{-4} \)[/tex] M
- Molar solubility of NaOH in 0.130 M solution = [tex]\( 3.32 \times 10^{-9} \)[/tex] M
1. Calculate the concentration of OH- ions in the 0.130 M NaOH solution:
The NaOH dissociates completely to produce Na+ and OH- ions.
Concentration of OH- ions from NaOH = 0.130 M
2. Determine the solubility of Mg(OH)₂ in the presence of 0.130 M NaOH:
According to the common ion effect, the presence of additional OH- ions from NaOH will shift the equilibrium of Mg(OH)2 dissolution. The solubility of Mg(OH)₂ will decrease due to the Le Chatelier's principle, which states that adding an ion that is involved in the equilibrium reaction will shift the equilibrium in the direction that reduces its effect.
Let's denote the solubility of Mg(OH)₂ in the presence of 0.130 M NaOH as [tex]\( S_{\text{NaOH}} \)[/tex].
Using the Ksp expression for Mg(OH)₂:
[tex]\[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \][/tex]
In pure water, [tex]\( [\text{OH}^-] = S_{\text{water}} \)[/tex] (molar solubility of Mg(OH)2 in pure water).
In the presence of 0.130 M NaOH, [tex]\( [\text{OH}^-] = 3.32 \times 10^{-9} \)[/tex]M (concentration of OH- from NaOH).
So, [tex]\( S_{\text{NaOH}} \)[/tex] can be found from:
[tex]\[ K_{sp} = [\text{Mg}^{2+}](3.32 \times 10^{-9})^2 \][/tex]
Solve for [tex]\( [\text{Mg}^{2+}] \)[/tex]:
[tex]\[ [\text{Mg}^{2+}] = \frac{K_{sp}}{(3.32 \times 10^{-9})^2} \][/tex]
Calculate [tex]\( [\text{Mg}^{2+}] \)[/tex]:
[tex]\[ [\text{Mg}^{2+}] = \frac{5.61 \times 10^{-11}}{(3.32 \times 10^{-9})^2} \][/tex]
[tex]\[ [\text{Mg}^{2+}] = \frac{5.61 \times 10^{-11}}{1.1 \times 10^{-17}} \][/tex]
[tex]\[ [\text{Mg}^{2+}] \approx 5.1 \times 10^{-4} \, \text{M} \][/tex]
3. Calculate the ratio of solubility in pure water to solubility in 0.130 M NaOH:
[tex]\[ \text{Ratio} = \frac{S_{\text{water}}}{S_{\text{NaOH}}} = \frac{2.41 \times 10^{-4}}{5.1 \times 10^{-4}} \][/tex]
[tex]\[ \text{Ratio} \approx 0.472 \][/tex]
Select the true statements about the electron transport chain. In the electron transport chain, a series of reactions moves electrons through carriers. The products of the electron transport chain are and either or . Coenzyme Q and cytochrome are components of the electron transport chain. Coenzyme A is a component of the electron transport chain. The electron transport chain operates independently of other metabolic processes.
Answer:
The true statements are given below.
Explanation:
1 In the electron transport chain a series of reactions moves electrons through carriers.
2 Coenzyme Q and cytochrome are the components of the electron transport chain.
During electron transport chain electrons are transported from one electron carrier have high reduction potential to another electron carrier having comparatively low reduction potential than the previous one by a series of redox reactions.
Coenzyme Q is a component of complex 2 whereas cytochrome is an important component of both complex 3 and complex 4.
The electron transport chain is a series of reactions that moves electrons through carriers and is located in the inner mitochondrial matrix in eukaryotic cells and the cytoplasmic membrane in prokaryotic cells. Its products are water and ATP. Coenzyme Q and cytochrome are components of the electron transport chain, but Coenzyme A is not. The electron transport chain operates independently of other metabolic processes.
Explanation:The electron transport chain (ETC) is a series of reactions that moves electrons through carriers. It is located in the inner mitochondrial matrix in eukaryotic cells and the cytoplasmic membrane in prokaryotic cells. The products of the electron transport chain are water and ATP. Coenzyme Q and cytochrome are components of the electron transport chain, but Coenzyme A is not. The electron transport chain operates independently of other metabolic processes.
Predict the sign of the entropy change, ΔS∘, for each of the reaction displayed.Drag the appropriate items to their respective bins: Positive, NegativeAg+(aq)+Br−(aq)→AgBr(s)CaCO3(s)→CaO(s)+CO2(g)2NH3(g)→N2(g)+3H2(g)2Na(s)+Cl2(g)→2NaCl(s)C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)I2(s)→I2(g)
The predicted sign of the entropy change in each reaction are as follows;
Ag+(aq). + Br−(aq) → AgBr(s). --NegativeCaCO3(s) → CaO(s)+CO2(g) --Positive2NH3(g) → N2(g)+3H2(g) --Positive2Na(s). + Cl2(g) → 2NaCl(s) --NegativeC3H8(g)+5O2(g)→3CO2(g)+4H2O(g) --PositiveI2(s) → I2(g). --PositiveTo solve this question, we must understand the meaning of entropy.
Entropy of a system or chemical entity is the degree of disorderliness in the substance.
The change in entropy, ΔS∘ is a measure of the difference between the entropy of products and the reactants and is given mathematically as;
ΔS∘ = S(products) - S(reactants).We must also know that the entropy of a gas is greater than a liquid and that in turn is greater than a solid.The statement above therefore provides a basis for predicting the sign of the entropy change; ΔS∘ in each reaction.Read more:
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In the first reaction entropy sign is negative while for the second reaction the sign on entropy is positive.
The Entropy is defined as degree of disorderliness in the system.
When entropy is negative the reaction will be non-spontaneous.When entropy id positive the reaction will be spontaneous.The entropy of the reaction can be calculated by,
[tex]\rm \bold{ \Delta S = S(products) - S(reactants)}[/tex]
Entropy for First reaction,
[tex]\rm \bold{ Ag(aq) + Br(aq) \rightarrow AgBr(s)}[/tex] will be negative
Entropy for second reaction,
[tex]\rm \bold{ CaCO_3(s) \rightarrow CaO(s)+CO_2(g) }[/tex] will be positive.
Hence, we can conclude that the the first reaction will be non- spontaneous while second reaction will spontaneous.
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A chemist prepares a solution of silver(I) nitrate (AgNO3) by measuring out 269. micromol of silver(I) nitrate into a 300. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mmol/L of the chemist's silver(I) nitrate solution. Be sure your answer has the correct number of significant digits.
Answer:
The concentration of the chemist's silver(I) nitrate solution is 0.897 mmol/L
Explanation:
Step 1: Data given
Number of moles AgNO3 = 269 micromol = 269 * 10^-6 mol
Volume of AgNO3 = 300 mL = 0.3 L
Molar mass of AgNO3 = 169.87 g/mol
Step 2: Calculate molarity of AgNO3
Molarity = number of moles per volume (in Liters)
Molarity AgNO3 = 269 *10^-6 mol / 0.3 L
Molarity AgNO3 = 8.97 * 10^-4 M
8.97 *10^-4 mol/L = 897 micromol/L = 0.897 mmol/L
The concentration of the chemist's silver(I) nitrate solution is 0.897 mmol/L
A basic solution contains the iodide and phosphate ions that are to be separated via selective precipitation. The I– concentration, which is 7.60×10-5 M, is 10,000 times less than that of the PO43– ion at 0.760 M . A solution containing the silver(I) ion is slowly added. Answer the questions below. Ksp of AgI is 8.30×10-17 and of Ag3PO4, 8.90×10-17.1.) Calculate the minimum Ag+ concentration required to cause precipitation of AgI.2.) Calculate the minimum Ag+ concentration required to cause precipitation of Ag3PO4.3.) Which salt will precipitate first?4.) Is the separation of the I– and PO43– ion complete (is the percentage of "remaining" ion less than 0.10% or its original value)?
Answer:
1) 1.092x10⁻¹² M
2) 4.89x10⁻⁶ M
3) Ag₃PO₄
4) Yes, it is.
Explanation:
1) The dissolution reaction of AgI is:
AgI(s) ⇄ Ag⁺(aq) + I⁻(aq)
The constant of equilibrium Kps is:
Kps = [Ag⁺]*[I⁻]
So, the minimum concentration of Ag⁺ is that one in equilibrium
8.30x10⁻¹⁷ = [Ag⁺]*7.60x10⁻⁵
[Ag+] = 1.092x10⁻¹² M
2) The dissolution reaction of Ag₃PO₄ is:
Ag₃PO₄(s) ⇄ 3Ag⁺(aq) + PO₄³⁻(aq)
The constant of equilibrium, Kps is:
Kps = [Ag⁺]³*[PO₄³⁻]
So, the minimum concentration of Ag⁺ is that one in equilibrium
8.90x10⁻¹⁷ = [Ag⁺]³*0.760
[Ag⁺]³ = 1.171x10⁻¹⁶
[Ag⁺] = 4.89x10⁻⁶ M
3) Because Ag₃PO₄ needs less Ag⁺ to precipitated, it will precipitate first.
4) To see if the separation will occur, let's verify is the ratio of the molar concentration of Ag⁺ for the substances is less then 0.1%:
(1.092x10⁻¹²)/(4.89x10⁻⁶) * 100% = 2.23x10⁻⁵ %
So, the separation is complete.
The Ksp is used to denote the extent of dissolution of compounds in water.
1) The equilibrium for the dissolution of AgI is set up as follows:
AgI(s) ⇄ Ag⁺(aq) + I⁻(aq)
Hence;
Ksp = [Ag⁺] [I⁻]
At equilibrium, the minimum concentration of Ag⁺ is
8.30x10⁻¹⁷ = [Ag⁺]*7.60x10⁻⁵
[Ag+] = 1.092x10⁻¹² M
2) The equilibrium for the dissolution of Ag₃PO₄ is set up as follows:
Ag₃PO₄(s) ⇄ 3Ag⁺(aq) + PO₄³⁻(aq)
Hence:
Ksp = [Ag⁺]³*[PO₄³⁻]
At equilibrium, the minimum concentration of Ag⁺ is
8.90x10⁻¹⁷ = [Ag⁺]³*0.760
[Ag⁺]³ = 1.171x10⁻¹⁶
[Ag⁺] = 4.89x10⁻⁶ M
3) The salt that is expected to precipitate first is Ag₃PO₄.
4) In order to determine if the separation is complete;
(1.092x10⁻¹²)/(4.89x10⁻⁶) * 100% = 2.23x10⁻⁵ %
This is less than 0.10% hence the separation is complete.
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Which of the following is NOT a standard state? Group of answer choices For a solid, it is 1 atm. For a solid, it is 25°C. For a liquid, it is 1 atm. For a liquid, it is 25°F. For a solution, it is 1 M.
Answer:
For a liquid, it is 25°F
Explanation:
The standard state for a liquid is 25°C
The state which is not a standard state for solid is 25 °C and for liquid is 25 F.
What are standard state?Every matter will behave idealy at a particular set of items and that set is known as standard state of that matter.
For solids:Solids will remain at the standard state at 1 atm pressure, but 25°C (298K) is not the standard temperature for solids as they will melt in this temperature.
For liquids:Standard state of liquids is 1 atm pressure and 25°C (298K) temperature, but in the question it is given 25F which is not true so.
For solution:Solution of 1 molariy will consider as the standard solution and it is define as the number of moles of solute present in per liter of the solution.
So, for solids and liquids 25°C and 25F respectively is not a standard state.
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Phosphoric acid (H3PO4) is important in the production of both fertilizers and detergents. It is prođuced in the following process Ca5 (PO4)3F+5H2SO4 -> 3H3PO4 +5 CaSO4 + HF. It is distributed commercially as a solution with a concentration of about 14.8 M. Approximately 2.00 x 109 gallons of this concentrated phosphoric acid solution are produced annually in this country alone. Assuming that all of this H3PO4 is produced in the reaction above, what mass (in g) of the mineral fluoroapatite, Ca5 (PO4)3F, would be required each year?
Final answer:
To calculate the mass of mineral fluoroapatite (Ca5(PO4)3F) required each year, we need to use stoichiometry. Given that approximately 2.00 x 10^9 gallons of concentrated phosphoric acid solution is produced annually containing 14.8 M concentration, we can convert this volume to moles of H3PO4 and then use stoichiometry to calculate the moles of Ca5(PO4)3F required. Finally, we can calculate the mass of Ca5(PO4)3F required, which is approximately 1.88 x 10^13 grams.
Explanation:
To calculate the mass of mineral fluoroapatite (Ca5(PO4)3F) required each year, we need to use stoichiometry.
Given that 2.00 x 109 gallons of concentrated phosphoric acid solution are produced annually, we need to convert this volume to moles of H3PO4 and then use the stoichiometric coefficients to calculate the moles of Ca5(PO4)3F required.
A balanced equation shows that for every 3 moles of H3PO4 produced, 1 mole of Ca5(PO4)3F is required.
Therefore, we can set up a ratio:
3 moles H3PO4 / 1 mole Ca5(PO4)3F
Now, we can calculate the moles of H3PO4:
14.8 M = 14.8 mol/L
2.00 x 109 gallons * 3.785 L/gallon = 7.57 x 109 L
7.57 x 109 L * 14.8 mol/L = 1.12 x 1011 mol H3PO4
Finally, we can use the ratio to calculate the moles of Ca5(PO4)3F:
1.12 x 1011 mol H3PO4 * (1 mol Ca5(PO4)3F / 3 mol H3PO4) = 3.73 x 1010 mol Ca5(PO4)3F
Since the molar mass of Ca5(PO4)3F is 504.05 g/mol, we can calculate the mass of Ca5(PO4)3F required:
3.73 x 1010 mol * 504.05 g/mol = 1.88 x 1013 g Ca5(PO4)3F
Therefore, approximately 1.88 x 1013 grams of mineral fluoroapatite (Ca5(PO4)3F) would be required each year.
Identify which sets of quantum numbers are valid for an electron. Each set is ordered (n,l,ml,ms).a. 3,1,1,-1/2b. 4,3,1,-1/2c. 2,0,0,-1/2d. 1,3.0,1/2e. 3,2,-1,-1/2f. 3,3,-1,1/2g. 3,2,1,-1h. 1,-1,-1,-1/2i. 3,3,1,1/2j. 3,0,0,1/2k. 4,3,4,-1/2l. 0,2,1,1/2
Answer:
The valid set of quantum numbers are:
a. 3,1,1,-1/2
b. 4,3,1,-1/2
c. 2,0,0,-1/2
e. 3,2,-1,-1/2
j. 3,0,0,1/2
Explanation:
Quantum numbers (n, l, mℓ, ms) are the set of numbers that describe the state of an electron in an atom.
The four quantum numbers and their rules are:
Principal: n ≥ 1 Azimuthal: ℓ ≤ (n-1)Magnetic: mℓ = (- ℓ) to (+ ℓ)Spin: ms = (- s), (+ s)Therefore,
a. 3,1,1,-1/2: Valid
b. 4,3,1,-1/2: Valid
c. 2,0,0,-1/2: Valid
d. 1,3,0,1/2: NOT Valid
Reason: ℓ ≤ (n-1). Therefore, ℓ can not be greater than n, .
e. 3,2,-1,-1/2: Valid
f. 3,3,-1,1/2: NOT Valid
Reason: ℓ ≤ (n-1). Therefore, ℓ can not be equal to n. (ℓ ≠n)
g. 3,2,1,-1: NOT Valid
Reason: the only allowed values of ms = (- s), (+ s) = (- 1/2), (+ 1/2)
h. 1,-1,-1,-1/2: NOT Valid
Reason: ℓ ≤ (n-1) and n ≥ 1. Therefore, ℓ can not be equal to n and also ℓ can't be negative.
i. 3,3,1,1/2: NOT Valid
Reason: ℓ ≤ (n-1). Therefore, ℓ can not be equal to n. (ℓ ≠n)
j. 3,0,0,1/2: Valid
k. 4,3,4,-1/2: NOT Valid
Reason: mℓ = (- ℓ) to (+ ℓ). Therefore, mℓ can't be greater than ℓ.
l. 0,2,1,1/2: NOT Valid
Reason: n ≥ 1. Therefore, value of n can't be 0 (n≠0)
Which statements accurately describe soap? Select one or more:
A. Soaps react with ions in hard water to create a precipitate.
B. Soaps are both hydrophobic and hydrophilic.
C. Soaps work in solutions of any pH.
D. Soaps should be weakly alkaline in solution.
E. Soaps are a mixture of fatty acid salts and glycerol.
Answer:
A. Soaps react with ions in hard water to create a precipitate.
B. Soaps are both hydrophobic and hydrophilic.
D. Soaps should be weakly alkaline in solution.
Explanation:
A. Hard water contains magnesium and calcium minerals like calcium and magnesium carbonates, sulfates and bicarbonates. As soon as these minerals come in contact with soap their ions like Mg²⁺ & Ca²⁺ form precipitates.
B. Soap are both hydrophilic and hydrophobic. They reason why they exhibit both the properties is really important for their functionality. The hydrophobic part of soap makes interaction with oil/dust particles while the hydrophilic part makes interaction with water. When the cloth is rinsed the dirt/soap particles are removed from the dirty clothes thereby making them clean.
C. Soaps have alkaline pH i.e. more than 7 that is why they have bitter taste.
Statements A, B, D, and E correctly describe soap: they react with hard water ions to create precipitate, are hydrophilic and hydrophobic, they are weakly alkaline in solution, and they are a mixture of fatty acid salts and glycerol. Statement C is incorrect, as soaps do not work well in all pH conditions.
Explanation:The following statements accurately describe soap:
A. Soaps react with ions in hard water to create a precipitate. Soap molecules react with the calcium and magnesium ions in hard water, forming an insoluble substance, or precipitate, which can leave residue.B. Soaps are both hydrophobic and hydrophilic. This means they have a part that is attracted to water (hydrophilic) and a part that is repelled by water (hydrophobic). This characteristic allows soaps to dissolve dirt and grease in water.D. Soaps should be weakly alkaline in solution. Most soaps are slightly alkaline, meaning they have a pH above 7. This helps to break down oils and fats on our skin.E. Soaps are a mixture of fatty acid salts and glycerol. Soap is typically a sodium or potassium salt of a fatty acid and is derived from glycerol (a type of alcohol), in a process called saponification.
Statement C. Soaps work in solutions of any pH, however, is not correct. Soap works best in slightly alkaline conditions.
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Be sure to answer all parts. A 0.365−mol sample of HX is dissolved in enough H2O to form 835.0 mL of solution. If the pH of the solution is 3.70, what is the Ka of HX? Be sure to report your answer to the correct number of significant figures.
The Ka of HX is mathematically given as
Ka = 9.11 *10^-8
What is the Ka of HX?
Generally, the equation for the molarity of HX is mathematically given as
M HX = moles HX / volume solution
Therefore
Molarity HX = 0.365 mol / 0.835 L
Molarity HX = 0.437 M
Therefore, ICE-chart
[H+] = [H3O+]
10^-3.70 = 10^-3.70 = 1.995 *10^-4
The concentration at the equilibrium is
[HX] = (0.437 - x)M
[H3O+] = 1.995*10^-4 M
x=1.995*10^-4
In conclusion
Ka = [X-]*[H3O+] / [HX]
Ka = ((1.995*10^-4)²)/ 0.437
Ka = 9.11 *10^-8
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Final answer:
The Ka of HX, given its concentration and the pH of solution, can be determined through a series of calculations leading to a Ka value of 9.17×10-8.
Explanation:
To find the Ka of HX given a 0.365-mol sample dissolved in 835.0 mL of solution with a pH of 3.70, we start by calculating the concentration of HX, then use the pH to find the concentration of H+ ions, which will help determine the acid dissociation constant (Ka).
First, convert volume from mL to L: 835.0 mL = 0.835 L. The concentration of HX (Molarity, M) is moles of solute (HX) divided by volume of solution in liters, which is 0.365 mol / 0.835 L = 0.4371 M.
Given pH = 3.70, we calculate the concentration of H+ ions ([H+]) as 10-pH = 10-3.70 = 2.00×10-4 M.
Assuming all H+ comes from the dissociation of HX and ignoring autoionization of water because HX is a weak acid, the Ka expression for HX is Ka = [H+][X-] / [HX]. Assuming the concentration of [X-] is equal to [H+] because each molecule of HX donates one H+, Ka becomes 2.00×10-4 M * 2.00×10-4 M / 0.4371 M = 9.17×10-8, after considering the initial concentration of HX and the change in concentration due to dissociation.
Nitrogen from a gaseous phase is to be diffused into pure iron at 700°C. If the surface concentration is maintained at 0.26 wt% N, what will be the concentration (in weight percent) 2.8 mm from the surface after 6.3 h? The diffusion coefficient for nitrogen in iron at 700°C is 2.2 × 10-10 m2/s.
Answer:
[tex]Cx = .376 \times 0.26 = 0.0978 wt% N[/tex]
Explanation:
Given details
concentration of surface is Cs 0.26 wt%
distance from surface is x 2.8 mm
time t = 3.6 h
original concentration is Co 0
we know that
[tex]\frac{Cx - Co}{Cs - Co} = 1 - erf(\frac{x}{2\sqrt{Dt}}[/tex]
[tex]\frac{Cx - 0}{0.26 - Co} = 1 - erf(\frac{2\times 10^{-3}}{2\sqrt{2.2\times 10^{-11} 5 hr (3600 /1 hr)}}[/tex]
[tex]\frac{Cx}{0.26} = 1 - erf(0.626)[/tex]
FROM TABLE OF ERROR FUNCTION WE HAVE
Z erf(z)
0.60 0.6039
0.65 0.6420
so by interpolation technique we have
for z = 0.626 we have value of erf(0.626) = 0.623
[tex]\frac{Cx}{0.26} = 1 - 0.623[/tex]
[tex]Cx = .376 \times 0.26 = 0.0978 wt% N[/tex]
The concentration in weight percent from the surface is 0.0962%.
Given the following data:
Temperature = 700°C.Surface concentration = 0.26 wt% N.Distance = 2.8 mm.Time = 6.3 hours.Diffusion coefficient for nitrogen = [tex]2.2 \times 10^{-10}\;m^2/s[/tex]Note: The original concentration of iron (Co) is 0.
To determine the concentration in weight percent:
How to calculate the concentration.In this exercise, we would apply Fick's second law and error function:
[tex]\frac{C_x - C_o}{C_s - C_o} =1-erf\frac{x}{2\sqrt{Dt} }[/tex]
Substituting the given parameters into the formula, we have;
[tex]\frac{C_x - 0}{0.26 - 0} =1-erf(\frac{2.8 \times 10^{-3}}{2\sqrt{2.2 \times 10^{-10} \times 22680} })\\\\\frac{C_x - 0}{0.26 - 0} =1-erf(\frac{2.8 \times 10^{-3}}{2\sqrt{4.99 \times 10^{-6}} })\\\\\frac{C_x }{0.26 } =1-erf(\frac{2.8 \times 10^{-3}}{2\times 0.0022 })\\\\\frac{C_x }{0.26} =1-erf(0.6364)\\\\\frac{C_x}{0.26 } =1-0.63\\\\\frac{C_x}{0.26 } =0.37\\\\C_x = 0.26 \times 0.37[/tex]
Cx = 0.0962%.
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The molar solubilities of the following compounds (in mol/L) are:
AgBr = 7.3 x 10-7
AgCN = 7.7 x 10-9
AgSCN = 1.0 x 10-6
When these compounds are arranged in order of decreasing Ksp values, what is the correct order?
AgCN > AgSCN > AgBr
AgBr > AgCN > AgSCN
AgSCN > AgBr > AgCN
AgCN > AgBr > AgSCN
Answer: The decreasing order of [tex]K_{sp}[/tex] is [tex]AgSCN>AgBr>AgCN[/tex]
Explanation:
For AgBr:The balanced equilibrium reaction for the ionization of silver bromide follows:
[tex]AgBr\rightleftharpoons Ag^{+}+Br^-[/tex]
s s
The expression for solubility constant for this reaction will be:
[tex]K_{sp}=[Ag^{+}][Br^-][/tex]
We are given:
[tex]s=7.3\times 10^{-7}mol/L[/tex]
Putting values in above equation, we get:
[tex]K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.3\times 10{-7})^2=5.33\times 10^{-13}[/tex]
Solubility product of AgBr = [tex]5.33\times 10^{-13}[/tex]
For AgCN:The balanced equilibrium reaction for the ionization of silver cyanide follows:
[tex]AgCN\rightleftharpoons Ag^{+}+CN^-[/tex]
s s
The expression for solubility constant for this reaction will be:
[tex]K_{sp}=[Ag^{+}][CN^-][/tex]
We are given:
[tex]s=7.7\times 10^{-9}mol/L[/tex]
Putting values in above equation, we get:
[tex]K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.7\times 10{-9})^2=5.93\times 10^{-17}[/tex]
Solubility product of AgCN = [tex]5.33\times 10^{-17}[/tex]
For AgSCN:The balanced equilibrium reaction for the ionization of silver thiocyanate follows:
[tex]AgSCN\rightleftharpoons Ag^{+}+SCN^-[/tex]
s s
The expression for solubility constant for this reaction will be:
[tex]K_{sp}=[Ag^{+}][SCN^-][/tex]
We are given:
[tex]s=1.0\times 10^{-6}mol/L[/tex]
Putting values in above equation, we get:
[tex]K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(1.0\times 10{-6})^2=1.0\times 10^{-12}[/tex]
Solubility product of AgSCN = [tex]1.0\times 10^{-12}[/tex]
The decreasing order of [tex]K_{sp}[/tex] follows:
[tex]AgSCN>AgBr>AgCN[/tex]
Which of the following atoms could have an expanded octet when it is the central atom in a covalent compound?a) B
b) O
c) Cl
d) F
Answer:
only chlorine can expand its octet.
Explanation:
An atom can expand its octet is it has empty d orbital
the electronic configuration of given elements will be:
B : 1s2 2s2 2p1 [Valence shell n =2 no d orbital]
O :1s2 2s2 2p4 [Valence shell n =2 no d orbital]
F : 1s2 2s2 2p5 [Valence shell n =2 no d orbital]
Cl :1s2 2s2 2p6 3s2 3p5 3d0 [Valence shell n =2 no d orbital]
Out of given elements only chlorine has empty d orbitals in its valence shell
Thus only chlorine can expand its octet.