Iron(III) oxide and hydrogen react to form iron and water, like this: Fe_2O_3(s) + 3H_2(g) rightarrow 2Fe(s) + 3H_2O(g) At a certain temperature, a chemist finds that a 5.4 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, iron, and water at equilibrium has the following composition: Calculate the value of the equilibrium constant K_c for this reaction. Round your answer to 2 significant digits.
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Complete Question
The complete question is shown on the first uploaded image
Answer:
The equilibrium constant is [tex]K_c= 2.8*10^{-4}[/tex]
Explanation:
From the question we are told that
The chemical reaction equation is
[tex]Fe_{2} O_{3}_{(s)} + 3H_{2}_{(g)} -----> 2Fe_{(s)} + 3H_{2} O_{(g)}[/tex]
The voume of the misture is [tex]V_m = 5.4L[/tex]
The molar mass of [tex]Fe_{2} O_{3}_{(s)}[/tex] is a constant with value of [tex]M_{Fe_{2} O_{3}_{(s)} } = 160g/mol[/tex]
The molar mass of [tex]H_{2}_{(g)}[/tex] is a constant with value of [tex]H_2 = 2g/mol[/tex]
The molar mass of [tex]H_{2}O[/tex] is a constant with value of [tex]H_2O = 18g/mol[/tex]
Generally the number of moles is mathematically given as
[tex]No \ of \ moles \ = \frac{mass}{molar\ mass}[/tex]
For [tex]Fe_{2} O_{3}_{(s)}[/tex]
[tex]No \ of\ moles = \frac{3.54}{160}[/tex]
[tex]= 0.022125 \ mols[/tex]
For [tex]H_{2}[/tex]
[tex]No \ of\ moles = \frac{3.63}{2}[/tex]
[tex]= 1.815 \ mols[/tex]
For [tex]H_{2}O[/tex]
[tex]No \ of\ moles = \frac{2.13}{18}[/tex]
[tex]= 0.12 \ mols[/tex]
Generally the concentration of a compound is mathematicallyrepresented as
[tex]Concentration = \frac{No \ of \ moles }{Volume }[/tex]
For [tex]Fe_{2} O_{3}_{(s)}[/tex]
[tex]Concentration[Fe_2 O_3] = \frac{0.222125}{5.4}[/tex]
[tex]= 4.10*10^{-3}M[/tex]
For [tex]H_{2}[/tex]
[tex]Concentration[H_2] = \frac{1.815}{5.4}[/tex]
[tex]= 0.336M[/tex]
For [tex]H_{2}O[/tex]
[tex]Concentration [H_2O] = \frac{0.12}{5.4}[/tex]
[tex]= 0.022M[/tex]
The equilibrium constant is mathematically represented as
[tex]K_c = \frac{[concentration \ of \ product]}{[concentration \ of \ reactant ]}[/tex]
Considering [tex]H_2O \ for \ product[/tex]
And [tex]H_2 \ for \ reactant[/tex]
At equilibrium the
[tex]K_c = \frac{0.022}{0.336}[/tex]
[tex]K_c= 2.8*10^{-4}[/tex]
Related Questions
Rock candy is formed when excess sugar is dissolved in hot water followed by crystallization. A student wants to make two batches of rock candy. He finds an unopened box of "cane sugar" in the pantry. He starts preparing batch A by dissolving sugar in of hot water (). He keeps adding sugar until no more sugar dissolves in the hot water. He cools the solution to room temperature. He prepares batch B by dissolving sugar in of water at room temperature until no more sugar is dissolved. He lets the solution sit at room temperature.It is likely that less rock candy will be formed in batch A. It is likely that no rock candy will be formed in either batch. I need more information to predict which batch is more likely to form rock candy.
Answers
Rock candy is formed through a chemical process known as crystallization which requires a supersaturated sugar solution. In the scenario, batch A prepared with hot water is more likely to form rock candy as it can dissolve more sugar creating a supersaturated solution. Batch B prepared at room temperature may not form as much rock candy due to lesser sugar dissolution.
Explanation:The question relates to the chemical process of crystallization, particularly in the formation of rock candy. When making rock candy, a supersaturated solution of sugar and water is required. This condition is achieved when as much sugar as possible is dissolved in hot water. Once cooled, the oversaturated solution starts to crystallize and forms rock candy. So in the given scenario, batch A is more likely to produce rock candy because it involves the preparation of a supersaturated solution through dissolving sugar in hot water. Batch B, prepared at room temperature, may not dissolve as much sugar as batch A, and thus, less or no rock candy might be formed.
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Batch A, which dissolved sugar in hot water that was then cooled, is more likely to form rock candy as the cooling process can lead to the crystallization of sugar from a supersaturated solution.
The process of making rock candy involves dissolving sugar in water to create a saturated solution from which sugar crystals can form upon cooling or evaporation of the solvent. The solubility of sugar increases with temperature, which means hot water can dissolve more sugar than room temperature water. Therefore, for batch A, dissolving sugar in hot water likely supersaturates the solution, and as it cools to room temperature, excess sugar will crystallize out.
Batch B, on the other hand, dissolves sugar at room temperature, potentially creating a saturated solution, but without the temperature change, it is less likely to form a supersaturated environment and thus may yield less crystallization compared to batch A. In summary, batch A is more likely to form rock candy due to the temperature-dependent solubility of sugar, and the cooling process allows crystals to form from the supersaturated solution.
A) Al3 (aq) is a stronger oxidizing agent than I2(s), and I-(aq) is a stronger reducing agent than Al(s).B) I2(s) is a stronger oxidizing agent than Al3 (aq), and Al(s) is a stronger reducing agent than I-(aq).C) Al(s) is a stronger oxidizing agent than I-(aq), and Al3 (aq) is a stronger reducing agent than I2(s).D) I-(aq) is a stronger oxidizing agent than Al(s), and I2(s) is a stronger reducing agent than Al3 (aq).
Answers
Answer:
B) I2(s) is a stronger oxidizing agent than Al3 (aq), and Al(s) is a stronger reducing agent than I-(aq).
Explanation:
An oxidizing agent accepts electrons in a redox reaction and become reduced while a reducing agent looses electrons and become oxidized.
Hence in a redox reaction, oxidizing agents are reduced while reducing agents are oxidized.
Looking at I2 and Al3+, I2 is a better oxidizing agent since it has a reduction potential of +0.54V compared to -1.66V for Al3+.
Given the statements above, the converse must be true, that is; All is a better reducing agent compared to I-
Which of the following statements is true? a. At equilibrium BOTH the rate of the forward reaction equals that of the reverse reaction AND the rate constant for the forward reaction equals that of the reverse. b. The equilibrium state is dynamic even though there is no change in concentrations. c. The equilibrium constant for a particular reaction is constant under all conditions. d. Starting with different initial concentrations will yield different individual equilibrium concentrations and a different relationship of equilibrium concentrations. e. None of these is true.
Answers
Answer:
a) True
Explanation:
a) From the definition of the equilibrium. When a reversible reaction is carried out in a closed vessel, a stage is reached when the forward and the backward reactions proceed with the same speed. This stage is known as chemical equilibrium.
Calculate the concentration of buffer components present in 287.00 mL of a buffer solution that contains 0.310 M NH4Cl and 0.310 M NH3 immediately after the addition of 1.50 mL of 6.00 M HNO3.
Answers
Final answer:
To determine the new concentrations of NH3 and NH4+ after the addition of HNO3, the moles of HNO3 added are calculated and the amounts of NH3 converted to NH4+ are accounted for. The final concentrations are found to be approximately 0.277 M for NH3 and 0.339 M for NH4+, after considering the change in total volume.
Explanation:
The student is asking to calculate the concentration of buffer components in a solution after the addition of a strong acid. To solve this, we have to determine how much the strong acid (HNO3) will neutralize the NH3 component of the buffer, and how it will change the concentrations of NH3 and NH4+ (the buffer components).
First, calculate the number of moles of HNO3 added:
1.50 mL × 6.00 M = 0.009 moles of HNO3Since NH3 and NH4+ are in a 1:1 molar ratio in the solution and they react with HNO3 in a 1:1 ratio, the added HNO3 will react with the NH3:
0.310 M × 0.28700 L = 0.08887 moles of NH3 (initial)0.08887 moles - 0.009 moles = 0.07987 moles of NH3 (after reaction with HNO3)And NH4+ will increase by 0.009 moles (since each mole of NH3 reacts to form one mole of NH4+):
0.310 M × 0.28700 L = 0.08887 moles of NH4+ (initial)0.08887 moles + 0.009 moles = 0.09787 moles of NH4+ (after reaction with HNO3)The new volume of the solution would be the initial volume of buffer solution plus the volume of HNO3 added, which totals 288.50 mL or 0.28850 L.
The new concentrations are therefore:
NH3 concentration = 0.07987 moles / 0.28850 L = approximately 0.277 MNH4+ concentration = 0.09787 moles / 0.28850 L = approximately 0.339 MThis change in concentrations due to the addition of HNO3 means the buffer will have adjusted to these new concentrations of NH3 and NH4+.
Complete the electron pushing mechanism for the formation of the following cyclic acetal under acidic conditions by adding any missing atoms, bonds, charges, non-bonding electrons, and curved arrows. Note the use of a generic base B: that represents any basic molecule in solution, in this case another ethylene glycol.
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Question:
The question is incomplete. See the attached file for the complete question and answer.
Explanation:
Find attached for explanation.
The first two pages is the additional question while the 3rd and last page is the answer .
Which statements are true about balancing chemical reactions?
Check all that apply
A. Atoms that are in only one of the reactants and only one of the
products should be done last.
B. Single atoms should be done last.
C. Balancing reactions involves trial and error
D. The final coefficients should be the biggest numbers possible
Answers
Final answer:
Balancing chemical reactions involves trial and error, with the aim of achieving the same number of atoms for each element on both sides of the equation. Atoms that are in only one of the reactants and only one of the products should be done last, and the final coefficients should be the biggest numbers possible.
Explanation:
There are several statements that are true about balancing chemical reactions:Atoms that are in only one of the reactants and only one of the products should be done last. This is because it is easier to balance elements that appear in multiple reactants and products first.The final coefficients should be the biggest numbers possible. This is because coefficient values should be integers and the aim is to attain the simplest whole number ratio.Balancing reactions involves trial and error. It requires adjusting the coefficients of the reactants and products until the number of atoms of each element on either side of the equation is balanced.What pressure is needed to reduce the volume of gases in a car’s cylinder from 48.0 cm3 at 102 kPa to 5.20 cm3?
Answers
Answer: The pressure that is needed is around 405kPa-755kPa
To summarize into 405PPM-755PPM
Explanation: In which 405 PPM-755 PPM which is about the same amount of pressure that water pressurises a car in water or a better example is that it's the same amount of pressure as if a penny was dropped from the sky towards a person holding a square piece of cardboard in which then the penny would directly go straight through the piece of cardboard.
The equation represents the decomposition of a generic diatomic element in its standard state. 12X2(g)⟶X(g) Assume that the standard molar Gibbs energy of formation of X(g) is 5.61 kJ·mol−1 at 2000. K and −52.80 kJ·mol−1 at 3000. K. Determine the value of K (the thermodynamic equilibrium constant) at each temperature. K at 2000. K= K at 3000. K= Assuming that ΔH∘rxn is independent of temperature, determine the value of ΔH∘rxn from this data. ΔH∘rxn=
Answers
Answer:
The equilibrium constant at 2000 K is 0.7139
The equilibrium constant at 3000 K is 8.306
ΔH = 122.2 kJ/mol
Explanation:
Step 1: Data given
the standard molar Gibbs energy of formation of X(g) is 5.61 kJ/mol at 2000 K
the standard molar Gibbs energy of formation of X(g) is -52.80 kJ/mol at 3000 K
Step 2: The equation
1/2X2(g)⟶X(g)
Step 3: Determine K at 2000 K
ΔG = -RT ln K
⇒R = 8.314 J/mol *K
⇒T = 2000 K
⇒K is the equilibrium constant
5610 J/mol = -8.314 J/molK * 2000 * ln K
ln K = -0.337
K = e^-0.337
K = 0.7139
The equilibrium constant at 2000 K is 0.7139
Step 4: Determine K at 3000 K
ΔG = -RT ln K
⇒R = 8.314 J/mol *K
⇒T = 3000 K
⇒K is the equilibrium constant
-52800 J/mol = -8.314 J/molK * 3000 * ln K
ln K = 2.117
K = e^2.117
K = 8.306
The equilibrium constant at 3000 K is 8.306
Step 5: Determine the value of ΔH∘rxn
ln K2/K1 = -ΔH/r * (1/T2 - 1/T1)
ln 8.306 /0.713 = -ΔH/8.314 * (1/3000 - 1/2000)
2.455 = -ΔH/8.314 * (3.33*10^-4 - 0.0005)
2.455 = -ΔH/8.314 * (-1.67*10^-4)
-14700= -ΔH/8.314
-ΔH = -122200 J/mol
ΔH = 122.2 kJ/mol
When The constant at 2000 K is 0.7139
Then The constant at 3000 K is 8.306
ΔH = 122.2 kJ/mol
What is Equilibrium?Step 1: Data is given
When the quality molar Gibbs effectiveness of the formation of X(g) is 5.61 kJ/mol at 2000 K
When the quality molar Gibbs effectiveness of the formation of X(g) is -52.80 kJ/mol at 3000 K
Step 2: The equation is:
[tex]1/2X2(g)⟶X(g)[/tex]
Step 3: Then Determine K at 2000 K
ΔG = -RT ln K
[tex]⇒R = 8.314 J/mol *K[/tex]
⇒[tex]T = 2000 K[/tex]
⇒ at that time K is that the constant
Then 5610 J/mol = [tex]-8.314 J/molK * 2000 * ln K[/tex]
ln K is = [tex]-0.337[/tex]
K is =[tex]e^-0.337[/tex]
K is = [tex]0.7139[/tex]
When The constant at 2000 K is 0.7139
Step 4: Then Determine K at 3000 K
ΔG = -RT ln K
⇒[tex]R = 8.314 J/mol *K[/tex]
⇒[tex]T = 3000 K[/tex]
⇒K is that the constant
[tex]-52800 J/mol = -8.314 J/mol K * 3000 * ln K[/tex]
ln [tex]K = 2.117[/tex]
K = e^2.117
K = 8.306
The constant at 3000 K is 8.306
Step 5: at the moment Determine the worth of ΔH∘rxn
ln K2/K1 = -ΔH/r * (1/T2 - 1/T1)
ln 8.306 /0.713 = -Δ[tex]H/8.314 * (1/3000 - 1/2000)[/tex]
2.455 = -ΔH/8.314 * (3.33*10^-4 - 0.0005)
2.455 = -ΔH/8.314 * [tex](-1.67*10^-4)[/tex]
-14700= -ΔH/8.314
-ΔH = [tex]-122200 J/mol[/tex]
Then ΔH = [tex]122.2 kJ/mol[/tex]
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Which statement describes solctices
Answers
Answer:funk
hot dog cat
Explanation:
uhughuhuhuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
Answer:
They occur when the sun reaches its highest or lowest point in the sky
Measurements show that the enthalpy of a mixture of gaseous reactants increases by 215. kJ during a certain chemical reaction, which is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that -155. kJ of work is done on the mixture during the reaction.Calculate the change in energy of the gas mixture during the reaction. Round your answer to 2 significant digits.Is the reaction exothermic or endothermic?
Answers
Answer: The change in energy of the gas mixture during the reaction is 60 kJ. The reaction is endothermic.
Explanation:
According to first law of thermodynamics:
[tex]\Delta E=q+w[/tex]
[tex]\Delta E[/tex]=Change in internal energy
q = heat absorbed or released
w = work done or by the system
w = work done on the system
w = -155 kJ
q = +215kJ {Heat absorbed by the system is positive}
[tex]\Delta E=+215+(-155)kJ=60kJ[/tex]
As the heat is absorbed and enthalpy increases, the reaction is endothermic.
The box below to the left represents ions in a certain volume of 0.10MHCl(aq) . In the box below to the right, draw a representation of ions in the same volume of 0.20MHCl(aq)
Answers
Complete question:
The student performs a second titration using the 0.10MNaOH(aq) solution again as the titrant, but this time with a 20.mL sample of 0.20MHCl(aq) instead of 0.10MHCl(aq).
1. The box below to the left represents ions in a certain volume of 0.10MHCl(aq). In the box below to the right, draw a representation of ions in the same volume of 0.20MHCl(aq). (Do not include any water molecules in your drawing.)
Answer:
The concentration of H+ and Cl- will be doubled.
Explanation:
See the attached photo for the representation of ions in the same volume.
Draw the organic product(s) of the reaction of p-methylbenzoic acid with CH3MgBr in dry ether, then H3O+ in the window below. If no new products are formed, tell OWL by drawing ethane, CH3CH3.
Answers
Answer:
Methane is produced as a new product.
Explanation:
[tex]CH_{3}MgBr[/tex] acts as a base toward -COOH group in p-methylbenzoic acid.
Hence an acid-base reaction occurs between p-methylbenzoic acid and [tex]CH_{3}MgBr[/tex] to produce methane and p-methylbenzoate.
[tex]H_{3}O^{+}[/tex]addition will convert p-methylbenzoate back to p-methylbenzoic acid.
Hence, methane is produce as a new product.
Reaction sequences are given below.
The concentration of a mixture can be increased in which of the following ways?
Answers
Answer:
C. adding more powder solute
Explanation:
The concentration of a mixture can be increased by removing solvent.
The concentration of a mixture can be increased in several ways. If we consider a distillery wanting to achieve a higher concentration of alcohol, they need to remove solvent, which in this case is water, from their product. This can be achieved through processes such as distillation, which separates alcohol from water due to their different boiling points. Another approach is the addition of a substance that reacts with the water, effectively removing water content from the mixture.
To increase the concentration of a solution in general, one can also add more solute (the substance being dissolved) into the solution or remove solvent (the substance dissolving the solute). It is important not to confuse the process of dilution, which involves adding solvent and decreases solute concentration, with the process of concentrating a solution, which involves removing solvent and increases solute concentration. Therefore, methods such as evaporation, reverse osmosis, or chemical reaction can be employed to increase the concentration of a solute in a solution.
The equipment used for aviation communications emits high‑frequency radiowave energy with a wavelength of 0.250 km. What is the energy of exactly one photon of this radiowave radiation?
Answers
Explanation:
Given that,
The wavelength of high‑frequency radio wave, [tex]\lambda=0.25\ km=250\ m[/tex]
We need to find the energy of exactly one photon of this radio wave radiation. It is given by :
[tex]E=\dfrac{nhc}{\lambda}[/tex]
Here, n = 1
[tex]E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{250}\\\\E=7.95\times 10^{-28}\ J[/tex]
So, the energy of exactly one photon of this radio wave radiation is [tex]7.95\times 10^{-28}\ J[/tex].
The energy of exactly one photon of this radio wave radiation is [tex]7.95 \times 10^{-28}J[/tex].
Given that,
The equipment used for aviation communications emits high‑frequency radio wave energy with a wavelength of 0.250 km i.e. = 250 m.Based on the above information, the calculation is as follows:
We know that
[tex]E = nhc \div \lambda\\\\= (6.63 \times 10^{-34} \times 3 \times 10^8) \div 250[/tex]
= [tex]7.95 \times 10^{-28}J[/tex]
Therefore we can conclude that the energy of exactly one photon of this radio wave radiation is [tex]7.95 \times 10^{-28}J[/tex].
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There are three competing factors at play here: 1. The effective nuclear charge 2. The size of the atom and the force of attraction according to Coulomb's law 3. A pair of electrons in an orbital The first ionization energy is the energy required to completely remove the first electron from the atom. The higher in energy an electron is to start with, the less additinal energy will be required to remove it, which translates to a lower ionization energy. By the same token, an electron arrangement which is lower in energy will require more energy to remove an electron.
Answers
Answer:
These three factors are required for ionization potential or ionization energy.
Explanation:
Ionization potential refers to the amount of energy which is required for the removal of outermost electron of the atom. If the atom size is big so the outermost electron is far from the nucleus and low energy is required for its removal due to lower force of attraction between nucleus and outermost electron. If the nuclear charge is higher, so the electron is tightly held by the nucleus and require more energy for its removal. Nuclear charge means number of protons present in the nucleus.
Benzene was treated with a set of unknown reagents. Using the spectral data below, draw the most likely product and select the most likely set of reagents. 1H NMR: See spectra below. 13C NMR: 142 ppm, 128 ppm, 128 ppm, 125 ppm, 38 ppm, 24 ppm, and 13 ppm. MS: There is no detectable M+2 peak. IR: Peaks observed at 3100 cm-1 and at 2900 cm-1. There is NO absorbance near 1700 cm-1.
Answers
Answer:
CHECK THE ATTACHMENT TO SEE THE STRUCTURAL DIAGRAM
Explanation:
The spectral data suggests the product is ethylbenzene, which can be synthesized from benzene using ethylchloride and a catalyst like AlCl3 via a Friedel-Crafts alkylation reaction.
Explanation:
The spectral data provided corresponds to the molecule ethylbenzene. The presence of a 13C NMR signal at 142 ppm indicates a carbon directly invested in a aromatic system. The signals at 38 ppm, 24 ppm and 13 ppm are indicative of the ethyl side chain. The IR peaks at 3100 cm-1 and 2900 cm-1 indicate C-H stretching of aromatic and aliphatic hydrogens respectively. The absence of an absorbance near 1700 cm-1 indicates absence of carbonyl group. The lack of a M+2 peak in MS data suggests no halogens are part of the structure. To obtain ethylbenzene from benzene, the reagents used would include ethylchloride and aluminum chloride (AlCl3) in a Friedel-Crafts alkylation reaction.Benzene, ethylchloride and AlCl3 are the key components to this reaction.
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A voltaic cell is constructed in which the following cell reaction occurs. The half-cell compartments are connected by a salt bridge. F2(g) + 2I-(aq) 2F-(aq) + I2(s)
The anode reaction is: + +
The cathode reaction is: + +
In the external circuit, electrons migrate the F-|F2 electrode the I-|I2 electrode. In the salt bridge, anions migrate the F-|F2 compartment the I-|I2 compartment.
Answers
Answer:
See explanation below
Explanation:
The anode reaction is :
2I^-(aq) -------> I2(g) +2e
Cathode reaction
F2(g) + 2e------> 2F^-(aq)
In the external circuit, electrons migrate from the I-|I2 electrode (anode) to the F-|F2 electrode (cathode)
In the salt bridge, anions migrate from the F-|F2
A voltaic cell is a type of cell in which a spontaneous redox reaction generates an electric current. This current is generated by the migration of electrons from the anode to the cathode in the external circuit, and the migration of anions in the salt bridge maintains electrical neutrality.
Explanation:A voltaic cell is a type of electrochemical cell where a spontaneous redox reaction generates an electric current. In this particular voltaic cell, the anode reaction is 2I-(aq) -> I2(s) + 2e-, where iodide ions are oxidized to solid iodine (losing electrons). The cathode reaction is F2(g) + 2e- -> 2F-(aq), where gaseous fluorine is reduced to fluoride ions (gaining electrons).
In the external circuit, electrons migrate from the anode (I-|I2 electrode) to the cathode (F-|F2 electrode). This migration of electrons generates the electric current. Lastly, in the salt bridge, anions migrate from the anode compartment (I-|I2) to the cathode compartment (F-|F2), allowing the cell to maintain electrical neutrality throughout the redox reaction.
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A flammable gas made up of only carbon and hydrogen is found to effuse through a porous barrier in 1.50 min. Under the same conditions of temperature and pressure, it takes an equal volume of bromine vapor 4.73 min to effuse through the same barrier. Calculate the molar mass of the unknown gas, and suggest what this gas might be.
Answers
Answer:
A. The molar mass of the unknown gas is 16g/mol
B. Compound is CH4 i.e methane
Explanation:
A. Step 1:
Representation:
Let t1 be time for unknown gas
Let t2 be the time for bromine vapor
Let M1 be molar mass of the unknown gas
Let M2 be the molar mass of bromine vapor
A. Step 2 :
Data obtained from the question.
Time for the unknown gas (t1) = 1.50 min
Time for Br2 (t2) = 4.73 min
Molar Mass of unknown gas (M1) =?
Molar Mass of Br2 (M2) = 80 x 2 = 160g/mol
A. Step 3:
Determination of the molar mass of the unknown gas.
Applying the equation:
t2/t1 = √(M2/M1)
The molar mass of can be obtained as follow:
t2/t1 = √(M2/M1)
4.73/1.5 = √(160/M1)
Take the square of both side
(4.73/1.5)^2 = 160/M1
9.94 = 160/M1
Cross multiply to express in linear form.
9.94 x M1 = 160
Divide both side by 9.94
M1 = 160/9.94
M1 = 16g/mol
Therefore, the molar mass of the unknown gas is 16g/mol
B. Identification of the gas.
The gas contains C and H only. From the calculations made above, the molar mass of the unknown gas is 16g/mol.
We know also that the molar mass of carbon is 12g/mol and that of Hydrogen is 1g/mol
Therefore,
C + H = 16
There would be only 1 atom of C in the compound since the molar mass of the compound is 16g/mol. With these understanding, let us determine the number of H atom in the compound. This is illustrated below :
C + H = 16
12 + H = 16
H = 16 - 12
H = 4
Divide by the molar mass of H i.e 1
H = 4/1 = 4
There are 4 atoms of H in the compound. Therefore, the compound is CH4 i.e methane
The molar mass of the unknown gas is 16 g/mol hence the unknown gas is methane.
We must note that the time taken for a gas to diffuse is directly proportional to the molar mass of the gas.
Hence;
t1/t2 = √M1/M2
Let t1 = time taken for the flammable gas to diffuse = 1.50 min
Let t2 = time taken for the bromine vapor to diffuse = 4.73 min
M1 = molar mass of the flammable gas = ?
M2 = molar mass of the bromine vapor = 160 g/mol
Substituting values;
1.50/4.73 = √M1/160
(1.50/4.73)^2 = M1/160
0.1006 = M1/160
M1 = 0.1006 × 160
M1 = 16 g/mol
The unknown gas is methane.
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Equation: 3Cu(s) 8HNO3(aq) --> 2NO(g) 3Cu(NO3)2(aq) 4H2O(l) In the above reaction, the element oxidized is ______, the reducing agent is ______ and the number of electrons transferred from reducing to oxidizing agent in the equation, as written, is ______.
Answers
Answer:
1. Cu
2. Cu
3. 2 electrons.
Explanation:
Step 1:
The equation for the reaction is given below:
3Cu(s) + 8HNO3(aq) -> 2NO(g) 3Cu(NO3)2(aq) + 4H2O(l)
Step 2:
Determination of the change of oxidation number of each element present.
For Cu:
Cu = 0 (ground state)
Cu(NO3)2 = 0
Cu + 2( N + 3O) = 0
Cu + 2(5 + (3 x -2)) =0
Cu + 2 (5 - 6) = 0
Cu + 2(-1) = 0
Cu - 2 = 0
Cu = 2
The oxidation number of Cu changed from 0 to +2
For N:
HNO3 = 0
H + N + 3O = 0
1 + N + (3 x - 2) = 0
1 + N - 6 = 0
N = 6 - 1
N = 5
NO = 0
N - 2 = 0
N = 2
The oxidation number of N changed from +5 to +2
The oxidation number of oxygen and hydrogen remains the same.
Note:
1. The oxidation number of Hydrogen is always +1 except in hydride where it is - 1
2. The oxidation number of oxygen is always - 2 except in peroxide where it is - 1
Step 3:
Answers to the questions given above
From the above illustration,
1. Cu is oxidize because its oxidation number increased from 0 to +2 as it loses electron.
2. Cu is the reducing agent because it reduces the oxidation number of N from +5 to +2.
3. The reducing agent i.e Cu transferred 2 electrons to the oxidising agent HNO3 because its oxidation number increase from 0 to +2 as it loses its electrons. This means that Cu transfer 2 electrons.
In the given chemical reaction, the element oxidized and the reducing agent is Copper (Cu). The number of electrons transferred from the reducing agent to the oxidizing agent is six.
Explanation:In the reaction equation, 3Cu(s) + 8HNO3(aq) --> 2NO(g) + 3Cu(NO3)2(aq) + 4H2O(l), the element oxidized is Cu (Copper), as Copper goes from an oxidation state of 0 in Cu(s) to +2 in Cu(NO3)2. Hence, the reducing agent is also Copper (Cu). Oxidation is the process of losing electrons, and in this chemical reaction, each copper atom loses 2 electrons (going from 0 to +2). Since the equation shows the reaction of 3 copper atoms, therefore, the total number of electrons transferred from reducing to oxidizing agent in the equation, as written, is 6.
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What is the density of N2 gas (molar mass: 28 g/mol), at 400 K and 2 atm?
Answers
Answer:
Density= 1.7g/dm3
Explanation:
Applying
P×M= D×R×T
P= 2atm, Mm= 28, D=? R= 0.082, T= 400K
2×28= D×0.082×400
D= (2×28)/(0.082×400)
D= 1.7g/dm3
What is the standard entropy change for the reaction below? 2 CO(g) + 2 NO(g) → N2(g) + 2 CO2(g) S o (CO(g)) = 197.7 J/(mol·K) S o (CO2(g)) = 213.8 J/(mol·K) S o (NO(g)) = 210.8 J/(mol·K) S o (N2(g)) = 191.6 J/(mol·K)
Answers
Answer:
The standard entropy change for the reaction is -197.8 J/mol*K
Explanation:
Step 1: Data given
S°(CO(g)) = 197.7 J/(mol*K)
S°(CO2(g)) = 213.8 J/(mol*K)
S°(NO(g)) = 210.8 J/(mol*K)
S°(N2(g)) = 191.6 J/(mol·K)
Step 2: The balanced equation
2 CO(g) + 2 NO(g) → N2(g) + 2 CO2(g)
Step 3: Calculate ΔS°
ΔS° = ∑S°(products) - ∑S°(reactants)
ΔS° = (191.6 + 2*213.8) - (2*210.8+2*197.7) J/mol*K
ΔS° = 619.2 J/mol*K - 817.0 J/mol *K
ΔS° = -197.8 J/mol* K
The standard entropy change for the reaction is -197.8 J/mol*K
119. In analytical chemistry, bases used for titrations must often be standardized; that is, their concentration must be precisely determined. Standardization of sodium hydroxide solutions can be accomplished by titrating potassium hydrogen phthalate (KHC8H4O4), also known as KHP, with the NaOH solution to be standardized. a. Write an equation for the reaction between NaOH and KHP. b. The titration of 0.5527 g of KHP required 25.87 mL of an NaOH solution to reach the equivalence point. What is the concentration of the NaOH solution
Answers
Answer:
0.1046M NaOH solution
Explanation:
a. KHP is a salt used as primary standard because allows direct standarization of bases solutions. The reaction of KHP with NaOH is:
KHP + NaOH → H₂O + KP⁻ + Na⁺
As you can see, KHP has 1 acid proton that reacts with NaOH.
Molar mass of KHP is 204.22g/mol; 0.5527g of KHP contains:
0.5527g KHP × (1mol / 204.22g) = 2.706x10⁻³moles of KHP. As 1 mole of KHP reacts per mole of NaOH, at equivalence point you must add 2.706x10⁻³moles of NaOH
As you spent 25.87mL of the solution, molarity of the solution is:
2.706x10⁻³moles of NaOH / 0.02587L = 0.1046M NaOH solution
Answer:
NaOH + KHC8H4O4 → NaKC8H4O4 + H2O
Concentration NaOH = 0.105 M
Explanation:
Step 1: Data given
Mass of KHP = 0.5527 grams
Molar mass KHP = 204.22 g/mol
Volume of NaOH = 25.87 mL
Step 2: The balanced equation
NaOH + KHC8H4O4 → NaKC8H4O4 + H2O
Step 3: Calculate moles KHP
Moles KHP = mass KHP / molar mass KHP
Moles KHP = 0.5527 grams / 204.22 g/mol
Moles KHP = 0.002706 moles
Step 4: Calculate moles NaOH
For 1 mol NaOH we need 1 mol KHP to react
For 0.002706 moles KHP we need 0.002706 moles NaOH
Step 5: Calculate concentration NaOH
Concentration = moles / volume
Concentration NaOH = 0.002706 moles / 0.02587 L
Concentration NaOH = 0.105 M
The reduced coenzymes generated by the citric acid cycle donate electrons in a series of reactions called the electron transport chain. The energy from the electron transport chain is used for oxidative phosphorylation. Which compounds donate electrons to the electron transport chain? H 2 O H2O FADH 2 FADH2 ADP ADP NADH NADH NAD + NAD+ O 2 O2 FAD FAD ATP ATP Which compound is the final electron acceptor? NADH NADH ATP ATP H 2 O H2O NAD + NAD+ FADH 2 FADH2 FAD FAD O 2 O2 ADP ADP Which compounds are the final products of the electron transport chain and oxidative phosphorylation? ADP ADP NADH NADH O 2 O2 NAD + NAD+ FADH 2 FADH2 ATP ATP H 2 O H2O FAD
Answers
Answer:
The reduced coenzymes generated by the citric acid cycle donate electrons in a series of reactions called the electron transport chain. The energy from the electron transport chain is used for oxidative phosphorylation.
a)The compounds that donate electrons to the electron transport chain are NADH and . FADH2
b) O2 is the final electron acceptor.
c) The final products of the electron transport chain and oxidative phosphorylation are NAD+, H2O, ATP and FAD
Explanation:
Which of these statements are true for a neutral, aqueous solution at 25 °C? pH = 7.00 [ H + ] = [ OH − ] pOH = 7.00 Which of these statements are true for a neutral, aqueous solution regardless of temperature? [ H + ] = [ OH − ] pH = 7.00 pOH = 7.00
Answers
The statement which are true for a neutral, aqueous solution at 25 °C are:
pH = 7.00 [ H + ] = [ OH − ] pOH = 7.00The statement which is/are true for a neutral, aqueous solution regardless of temperature is;
[ H + ] = [ OH − ]We must know that at standard temperature, 25°C, a neutral, aqueous solution has it's pH = pOH = 7. Additionally, the hydrogen ion concentration, [H+] is equal to its hydroxide ion concentration, [OH-]
pH is slightly affected by change in temperature as it decreases with increase in temperature. In a neutral aqueous solution, there are always the same concentration of hydrogen ions,[H+] and hydroxide ions, [OH-] and hence, the solution is still neutral (even if its pH changes).Ultimately, the pH of a solution decreases with increase in temperature and vice versa.
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Use the Ideal Gas Law to calculate the number of moles (n) of helium in a 4000 Liter weather balloon near the top of Mt. Rainier with a pressure of 0.6 atm and temperature of 260K.
Answers
Answer: 112.5moles
Explanation:
P= 0.6atm, V= 4000L, R= 0.082, T= 260K
Applying PV = nRT
0.6×4000= n×0.082×260
Simplify n= (0.6× 4000)/(0.082×260)
n= 112.5moles
Using the Ideal Gas Law, we calculated that the weather balloon near the top of Mt. Rainier, with given conditions, contains approximately 113.1 moles of helium.
The Ideal Gas Law formula is PV = nRT, where:
P = pressure (in atm)V = volume (in liters)n = number of molesR = ideal gas constant (0.0821 L·atm/mol·K)T = temperature (in Kelvin)We can use this formula to calculate the number of moles of helium in the weather balloon.
Given:
Pressure (P) = 0.6 atmVolume (V) = 4000 LTemperature (T) = 260KUsing the Ideal Gas Law:
PV = nRT[tex]n = \frac{PV}{RT}[/tex]Plugging in the given values:
[tex]n = \frac{0.6 \, \text{atm} \times 4000 \, \text{L}}{0.0821 \, \text{L} \cdot \text{atm/mol} \cdot \text{K} \times 260 \, \text{K}}[/tex]n ≈ 113.1 molesThus, the number of moles of helium in the weather balloon is approximately 113.1 moles.
You add 100.0 g of water at 52.0 °C to 100.0 g of ice at 0.00 °C. Some of the ice melts and cools the water to 0.00 °C. When the ice and water mixture reaches thermal equilibrium at 0 °C, how much ice has melted? (The specific heat capacity of liquid water is 4.184 J/g ⋅ K. The enthalpy of fusion of ice at 0 °C is 333 J/g.) Mass of ice = g
Answers
Answer:
[tex]m_{ice} = 65.336\,g[/tex]
Explanation:
Accoding to the First Law of Thermodynamics, the heat released by the water melts a portion of ice. That is to say:
[tex]Q_{water} = Q_{ice}[/tex]
[tex](100\,g)\cdot \left(4.184\,\frac{J}{kg\cdot ^{\textdegree}C}\right)\cdot (52\,^{\textdegree}C - 0\,^{\textdegree}C) = m_{ice}\cdot \left(333\,\frac{J}{g} \right)[/tex]
The amount of ice that is melt is:
[tex]m_{ice} = 65.336\,g[/tex]
Draw the mechanism arrows for both propagation steps for the radical addition of HBr to the alkene. When drawing single-headed radical arrows, this software requires that they meet at one atom (not in space between atoms like you may do in class). In the second box you will need to draw the first product and another reactant. In the last box you will need to draw an additional product.
Answers
Answer:
See the attached file for the structure
Explanation:
See the attached file for the explanation
uestion 174 Which sequence of reactions is expected to produce the product below as the final, and major, organic product? I 1. HC≡CH, NaNH2; 2. (CH3)2CHCH2Br 3. Aqueous H2SO4, HgSO4 II 1. CH3C≡CH, NaNH2; 2. (CH3)2CHBr; 3. Disisamylborane; 4. H2O2, NaOH III 1. (CH3)2CHBr, NaNH2; 2. CH3C≡CH; 3. O3; 4. H2O IV 1. CH3C≡CH, NaNH2: 2. (CH3)2CHCH2Br; 3. BH3·THF; 4. H2O2, NaOH V 1. (CH3)2CHCH2Br, NaNH2; 2. HC≡CH; 3. 9-BBN; 4. H2O2, NaOH I II III IV V
Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
The correction option is is [tex]I[/tex]
Explanation:
The mechanism of the reaction is show on the second uploaded image
Draw the line‑bond structure of oleic acid (cis‑9‑octadecenoic acid), CH 3 ( CH 2 ) 7 CH = CH ( CH 2 ) 7 COOH , at physiological pH. Hydrogen atoms attached to carbon atoms do not need to be drawn.
Answers
Answer : The line-bonds structure of oleic acid (cis‑9‑octadecenoic acid) is shown below.
Explanation :
In line-bonds formula, the organic structures of the compound are represented in such a way that covalent bonds are represented by line and frame the structure by zig-zag straight lines which emits all the hydrogen atom.
The terminals of the line and vertex represents carbon atoms whose valences are satisfied by formation of single bonds with H atom.
The given compound is, (cis‑9‑octadecenoic acid)
In this compound, the parent chain is 18 membered and the carboxylic acid functional group is attached to it.
Huan wants to enter the science fair at his school. He has a list of ideas for his project. Which questions could be
answered through scientific investigation? Check all that apply.
Does pressure have an effect on the volume of a gas?
Which physicist was the smartest?
Is the information on the periodic table difficult to understand?
Which brand of soap is the best for cleaning grease off dishes?
Which laboratory experiment is the most fun to perform?
Answers
Correct answer:
•Does pressure have an effect on the volume of a gas?
•Which brand of soap is the best for cleaning grease off dishes?
The questions that can be answered by science must be empirical, that is they must be answerable by experiments.
The question that can be answered by science is; Which brand of soap is the best for cleaning grease off dishes?
This question can be answered by using various brands of soap to clean the same type of dish with the same type of grease and comparing the results. The answer to this question is pure empirical.
The other questions listed can not be answered by experiment. Their answer may vary from person to person therefore they are not scientific questions.
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