Answer:
[tex]\boxed{\text{c. The concentration of reactants is much greater than that of products.}}[/tex]
Explanation:
NO₂ + NO₃ ⇌ N₂O₅; K = 2.1 × 10⁻²⁰
We often write K as
[tex]K = \dfrac{[\text{Products}]}{[\text{Reactants}]}[/tex]
If K is large, more of the molecules exist as products.
If K is small, more of the molecules exist as reactants.
[tex]\text{Since K is small, }\\\boxed{\textbf{the concentration of reactants is much greater than that of products.}}[/tex]
a. b., and d. are wrong. The concentration of reactants is greater.
Final answer:
The equilibrium constant of K = 2.1 × 10−20 suggests that at equilibrium, the reactants' concentration significantly surpasses the products', confirming option (c) as correct.
Explanation:
The equilibrium constant for the reaction NO2(g) + NO3(g) ⇌ N2O5(g) is K = 2.1 × 10−20. An equilibrium constant of such a small magnitude indicates that, at equilibrium, the concentration of reactants (NO2 and NO3) will be much greater than the concentration of products (N2O5). This means that option (c) is correct: At equilibrium the concentration of reactants is much greater than that of products. It does not mean reactions stop; both the forward and reverse reactions continue occurring at equal rates, maintaining the equilibrium concentrations of reactants and products.
A sample of oxygen gas was collected via water displacement. Since the oxygen was collected via water displacement, the sample is saturated with water vapor. If the total pressure of the mixture at 26.4 °C is 805 torr, what is the partial pressure of oxygen? The vapor pressure of water at 26.4 °C is 25.81 mm Hg.
Answer : The partial pressure of oxygen is, 799.19 torr
Solution :
According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gasses.
[tex]P_T=p_{H_2O}+p_{O_2}[/tex]
where,
[tex]P_T[/tex] = total partial pressure = 805 torr
[tex]P_{O_2}[/tex] = partial pressure of oxygen gas = ?
[tex]P_{H_2O}[/tex] = partial pressure of water = 25.81 mm Hg = 25.81 torr
Now put all the given values is expression, we get the partial pressure of the oxygen gas.
[tex]805\text{ torr}=25.81\text{ torr}+p_{O_2}[/tex]
[tex]p_{O_2}=779.19\text{ torr}[/tex]
Therefore, the partial pressure of oxygen gas is, 799.19 torr
The reform reaction between steam and gaseous methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and dihydrogen gas. Synthesis gas is one of the most widely used industrial chemicals, and is the major industrial source of hydrogen.
Suppose a chemical engineer studying a new catalyst for the reform reaction finds that 924. liters per second of methane are consumed when the reaction is run at 261.°C and 0.96atm. Calculate the rate at which dihydrogen is being produced. Give your answer in kilograms per second. Round your answer to 2 significant digits.
Answer:
The rate at which dihydrogen is being produced is 0.12 kg/sec.
Explanation:
[tex]CH_4+H_2O\rightarrow CO+3H_2[/tex] Haber reaction
Volume of methane consumed in a second = 924 L
Temperature at which reaction is carried out,T= 261°C = 538.15 K
Pressure at which reaction is carried out, P = 0.96 atm
Let the moles of methane be n.
Using an Ideal gas equation:
[tex]PV=nRT[/tex]
[tex]n=\frac{PV}{RT}=\frac{0.96 atm\times 924 L}{0.0821 atm l/mol K\times 538.15 K}=20.0769 mol[/tex]
According to reaction , 1 mol of methane gas produces 3 moles of dihydrogen gas.
Then 20.0769 mol of dihydrogen will produce :
[tex]\frac{3}{1}\times 20.0769 mol=60.2307 mol[/tex] of dihydrogen
Mass of 24.3194 moles of ammonia =24.3194 mol × 2 g/mol
=120.46 g=0.12046 kg ≈ 0.12 kg
924 L of methane are consumed in 1 second to produce 0.12 kg of dihydrogen in 1 second. So the rate at which dihydrogen is being produced is 0.12 kg/sec.
You can identify a metal by carefully determining its density. A 8.44 g piece of an unknown metal is 1.25 cm long, 2.50 cm wide, and 1.00 cm thick. What is a possible identity of the element?Aluminum, 2.70 g/cm3Titanium, 4.54 g/cm3Zirconium, 6.51 g/cm3Chromium, 7.20 g/cm3Nickel, 8.90 g/cm3
Answer: Aluminum, [tex]2.70 g/cm^3[/tex]
Explanation:
Density is defined as the mass contained per unit volume. It is characteristic of a substance.
[tex]Density=\frac{mass}{Volume}[/tex]
Given : Mass of object = 8.44 grams
Volume of object=[tex]length\times breadth\times height=1.25cm\times 2.50cm\times 1.0cm=3.125cm^3[/tex]
Putting in the values we get:
[tex]Density=\frac{8.44g}{3.125cm^3}=2.70g/cm^3[/tex]
Thus density of the object will be [tex]2.70 g/cm^3[/tex] which matches that of aluminium.
Answer: Nickel
Explanation:
They have the same density as each other
Lithium hydrogen carbonate, LiHCO3, decomposes when heated to form Li2O, CO2, and H2O. How many moles of H2O are formed when 0.50 mol LiHCO3 decomposes?
Answer:
0.25 mol of water
Explanation:
The starting point is writing out a balanced chemical equation
[tex]2LiHCO_{3} --->Li_{2}O +2CO_{2} +H_{2}O[/tex]
From here we can see that 2 moles of Lithium hydrogen carbonate produce 1 mole of water. It follows that 0.5 moles of Lithium hydrogen carbonate will produce half of that too, which is 0.25 moles. The ratio is always mantained.
Final answer:
When 0.50 mol of lithium hydrogen carbonate (LiHCO3) decomposes, it produces 0.25 mol of water (H2O).
Explanation:
You asked how many moles of H2O are formed when 0.50 mol LiHCO3 decomposes. The decomposition reaction for LiHCO3 to produce Li2O, CO2, and H2O is:
2 LiHCO3(s) → Li2O(s) + 2 CO2(g) + H2O(g)
According to the balanced equation, 2 moles of LiHCO3 produce 1 mole of H2O. Therefore, when 0.50 mole of LiHCO3 decomposes, you would get half of that ratio in moles of water:
(0.50 mol LiHCO3) / (2 mol LiHCO3) = 0.25 mol H2O
Therefore, the decomposition of 0.50 mol of LiHCO3 will produce 0.25 mol of H2O.
Which of the following assumptions appears reasonable for the isothermal process? H20 (liq, 1 bar) → H2O (liq, 1300 bar), T = 20°C. A. AU = 0, AH 0 B. AU = 0, AH 0 C. AU #0, AH = 0 D. None of the above Hinn nf diethyl ether using the Chen's rule is
Answer : The assumptions appears reasonable for the isothermal process is, [tex]\Delta U=0[/tex] and [tex]\Delta H=0[/tex]
Explanation :
First law of thermodynamic : It states that the energy can not be created or destroyed, it can only change or transfer from one state to another state.
As per first law of thermodynamic,
[tex]\Delta U=q+w[/tex]
The expression for internal energy is:
[tex]\Delta U=nC_vdt[/tex]
The expression for enthalpy is:
[tex]\Delta H=nC_pdt[/tex]
where,
[tex]\Delta U[/tex] = internal energy
q = heat
w = work done
n = number of moles
[tex]C_v[/tex] = specific heat capacity at constant volume
[tex]C_p[/tex] = specific heat capacity at constant pressure
[tex]dt[/tex] = change in temperature
As we know that, the term internal energy and enthalpy is the depend on the temperature and the process is isothermal that means at constant temperature.
T = constant
[tex]\dt[/tex] = 0
So, at constant temperature the internal energy and enthalpy is equal to zero. That means,
[tex]\Delta U=0[/tex] and [tex]\Delta H=0[/tex]
For an isothermal process in which the pressure changes significantly (as in the one outlined in the question, from 1 bar to 1300), the change in internal energy (ΔU) will be zero but a change in enthalpy (ΔH) will be non-zero, since the work done on the system is not constant. This corresponds to option B in the question.
Explanation:In order to understand the question, it's crucial to define two key terms: isothermal and internal energy. An isothermal process is a change of a system, in which the temperature remains constant: ΔT = 0. For an ideal gas, the internal energy is a function of temperature only. Hence, in an isothermal process, the change in internal energy ΔU = 0. The enthalpy H of a system is defined as H = U + PV where U is the internal energy of the system, P is the pressure, and V is the volume.
Given that the process we're examining here involves a change in pressure, there is work done on the system (from 1 bar to 1300 bar), so the enthalpy H = U + PV will not be zero as P and V do not remain constant. Therefore, the most reasonable assumption for this process is ΔU = 0 and ΔH ≠ 0, which would correspond to option B.
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A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4 will be needed to produce 375 mL of a solution that has a concentration of Na+ ions of 1.50 M ?
The mass of sodium phosphate needed to make 375 mL of a solution with a molarity of sodium ions of 0.900 M is indeed 18.45 grams.
Given:
Volume of solution (V) = 375 mL = 0.375 L
Molarity of sodium ions (c) = 0.900 M
Using the formula for molarity: c = n/V
Since each mole of sodium phosphate produces 3 moles of sodium ions:
Effective molarity of sodium phosphate solution = c / 3 = 0.300 M
Number of moles of sodium phosphate (n) = c * V = 0.300 M * 0.375 L = 0.1125 mol
Molar mass of sodium phosphate (M) = 164 g/mol
Mass of sodium phosphate (m) = n * M = 0.1125 mol * 164 g/mol = 18.45 g
Therefore, the mass of sodium phosphate needed to make 375 mL of a solution with a molarity of sodium ions of 0.900 M is indeed 18.45 grams.
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To make a solution of tribasic sodium phosphate with a concentration of 1.50 M Na+ ions in 375 mL, you will need 276.38 grams of Na3PO4.
Explanation:To calculate the grams of Na3PO4 needed, we need to use the equation relating moles, volume, and concentration:
Moles of Na+ ions = Molarity of Na+ ions x Volume of solution (in L)
Moles of Na3PO4 = 3 x Moles of Na+ ions
First, convert the volume in mL to L (375 mL = 0.375 L). Then, substitute the values into the equation:
Moles of Na+ ions = 1.50 M x 0.375 L = 0.5625 mol
Moles of Na3PO4 = 3 x 0.5625 mol = 1.6875 mol
Finally, convert moles to grams using the molar mass of Na3PO4 (163.94 g/mol):
Mass of Na3PO4 = 1.6875 mol x 163.94 g/mol = 276.38 g
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Which of the following statements are true with respect to enzyme activity? Check all that apply. View Available Hint(s) Check all that apply. Enzymes speed up the reaction rate. The activation energy of a reaction increases when an enzyme is used to catalyze the reaction. Enzyme reactivity is not affected by change in pH and temperature. Enzymes are regenerated when the reaction is completed. Enzymes affect the reaction pathway by forming an enzyme-substrate complex.
Enzymes speed up reactions, are regenerated after the reaction, and form enzyme-substrate complexes. They do not increase activation energy or remain unaffected by pH and temperature changes.
Explanation:The following statements are true regarding enzyme activity:
Enzymes speed up the reaction rate. They do this by providing an alternative pathway for a reaction with a lower activation energy.Enzymes are regenerated when the reaction is completed. After an enzyme has bound to its substrate and completed its catalytic function, it's released intact and can participate in future reactions.Enzymes affect the reaction pathway by forming an enzyme-substrate complex. The active site of an enzyme accommodates the substrate molecule, leading to a temporary enzyme-substrate complex that allows the reaction to proceed more easily.However, the statement that "The activation energy of a reaction increases when an enzyme is used to catalyze the reaction" is incorrect. Enzymes lower the activation energy of a reaction rather than increasing it. Also false is "Enzyme reactivity is not affected by change in pH and temperature." Enzyme activity is very sensitive to changes in pH and temperature.
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What mass of Cu(s) would be produced if 0.40 mol of Cu2O(s) was reduced completely with excess H2(g) ?
Answer:
50.8g
Explanation:
Given parameters:
Nubmer of mole of Cu₂O = 0.4mol
Unknown:
Mass of Cu produced = ?
Solution:
The balanced reaction equation:
Cu₂O + H₂ ⇄ 2Cu + H₂O
From the reaction, we know that:
1 mole of Cu₂O produced 2 moles of Cu
0.4 mol of Cu₂O would give 0.8mol of Cu
Now we know the number of moles of Cu produced because the limiting reactant here is Cu₂O. It was used up in the reaction in which the hydrogen gas was in excess.
To find the mass of Cu produced, we use the equation below :
Mass of Cu = Number of moles of Cu x molar mass of Cu
Molar mass of Cu = 63.5gmol⁻¹
Mass of Cu = 0.8 x 63.5 = 50.8g
If [tex]0.40[/tex] mol of [tex]Cu_2O(s)[/tex] is reduced completely with excess [tex]H_2(g)[/tex], approximately [tex]50.84[/tex] grams of [tex]Cu(s)[/tex] would be produced.
The balanced chemical equation for the reduction of [tex]Cu_2O(s)[/tex] with [tex]H_2(g)[/tex] is:
[tex]\[ \text{Cu}_2\text{O}(s) + \text{H}_2(g) \rightarrow \text{Cu}(s) + \text{H}_2\text{O}(l) \][/tex]
From the balanced equation, we can see that [tex]1[/tex] mole of [tex]Cu_2O[/tex] produces [tex]2[/tex] moles of [tex]Cu[/tex].
Given that we have [tex]0.40[/tex] mol of [tex]Cu_2O[/tex], we can use the stoichiometry of the reaction to find the amount of [tex]Cu[/tex] produced:
[tex]\[ \text{mol of Cu} = 0.40 \, \text{mol Cu}_2\text{O} \times \frac{2 \, \text{mol Cu}}{1 \, \text{mol Cu}_2\text{O}} \][/tex]
[tex]\[ \text{mol of Cu} = 0.40 \times 2 \][/tex]
[tex]\[ \text{mol of Cu} = 0.80 \, \text{mol} \][/tex]
Now, we need to convert the moles of [tex]Cu[/tex] to grams using the molar mass of copper ([tex]Cu[/tex]), which is approximately [tex]63.55 g/mol[/tex]
[tex]\[ \text{mass of Cu} = \text{mol of Cu} \times \text{molar mass of Cu} \][/tex]
[tex]\[ \text{mass of Cu} = 0.80 \, \text{mol} \times 63.55 \, \text{g/mol} \][/tex]
[tex]\[ \text{mass of Cu} = 50.84 \, \text{g} \][/tex]
Which of the following has the greatest number of particles? Which of the following has the greatest number of particles? 22.990 g of Na 1 mole of Na 9.012 g of Be 1 mole of Be All are the same.
Answer: All the given options have the same number of particles.
Explanation:
For the given options:
Option 1: 22.990 g of NaAccording to mole concept:
1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of atoms.
If, 22.99 g of sodium metal contains [tex]6.022\times 10^{23}[/tex] number of atoms.
So, 22.990 g of sodium metal will contain [tex]\frac{6.022\times 10^{23}}{22.99}\times 22.990=6.022\times 10^{23}[/tex] number of atoms.
Number of particles = [tex]6.022\times 10^{23}[/tex]
Option 2: 1 mole of NaAccording to mole concept:
1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of atoms.
So, 1 mole of sodium element will also contain [tex]6.022\times 10^{23}[/tex] number of atoms.
Number of particles = [tex]6.022\times 10^{23}[/tex]
Option 3: 9.012 g of BeAccording to mole concept:
1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of atoms.
If, 9.01 g of beryllium metal contains [tex]6.022\times 10^{23}[/tex] number of atoms.
So, 9.012 g of beryllium metal will contain [tex]\frac{6.022\times 10^{23}}{9.01}\times 9.012=6.022\times 10^{23}[/tex] number of atoms.
Number of particles = [tex]6.022\times 10^{23}[/tex]
Option 4: 1 mole of BeAccording to mole concept:
1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of atoms.
So, 1 mole of beryllium element will also contain [tex]6.022\times 10^{23}[/tex] number of atoms.
Number of particles = [tex]6.022\times 10^{23}[/tex]
Hence, all the options have same number of particles which is [tex]6.022\times 10^{23}[/tex]
Both 1 mole of sodium (Na) and 1 mole of beryllium (Be) contain the greatest number of particles, which is Avogadro's Number or 6.02×1023 particles. The other listed quantities, 22.990 g of Na and 9.012 g of Be, contain fewer particles as they are less than one mole of their respective elements.
Explanation:To determine which of the given options has the greatest number of particles, we must refer to Avogadro's Number and the concept of molar mass. Avogadro's Number, which is 6.02×1023 mol-1, represents the number of particles in one mole of a substance. The molar mass is the mass of one mole of representative particles of a substance and is equivalent numerically to the atomic mass of an element in grams. For instance, the atomic mass of sodium (Na) is approximately 23 g/mol, which means that 1 mole of Na has a mass of approximately 23 grams and contains 6.02×1023 particles of Na.
Therefore, 1 mole of any substance, whether Na or beryllium (Be), will always contain 6.02×1023 particles. Comparing the options given:
22.990 g of Na - This is less than 1 mole of Na, so it contains fewer than 6.02×1023 particles.1 mole of Na - This contains 6.02×1023 particles.9.012 g of Be - This is less than 1 mole of Be (given that Be has a molar mass around 9.012 g/mol), so it also has fewer than 6.02×1023 particles.1 mole of Be - This too contains 6.02×1023 particles.All are the same - This is incorrect since the options with amounts less than 1 mole (22.990 g of Na and 9.012 g of Be) have fewer particles.Both 1 mole of Na and 1 mole of Be have the greatest number of particles, which is Avogadro's Number of particles. The quantities that are less than one mole have fewer particles in comparison.
The ionization energies of rubidium and silver are 4.18 and 7.57 eV, respectively. Calculate the ionization energies of an H atom with its electron in the same outermost orbitals as in these two atoms and account for the differences in values in these different elements.
Answer:
0.544 eV; Ag is a smaller atom.
Explanation:
1. Ionization energy of hydrogen
The outermost electrons in Rb and Ag are in 5s orbitals.
The formula for the energy of an electron in a hydrogen atom is
E = -13.6/n² eV
For a hydrogen atom in a 5s orbital,
E = -13.6/5² = -13.6/25 = -0.544 eV
The ionization energy would be 0.544 eV.
2. Rb vs Ag
The first electrons to be removed from Rb and Ag are in 5s orbitals.
The atomic radius of Ag is less than that of Rb because, as you go from left to right across the Row, you are adding 10 protons to the nucleus and 10 electrons to the outer shell.
The added electrons do not effectively shield each other from the attraction of the nucleus, so the 5s electron of Ag is closer in.
It takes more energy to remove the electron from silver, so the ionization energy of Ag is greater than that of Rb.
Rubidium's and silver's higher ionization energies arise from effective nuclear charge, quantum defect, and relativistic effects.
The ionization energies of rubidium and silver are 4.18 and 7.57 eV, respectively. To calculate the ionization energies of an H atom with its electron in the same outermost orbitals as in these two atoms, we must consider the effective nuclear charge experienced by the outermost electron in each atom.
For hydrogen, the ionization energy is given by the formula:
[tex]\[ IE = \frac{13.6 \text{ eV}}{n^2} \][/tex]
where n is the principal quantum number of the electron's orbital.
For rubidium (Rb), the outermost electron is in the 5s orbital, which corresponds to n = 5. Therefore, the ionization energy for a hydrogen-like atom with an electron in the 5s orbital is:
[tex]\[ IE_{\text{H, Rb-like}} = \frac{13.6 \text{ eV}}{5^2} = \frac{13.6 \text{ eV}}{25} \approx 0.544 \text{ eV} \][/tex]
For silver (Ag), the outermost electron is in the 5s orbital as well, which also corresponds to n = 5. Thus, the ionization energy for a hydrogen-like atom with an electron in the 5s orbital is the same as for the rubidium-like case:
[tex]\[ IE_{\text{H, Ag-like}} = \frac{13.6 \text{ eV}}{5^2} = \frac{13.6 \text{ eV}}{25} \approx 0.544 \text{ eV} \][/tex]
However, the actual ionization energies of rubidium and silver are much higher than the calculated values for hydrogen-like atoms because of the following reasons:
1. Penetration Effect:
The inner electrons shield the outermost electron from the nucleus. In a multi-electron atom, the effective nuclear charge [tex](\( Z_{\text{eff}} \))[/tex] is less than the actual nuclear charge due to this shielding.
The outermost electron in rubidium and silver experiences a [tex]\( Z_{\text{eff}} \)[/tex] that is less than the full nuclear charge.
2. Quantum Defect:
The quantum defect accounts for the difference in energy levels between hydrogen-like atoms and many-electron atoms due to the presence of inner electrons.
This defect is more significant for atoms with higher atomic numbers, like silver.
3. Relativistic Effects:
For heavier atoms like silver, relativistic effects become significant.
The mass of the electron increases as it moves at a significant fraction of the speed of light, which increases the electron's binding energy to the nucleus, thus increasing the ionization energy.
4. Full d Subshell:
Silver has a full 4d subshell, which provides additional stability and pulls the 5s orbital closer to the nucleus, increasing the ionization energy.
Due to these effects, the actual ionization energies of rubidium and silver are much higher than those calculated for a hydrogen-like atom.
The differences in values between rubidium and silver are primarily due to the increased nuclear charge, the presence of filled d subshells in silver, and relativistic effects, which are more pronounced in silver due to its higher atomic number.
What volume (in liters) of carbon monoxide gas, measured at a temperature of 212 K and a pressure of 676 mm Hg, is required to synthesize 19.3 g of methanol. How many liters of oxygen (at STP) are required to form 12.5 g of H2O ? Show your work
Answer:
For 1: The volume of carbon monoxide required to produce given amount of methanol is 11.9 L.
For 2: The volume of oxygen required to form given amount of water is 7.77 L.
Explanation:
For 1: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 methanol:
Given mass of methanol = 19.3 g
Molar mass of methanol = 32 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of methanol}=\frac{19.3g}{32g/mol}=0.603mol[/tex]
The chemical reaction of formation of methanol from carbon monoxide follows:
[tex]CO+2H_2\rightarrow CH_3OH[/tex]
By Stoichiometry of the reaction:
1 mole of methanol is produced from 1 mole of carbon monoxide.
so, 0.603 moles of methanol will be produced from = [tex]\frac{1}{1}\times 0.603=0.603mol[/tex] of carbon monoxide.
To calculate the volume of carbon monoxide, we use the equation given by ideal gas:
PV = nRT
where,
P = Pressure of carbon monoxide = 676 mmHg
V = Volume of carbon monoxide = ? L
n = Number of moles of carbon monoxide = 0.603 mol
R = Gas constant = [tex]62.3637\text{ L.mmHg }mol^{-1}K^{-1}[/tex]
T = Temperature of carbon monoxide = 212 K
Putting values in above equation, we get:
[tex]676mmHg\times V=\frac{19.3}{32g/mol}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 212K\\\\V=11.79L[/tex]
Hence, the volume of carbon monoxide required to produce given amount of methanol is 11.9 L.
For 2:Calculating the moles of water by using equation 1, we get:
Given mass of water = 12.5 g
Molar mass of water = 18 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of water}=\frac{12.5g}{18g/mol}=0.694mol[/tex]
The chemical reaction of formation of water from oxygen and hydrogen follows:
[tex]O_2+2H_2\rightarrow 2H_2O[/tex]
By Stoichiometry of the reaction:
2 moles of water is produced from 1 mole of oxygen gas.
so, 0.694 moles of water will be produced from = [tex]\frac{1}{2}\times 0.694=0.347mol[/tex] of oxygen gas.
At STP:
1 mole of a gas occupies 22.4 L of volume.
So, 0.347 moles of oxygen gas will occupy = [tex]\frac{22.4L}{1mol}\times 0.347mol=7.77L[/tex]
Hence, the volume of oxygen required to form given amount of water is 7.77 L.
Which statements are true? The oxidation number for Cu(s) is +2. An oxidizing agent gains electrons. Na+ is formed from the reduction of Na(s) . The oxidation number for Hg(????) is 0. A reducing agent gains electrons. Zn2+ is formed from the oxidation of Zn(s) .
Final answer:
The true statements are: an oxidizing agent gains electrons, a reducing agent loses electrons, and Zn2+ is formed from the oxidation of Zn(s).
Explanation:
Let's address each statement in the question about oxidation numbers and redox reactions:The oxidation number for Cu(s) is 0, not +2, because the oxidation number of a free element is always 0.An oxidizing agent gains electrons during a redox reaction.Na+ is formed from the oxidation of Na(s), not reduction, because sodium loses an electron to form Na+.The oxidation number for elemental Hg(l) is 0 since the oxidation state of a free element is 0.A reducing agent loses electrons in a redox reaction.Zn2+ is formed from the oxidation of Zn(s) since zinc loses electrons in the process.Ethanol has a Kb of 1.22 Degrees C/m and usually boils at 78.4 Degrees Celcius. How many mol of an nonionizing solute would need to be added to 47.84 g ethanol in order to raise the boiling point to 86.30?
Answer:
0.3097 moles of an nonionizing solute would need to be added.
Explanation:
Molal elevation constant = [tex]k_b=1.22^oC/m[/tex]
Normal boiling point of ethanol = [tex]T_o=78.4^oC[/tex]
Boiling of solution =[tex]T_b=86.30^oC[/tex]
Moles of nonionizing solute = n
Mass of ethanol (solvent) = 47.84 g
Elevation boiling point:
[tex]\Delta T_b=T_b-T_o[/tex]
[tex]\Delta T_b=86.30^oC-78.4^oC=7.9^oC[/tex]
[tex]\Delta T_b=K_b\times m[/tex]
[tex]m=\frac{\text{Moloes of solute}}{\text{Mass of solvent(kg)}}[/tex]
[tex]7.9^oC=1.22^oC/m\times \frac{n}{0.04784 kg}[/tex]
n = 0.3097 mol
0.3097 moles of an nonionizing solute would need to be added.
Final answer:
To raise the boiling point of ethanol to 86.30°C, 0.310 moles of a nonionizing solute need to be added to 47.84 g of ethanol, using the boiling point elevation formula and the given Kb of ethanol.
Explanation:
To calculate the number of moles of a nonionizing solute needed to raise the boiling point of ethanol to 86.30°C, we use the boiling point elevation formula: ΔT = Kb × m, where ΔT is the change in boiling point, Kb is the ebullioscopic constant of ethanol, and m is the molality of the solution.
First, determine the change in boiling point (ΔT): ΔT = final boiling point - initial boiling point = 86.30°C - 78.4°C = 7.9°C.
Use the given Kb for ethanol, 1.22°C/m, and solve for molality (m): m = ΔT / Kb = 7.9°C / 1.22°C/m = 6.48 m.
To find the number of moles of solute, relate molality to the mass of solvent in kilograms: molality (m) = moles of solute/kg of solvent. Therefore, moles of solute = m × kg of solvent. The mass of ethanol is 47.84 g or 0.04784 kg, thus moles of solute = 6.48 × 0.04784 kg = 0.310 moles.
4.2 g of 1,4-di-t-butyl-2,5-dimethoxybenzene (250.37 g/mol) were synthesized by reacting 10.4 mL of t-butyl alcohol (MW 74.12 g/mol, D 0.79 g/mL), 25 mL of concentrated sulfuric acid (MW 98.08 g/mol, D 1.84 g/mL), and 5.6 g of 1,4-dimethoxybenzene (MW 138.17 g/mol) together. Calculate the percent yield of this reaction.
Answer:
Percentage yield of 1,4-di-t-butyl-2,5-dimethoxybenzene is 41.40%.
Explanation:
Here, in the reaction sulfuric acid is playing the role of catalyst by donating its proton in initial stage of the reaction and in the end of the reaction the proton is returned back to sulfuric acid.
Mass = Density × Volume
Mass of t-butyl alcohol = [tex]0.79 g/mL\times 10.4 mL=8.219 g[/tex]
Moles of t-butyl alcohol =[tex]\frac{8.219 g}{74.12 g/mol}=0.11084 mol[/tex]
Moles of 1,4-dimethoxybenzene = [tex]\frac{5.6 g}{138.17 g/mol}=0.04052 mol[/tex]
According to reaction 2 mol of t-butyl alcohol reacts with 1 mol of 1,4-dimethoxybenzene.
Then 0.11084 moles of t-butyl alcohol will react with :
[tex]\frac{1}{2}\times 0.11084 mole=0.05542 mol[/tex] of 1,4-dimethoxybenzene.
This means that moles of 1,4-dimethoxybenzene are limited and moles of t-butyl alcohol are in excess.So, the moles of product will depend upon the moles of 1,4-dimethoxybenzene.
According top reaction 1 mol of 1,4-dimethoxybenzene gives 1 mol of 1,4-di-t-butyl-2,5-dimethoxybenzene.
Then 0.04052 moles of 1,4-di-t-butyl-2,5-dimethoxybenzene will give:
[tex]\frac{1}{1}\times 0.04052 mol= 0.04052 mol[/tex] of 1,4-di-t-butyl-2,5-dimethoxybenzene.
Mass of 0.04052 moles of 1,4-di-t-butyl-2,5-dimethoxybenzene:
0.04052 mol × 250.37 g/mol = 10.144 g
Percentage yield:
[tex]\frac{Experimental}{Theoretical}\times 100[/tex]
Percentage yield of 1,4-di-t-butyl-2,5-dimethoxybenzene:
Experimental yield = 4.2 g
Theoretical yield = 10.144 g
[tex]\frac{4.2 g}{10.144 g}\times 100=41.40\%[/tex]
The percent yield of the reaction is approximately 41.52%.
The correct format for the answer is as follows:
First, we need to calculate the moles of the limiting reactant to determine the theoretical yield. The balanced chemical equation for the synthesis of 1,4-di-t-butyl-2,5-dimethoxybenzene is not provided, but we can assume that one mole of 1,4-dimethoxybenzene reacts with two moles of t-butyl alcohol to produce one mole of the product.
Let's calculate the moles of each reactant:
For t-butyl alcohol:
[tex]\[ \text{Moles of t-butyl alcohol} = \frac{\text{Volume (mL)} \times \text{Density (g/mL)}}{\text{Molecular Weight (g/mol)}} \] \[ \text{Moles of t-butyl alcohol} = \frac{10.4 \text{ mL} \times 0.79 \text{ g/mL}}{74.12 \text{ g/mol}} \] \[ \text{Moles of t-butyl alcohol} = \frac{8.216 \text{ g}}{74.12 \text{ g/mol}} \] \[ \text{Moles of t-butyl alcohol} \approx 0.1108 \text{ mol} \][/tex]
For 1,4-dimethoxybenzene:
[tex]\[ \text{Moles of 1,4-dimethoxybenzene} = \frac{\text{Mass (g)}}{\text{Molecular Weight (g/mol)}} \] \[ \text{Moles of 1,4-dimethoxybenzene} = \frac{5.6 \text{ g}}{138.17 \text{ g/mol}} \] \[ \text{Moles of 1,4-dimethoxybenzene} \approx 0.0405 \text{ mol} \][/tex]
Since the reaction requires two moles of t-butyl alcohol for every mole of 1,4-dimethoxybenzene, t-butyl alcohol is in excess, and 1,4-dimethoxybenzene is the limiting reactant.
Now, we calculate the theoretical yield of 1,4-di-t-butyl-2,5-dimethoxybenzene:
[tex]\[ \text{Theoretical yield (g)} = \text{Moles of limiting reactant} \times \text{Molecular Weight of product} \] \[ \text{Theoretical yield (g)} = 0.0405 \text{ mol} \times 250.37 \text{ g/mol} \] \[ \text{Theoretical yield (g)} \approx 10.11955 \text{ g} \]The actual yield is given as 4.2 g.[/tex]
Finally, we calculate the percent yield:
[tex]\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield (g)}}{\text{Theoretical Yield (g)}} \right) \times 100\% \] \[ \text{Percent Yield} = \left( \frac{4.2 \text{ g}}{10.11955 \text{ g}} \right) \times 100\% \] \[ \text{Percent Yield} \approx 41.52\% \][/tex]
The atomic mass of silicon is approximately 28.055. Which of the following would indicate the number closest to 28.055?
1. atomic number
2. mass number
3. number of electrons + neutrons
4. number of protons + electrons
Answer:
2. mass number
Explanation:
Mass number is equal to the sum of protons + neutrons in an element. For silicon is 14 + 14 = 28 (for the silicon 14 isotope).
Answer:
2. mass number
Explanation:
The atomic mass is concentrated in the tiny nucleus which contains protons and neutrons. These particles are called the nucleons. Nucleons dictate the mass of an atom.
The number of protons and neutrons in a nucleus is the atomic mass. Also, when we add the protons and neutrons together, we are calculating the mass number. The mass number also gives the atomic mass.
Atomic number on the other hand is the number of electrons or protons in a neutral atom.
2Al(s)+Fe2O3(s)−→−heatAl2O3(s)+2Fe(l) 2Al(s)+Fe2O3(s)→heatAl2O3(s)+2Fe(l) If 26.1 kg Al26.1 kg Al reacts with an excess of Fe2O3,Fe2O3, how many kilograms of Al2O3Al2O3 will be produced?
Answer : The mass of [tex]Al_2O_3[/tex] produced will be, 49.32 Kg
Explanation : Given,
Mass of [tex]Al[/tex] = 26.1 Kg = 26100 g
Molar mass of [tex]Al[/tex] = 26.98 g/mole
Molar mass of [tex]Al_2O_3[/tex] = 32 g/mole
First we have to calculate the moles of [tex]Al[/tex].
[tex]\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}=\frac{26100g}{26.98g/mole}=967.38moles[/tex]
Now we have to calculate the moles of [tex]Al_2O_3[/tex].
The balanced chemical reaction is,
[tex]2Al+Fe_2O_3\rightarrow Heat+Al_2O_3+2Fe[/tex]
From the balanced reaction we conclude that
As, 2 moles of [tex]Al[/tex] react to give 1 mole of [tex]Al_2O_3[/tex]
So, 967.38 moles of [tex]Al[/tex] react to give [tex]\frac{967.38}{2}=483.69[/tex] moles of [tex]Al_2O_3[/tex]
Now we have to calculate the mass of [tex]Al_2O_3[/tex].
[tex]\text{Mass of }Al_2O_3=\text{Moles of }Al_2O_3\times \text{Molar mass of }Al_2O_3[/tex]
[tex]\text{Mass of }Al_2O_3=(483.69mole)\times (101.96g/mole)=49317.0324g=49.32Kg[/tex]
Therefore, the mass of [tex]Al_2O_3[/tex] produced will be, 49.32 Kg
In the reaction of 26.1 kg of Al with an excess of Fe₂O₃, whose equation is 2Al(s) + Fe₂O₃(s) → heat + Al₂O₃(s) + 2Fe(l), will be produced 49.31 kilograms of Al₂O₃.
The reaction is:
2Al(s) + Fe₂O₃(s) → heat + Al₂O₃(s) + 2Fe(l) (1)
To find the mass of Al₂O₃ produced, we need to find the number of moles of Al since it is the limiting reactant (Fe₂O₃ is in excess).
[tex] n_{Al} = \frac{m_{Al}}{A_{Al}} [/tex] (2)
Where:
[tex]m_{Al}[/tex]: is the mass of Al = 26.1 kg = 26100 g
[tex]A_{Al}[/tex]: is the atomic mass of Al = 26.982 g/mol
The number of moles of Al is (eq 2):
[tex]n_{Al} = \frac{m_{Al}}{A_{Al}} = \frac{26100 g}{26.982 g/mol} = 967.31 \:moles[/tex]
From equation (1) we have that 2 moles of Al react with 1 mol of Fe₂O₃ to form 1 mol of Al₂O₃(s), so the number of moles of Al₂O₃ produced is:
[tex]n_{Al_{2}O_{3}} = \frac{1 \:mol \:Al_{2}O_{3}}{2 \:moles \:Al}*967.31 \:moles \:Al = 483.66 \:moles[/tex]
Finally, the mass of Al₂O₃ in kilograms is:
[tex]m = n_{Al_{2}O_{3}}*MM = 483.66 \:moles*101.96 \:g/mol = 49313.5 g = 49.31 kg[/tex]
Therefore, will be produced 49.31 kilograms of Al₂O₃.
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Excess Ca(OH)2 is shaken with water to produce a saturated solution. The solution is filtered, and a 50.00 mL sample titrated with HCl requires 11.22 mL of 0.0983 M HCl to reach the end point. Part A Calculate Ksp for Ca(OH)2.
Answer: The [tex]K_{sp}[/tex] for calcium hydroxide is [tex]5.324\times 10^{-6}[/tex]
Explanation:
To calculate the concentration of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HCl[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ca(OH)_2[/tex]
We are given:
[tex]n_1=1\\M_1=0.0983M\\V_1=11.22mL\\n_2=2\\M_2=?M\\V_2=50mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.0983\times 11.22=2\times M_2\times 50\\\\M_2=0.011M[/tex]
The concentration of [tex]Ca(OH)_2[/tex] comes out to be 0.011 M.
The balanced equilibrium reaction for the ionization of calcium hydroxide follows:
[tex]Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-[/tex]
The expression for solubility constant for this reaction follows:
[tex]K_{sp}=[Ca^{2+}][OH^-]^2[/tex]
Putting the values in above equation, we get:
[tex]K_{sp}=(0.011)\times (2\times 0.11)^2[/tex]
[tex]K_{sp}=5.324\times 10^{-6}[/tex]
Hence, the [tex]K_{sp}[/tex] for calcium hydroxide is [tex]5.324\times 10^{-6}[/tex]
Answer:
[tex]5.2*10^{-6}[/tex]
Explanation:
The balanced chemical equation of the reaction is :
Ca(OH)2 + 2HCl → CaCl2 + 2 H20.
Ksp can be calculated by the following formula:
Ksp = [Ca^{2+} ]+ [OH^{2-}].
Moles of HCl = Molarity × Volume of solution ( liters).
Moles of HCl can be calculated by multiplying 0.01122 (liters) ×0.0983
Moles of HCl = 0.0011 or [tex]1.0*10^{-3}[/tex]
The calculation of the concentration of Calcium hydroxide ( as starting with 50 ml) is :
[tex]Ca(OH)_2 = \frac{1/2 * 0.0011}{0.05 (liters)}[/tex]
[tex]Ca(OH)_2 =0.011[/tex].
[tex]Ca(OH)_2 = 1.1 \times 10^{-2}[/tex].
Ksp = [Ca^{2+} ]+ [2OH^{2-}].
Ksp = [tex]1.1 \times 10^{-2}* (2.2 \times 10^{-2})^2[/tex]
Ksp = [tex]5.2*10^{-6}[/tex]
Hence, the Ksp of calcium hydroxide is [tex]5.2*10^{-6}[/tex]
The normal boiling point of a certain liquid X is 120.7°C , but when 58.66g of urea NH22CO are dissolved in 950.g of X , it is found that the solution boils at 123.0°C instead. Use this information to calculate the molal boiling point elevation constant Kb of X .
The molal boiling point elevation constant, Kb, of the liquid X in the presence of the given amount of urea is calculated to be approximately 2.239 °C kg/mol.
Explanation:The question requires us to calculate the molal boiling point elevation constant, Kb, of a fluid X. The molal boiling point elevation constant is a measure of how much the boiling point of a solution increases when a solute is added. This is represented by the formula: ∆TB = Kb*m, where ∆TB is the change in boiling point and m is the molal concentration of the solute. Given that
∆TB = TB(solution) - TB(pure solvent) = 123.0°C - 120.7°C = 2.3°C,
and the molality m = moles of solute/kg of solvent = (58.66g urea / 60.056 g/mol urea) / (950g solvent / 1000g/kg) = 1.027 mol/kg,
The molal boiling point elevation constant, Kb, can be calculated by rearranging ∆TB = Kb*m to Kb = ∆TB / m = 2.3°C / 1.027 mol/kg = 2.239 °C kg/mol.
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Based on boiling point elevation, mole number, and molality, the molal boiling point elevation constant (Kb) of a liquid X containing 58.66g of urea is computed as 2.23 °C/m.
Step-by-Step Solution:
Determine the boiling point elevation (∆Tb):
[tex]\Delta T_b = 123.0\textdegree C - 120.7\textdegree C = 2.3\textdegree C[/tex]
Calculate the number of moles of urea (NH₂₂CO):
[tex]Molar\ mass\ of\ urea\ = 14 + 1\times 4 + 12 + 16 = 60\ g/mol[/tex]
Urea moles are equal to 58.66 g / 60 g/mol, or 0.978 moles.
Calculate the molality (m) of the solution:
Molality (m) is calculated as solute moles per kilogram of solvent.
[tex]m = 0.978\ moles / 0.950\ kg = 1.03\ m[/tex]
Use the boiling point elevation formula to calculate Kb:
[tex]\Delta T_b = K_b \times m[/tex]
[tex]K_b = \Delta T_b / m = 2.3\textdegree C / 1.03\ m = 2.23 \textdegree C/m[/tex]
As a result, liquid X's molal boiling point elevation constant, or Kb, is 2.23 °C/m.
Antacids, such as Alka-Seltzer, use the reaction of sodium bicarbonate with citric acid in water solution to produce a fizz as follows: 3NaHCO3 + C6H8O7 → 3CO2 + 3H2O + Na3C6H5O7 If 4.11 g of the citric acid (C6H8O7, MW = 192 g/mol) react with excess sodium bicarbonate (NaHCO3), how many grams of carbon dioxide (CO2, MW = 44 g/mol) are formed as the solution fizzes?
When 4.11 g of citric acid reacts with sodium bicarbonate, 2.82 g of carbon dioxide gas is produced according to the provided chemical reaction.
Explanation:In the given chemical reaction, 3NaHCO3 + C6H8O7 → 3CO2 + 3H2O + Na3C6H5O7, specifically between sodium bicarbonate and citric acid, it's clear that for every one molecule of citric acid, three molecules of carbon dioxide are formed. Given that you have 4.11 g of citric acid, and knowing that the molecular weight of citric acid is 192 g/mol, first we calculate the number of moles of citric acid by dividing the mass by the molecular weight: 4.11 g/192 g/mol = 0.0214 moles. Since each mole of citric acid react to produce 3 moles of carbon dioxide, then 0.0214 moles of citric acid would produce 0.0214 moles x 3 = 0.0642 moles of CO2. To find the mass of CO2 formed (since its molar mass is 44 g/mol), multiply the resulting moles of CO2 by its molar mass: 0.0642 moles x 44 g/mol = 2.82 g. Therefore, when 4.11 g of citric acid reacts with sodium bicarbonate, 2.82 g of carbon dioxide will form.
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hotosynthesis can be represented by 6CO2(g)+6H2O(l)⇌C6H12O6(s)+6O2(g) Which of the following will be false when the photosynthesis reaction reaches equilibrium?
a.The rate of formation of O2 is equal to the rate of formation of CO2.
b.The concentration of O2 will begin decreasing.
c.The concentrations of CO2 and O2 will not change.
d.The concentrations of CO2 and O2 will be equal.
Answer: Option (b) and (d) are correct.
Explanation:
An equilibrium reaction is defined as the reaction in which rate of forward reaction equals rate of backward reaction.
A photosynthesis reaction is the reaction in which plants in the presence of sunlight, water, and carbon dioxide make their own food.
The amount of carbon dioxide consumed will be equal to the rate of formation of oxygen into the atmosphere.Hence, the statement rate of formation of [tex]O_{2}[/tex] is equal to the rate of formation of [tex]CO_{2}[/tex] is true.
Concentration of oxygen may remain constant but it will not decrease. Hence, the statement concentration of [tex]O_{2}[/tex] will begin decreasing, is false.As both oxygen and carbon dioxide are present in an equilibrium reaction, So, their concentration will not change.Hence, the statement concentrations of [tex]CO_{2}[/tex] and [tex]O_{2}[/tex] will not change, is true.
Concentration of carbon dioxide and oxygen will also depend on the limiting reagent. So, it is not necessary that the concentrations of [tex]CO_{2}[/tex] and [tex]O_{2}[/tex] will be equal.Some amount of carbon dioxide might escape out into the air.
Hence, the statement concentrations of [tex]CO_{2}[/tex] and [tex]O_{2}[/tex] will be equal, is false.
The equilibrium constant, Kc, for the following reaction is 77.5 at 600 K. CO(g) + Cl2(g) COCl2(g) Calculate the equilibrium concentrations of reactant and products when 0.442 moles of CO and 0.442 moles of Cl2 are introduced into a 1.00 L vessel at 600 K. [CO] = M [Cl2] = M [COCl2] = M
To calculate the equilibrium concentrations of reactant and products, we use the equilibrium constant (Kc) and the initial concentrations of the reactants and products. Using the stoichiometry of the reaction, we can determine the change in concentrations. Solve for x in the equilibrium constant expression to find the equilibrium concentrations.
Explanation:To calculate the equilibrium concentrations of reactant and products, we need to use the equilibrium constant (Kc) and the initial concentrations of the reactants and products. In this case, we have 0.442 moles of CO and 0.442 moles of Cl2 in a 1.00 L vessel at 600 K. Using the stoichiometry of the reaction, we can determine the change in concentrations. Let's denote the change in concentration of COCl2 as x.
The initial concentration of CO is [CO] = 0.442 M, so the change in concentration is -x (due to the decrease in CO).
The initial concentration of Cl2 is [Cl2] = 0.442 M, so the change in concentration is -x (due to the decrease in Cl2).
The initial concentration of COCl2 is [COCl2] = 0 M, so the change in concentration is +x (due to the formation of COCl2).
Using the equilibrium constant expression Kc = [COCl2] / ([CO][Cl2]), we can set up the following equation: 77.5 = x / (0.442 * 0.442)
Solving for x, we get x = 0.442 * 0.442 * 77.5 = 15.558 M. Therefore, the equilibrium concentrations are: [CO] = 0.442 - 15.558 = -15.116 M (neglecting the negative sign), [Cl2] = 0.442 - 15.558 = -15.116 M (neglecting the negative sign), and [COCl2] = 15.558 M.
In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for the following redox reaction that occurs in an electrochemical cell having two electrodes: a cathode and an anode. The two half-reactions that occur in the cell are Cu2 (aq) 2e−→Cu(s) and Co(s)→Co2 (aq) 2e− The net reaction is Cu2 (aq) Co(s)→Cu(s) Co2 (aq) Use the given standard reduction potentials in your calculation as appropriate.
Determine the standard cell potential for the reaction from given standard reduction potentials. Use this in the Nernst equation to calculate the equilibrium constant (Keq). From Keq, interpret the system at equilibrium.
Explanation:The calculation involved in this chemistry question is related to electrochemistry, specifically determining the equilibrium constant (Keq) for a redox reaction involving copper and cobalt in an electrochemical cell. This calculation can be performed using standard reduction potentials and the Nernst equation.
Based on the given half-reactions, the net reaction is Cu2+ (aq) + Co(s) → Cu(s) + Co2+ (aq). You would first determine the E°cell (standard cell potential) for the reaction by subtracting the standard reduction potential for the oxidation half-reaction (occurring at the anode) from that for the reduction half-reaction (at the cathode).
Assuming we have the standard reduction potential values for the cathode and anode processes, E.cell can be calculated. Note that at equilibrium, E.cell equals to 0, so from the Nernst equation we derive the formula for Keq. Remember that n is the number of moles of electrons transferred in the reaction, which is equivalent to 2 in this reaction. From this calculated Keq value, you will be able to infer changes in the system at equilibrium.
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The following reaction was carried out in a 2.00 L reaction vessel at 1100 K: C(s)+H2O(g)⇌CO(g)+H2(g) If during the course of the reaction, the vessel is found to contain 8.75 mol of C, 12.9 mol of H2O, 4.00 mol of CO, and 7.20 mol of H2, what is the reaction quotient Q?
The ratio of products to reactants in a process at a particular time is measured by the reaction quotient Q. The reaction quotient Q is 0.2552.
What is the reaction quotient ?The reactivity factor Q measures the proportional amounts of reactants and products present in a reaction at a specific time. A reaction's direction of shift toward equilibrium can be predicted using Q.
A reaction will go forward and change reactants into products if K > Q. The reaction will go in the opposite direction, turning reactants into products, if K Q. The system is already in equilibrium if Q = K.
Volume of container = 2 L
Concentration = mass / volume
Moles of carbon = 8.75 mol
Concentration of carbon = 8.75 / 2
= 4.375 mol
Moles of H₂O = 12.9 mol
Concentration of water = 12.9 /2
= 6.45 mol
Moles of CO = 4
Concentration of CO = 4 / 2
= 2
Moles of Hydrogen = 7.20 mol
Concentration of Hydrogen is 7.20 / 2
= 3.6 mol
Q is given as
Q = ( H₂ ) ( O₂ ) / ( C ) ( H₂O )
= 7.20 / 28.21
= 0.2552
Thus, The reaction quotient Q is 0.2552.
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Final answer:
The reaction quotient Q for C(s) + H2O(g) ⇌ CO(g) + H2(g), given the amounts of reactants and products and the volume of the container, is calculated to be 1.12.
Explanation:
The reaction quotient, Q, expresses the relative amounts of reactants and products during the progress of a reaction at a particular moment in time. It is calculated using the same expression as the equilibrium constant, Kc, but with the concentrations (or in the case of gases, the partial pressures) that are currently present, rather than at equilibrium. For the reaction given:
C(s) + H₂O(g) ⇌ CO(g) + H₂(g)
The reaction quotient, Q, can be expressed as:
Q = ([CO][H₂]) / [H₂O]
The concentrations can be found by dividing the number of moles of each gas by the volume of the container. Since carbon is a solid, its amount does not factor into the expression for Q. In a 2.00 L container:
[H₂O] = 12.9 mol / 2.00 L = 6.45 M
[CO] = 4.00 mol / 2.00 L = 2.00 M
[H₂] = 7.20 mol / 2.00 L = 3.60 M
To calculate Q:
Q = (2.00 M × 3.60 M) / 6.45 M
Q = 7.20 M2 / 6.45 M
Q = 1.12
Thus, the reaction quotient Q for the given reaction under the specified conditions is 1.12.
A 60.0 mL60.0 mL aliquot of a 1.80 M1.80 M solution is diluted to a total volume of 218 mL.218 mL. A 109 mL109 mL portion of that solution is diluted by adding 107 mL107 mL of water. What is the final concentration? Assume the volumes are additive
Answer:
The final concentration is 0.2497 M.
Explanation:
1. Molarity of the starting solution = [tex]M_1=1.80 M[/tex]
Volume of the starting solution = [tex]V_1=60.0 mL[/tex]
After dilution of the solution to 219 mL.
Molarity of solution after dilution = [tex]M_2[/tex]
Volume of the new solution = [tex]V_2=218 mL[/tex]
[tex]M_1\times V_1=M_2\times V_2[/tex] (Dilution law)
[tex]M_2=\frac{1.80 M\times 60.0 mL}{218 mL}=0.495 M[/tex]
2. Now,109 mL of 0.495 M solution was diluted by 107 mL
Molarity of the solution = [tex]M_1=0.495 M[/tex]
Volume of the solution = [tex]V_1=109 mL[/tex]
Molarity of solution after dilution = [tex]M_2[/tex]
Volume of the new solution(107 mL of water is added) = [tex]V_2=109 mL+107 mL = 216 mL[/tex]
[tex]M_1\times V_1=M_2\times V_2[/tex] (Dilution law)
[tex]M_2=\frac{0.495 M\times 109 mL}{216 mL}=0.2497 M[/tex]
The final concentration is 0.2497 M.
To calculate the final concentration after multiple dilutions, use the dilution formula (M1V1 = M2V2) for each dilution step. The first dilution involves diluting a 60.0 mL sample of a 1.80 M solution to 218 mL. The second dilution involves diluting a 109 mL sample of this diluted solution by adding 107 mL of water.
Explanation:The calculation involved here relies on the formula used for dilution which states that M1V1 = M2V2, where M represents molarity and V represents volume. Initially, we have a 1.80 M solution and we take a 60.0 mL aliquot. After dilution, this becomes a 218 mL solution. Hence, the new molarity (M2) can be calculated as (1.80 M * 60.0 mL) / 218 mL.
From that diluted solution, we take a 109 mL sample and add 107mL of water to it, again diluting this solution further. This process is again similar to the first step of dilution. To find the final concentration, use the dilution formula again with the newly calculated molarity as M1 and 109 mL as V1. The final volume (V2) would be 109 mL + 107 mL. So the final concentration will be given by: (M1 * 109 mL) / (109 mL + 107 mL).
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What are the resulting coefficients when you balance the chemical equation for the combustion of ethane, C2H6? In this reaction, ethane is burned in the presence of oxygen (O2) to form carbon dioxide (CO2) and water (H2O). ____C2H6(g)+____O2(g)→____CO2(g)+____H2O(g) Recall that the coefficients of the final balanced equation should be whole numbers. Thus, you might need to multiply through the equation by a factor of two to obtain whole numbers in your last step. If you have trouble balancing the equation below, use the first hint to view a video of a similar equation being balanced. Then, use the rest of the hints to help you balance the equation, step-by-step. Express the coefficients as integers separated by commas. View Available Hint(s)
Answer: The coefficients are 2, 7, 4 and 6.
Explanation:
Every balanced chemical equation follows Law of conservation of mass.
This law states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form. This means that total mass on the reactant side is equal to the total mass on the product side.
This also means that the total number of individual atoms on the reactant side will be equal to the total number of individual atoms on the product side.
For the given chemical reaction, the balanced equation follows:
[tex]2C_2H_6(g)+7O_2(g)\rightarrow 4CO2(g)+6H_2O(g)[/tex]
On reactant side:
Number of carbon atoms = 4
Number of hydrogen atoms = 12
Number of oxygen atoms = 14
On product side:
Number of carbon atoms = 4
Number of hydrogen atoms = 12
Number of oxygen atoms = 14
Hence, the coefficients are 2, 7, 4 and 6.
To balance the combustion of ethane (C2H6), the correct coefficients are 2 for C2H6, 7 for O2, 4 for CO2, and 6 for H2O resulting in the equation 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O.
Explanation:The chemical equation for the combustion of ethane to form carbon dioxide and water is C₂H₆ + O₂ → CO₂ + H₂O. After balancing the number of atoms of each element on both sides of the equation, we arrive at the balanced chemical equation: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O. Thus, the resulting coefficients are 2, 7, 4, and 6.
Firstly, the carbon (C) atoms are balanced. As we have 2 carbon atoms on the left side, we can put the coefficient 2 in front of CO₂ on the right side.Next, we balance the hydrogen (H) atoms. Since we have 6 hydrogen atoms on the left side, we put the coefficient 3 in front of H₂O on the right side.Finally, we balance the oxygen (O) atoms. Counting the oxygen atoms on both sides, we see that we have 7 oxygen atoms on the right (4 from CO₂ and 3 from H₂O), meaning we need 7 O₂ molecules on the left side.Learn more about Chemical reaction balancing here:https://brainly.com/question/32834309
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The Heat transferred through wave energy (electromagnetic waves) is: A. Sublimation B. Conduction C. Radiation D. Nuclear Fusion
Answer:
C. Radiation
Explanation:
The Heat transferred through wave energy (electromagnetic waves) is Radiation.
The Heat transferred through wave energy (electromagnetic waves) is Radiation.
A mixture of helium and nitrogen gases at a total pressure of 641 mm mm Hg. If the gas mixture contains 0.399 grams of helium, how many grams of nitrogen are present? Hg contains helium at a partial pressure of 231 g
Answer : The mass of nitrogen present are, 4.965 grams.
Explanation :
According to the Raoult's law,
[tex]p_{He}=X_{He}\times p_T[/tex]
where,
[tex]p_{He}[/tex] = partial pressure of gas = 231 mmHg
[tex]p_T[/tex] = total pressure of gas = 641 mmHg
[tex]X_{He}[/tex] = mole fraction of helium gas = ?
Now put all the given values in this formula, we get the mole fraction of helium gas.
[tex]231mmHg=X_{He}\times 641mmHg[/tex]
[tex]X_{He}=0.36[/tex]
Now we have to calculate the mole fraction of nitrogen gas.
[tex]X_{He}+X_{N_2}=1[/tex]
[tex]X_{N_2}=1=0.36=0.64[/tex]
Now we have to calculate the mass nitrogen gas.
[tex]\frac{X_{He}}{X_{N_2}}=\frac{n_{He}}{n_{N_2}}[/tex]
[tex]\frac{X_{He}}{X_{N_2}}=\frac{\frac{w_{He}}{M_{He}}}{\frac{w_{N_2}}{M_{N_2}}}[/tex]
where,
n = moles, w = mass, M = molar mass
Now put all the given values in this expression, we get:
[tex]\frac{0.36}{0.64}=\frac{\frac{0.399}{4}}{\frac{w_{N_2}}{28}}[/tex]
[tex]w_{N_2}=4.965g[/tex]
Therefore, the mass of nitrogen present are, 4.965 grams.
Assume the diameter of a neutral helium atom is 1.40 × 102 pm. Suppose that we could line up helium atoms side by side in contact with one another. How many atoms would it take to make the distance 6.20 cm from end to end?
Answer:
[tex]4.4285\times 10^8 [/tex] many atoms would it take to make the distance 6.20 cm from end to end.
Explanation:
Diameter of the helium atom = [tex]d=1.40\times 10^2 pm[/tex]
Let the number of atoms required to make the distance 6.20 cm be n.
[tex]d\times n= 6.20 cm =6.20\times 10^{10} pm[/tex]
[tex]1.40\times 10^2 pm\times n=6.20\times 10^{10} pm[/tex]
[tex]n=\frac{6.20\times 10^{10} pm}{1.40\times 10^2 }=4.4285\times 10^8 [/tex]
[tex]4.4285\times 10^8 [/tex] many atoms would it take to make the distance 6.20 cm from end to end.
To find the number of helium atoms needed to span 6.20 cm, convert the atom's diameter to centimeters and divide 6.20 cm by the atom's diameter. After calculation, you will get the number of helium atoms required.
We first need to convert the diameter of a helium atom from picometers to centimeters: 1.40 × 102 pm is equivalent to 1.40 × 10-10 cm. Then, we divide the total length required, 6.20 cm, by the diameter of one helium atom in centimeters to find the number of atoms required:
Number of atoms = Total distance / Diameter of one atom = 6.20 cm / (1.40 × 10-10 cm)
After calculating this expression, we would arrive at the number of helium atoms needed to span the distance of 6.20 cm.
The depletion of ozone (O3) in the stratosphere has been a matter of great concern among scientists in recent years. It is believed that ozone can react with nitric oxide (NO) that is discharged from the high-altitude jet plane, the SST. The reaction is O3 + NO longrightarrow O2 + NO2 If 0.781 g of O3 reacts with 0.589 g of NO, how many grams of NO2 will be produced?
Answer: 0.736 g
Explanation:
[tex]O_3+NO\rightarrow O_2+NO2[/tex]
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
[tex]\text{Number of moles of}O_3=\frac{0.781g}{48g/mol}=0.016moles[/tex]
[tex]\text{Number of moles of}NO=\frac{0.589g}{30g/mol}=0.019moles[/tex]
By Stoichiometry:
1 mole of ozone [tex]O_3[/tex] reacts with 1 mole of nitric oxide [tex]NO[/tex] to form 1 mole of nitrogen dioxide [tex]NO_2[/tex]
0.016 moles of ozone reacts with=[tex]\frac{1}{1}\times 0.016=0.016moles[/tex] of nitric oxide to form 0.016 mole of [tex]NO_2[/tex]
Thus ozone is the limiting reagent as it limits the formation of products and nitric oxide is the excess reagent as (0.019-0.016) g= 0.003 g remains as such.
Mass of [tex]NO_2=moles\times {\text{Molar mass}}=0.016\times 46=0.736g[/tex]
0.736 g of [tex]NO_2[/tex] will be produced.
0.748 grams of [tex]\( \text{NO}_2 \)[/tex] will be produced when 0.781 grams of [tex]\( \text{O}_3 \)[/tex] reacts with 0.589 grams of NO, as [tex]\( \text{O}_3 \)[/tex] is the limiting reactant.
To determine how many grams of [tex]\( \text{NO}_2 \)[/tex] will be produced from the reaction of [tex]\( \text{O}_3 \)[/tex] with NO, we need to follow these steps:
1. Write the balanced chemical equation:
[tex]\text{O}_3 + \text{NO} \longrightarrow \text{O}_2 + \text{NO}_2[/tex]
2. Calculate the moles of reactants:
Moles of [tex]\( \text{O}_3 \)[/tex]:
[tex]\text{Molar mass of } \text{O}_3 = 3 \times 16.00 \, \text{g/mol} = 48.00 \, \text{g/mol}[/tex]
[tex]\text{Moles of } \text{O}_3 = \frac{0.781 \, \text{g}}{48.00 \, \text{g/mol}} = 0.01627 \, \text{mol}[/tex]
Moles of NO:
[tex]\text{Molar mass of } \text{NO} = 14.01 \, \text{g/mol} + 16.00 \, \text{g/mol} = 30.01 \, \text{g/mol}[/tex]
[tex]\text{Moles of } \text{NO} = \frac{0.589 \, \text{g}}{30.01 \, \text{g/mol}} = 0.01963 \, \text{mol}[/tex]
3. Determine the limiting reactant:
The balanced equation shows that 1 mole of [tex]\( \text{O}_3 \)[/tex] reacts with 1 mole of NO.
Compare the moles of each reactant:
[tex]0.01627 \, \text{mol} \, \text{O}_3 \quad \text{and} \quad 0.01963 \, \text{mol} \, \text{NO}[/tex]
Since [tex]\( \text{O}_3 \)[/tex] has fewer moles, it is the limiting reactant.
4. Calculate the moles of [tex]\( \text{NO}_2 \)[/tex] produced:
From the balanced equation, 1 mole of [tex]\( \text{O}_3 \)[/tex] produces 1 mole of [tex]\( \text{NO}_2 \)[/tex].
Therefore, moles of [tex]\( \text{NO}_2 \)[/tex] produced = moles of [tex]\( \text{O}_3 \)[/tex] reacted:
[tex]\text{Moles of } \text{NO}_2 = 0.01627 \, \text{mol}[/tex]
5. Convert moles of [tex]\( \text{NO}_2 \)[/tex] to grams:
Molar mass of [tex]\( \text{NO}_2 \)[/tex]:
[tex]\text{Molar mass of } \text{NO}_2 = 14.01 \, \text{g/mol} + 2 \times 16.00 \, \text{g/mol} = 46.01 \, \text{g/mol}[/tex]
Grams of [tex]\( \text{NO}_2 \)[/tex]:
[tex]\text{Mass of } \text{NO}_2 = 0.01627 \, \text{mol} \times 46.01 \, \text{g/mol} = 0.748 \, \text{g}[/tex]
A more experienced colleague mentions to you that you should aim for a dissolved oxygen concentration around 1.0 mg/L at day 5 of the experiment. They show you data from a similar wastewater spill that happened last year, in which the BOD5 of the stream water was 60 mg/L. From your previous experiment, you found that the dissolved oxygen concentration at day 0 was 10 mg/L. Based on this information, what dilution factor, P, should you try in your reactor?
Answer:
3
Explanation:
Lt= Loe^(-kt)
Data:
Lo = 10 mg/mL
Assume k = 0.23/da
1. Calculate L5
L5 = 10e^(-5×0.23) = 10e^-1.15 = 10 × 0.317 = 3.17 mg/mL
2. Calculate the dilution factor
You expect to find L5 to be about 3
You want L5 to be about 1.
You should use a dilute your sample by a factor of 3.
P = 3