a)Exothermic
b) it goes up
c) The piston must move out
d)The energy is released by the system
e)ΔE = -616 kJ
Explanation:
sign convention for heat flow:
if the heat flows into the system it is marked with positive sign
when the heat flows out the system it is marked with negative sign
sign convention for work
if work is done on the system it is marked positive
if work is done by the system it is marked negative
sign convention for internal energy
if energy absorbed by system mark positive
if energy released by system mark negative
a)
the heat flow flowing out of the system (q= -300 kJ)
thus the reaction is Exothermic
b)
In exothermic reaction, heat flows from the system to surroundings.
surrounding here is water bath.
As the system is exothermic, hence the temperature of water bath goes up
c)
It is given that 316 kJ of work is done by the system, applying expansion.
To maintain constant pressure of 1 atm the decreased pressure can be compensated with an increase on volume . To increase volume the piston must move out
d)
According to 1st law of thermodynamics,
ΔE =q +w ⇒1
here,
ΔE = q +w
ΔE = -300 kJ -316 kJ
ΔE = -616 kJ ⇒2
the negative sign denotes the energy is released by the system
e)
from ΔE = -616 kJ the amount of energy released by the reaction is ΔE = -616kJ
Which of the following reagents convert(s) benzoyl chloride to phenyl propyl ketone?a. CH3CH2CH2MgBrb. CH3CH2CH2Lic. (CH3CH2CH2)2CuLid. Both a and b
Answer:C
(CH3CH2CH2)2CuLi
Explanation:
The reaction between acyl halides and Grignard reagents or acyl halides and organolithium compounds does not form ketones because the reagents ( Grignard reagent and organo lithium compounds) are too reactive hence the ketone intermediate reacts further to form tertiary alcohols hence the ketone cannot be isolated.
I have a cup of hot coffee at 140 oC but I want to cool it to 110 oC. My cup holds about 0.3 kg of coffee. Fortunately, I have a bunch of aluminum cubes in the freezer that I can drop into my hot coffee to cool it down.
If each aluminum cube has a mass of 1 g (not 1 kg!) and my freezer keeps its contents at a temperature of –10 oC, how many cubes do I have to drop into my coffee? The specific heat of water is around 4000 joules/kg/oC and aluminum is about 900 joules/kg/oC. (Pick the answer closest to the true value and ignore any thermal losses to surroundings.)
A. 200
B. 330
C. 400
D. 110
E. 88
Answer:
The correct answer is option B.
Explanation:
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
[tex]q_1=-q_2[/tex]
[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]
where,
[tex]C_1[/tex] = specific heat of metal = [tex]900 J/kg^oC[/tex]
[tex]C_2[/tex] = specific heat of coffee= [tex]4000 J/kg^oC[/tex]
[tex]m_1[/tex] = mass of metal = x
[tex]m_2[/tex] = mass of coffee = 0.3 kg
[tex]T_f[/tex] = final temperature of aluminum metal= [tex]110^oC[/tex]
[tex]T_1[/tex] = initial temperature of aluminum metal = [tex]-10^oC[/tex]
[tex]T_2[/tex] = initial temperature of coffee= [tex]140^oC[/tex]
Now put all the given values in the above formula, we get
[tex]x\times 900 J/kg^oC\times (110-(-10))^oC=-(0.3 kg\times 4000 J/kg^oC\times (110-140)^oC[/tex]
[tex]x=0.333 kg[/tex]
Mass of aluminum cubes = 0.3333 kg = 333.3 g
If mass of 1 cube is 1 gram, then numbers of cubes in 333.3 grams will be:
[tex]=\frac{333.3 g}{1 g}=333.3\approx 330[/tex]
330 cubes of aluminum cubes will be required.
Scientists plan to place detection devices outside Earth's atmosphere to detect radiations from a star. Which of the following best explains the type of detection device that the scientists would use.a. A radar, because high frequency microwaves are limited to the region outside Earth's atmosphereb. Special cameras, because high frequency UV rays are emitted by objects at very high temperaturesc. Special cameras, because high energy UV rays are emitted by objects having very high massd. A radar, because high energy microwaves are produced from very hot objects
Answer: the correct option is A (A radar, because high frequency microwaves are limited to the region outside Earth's atmosphere)
Explanation:
A radar can be used for detection of devices outside Earth's atmosphere to detect radiations from a star. A radar system consists of a transmitter producing electromagnetic waves in the radio or microwaves domain, a transmitting antenna, a receiving antenna (often the same antenna is used for transmitting and receiving) and a receiver and processor to determine properties of the object(s) example the stars
Answer:
A radar, because high frequency microwaves are limited to the region outside Earth's atmosphere
Explanation:
I took the test and its correct
What compound is produced when cyclohexene is treated with concentrated KMnO4?
A) hexanoic acid
B) adipic acid
C) cyclohexanecarboxylic acid
D) benzoic acid
E) succinic acid
Final answer:
The reaction of cyclohexene with concentrated KMnO₄ cleads to the formation of adipic acid due to oxidative cleavage of the carbon-carbon double bond.
Explanation:
When cyclohexene is treated with concentrated KMnO₄ (potassium permanganate), the reaction is an oxidative cleavage, and it produces a compound called adipic acid. This is because KMnO₄ is a strong oxidizing agent, which will typically cleave the double bond in the cyclohexene and oxidize the resulting fragments into carboxylic acids. Given that cyclohexene has a six-carbon ring, the oxidative cleavage will produce a six-carbon diacid, which is known as adipic acid.
Therefore, the correct answer to what compound is produced when cyclohexene is treated with concentrated KMnO₄ is B) adipic acid.
Write Lewis structural formulas for the following molecules:
NH3 SO2
CH3OH HNO2
N2 CH2O
Answer : The Lewis-dot structure of the following molecules are shown below.
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
(1) The given molecule is, [tex]NH_3[/tex]
As we know that hydrogen has '1' valence electron, nitrogen has '5' valence electrons.
Therefore, the total number of valence electrons in [tex]NH_3[/tex] = 5 + 3(1) = 8
According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.
(2) The given molecule is, [tex]SO_2[/tex]
As we know that sulfur and oxygen has '6' valence electrons.
Therefore, the total number of valence electrons in [tex]SO_2[/tex] = 6 + 2(6) = 18
According to Lewis-dot structure, there are 8 number of bonding electrons and 10 number of non-bonding electrons.
(3) The given molecule is, [tex]CH_3OH[/tex]
As we know that carbon has '4' valence electrons, hydrogen has '1' valence electron and oxygen has '6' valence electrons.
Therefore, the total number of valence electrons in [tex]CH_3OH[/tex] = 4 + 4(1) + 6 = 14
According to Lewis-dot structure, there are 10 number of bonding electrons and 4 number of non-bonding electrons.
(4) The given molecule is, [tex]HNO_2[/tex]
As we know that hydrogen has '1' valence electron, nitrogen has '5' valence electrons and oxygen has '6' valence electrons.
Therefore, the total number of valence electrons in [tex]HNO_2[/tex] = 1 + 5 + 2(6) = 18
According to Lewis-dot structure, there are 8 number of bonding electrons and 10 number of non-bonding electrons.
(5) The given molecule is, [tex]N_2[/tex]
As we know that nitrogen has '5' valence electrons.
Therefore, the total number of valence electrons in [tex]N_2[/tex] = 2(5) = 10
According to Lewis-dot structure, there are 6 number of bonding electrons and 4 number of non-bonding electrons.
(6) The given molecule is, [tex]CH_2O[/tex]
As we know that carbon has '4' valence electrons, hydrogen has '1' valence electron and oxygen has '6' valence electrons.
Therefore, the total number of valence electrons in [tex]CH_2O[/tex] = 4 + 2(1) + 6 = 12
According to Lewis-dot structure, there are 8 number of bonding electrons and 4 number of non-bonding electrons.
Acetylene (C2H2) has a tendency to lose two protons (H1) and form the carbide ion (C2 22), which is present in a number of ionic compounds, such as CaC2 and MgC2. Describe the bonding scheme in the C2 22 ion in terms of molecular orbital theory. Compare the bond order in C2 22 with that in C2.
Answer:
The C22- ion is stable
Explanation:
The C22- ion is a stable ion having a bond order of three. It has a favourable stabilization energy of 6∆ compared to 4∆ in C2. In the carbide ion carbon firms three bonds rather than two in dicarbon, hence the formation of the carbide ion is preferred. The molecular orbital configuration of the carbide ion is shown in the image attached.
Final answer:
The carbide ion (C2 2-) is formed from acetylene (C2H2) by loss of two protons and has a bonding scheme where the triple bond of acetylene is reduced to a double bond in the ion, yielding a bond order of 2 for the carbide ion compared to 3 for acetylene, according to molecular orbital theory.
Explanation:
The carbide ion (C2 2-), formed when acetylene (C2H2) loses two protons, has an interesting bonding scheme described by molecular orbital theory. In acetylene, there is a triple bond between the two carbon atoms consisting of one sigma (σ) bond and two pi (π) bonds. This triple bond is the result of the overlapping of sp hybrid orbitals for the σ bond and the side-by-side overlap of unhybridized p orbitals for the π bonds. When the two protons are removed to form the carbide ion, two additional electrons are added to the system which occupy the antibonding π orbitals (π*). In molecular orbital theory, the bond order is equal to the difference between the number of bonding electrons and antibonding electrons divided by two. Thus, in the carbide ion, the triple bond of acetylene is reduced to a double bond due to the additional electrons in the antibonding orbitals, which gives a bond order of 2 in C2 2- as compared to a bond order of 3 in C2.
Be sure to answer all parts. The sulfate ion can be represented with four S―O bonds or with two S―O and two S═O bonds. (a) Which representation is better from the standpoint of formal charges? four S―O bonds two S―O bonds and two S═O bonds (b) What shape is the sulfate ion, and what hybrid orbitals of S are postulated for the σ bonding? bent tetrahedral trigonal planar trigonal pyramidal spmdn where m = and n = (c) In view of the answer to part (b), what orbitals of S must be used for the π bonds? What orbitals of O? sulfur: 3d 3p 2p 3s oxygen: 3p 2p 2s 3s
Answer:
a) Representation - (in attachment)
b) Tetrahedral geometry and
c) [tex]sp^{3}[/tex] hybrid orbitals invovle sigma bonding.
[tex]\pi[/tex] orbitals are formed by the overlapping of d-orbitals of sulfur with p-orbitals of oxygen.
Explanation:
a)
Representation in attachment.
The overall charge in the both structures are -2. The structure (b) is favored by the resonance structures since the formal charge with in the species are remained.
Therefore, structure (b) is the better representation of sulfate ion.
b)
In sulfate ion, sulfur atom is attached with four different oxygen atoms. According to the VSEPR theory , the sufate ion has tetrahedral geometry.
There is four sigma bonds and zero lone pairs present on the central metal atom.
Hence, the hybridization of the sulfur atom is [tex]sp^{3}[/tex]
c)
The s and p orbitals of sulfur invovles hybridization and form sigma bonds.The orbitals of sulfur involves [tex]\pi[/tex] bonding.
Therefore, [tex]\pi[/tex] bonds are formed by the overlapping of d-orbitals of sulfur with p-orbitals of oxygen.
The sulfate ion is best represented with four S―O single bonds maintaining a tetrahedral geometry and sulfurs' sp³ hybridization. The pi bonds are formed due to the overlap of sulfur's 3d and oxygen's 2p orbitals.
Explanation:The sulfate ion can be represented better with four S―O single bonds. This equalizes the formal charges, since every oxygen atom carries a charge of -2 while sulfur carries a charge of +6.
The geometry shape of the sulfate (SO4²-) ion is tetrahedral, and the hybridization of the sulfur (S) atom is sp³, based on the fact that there are four regions of electron density around sulfur in the sulfate ion. These regions come from the four sigma bonds formed between sulfur and oxygen.
In view of the tetrahedral shape of the sulfate ion, the sulfur atom uses its 3p orbitals and 3s orbital to form sp³ hybrid orbitals for sigma bonds with the oxygen atom's 2p orbitals. The double bonds observed in some resonance structures of sulfate ion arise due to the overlap of the 3d orbitals of sulfur with the 2p orbitals of oxygen to form pi bonds.
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On top of one of the peaks in rocky mountain national park the pressure of the atmosphere is 550 torr determine the boiling point of water at this location.
Answer:
The boiling point of water at 550 torr will be 91 °C or 364 Kelvin
Explanation:
Step 1: Data given
Pressure = 550 torr
The heat of vaporization of water is 40.7 kJ/mol.
Step 2: Calculate boiling point
⇒ We'll use the Clausius-Clapeyron equation
ln(P2/P1) = (ΔHvap/R)*(1/T1-1/T2)
ln(P2/P1) = (40.7*10^3 / 8.314)*(1/T1 - 1/T2)
⇒ with P1 = 760 torr = 1 atm
⇒ with P2 = 550 torr
⇒ with T1 = the boiling point of water at 760 torr = 373.15 Kelvin
⇒ with T2 = the boiling point of water at 550 torr = TO BE DETERMINED
ln(550/760) = 4895.4*(1/373.15 - 1/T2)
-0.3234 = 13.119 - 4895.4/T2
-13.4424= -4895.4/T2
T2 = 364.2 Kelvin = 91 °C
The boiling point of water at 550 torr will be 91 °C or 364 Kelvin
The boiling point of water decreases with altitude due to reduced atmospheric pressure. At an atmospheric pressure of 550 torr, which one might find on a peak in the Rocky Mountain National Park, the boiling point of water would be slightly higher than 90°C, but less than 100°C.
Explanation:The boiling point of water is impacted by the atmospheric pressure. At higher altitudes, like on a peak in Rocky Mountain National Park where the pressure is 550 torr, the atmospheric pressure is lower and results in a lower boiling point compared to sea level. This can be confirmed with the use of a vapor pressure curve.
Deciphering the vapor pressure curve: Usually, 500 torr corresponds to a temperature around 80-90°C. So, it can be inferred that the boiling point of water at 550 torr would be slightly higher than 90°C, but still less than 100°C, which is the boiling point of water at sea level at standard atmospheric pressure.
Summarizing the concept, the boiling point of a liquid increases as pressure increases and decreases as pressure decreases. In other words, the boiling point of water decreases with altitude as the atmospheric pressure decreases.
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What is the Molecule Geometry of a ABE3 molecule?
A. linear
B. trigonal bipyramidal
C. trigonal planar
D. tetrahedral
The Molecule Geometry of an ABE3 molecule, with three electron groups and no lone pairs, is trigonal planar, with groups arranged 120° apart in a plane.
Explanation:The Molecule Geometry of a molecule with the designation ABE3, where 'A' represents the central atom and 'E' represents electron groups (or bonding domains) surrounding the central atom, relates to how these groups are spatially distributed around the central atom. In an ABE3 molecule configuration, with three electron groups around the central atom and no lone pairs (indicated by the '3'), all three groups are arranged to be as far apart from one another as possible. This arrangement forms trigonal planar geometry, where the groups adopt the positions at the corners of an equilateral triangle, each 120° apart and all in the same plane.
The solubility of oxygen in lakes high in the Rocky Mountains is affected by the altitude. If the solubility of O 2 from the air is 2.67 × 10 –4 M at sea level and 25°C, what is the solubility of O 2 at an elevation of 12,000 ft where the atmospheric pressure is 0.657 atm?
Assume the temperature is 25°C, and that the mole fraction of O 2 in air is 0.209 at both 12,000 ft and at sea level. 1.75 × 10–4 M 2.67 × 10–4 M 3.66 × 10–5 M 4.06 × 10–4 M
Answer:
S = 1,75x10⁻⁴M
Explanation:
It is possible to answer this question using Henry's law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid. The formula is:
S = k×Pi
Where S is solubility, K is Henry's constant and Pi is partial pressure.
At sea level, Pi of O₂ is:
1atm×0,209 = 0,209 atm. Replacing:
2,67x10⁻⁴M = k×0,209atm
k = 1,28x10⁻³M/atm
At 12,000ft Pi of O₂ is:
0,657atm×0,209 = 0,137 atm. Replacing:
S = 1,28x10⁻³M/atm×0,137atm
S = 1,75x10⁻⁴M
I hope it helps!
Olympic cyclist fill their tires with helium to make them lighter. Calculate the mass of air in an air filled tire and the mass of helium in a helium filled tire. What is the mass difference between the two? Assume that the volume of the tire is 855ml that is filled with a total pressure of 125psi, and that the temperature is 25 degrees Celsius. Also, assume an average molar mass for air of 28.8g/mol
Answer: The mass difference between the two is 7.38 grams.
Explanation:
To calculate the number of moles, we use the equation given by ideal gas follows:
[tex]PV=nRT[/tex]
where,
P = pressure = 125 psi = 8.50 atm (Conversion factor: 1 atm = 14.7 psi)
V = Volume = 855 mL = 0.855 L (Conversion factor: 1 L = 1000 mL)
T = Temperature = [tex]25^oC=[25+273]K=298K[/tex]
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles = ?
Putting values in above equation, we get:
[tex]8.50atm\times 0.855L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\n=\frac{8.50\times 0.855}{0.0821\times 298}=0.297mol[/tex]
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
For air:Moles of air = 0.297 moles
Average molar mass of air = 28.8 g/mol
Putting values in equation 1, we get:
[tex]0.297mol=\frac{\text{Mass of air}}{28.8g/mol}\\\\\text{Mass of air}=(0.297mol\times 28.8g/mol)=8.56g[/tex]
Mass of air, [tex]m_1[/tex] = 8.56 g
For helium gas:Moles of helium = 0.297 moles
Molar mass of helium = 4 g/mol
Putting values in equation 1, we get:
[tex]0.297mol=\frac{\text{Mass of helium}}{4g/mol}\\\\\text{Mass of helium}=(0.297mol\times 4g/mol)=1.18g[/tex]
Mass of helium, [tex]m_2[/tex] = 1.18 g
Calculating the mass difference between the two:
[tex]\Delta m=m_1-m_2[/tex]
[tex]\Delta m=(8.56-1.18)g=7.38g[/tex]
Hence, the mass difference between the two is 7.38 grams.
Final answer:
The mass of air in the tire calculated using the ideal gas law is 8.4752 grams, and the mass of helium is 1.1778 grams; the mass difference is 7.2974 grams, with helium being the lighter gas.
Explanation:
To calculate the mass of air and helium in a bike tire, we will use the ideal gas law which relates pressure (P), volume (V), temperature (T), and the number of moles of gas (n) using the equation PV = nRT where R is the gas constant. For this scenario, we are given that the volume (V) of the tire is 0.855 L (converted from 855 ml), the pressure (P) is 125 psi (which we must convert to atmospheres, since the gas constant uses atmospheres), and the temperature (T) is 25 degrees Celsius (which we must convert to Kelvin).
First, let's convert the pressure to atmospheres. Since 1 atm = 14.7 psi, the pressure in atmospheres is 125 psi / 14.7 psi/atm = 8.5034 atm. Next, let's convert the temperature to Kelvin: T = 25 °C + 273.15 = 298.15 K.
Using the ideal gas law equation PV = nRT, and solving for n (the number of moles), we get n = PV / RT. Substituting the known values for helium (R = 0.0821 L atm/mol K) and air (assuming average R = same as for helium, as there are no significant differences for the purposes of this problem):
For Air: n = (8.5034 atm) * (0.855 L) / (0.0821 L atm/mol K × 298.15 K) = 0.294 moles of air
For Helium: We use the same calculation as for air, since the volume, pressure, and temperature are the same.
Now, using the molar mass, we calculate the mass of each gas:
The mass of air = n (number of moles of air) × molar mass of air = 0.294 moles × 28.8g/mol = 8.4752 g
The molar mass of helium = 4.0026 g/mol, so the mass of helium = n (number of moles of helium) × molar mass of helium = 0.294 moles × 4.0026 g/mol = 1.1778 g
The mass difference between air-filled and helium-filled tires = mass of air - mass of helium = 8.4752 g - 1.1778 g = 7.2974 g. Thus, filling tires with helium instead of air would make them lighter by 7.2974 g.
An archeological artifact has a carbon-14 decay rate of 2.75 dis/min·gC. If the rate of decay of a living organism is 15.3 dis/min·gC, how old is this artifact? Assume that t1/2 for carbon-14 is 5730 yr.1.42 × 104 yr1.209 × 10−4 yr6.16 × 103 yr1.03 × 103 yr
The artifact has an approximate age of 6146 years.
Explanation:To determine the age of the artifact, we can use the formula for radioactive decay. The decay rate of the artifact is given as 2.75 dis/min·gC, and the decay rate of a living organism is 15.3 dis/min·gC. Using the formula:
t = (ln(N0/N))/(k)
where t is the time in years, N0 is the initial amount of carbon-14, N is the current amount of carbon-14, and k is the decay constant, we can calculate the age of the artifact.
Substituting the given values:
t = (ln(15.3/2.75))/(k)
where k = 0.693/t1/2
Using the half-life of carbon-14 (5730 years), we can calculate k:
t = (ln(15.3/2.75))/(0.693/5730)
Calculating this expression gives us an answer of approximately 6146 years, so the artifact is approximately 6146 years old.
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If 25 grams of NaI is mixed with excess amount Pb(NO3)2, what would be the product?
Answer:
If 25 grams of NaI is mixed with an excess amount of Pb(NO3)2 we get 38.45 grams of PbI2 and 14.18 grams of NaNO3
Explanation:
Step 1: Data given
Mass of NaI = 25.00 grams
Pb(NO3)2 is in excess
Step 2: The balanced equation
2NaI + Pb(NO3)2 → PbI2 + 2NaNO3
Step 3: Calculate moles of NaI
Moles NaI = mass NaI/ molar mass NaI
Moles NaI = 25.00/ 149.89 g/mol
Moles NaI = 0.1668 moles
Step 4: Calculate moles of PbI2 and NaNO3
For 2 moles of NaI we need 1 mol of Pb(NO3)2 to produce 1 mol of PbI2 and 2 moles of NaNO3
For 0.1668 moles of of NaI we will have 0.1668/2 = 0.0834 moles of PbI2 and 0.1668 moles of NaNO3
Step 5: Calculate mass of the products
Mass PbI2 = 0.0834 *461.01 g/mol
Mass PbI2 = 38.45 grams
Mass NaNO3 = 0.1668 mol * 84.99 g/mol
Mass NaNO3 = 14.18 grams
If 25 grams of NaI is mixed with an excess amount of Pb(NO3)2 we get 38.45 grams of PbI2 and 14.18 grams of NaNO3
Lithium diisopropylamide is a strong, nonnucleophilic base. It is often freshly prepared by treating a certain reactant with n-butyllithium (n-BuLi). Draw the starting material and draw the product (lithium diisopropylamide). Include any charges, but you do not need to draw electron pairs.
Answer:
Starting material is Diisopropylamine.
Explanation:
By reacting Diisopropylamine with n-Butyllithium in dry cold conditions with Tetrahydrofuran as solvent, Lithium diisopropylamide is prepared. Please see the attached image for reference.
5 Fe2+ + MnO4− + 8 H+ ⇄ 5 Fe3+ + Mn2+ + 4 H2O In a titration experiment based on the equation above, 25.0 milliliters of an acidified Fe2+ solution requires 14.0 milliliters of standard 0.050-molar MnO4− solution to reach the equivalence point. The concentration of Fe2+ in the original solution is…
(A) 0.0010 M
(B) 0.0056 M
(C) 0.028 M
(D) 0.090 M
(E) 0.14 M
Answer:
0.14 M
Explanation:
Volume of Fe2+ required for titration = 25 mL
Volume of MnO4- required for titration = 14 mL
Concentration of MnO4- = 0.05 M
We have to find the concentration of Fe2+.
Let it be x.
In a redox reaction:
Number of equivalence of oxidising agent = Number of equivalence of reducing agent.
Formula for number of equivalence: n*M*V
n = n-factor (number of electrons gained or lost by 1 mole of reducing or oxidising agent)
M = molarity
V = volume
n-factor of Fe2+ is 1 as it changes to Fe3+ by loosing 1 electron
n-factor of MnO4- is 5 as it changes to Mn2+ by gaining 5 electrons.
Number of equivalence of Fe2+ = Number of equivalence of MnO4-
1*(x)*25 = 5*0.05*14
x = 0.14 M
The concentration of Fe2+ in the original solution is 0.14 M. This is calculated using stoichiometry and the molarity of the MnO4− solution used in the reaction.
Explanation:The given question is about stoichiometry and molarity in a chemistry experiment involving a redox reaction of Fe2+ and MnO4−. To determine the concentration of Fe2+ in the original solution, we will use the reaction stoichiometry and the information given about the volume and concentration of the MnO4− solution. From the balanced equation, 5 moles of Fe2+ react with 1 mole of MnO4−. We know that the number of moles can be calculated as the product of volume (in litres) and molarity. Hence the moles of MnO4− is 14.0 mL (or 0.014 L) * 0.050 mol/L = 0.0007 mol. Therefore, the moles of Fe2+ = 5 * moles of MnO4− = 5 * 0.0007 mol = 0.0035 mol. The volume of Fe2+ solution is 25.0 mL or 0.025 L. Therefore, the concentration (molarity) of Fe2+ in the original solution = moles/volume = 0.0035 mol/0.025 L = 0.14 M. Therefore, the answer is (E) 0.14 M.
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Phosphorus is present in seawater to the extent of 0.07 ppm by mass. You may want to reference (Page) Section 18.3 while completing this problem. Part A Assuming that the phosphorus is present as dihydrogenphosphate, H2PO43−, calculate the corresponding molar concentration of phosphate in seawater. The density of seawater is 1.025 g/mL.
The molar concentration of phosphate (present as dihydrogen phosphate) in seawater can be calculated by converting the given mass concentration of phosphate (0.07 ppm) into moles using the molar mass of phosphate. After this conversion, the phosphate concentration is approximately 7.14 x 10^-4 mol/L.
Explanation:The question asks us to calculate the molar concentration of the dihydrogen phosphate ion in seawater. Remember that molar concentration is the amount of a solute (in this case phosphorus as dihydrogen phosphate) divided by the volume of the solution (seawater). We are given that the concentration of phosphorus is 0.07 ppm by mass, which, converted into grams per liter (g/L), yields 0.07 g/L. Additionally, the molar mass of H₂PO₄⁻ is approximately 97.994 g/mol.
Step-by-step calculation:
First, convert the given mass concentration of phosphorus into moles by using the molar mass of phosphorus: (0.07 g/L) / (97.994 g/mol) = 7.14 x 10⁻⁴ mol/L.Since the phosphorus was assumed to be in the form of dihydrogen phosphate, we don't need any further calculations as the molar concentration of dihydrogen phosphate is equal to that of phosphorus.Therefore, the molar concentration of phosphate in seawater is approximately 7.14 x 10⁻⁴ mol/L.
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A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 3.70kg of water at 30.5 C. During the reaction 97.1 kJ of heat flows out of the flask and into the bath.
Calculate the new temperature of the water bath. You can assume that the specific heat capacity of water under these conditions is 4.18J / g*K. Round to the 3 significant digit.
Answer:
The new temperature of the water bath is 36.8°C
Explanation:
A Calorimety problem to apply this formula:
Q = m . C . ΔT
Where Q is heat
m is mass
C is specific heat capacity
ΔT = T° final - T° initial
Let's make some conversion before
97.1 kJ .1000 = 97100 Joules
3.70 kg . 1000 = 3700 g
This because the specific heat capacity units
ΔT it's a difference and mathematically, in numbers, it is the same value in °C as in K
97100 J = 3700 g . 4.18J / g°C ( T°final - 30.5°C)
97100 J = 15466 J/°C (T°final - 30.5°C)
97100 J / 15466 °C/J = (T°final - 30.5°C)
6.28°C = T °final - 30.5°C
6.28° C + 30.5°C = T°final
36.8°C = T° final
A copper cube with a mass of 100 grams is heated in a boiling water bath to 100.0 Celsius. The cube is removed from the bath and placed on a very large block of ice which is at a temperature of 0.00 Celsius. What is the maximum mass of ice that could conceivably be melted by the heat exchange, assuming all the thermal energy is used to melt the ice? The specific heat of copper is 0.39 J/g-C and the enthalpy of fusion for water is 334 J/g.
Answer:
The copper cube has a mass of 11.68 grams
Explanation:
Step 1: Data given
Mass of the copper cube = 100 grams
Temperature of water = 100°C
Temperature of ice = 0°C
Specific heat of copper = 0.39J/g°C
Enthalpy of fusion of water = 334 J/g
Step 2: Calculate the energy for cooling the copper
energy in the copper as it cools from 100º to 0º = 0.39 J/g°C * 100g *100°C = 3900 J
Step 3: Calculate the mass of the copper cube
3900 J / 334 J/g = 11.68 grams
The copper cube has a mass of 11.68 grams
M = 3900/334 = 11.68 g
change is 3900 J
The maximum mass of ice that could be melted by the heat exchange is approximately 11.68 grams.
Explanation:To find the maximum mass of ice that could be melted by the heat exchange, we need to calculate the amount of thermal energy transferred from the copper cube to the ice. We can use the equation Q = m * c * ΔT, where Q is the thermal energy, m is the mass, c is the specific heat, and ΔT is the change in temperature.
First, let's calculate the thermal energy transferred by the copper cube in the boiling water bath. The cube's mass is 100 grams, the specific heat of copper is 0.39 J/g-C, and the temperature change is 100.0 - 0.0 = 100.0 Celsius. Therefore, Q1 = (100g) * (0.39 J/g-C) * (100.0 C) = 3900 J.
Next, let's calculate the thermal energy required to melt the ice. The enthalpy of fusion for water is 334 J/g. Assuming that all the thermal energy is used to melt the ice, we can calculate the maximum mass of ice that could be melted using the equation Q2 = m * ΔHf, where Q2 is the thermal energy, m is the mass, and ΔHf is the enthalpy of fusion. Rearranging the equation, we have m = Q2 / ΔHf.
Since the thermal energy transferred by the copper cube is equal to the thermal energy required to melt the ice, we can equate Q1 and Q2: 3900 J = m * (334 J/g). Solving for m, we get m = 3900 J / 334 J/g = 11.68 g.
Therefore, the maximum mass of ice that could be melted by the heat exchange is approximately 11.68 grams.
A sample has a C614 activity of 0.0015 Bq per gram of carbon.
(a) Find the age of the sample, assuming that the activity per gram of carbon in a living organism has been constant at a value of 0.23 Bq.
(b) Evidence suggests that the value of 0.23 Bq might have been as much as 45% larger. Repeat part (a), taking into account this 45% increase.
Answer:
[tex]\large \boxed{\text{(a) 42 000 yr; (b) 45 000 yr}}[/tex]
Explanation:
Two important equations in radioactive decay are
[tex]\ln \dfrac{N_{0} }{N_{t}} = kt\\\\t_{\frac{1}{2}} = \dfrac{\ln2}{k }[/tex]
We use them for carbon dating.
(a) Initial activity = 0.23 Bq
(i) Calculate the rate constant
The half-life of ¹⁴C is 5730 yr.
[tex]\begin{array}{rcl}t_{\frac{1}{2}}& = &\dfrac{\ln2}{k }\\\\k& = &\dfrac{\ln2}{t_{\frac{1}{2}}}\\\\ & = & \dfrac{\ln2}{\text{5730 yr}}\\\\ & = & 1.210 \times 10^{-4}\text{ yr}^{-1}\\\end{array}[/tex]
(ii) Calculate the age of the sample
[tex]\begin{array}{rcl}\ln \dfrac{N_{0} }{N_{t}} & = & kt\\\\\ln \dfrac{0.23 }{0.0015} & = & k\times 1.210 \times 10^{-4}\text{ yr}^{-1}\\\\\ln 153 & = & 1.210 \times 10^{-4}k \text{ yr}^{-1}\\5.03 & = & 1.210 \times 10^{-4}k \text{ yr}^{-1}\\k & = & \dfrac{5.03}{1.210 \times 10^{-4} \text{ yr}^{-1}}\\\\ & = & \textbf{42 000 yr}\\\end{array}\\\text{The age of the sample is $\large \boxed{\textbf{42 000 yr}}$}[/tex]
(b) Initial activity = 45 % larger
N₀ = 1.45 × 0.230 Bq = 0.334 Bq
[tex]\begin{array}{rcl}\ln \dfrac{N_{0} }{N_{t}} & = & kt\\\\\ln \dfrac{0.334 }{0.0015} & = & k\times 1.210 \times 10^{-4}\text{ yr}^{-1}\\\\\ln 222 & = & 1.210 \times 10^{-4}k \text{ yr}^{-1}\\5.40 & = & 1.210 \times 10^{-4}k \text{ yr}^{-1}\\k & = & \dfrac{5.40}{1.210 \times 10^{-4} \text{ yr}^{-1}}\\\\ & = & \textbf{45 000 yr}\\\end{array}\\\text{The age of the sample is $\large \boxed{\textbf{45 000 yr}}$}[/tex]
Which processes of the water cycle involve the movement of water in the liquid state?
Question 6 options:
evaporation and precipitation
surface runoff and condensation
precipitation and surface runoff
condensation and evaporation
Answer:
Surface runoff and condensation
Explanation:
Let's define each of the given processes in order to understand them better:
evaporation is a process in which liquid phase transforms into a gas phase;precipitation is a process in which we produce a solid phase, usually this is the case when we precipitate a salt out of a solution, analogy of precipitation for water would be transformation from a liquid to a solid phase, such as freezing;surface runoff is a process in which water flows over the surface of a land without any change in its phase;condensation is a process in which a gas transforms into a liquid.All in all, notice that surface runoff keeps water in its liquid state, while all the other three options consider phase change. The only phase change of interest is condensation: we produce liquid water from water vapor and then we can analyze its movement in the liquid state.
Combustion of hydrocarbons such as propane ( C3H8 ) produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's atmosphere can trap the Sun's heat, raising the average temperature of the Earth. For this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide.
1. Write a balanced chemical equation, including physical state symbols, for the combustion of gaseous propane into gaseous carbon dioxide and gaseous water.
2. Suppose 0.150kg of propane are burned in air at a pressure of exactly 1atm and a temperature of 12.0°C . Calculate the volume of carbon dioxide gas that is produced. Round your answer to 3 significant digits. L
Answer:
239 L
Explanation:
1. Propane reacts with oxygen to produce carbon dioxide and water:
[tex]C_3H_8 (g) + O_2 (g)\rightarrow CO_2 (g) + H_2O (l)[/tex]
Firstly, 3 carbon atoms are required on the right:
[tex]C_3H_8 (g) + O_2 (g)\rightarrow 3 CO_2 (g) + H_2O (l)[/tex]
Secondly, 8 hydrogens in total (4 water molecules) are required on the right:
[tex]C_3H_8 (g) + O_2 (g)\rightarrow 3 CO_2 (g) + 4 H_2O (l)[/tex]
On the right, we have a total of 10 oxygen atoms, this implies we need 5 oxygen molecules on the left:
[tex]C_3H_8 (g) + 5 O_2 (g)\rightarrow 3 CO_2 (g) + 4 H_2O (l)[/tex]
2. Calculate moles of propane using the the ratio of mass to molar mass:
[tex]n_1 = \frac{m_1}{M_1} = \frac{150 g}{44.1 g/mol} = 3.40 mol[/tex]
According to the stoichiometry, we have 3 times greater amount of carbon dioxide:
[tex]n_2 = 3n_2 = 3\cdot 3.40 mol = 10.2 mol[/tex]
Use the ideal gas law to solve for volume:
[tex]pV = nRT\therefore V = \frac{nRT}{p} = \frac{10.2 mol\cdot 0.08206 \frac{L atm}{mol K}\cdot 285.15 K}{1 atm} = 239 L[/tex]
An analytical chemist weighs out of an unknown diprotic acid into a volumetric flask and dilutes to the mark with distilled water. He then titrates this solution with solution. When the titration reaches the equivalence point, the chemist finds he has added of solution. Calculate the molar mass of the unknown acid. Be sure your answer has the correct number of significant digits.
The molar mass of an unknown diprotic acid is determined through a titration procedure. Using the volume and molarity of NaOH, we calculate the moles of NaOH, then divide by 2 to get the moles of diprotic acid. Dividing the mass of the acid by the moles gives us the molar mass.
Explanation:In this problem, the student is asked to calculate the molar mass of an unknown diprotic acid. The task requires the understanding of acids, stoichiometry, and titration concepts in analytical chemistry.
A diprotic acid is an acid that can donate two protons, or hydrogen ions, per molecule in solution. As the question doesn't provide the exact values, I'll explain conceptually. The molar mass of the unknown acid is calculated by using the volume of NaOH needed to reach the equivalence point and the concentration of NaOH. We'd start by calculating the moles of NaOH used (moles = volume x molarity), then because it's a diprotic acid, for every mole of acid, two moles of NaOH are needed, so we'd divide the moles of NaOH by 2 to get moles of unknown acid (moles acid = moles NaOH / 2). Lastly to find the molar mass, divide the mass of the acid used by the moles of acid calculated. So, the molar mass = mass of acid / moles of acid.
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The titration analysis provides a method to calculate the molar mass of an unknown acid using the volume of a known solution required to reach the equivalence point of the reaction. The process involves incrementally adding the known solution to the unknown, and using the reaction stoichiometry to find the unknown concentration. The molar mass is then calculated from this concentration, the initial volume of solution, and the molecular formula of the reaction products.
Explanation:Based on the information provided, we need further details to accurately calculate the molar mass of the unknown diprotic acid. However, we can understand the general process involved in calculating such values using titration data.
The process is generally as follows:
Titration analysis is conducted by incrementally adding a solution of known concentration (the titrant) to the unknown solution (analyte).The titrant and analyte undergo a chemical reaction of known stoichiometry. The volume of titrant required to reach the 'equivalence point' of the reaction - where both the titrant and analyte have been completely reacted - allows us to calculate the concentration of the unknown analyte solution. The molar mass of the unknown acid can then be calculated using the resulting concentration, the initial volume of solution, and the defined molecular formula and mass of the reaction products. Learn more about Titration here:https://brainly.com/question/31271061
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A voltaic cell is constructed in which the anode is a Zn|Zn2+ half cell and the cathode is a Fe2+|Fe3+ half cell. The half-cell compartments are connected by a salt bridge.(Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.)The anode reaction is: ____+_____ yeilds _____ +______The cathode reaction is: _____+_______ yeilds_____ +________The net cell reaction is: _____+ _______yeilds _______+______In the external circuit, electrons migrate _______the Fe2+|Fe3+ electrode _______the Zn|Zn2+ electrode.In the salt bridge, anions migrate ________ the Fe2+|Fe3+ compartment ________ the Zn|Zn2+ compartment.
In a voltaic cell set up with Zn and Fe half cells, the anode reaction is Zn (s) → Zn2+ (aq) + 2e-, the cathode reaction is Fe3+ (aq) + e- → Fe2+ (aq), and the net cell reaction is Zn (s) + Fe3+ (aq) → Zn2+ (aq) + Fe2+ (aq). Electrons move from the Zn electrode to the Fe electrode and anions in the salt bridge migrate from the Zn compartment to the Fe compartment.
Explanation:In a voltaic cell constructed in which the anode is a Zn|Zn2+ and the cathode is a Fe2+|Fe3+ half cell, the reactions would look something like this:
Anode reaction: Zn (s) → Zn2+ (aq) + 2e-
Cathode reaction: Fe3+ (aq) + e- → Fe2+ (aq)
Net cell reaction: Zn (s) + Fe3+ (aq) → Zn2+ (aq) + Fe2+ (aq)
In the external circuit, electrons migrate from the Zn|Zn2+ electrode to the Fe2+|Fe3+ electrode. In the salt bridge, anions migrate from the Zn|Zn2+ compartment to the Fe2+|Fe3+ compartment.
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A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard reduction potential (s)(aq)(aq)(l) (aq)(aq) Answer the following questions about this cell. Write a balanced equation for the half-reaction that happens at the cathode. Write a balanced equation for the half-reaction that happens at the anode. Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written. Do you have enough information to calculate the cell voltage under standard conditions
Answer:
Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)
Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻
Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)
E°cell = 1.10 V
Explanation:
The half-reactions are missing, but I will propose some to show you the general procedure and then you can apply it to your equations.
Suppose we have the following half-reactions.
Cu²⁺(⁺aq) + 2 e⁻ → Cu(s) E°red = 0.34 V
Zn²⁺(⁺aq) + 2 e⁻ → Zn(s) E°red = -0.76 V
To identify how to make a spontaneous cell, we need to consider the standard reduction potentials (E°red). The half-reaction with the higher E°red will occur as a reduction (in the cathode), whereas the one with the lower E°red will occur as an oxidation (in the anode).
Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s) E°red = 0.34 V
Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻ E°red = -0.76 V
To get the overall equation we add both half-reactions.
Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)
The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E°cell = E°red, cat - E°red, an
E°cell = 0.34 V - (-0.76 V) = 1.10 V
Since E°cell > 0, the reaction is spontaneous.
With the provided half-reactions information, one could write the reduction and oxidation reactions at the cathode and anode, respectively, and determine the overall balanced equation for the spontaneous redox reaction in a galvanic cell. The positive and negative electrodes, identifying the cathode and anode, play a key role in this determination. However, without the specific details of the half-reactions, it's not possible to write the balanced equations or calculate the cell voltage.
Explanation:To answer questions about the galvanic cell, we require specifics about the half-reactions involved. The half-reaction that occurs at the cathode is the reduction process, and the half-reaction that occurs at the anode is the oxidation process. However, without the actual half-reactions or the substances involved, it is impossible to write the precise equations.
In general terms, if the half-reactions and their respective standard reduction potentials are known, we can write the reactions as follows:
Cathode (Reduction): A⁺⁺+(aq) + ne⁻ → A(s) (where n is the number of electrons transferred)Anode (Oxidation): B(s) → B⁺⁺+(aq) + ne⁻The electrode at which reduction occurs, the cathode, is the positive electrode, while the electrode at which oxidation occurs, the anode, is the negative electrode. For the overall balanced equation that powers the cell, simply combine the two half-reactions to get the full-cell reaction.
To calculate the cell voltage under standard conditions, the difference in standard electrode potentials between the cathode and anode is taken (E°cathode - E°anode). The overall reaction should be spontaneous if the cell voltage is positive.
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For which of the following processes will \DeltaΔS be negative?PbCl2(s) = Pb2+(aq) + 2 Cl-(aq)MgO(s) + CO2(g) = MgCO3(s)CO2(aq) = CO2(g)C5H12(l) + 8 O2(g) = 5 CO2(g) + 6 H2O(g)
Answer
a) [tex]PbCl_2(s)\rightarrow Pb^{2+}(aq)+2Cl^-(aq)[/tex]: [tex]\Delta S[/tex] = +ve
b) [tex]MgO(s)+CO_2(g)\rightarrow MgCO_3(s)[/tex]: [tex]\Delta S[/tex] = -ve
c) [tex]CO_2(aq)\rightarrow CO_2(g)[/tex]: [tex]\Delta S[/tex]= +ve
d) [tex]C_5H_{12}(l)+8O_2(g)\rightarrow 5CO_2(g)+6H_2O(g)[/tex]: [tex]\Delta S[/tex] = +ve
Explanation:
Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.
a) [tex]PbCl_2(s)\rightarrow Pb^(2+)(aq)+2Cl^-(aq)[/tex]
As solid is moving to ions form , randomness increases and thus sign of [tex]\Delta S[/tex] is positive.
b)[tex]MgO(s)+CO_2(g)\rightarrow MgCO_3(s)[/tex]
As gaseous reactants are converted to solid products , randomness decreases and thus sign of [tex]\Delta S[/tex] is negative.
c) [tex]CO_2(aq)\rightarrow CO_2(g)[/tex]
As liquid is changing to gas randomness increases and thus sign of [tex]\Delta S[/tex] is positive.
d) [tex]C_5H_{12}(l)+8O_2(g)rightarrow 5CO_2(g)+6H_2O(g)[/tex]
As 8 moles of gaseous reactants are converted to 11 moles of gaseous products , randomness increases and thus sign of [tex]\Delta S[/tex] is positive.
How do you read beakers to the correct amount of significant figures?
Please help
Final answer:
To read beakers accurately, identify the smallest increment, estimate one decimal place beyond it, align your eye level with the meniscus, and report the measurement with the correct number of significant figures and units.
Explanation:
To read beakers to the correct amount of significant figures, you should:
Look at the measurement scale of the beaker and identify the smallest increment.Determine the value of one division by the smallest increment. This informs you of the degree of precision of the beaker.Record measurements by estimating one decimal place beyond the smallest marked increment on the beaker. For example, if the beaker has markings every 10 mL, you should estimate and record the volume to the nearest 1 mL.Make sure your eye level is at the same height as the meniscus of the liquid to avoid parallax error when taking the reading.Perform calculations using the measured volume, maintaining the correct number of significant figures throughout the process.When reporting your final measurement, round appropriately according to the number of significant figures your equipment allows, and apply the correct units.For instance, if you measure a volume that reads between 50 mL and 60 mL and the meniscus is halfway between the two, you would report a volume of 55 mL. If you are using a more precise beaker with 1 mL increments, you might report 55.5 mL where the 5 is an estimate.
It is crucial to ensure that all your reported measurements and calculated results have the proper number of significant figures and proper units, maintaining precision and accuracy in your experiments.
Which of the following conditions remain constant in Charles's law?
Volume and pressure
Density and temperature
Pressure and number of moles
Temperature and number of moles
Answer:
Pressure and number of moles.
Explanation:
According to Charles law,
For an ideal gas, volume is directly proportional to temperature at constant pressure for a fixed amount of gas.
V ∝ T
[tex]\frac{V}{T}=constant[/tex]
From the law we can say that the two conditions are:
Pressure remains constant.Number of moles do not change (As the amount of gas is fixed).The condition that remains constant in Charles's law is Pressure and number of moles.
What is Charles's law?Charles's law states that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature, if the pressure remains constant.
The volume of a gas is directly proportional to it's number of moles.
Therefore, the condition that remains constant in Charles's law is Pressure and number of moles.
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There are 7.11 x1024molecules in 100.0 cm3of a certain substance.a. What is the number of molecules in 1.09 cm3of the substance?7.75 x 1022moleculesb.What is the number of molecules in 2.24 x 104cm3of the substance?1.59 x 10 27moleculesc. What number of molecules would be in 9.01 x 10-6cm3?6.41 x 1017molecules
Answer:
a. 7.75 × 10²² molecules
b. 1.59 × 10²⁷ molecules
c. 6.41 × 10¹⁷ molecules
Explanation:
There are 7.11 × 10²⁴ molecules in 100.0 cm³ of a certain substance. We will use this ratio in our conversion fractions.
a. What is the number of molecules in 1.09 cm³ of the substance?
1.09 cm³ × (7.11 × 10²⁴ molecules/100.0 cm³) = 7.75 × 10²² molecules
b. What is the number of molecules in 2.24 × 10⁴ cm³ of the substance?
2.24 x 10⁴ cm³ × (7.11 × 10²⁴ molecules/100.0 cm³) = 1.59 × 10²⁷ molecules
c. What number of molecules would be in 9.01 × 10⁻⁶ cm³?
9.01 × 10⁻⁶ cm³ × (7.11 × 10²⁴ molecules/100.0 cm³) = 6.41 × 10¹⁷ molecules
Calculating the number of particles in a given volume involves using Avogadro's number and proportionality between the volume and the number of particles, assuming constant density.
Explanation:The question involves concepts from chemistry, specifically dealing with Avogadro's number, the mole, and calculations of the number of molecules in a given volume. Avogadro's number is defined as the number of atoms or molecules in one mole of a substance, which is 6.022 × 10²23. When given the number of molecules in a specific volume, you can calculate the number of molecules in a different volume by setting up a proportional relationship. This is because the number of molecules is directly proportional to the volume if the density remains the same.
For example, if there are 7.11 × 10²24 molecules in 100.0 cm³ of a substance, to find the number of molecules in 1.09 cm³:
Set up a proportion: (7.11 × 10²24 molecules / 100 cm³) = (x molecules / 1.09 cm³)Solve for x to find the number of molecules in 1.09 cm³.Similarly, you would use the same approach for other volumes to determine the number of molecules in 2.24 × 10²4 cm³ and 9.01 × 10²²6 cm³ respectively.
According to the following reaction, how many grams of potassium phosphate will be formed upon the complete reaction of 29.6 grams of phosphoric acid with excess potassium hydroxide? 3KOH(aq)+H3PO4(aq) <=>K3PO4(aq)+3H2O(l)
Upon the complete reaction of 29.6 grams of phosphoric acid with excess potassium hydroxide, approximately 64 grams of potassium phosphate will be produced.
Explanation:To determine the amount of potassium phosphate that will be produced, we first need to find the molar mass of phosphoric acid (H3PO4), which is approximately 98 g/mol. So, 29.6 grams of phosphoric acid is equal to 0.302 moles (29.6g / 98g/mol).
The balanced equation shows that each mole of phosphoric acid will produce one mole of potassium phosphate (K3PO4) in the reaction. Therefore, 0.302 moles of phosphoric acid will produce 0.302 moles of potassium phosphate.
Next, we need to convert this amount from moles to grams. The molar mass of potassium phosphate is roughly 212 g/mol. So, 0.302 moles of K3PO4 is equal to approximately 64 grams (0.302 moles * 212 g / mol).
Therefore, upon the complete reaction of 29.6 grams of phosphoric acid with excess potassium hydroxide, approximately 64 grams of potassium phosphate will be formed.
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The standard reaction enthalpy of Zn(s) + H2O(g) →ZnO(s) + H2(g) is known to be HR 0 = 224 kJ and is approximately constant from 920 K up to 1280 K. The standard reaction free energy is +33 kJ at 1280 K. Calculate the equilibrium constant at 1280 K and then calculate the temperature at which the equilibrium constant becomes greater than 1.
Answer:
K = 0.0450, T > 1501 K
Explanation:
We may apply the Gibbs free energy equation relating the Gibbs free energy to an equilibrium constant of a reaction. The relationship is described by the following equation:
[tex]\Delta G^o = -RT ln(K)[/tex]
Rearrange the equation for the equilibrium constant:
[tex]ln(K) = -\frac{\Delta G^o}{RT}\therefore K = e^{-\frac{\Delta G^o}{RT}}[/tex]
Given the temperature [tex]T = 1280 K[/tex] and the ideal gas law constant [tex]R = 8.314 \frac{J}{K mol}[/tex], we obtain:
[tex]K = e^{-\frac{33\cdot 10^3 J}{8.314 \frac{J}{K mol}\cdot 1280 K}} = 0.0450[/tex]
Now notice if [tex]K > 1[/tex], then [tex]ln(K) > 1[/tex] and [tex]\Delta G^o < 0[/tex].
We may firstly solve for the entropy change of this reaction using the following equation:
[tex]\Delta G^o = \Delta H^o - T\Delta S^o \therefore \Delta S^o = \frac{\Delta H^o - \Delta G^o}{T} = \frac{224\cdot 10^3 J - 33\cdot 10^3 J}{1280 K} = 149.2 \frac{J}{K}[/tex]
Using the same equation, solve when the change in the Gibbs free energy is negative:
[tex]\Delta H^o - T\Delta S^o < 0\therefore T > \frac{\Delta H^o}{\Delta S^o} = \frac{224\cdot 10^3 J}{149.2 \frac{J}{K}} = 1501 K[/tex]
Final answer:
The equilibrium constant (K) at 1280 K for the reaction Zn(s) + H2O(g) → ZnO(s) + H2(g) is calculated using the standard reaction free energy and the relationship ΔG0 = -RTln(K). K is found to be e⁻³.152, which is less than 1. The temperature at which K becomes greater than 1 is approximately 1501 K.
Explanation:
The student's question is about calculating the equilibrium constant (K) at 1280 K for the reaction Zn(s) + H2O(g) → ZnO(s) + H2(g) given the standard reaction enthalpy (ΔHR0 = 224 kJ) and the standard reaction free energy (ΔG0 = +33 kJ) at this temperature. To calculate K, we can use the relationship between free energy and the equilibrium constant given by the equation ΔG0 = -RTln(K), where R is the gas constant (8.314 J/mol·K) and T is the temperature in kelvins. Rearranging and solving for K yields:
K = e(-ΔG0)/(RT)
At 1280 K:
K = e(-33000 J/mol)/(8.314 J/mol·K × 1280 K)
K ≈ e-3.152
The equilibrium constant K at 1280 K is approximately e-3.152, which is less than 1.
To find the temperature at which K becomes greater than 1, we need to find the temperature at which ΔG0 becomes negative. Since the relationship between ΔH0, ΔG0, and ΔS0 (the standard entropy change) is given by ΔG0 = ΔH0 - TΔS0, we can set ΔG0 to 0 and solve for T, assuming ΔS0 remains constant.
0 = ΔH0 - TΔS0
T = ΔH0 / ΔS0
We don't have ΔS0, but using the provided ΔH0 and ΔG0 values at a known temperature, we can calculate it:
ΔS0 = (ΔH0 - ΔG0) / T
ΔS0 = (224000 J/mol - 33000 J/mol) / 1280 K
ΔS0 = 149.22 J/(mol·K)
Now, substituting ΔS0 back into the original equation to solve for T when K becomes greater than 1:
0 = ΔH0 - TΔS0
T = ΔH0 / ΔS0
T = 224000 J/mol / 149.22 J/(mol·K)
T ≈ 1501 K
Therefore, the equilibrium constant becomes greater than 1 at a temperature of approximately 1501 K.