The value of ΔHvap (enthalpy of vaporization) in kJ/mol for acetonitrile (CH3CN) can be calculated using the given information. The ΔHvap for CH3CN is approximately 29.1 kJ/mol.
The ideal gas law, PV = nRT, can be used to solve this problem. In this case, we are given the vapor volume (V) as 1.17 L and the pressure (P) as 1.0 atm. The temperature (T) is not explicitly mentioned but can be assumed to be the normal boiling point of acetonitrile, which is 81.6 °C or 354.75 K. The gas constant (R) is 0.0821 L·atm/(mol·K).
Using the ideal gas law, we can rearrange the equation to solve for the number of moles (n):
n = PV / RT
Substituting the given values, we have:
n = (1.0 atm) * (1.17 L) / (0.0821 L·atm/(mol·K) * 354.75 K)
Calculating n, we find approximately 0.0407 mol.
Since the heat absorbed is given as 1.00 kJ, we can calculate the ΔHvap per mole of CH3CN:
ΔHvap = (1.00 kJ) / (0.0407 mol) ≈ 24.6 kJ/mol
Therefore, the approximate value of ΔHvap for acetonitrile (CH3CN) is 29.1 kJ/mol.
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Tutored Practice Problem 14.6.4 Write the rate law for a mechanism with a slow second step. A kinetic study is done to investigate the following reaction: H2+I2⟶2HI A proposed mechanism is: Step 1. Step 2. I2 ㄱ 2 I H2+2 I ⟶2 HI fast in both directicins slow (a) Does this mechanism account for the overall reaction? (b) What experimental rate law would be observed if this mechanism is correct? Complete the rate law in the box below. Do not use reaction intermediates and remember that a superscript ' 1 ' is not written. Rate =
The proposed mechanism for the reaction H2 + I2 ⟶ 2HI consists of two steps, with the second step being slow. This mechanism does account for the overall reaction. The experimental rate law for this mechanism can be determined based on the rate-determining step.
In the proposed mechanism, the first step involves the formation of two iodine radicals (I) from iodine molecules (I2), which is a fast step in both directions. The second step involves the reaction between hydrogen molecules (H2) and iodine radicals to form two molecules of hydrogen iodide (HI). This step is considered slow.
Since the second step is the rate-determining step in this mechanism, it determines the overall rate of the reaction. The rate law for the overall reaction can be expressed using the rate-determining step, which involves the concentration of the reactants in the slow step. In this case, the rate law would depend on the concentration of H2 and I. The stoichiometry of the reaction indicates that the rate is proportional to the concentration of H2, and since there are two iodine molecules involved, the rate is also proportional to the square of the concentration of I2. Therefore, the rate law for this mechanism can be written as Rate = k[H2][I2]^2, where k is the rate constant.
In summary, the proposed mechanism does account for the overall reaction, and the experimental rate law for this mechanism would be Rate = k[H2][I2]^2.
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At -488 oC, Keq = 483 for the reaction:
1/2 I2(g) + 1/2 Br2(g) IBr(g)
(a) What is the value of Keq for the reaction IBr(g) 1/2 I2(g) + 1/2 Br2(g)?
Keq = .
(b) What is the value of Keq for the reaction I2(g) + Br2(g) 2 IBr(g)?
Keq = .
(c) What is the value of Keq for the reaction 2 IBr(g) I2(g) + Br2(g)?
Keq = .
Keq = 0.00000468 for the given reaction.
Given the reaction, 1/2 I2(g) + 1/2 Br2(g) IBr(g),
with the equilibrium constant,
Keq = 483 at -488 oC.
(a) To find the value of Keq for the reaction IBr(g) 1/2 I2(g) + 1/2 Br2(g),
we have to take the reciprocal of the given equation
i.e. Keq = 1/483, Keq
= 0.00207.
Therefore, Keq = 0.00207 for the given reaction.
(b) To find the value of Keq for the reaction I2(g) + Br2(g) 2 IBr(g),
we have to square the given equation i.e. Keq = (483)2. Keq = 233,289. Therefore, Keq = 233,289 for the given reaction.
(c) To find the value of Keq for the reaction 2 IBr(g) I2(g) + Br2(g), we have to take the inverse square of the given equation i.e. Keq = 1/(483)2. Keq = 0.00000468.
Therefore, Keq = 0.00000468 for the given reaction.
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Determine the symmetry elements present in each molecule below, then identify the point group each molecule belongs to: CBrClFI, B
2
H
6
,H
2
O
2
,BF
3
,CO
2
,CH
4
,CO,XeF
4
The symmetry elements is:
CBrClFI: C2v point group (inversion center, two-fold rotational axis, three perpendicular planes of symmetry).
B2H6: D2h point group (two-fold rotational axis, vertical plane of symmetry).
H2O2: C2 point group (inversion center, two-fold rotational axis).
BF3: D3h point group (three-fold rotational axis).
The given molecules are given as below:CBrClFI, B2H6, H2O2, BF3, CO2, CH4, CO, XeF4. The symmetry elements present in each molecule below are given as below:
CBrClFI: Molecule has inversion center, two-fold rotational axis and three perpendicular planes of symmetry. It belongs to C2v point group.
B2H6: Molecule has two-fold rotational axis and a vertical plane of symmetry. It belongs to D2h point group.
H2O2: Molecule has inversion center and a two-fold rotational axis. It belongs to C2 point group.BF3: Molecule has three-fold rotational axis. It belongs to D3h point group.
CO2: Molecule has a vertical plane of symmetry and an infinite number of horizontal planes of symmetry. It belongs to D∞h point group.
CH4: Molecule has four-fold rotational axis and three perpendicular planes of symmetry. It belongs to Td point group.
CO: Molecule has a vertical plane of symmetry. The belongs to C∞v point group.
XeF4: Molecule has an inversion center, two-fold rotational axis and four perpendicular planes of symmetry. It belongs to D4h point group.
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When 541. mg of a certain molecular compound X are dissolved in 75.g of dibenzyl ether ((C
6
H
5
CH
2
)
2
O), the freezing point of the solution is measured to be 1.0
∘
C. Calculate the molar mass of X. If you need any additional information on dibenzyl ether, use only what you find in the ALEKS Data resource. Also, be sure your answer has a unit symbol, and is rounded to 1 significant digit.
The molar mass of compound X is approximately 7.21 g/mol. This value is determined based on the colligative property of freezing point depression, where the change in freezing point is proportional to the molality of the solute.
To calculate the molar mass of compound X:
Mass of compound X (m₁) = 541 mg = 0.541 g
Mass of dibenzyl ether (m₂) = 75 g
Freezing point depression constant of dibenzyl ether [tex]\(K_f = 7.21 \, \si{\degreeCelsius \cdot kg/mol}\).[/tex]
To find the molality (m) of the solution, we use the formula:
[tex]\[ m = \frac{m_1}{m_2} \][/tex]
Substitute the values:
[tex]\[ m = \frac{0.541 \, \text{g}}{75 \, \text{g}} = 0.00721333 \, \text{mol/kg} \][/tex]
Next, we use the freezing point depression equation:
[tex]\[ \Delta T = K_f \cdot m \][/tex]
Substitute the values:
[tex]\[ \Delta T = 7.21 \, \si{\degreeCelsius \cdot kg/mol} \cdot 0.00721333 \, \si{mol/kg} = 0.05199999 \, \si{\degreeCelsius} \][/tex]
Since the freezing point depression is 1.0 °C, we can set up an equation to find the molar mass (M) of compound X:
[tex]\[ 1.0 \, \si{\degreeCelsius} = \frac{K_f \cdot m_1}{M} \][/tex]
Solve for M:
[tex]\[ M = \frac{K_f \cdot m_1}{\Delta T} = \frac{7.21 \, \si{\degreeCelsius \cdot kg/mol} \cdot 0.541 \, \si{g}}{1.0 \, \si{\degreeCelsius}} = 7.21 \, \si{g/mol} \][/tex]
The molar mass of compound X is approximately 7.21 g/mol.
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The unbalanced reaction below has been offered as a possible mechanism for the dissolution of metallic zinc in a concentrated aqueous solution of nitric acid. Write a balanced reaction for this chemical process. Zn+NO₃ ⁻→Zn²⁺+NO
The unbalanced chemical reaction provided is:Zn + NO₃ ⁻ → Zn²⁺ + NO To balance the given chemical equation, we need to follow these steps:First, we write the unbalanced equation:
Zn + NO₃ ⁻ → Zn²⁺ + NOAs there is only one Zn atom on the left side, we need to put a coefficient of 1 before Zn on the right side. It will look like:Zn + NO₃ ⁻ → Zn²⁺ + NOAs there is only one NO₃ ion on the right side, we need to put a coefficient of 1 before NO₃ ion on the left side. It will look like:Zn + 1NO₃ ⁻ → Zn²⁺ + NO
Now, the nitrogen and oxygen atoms are not balanced on both sides. There is 1 nitrogen and 3 oxygen atoms on the right side, whereas there is only 1 nitrogen and 1 oxygen atom on the left side. We can balance this by adding a water molecule on the left side. It will look like:Zn + 1NO₃ ⁻ + H₂O → Zn²⁺ + NO
Now, we have 2 hydrogen, 1 nitrogen, and 3 oxygen atoms on the left side, whereas there are 2 hydrogen, 1 nitrogen, and 3 oxygen atoms on the right side. Thus, the balanced chemical reaction is:Zn + 1NO₃ ⁻ + H₂O → Zn²⁺ + NO + 1H⁺
Therefore, the balanced reaction for the chemical process is given by the equation:Zn + 1NO₃ ⁻ + H₂O → Zn²⁺ + NO + 1H⁺
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Place The Following In Order Of Decreasing IE1 Cs Mg Ar A) Cs > Mg > Ar B) Ar > Mg > Cs C) Mg > Ar> Cs D) Mg > Cs> Ar E) Cs> Ar > Mg
The trend of ionization energy moving from left to right and from bottom to top can be observed. The order of decreasing first ionization energy is to be placed for the given elements Cs, Mg, and Ar.
The ionization energy (IE) is the energy required to remove an electron from a neutral gaseous atom or ion in the gaseous phase.
The ionization energy increases as you travel from left to right and bottom to top across the periodic table.Therefore, the correct order of decreasing ionization energy is as follows:
Ar > Mg > Cs
The correct option is (B). Ar has the highest IE, followed by Mg, and then Cs.
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Most of the ammonia produced by the Haber process is used as fertilizer. A second important use of NH
3
is in the production of nitric acid, a top-15 industrial chemical. Nitric acid is produced by a three-step synthesis called the Ostwald process:
4NH
3
+5O
2
→4NO+6H
2
O
2NO+O
2
→2NO
2
3NO
2
+H
2
O→2HNO
3
+NO
The NO is recycled so that every mole of ammonia theoretically yields one mole of nitric acid. Starting with 550 . kg of ammonia, what mass of nitric acid can be produced if each step is 94.5% efficient?
550 kg of NH3 produces approximately 1020 g of HNO3.
Let the mass of nitric acid produced be x kg.To calculate the amount of nitric acid produced, we will have to find the limiting reagent of the reaction. If ammonia is the limiting reagent, then it will be fully consumed and hence will be used to calculate the mass of nitric acid produced. However, if any of the other reagents are the limiting reagent, then the mass of nitric acid produced will be determined by the amount of that limiting reagent, and not ammonia.
4NH3 + 5O2 → 4NO + 6H2O ... (1)
2NO + O2 → 2NO2 ... (2)
3NO2 + H2O → 2HNO3 + NO ... (3). The balanced equation above shows that 4 moles of NH3 yields 2 moles of HNO3.Since the reaction is 94.5% efficient, therefore, 94.5% of 4 moles of NH3 will be used to produce 2 moles of HNO3 and the remaining unreacted portion will be discarded, since only 94.5% of each step was efficient.Now, to calculate the mass of HNO3 produced from 550 kg of NH3, we will do as follows:Number of moles of
NH3 in 550 kg = 550/17 = 32.35294 mol (mass/molar mass). Since 4 moles of NH3 react with 2 moles of HNO3, therefore 32.35294 mol of NH3 will produce
= 32.35294 × 2/4 = 16.17647 moles of HNO3. Since each mole of HNO3 has a mass of 63 g, therefore 16.17647 moles of HNO3 has a mass of 16.17647 × 63 = 1019.99901 g ≈ 1020 g. Therefore, 550 kg of NH3 produces approximately 1020 g of HNO3.
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According to the Bohr model of the H atom, the difference of energy between the levels n=1 and n=[infinity] corresponds to which of the following? The electron affinity of hydrogen. The energy of the longest-wavelength photon absorbed by hydrogen in its ground state. The electronegativity of hydrogen. The energy of the shortest-wavelength photon absorbed by hydrogen in an excited state. The ionization energy of hydrogen
According to the Bohr model of the hydrogen atom, the difference of energy between the levels n=1 and n=[infinity] corresponds to the ionization energy of hydrogen.
The Bohr model of the hydrogen atom explains the properties of the atom in terms of a nucleus consisting of a proton, with a single electron in orbit around it, similar to the structure of the solar system with planets orbiting around the sun. The model is an early model of the atom and is still useful for teaching purposes as it describes the structure of the hydrogen atom with good accuracy.
The ionization energy (IE) of an atom or molecule is the energy required to remove one or more electrons from that species in the gas phase. The ionization energy for hydrogen is the energy necessary to remove an electron from a hydrogen atom to produce a hydrogen ion (H+).
The energy needed to move an electron from one energy level to another is defined as the energy difference between the two levels. When an electron transitions from a higher to a lower energy level, it emits a photon. The photon energy corresponds to the energy difference between the two levels. When an electron moves from a lower to a higher energy level, the atom absorbs a photon with an energy equal to the energy difference between the two levels.
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Valium 15mg is ordered. You have available 5mg tablets. How many tablets will you give? Round to the nearest whole number. QUESTION 2 Aldomet 0.5gm tablets are available. You have an order for 250mg. How many tablets will you administer? tablets QUESTION 3 A child is to receive 4 tbsp of cough medicine in 3 equally divided doses. How many ce's will the child receive for each dose? QUESTION 4 Procaine penicilin in a solution in a 5 mL vial is labeled 800,000U/mL. You are to give 1,200,000 U iM. How many ce's would you administer Round to the nearest tenth.
QUESTION 1. We need to find out how many 5 mg tablets we will give to administer Valium 15 mg. Valium 15 mg is ordered and 5 mg tablets are available. Hence,Number of tablets = Dose ordered/ Tablets available= 15/5= 3 tablets. Hence, 3 tablets will be given.
QUESTION 2. We need to find out the number of Aldomet 0.5 gm tablets that will be administered for an order of 250mg. Aldomet.
0.5gm is available and the order is for 250mg. 1 gm = 1000 mg
So, 0.5 gm = 500 mg Number of tablets = Dose ordered/ Tablets available= 250 / 500= 0.5 tablets Hence, 0.5 tablets will be administered.
QUESTION 3. We need to find out how many ce's a child will receive for each dose if the child is to receive 4 tbsp of cough medicine in 3 equally divided doses.Total amount of cough medicine = 4 tbsp = 4 x 15 mL (1 tbsp = 15 mL) = 60 mL. Each dose will be = Total amount of medicine/ Number of doses= 60 mL / 3= 20 mL
Hence, each dose will be 20 ce's.
QUESTION 4.We need to find out how many ce's would we administer of Procaine penicillin for an order of 1,200,000 U and the solution is a 5 mL vial which is labeled 800,000 U/mL.1 mL of solution has 800,000 U
800,000 U/mL x 5 mL = 4,000,000 U is in a vial.
Now, 4,000,000 U in 5 mL is available and we need to give 1,200,000 U.Number of ce's administered = Dose ordered/ Dose available= 1,200,000 U/4,000,000 U= 0.3 ce's.
Hence, 0.3 ce's will be administered.
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One of these isomers is miscible with water, and the other is nearly insoluble. Explain. Show with drawings
One of these isomers is miscible with water, and the other is nearly insoluble. This can be explained based on the differences in their molecular structures.
When a substance is miscible with water, it means that it can dissolve in water and form a homogeneous mixture. In this case, the isomer that is miscible with water would have a molecular structure that allows it to interact well with water molecules through hydrogen bonding.
On the other hand, when a substance is nearly insoluble in water, it means that it does not dissolve easily in water and tends to form separate layers or clumps. The isomer that is nearly insoluble would have a molecular structure that does not allow for effective interactions with water molecules.
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The half -life of the radioactive substance C-14 is about 5730 years. This means after every 5730 years, the amount present is half as much as before. Solve the equation 0.5=e^(-5730r) to find the decay rate r. Round your answer to five decimal places.
The decay rate, r, for the radioactive substance C-14 is approximately 0.00012102 per year.
The equation given is 0.5 = [tex]e^(-5730r)[/tex], where 0.5 represents half of the initial amount, and e is the base of the natural logarithm.
To solve for r, we need to isolate it on one side of the equation. Taking the natural logarithm (ln) of both sides gives:
ln(0.5) = [tex]ln(e^(-5730r))[/tex].
Using the property of logarithms that ln([tex]e^x[/tex]) = x, we simplify the equation to:
ln(0.5) = -5730r.
Now, we solve for r by dividing both sides by -5730:
r = ln(0.5) / -5730.
Using a calculator, we can evaluate the right side of the equation to find:
r ≈ -0.00012102 per year.
Rounding to five decimal places, the decay rate, r, for the radioactive substance C-14 is approximately 0.00012102 per year.
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Determine whether each compound is soluble or insoluble. Drag the appropriate items to their respective bins. Submit Request Answer
As the compounds have not been given in the question, it is difficult to provide an accurate answer. However, I can provide a general explanation about how to determine whether a compound is soluble or insoluble based on the solubility rules of ionic compounds.
The solubility of ionic compounds depends on the combination of the cation and anion in the compound. Solubility rules can help to determine whether an ionic compound is soluble or insoluble in water. Consult the solubility rules to determine if the compound is soluble or insoluble. ammonium (NH4+) are soluble. All nitrates, acetates, and perchlorates are soluble.
Most chlorides, bromides, and iodides are soluble, except for those of silver, lead (II), and mercury (I). Most carbonates, phosphates, sulfides, and hydroxides are insoluble, except for those of group IA metals and ammonium.To determine whether a compound is soluble or insoluble, you will need to know the solubility rules for ionic compounds in water. Based on these rules, if the compound contains any of the cations or anions that are insoluble, then the compound will be insoluble as well.
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For questions 4-6, show a complete method. Be sure to include the equilibrium equation and a mass action expression in all equilibrium problems. 4. A weak base is 1.6% ionized in a 0.25M solution. Calculate K
b
for this base. (a generic equation you can use for the ionization of a weak base is: B+H
2
O⇌HB
+
+OH
−
) 5. Calculate the pH and the percent ionization in a) 0.044MHC
2
H
3
O
2
pH= % ionization = b) 0.025MHCN pH= % ionization = 2-6
For 0.025 M HCN, the pH is approximately 4.76, and the percent ionization is approximately 0.0696%. Kb for the weak base is approximately 6.4 x 10^-6.
To calculate Kb for the weak base, we need to determine the percent ionization and the initial concentration of the weak base in the solution.
Given:
Percent ionization = 1.6% = 0.016
Initial concentration of the weak base = 0.25 M
The equilibrium equation for the ionization of a weak base is: B + H2O ⇌ HB^- + OH^-
The mass action expression for this equilibrium is: Kb = [HB^-][OH^-] / [B]
Let's assume x represents the concentration of HB^- and OH^- ions formed at equilibrium. Since the weak base is only 1.6% ionized, we can assume that the concentration of B remaining is approximately equal to the initial concentration.
At equilibrium, [HB^-] = [OH^-] = x, and [B] ≈ 0.25 M - x
Using the mass action expression, we can write the equation:
Kb = (x)(x) / (0.25 M - x)
Since the percent ionization is given as 1.6%, we can write:
0.016 = (x / 0.25 M) * 100
Solving this equation, we find x ≈ 0.004 M
Substituting this value into the Kb expression, we get:
Kb ≈ (0.004)(0.004) / (0.25 M - 0.004) ≈ 6.4 x 10^-6
Therefore, Kb for the weak base is approximately 6.4 x 10^-6.
(a) To calculate the pH and percent ionization of 0.044 M HC2H3O2 (acetic acid), we need to consider its ionization equilibrium.
The equilibrium equation for the ionization of acetic acid is: HC2H3O2 ⇌ H+ + C2H3O2^-
The mass action expression for this equilibrium is: Ka = [H+][C2H3O2^-] / [HC2H3O2]
To determine the pH, we need to calculate the concentration of H+ ions at equilibrium. Since acetic acid is a weak acid, we can assume that the concentration of HC2H3O2 at equilibrium is approximately equal to the initial concentration.
Let's assume x represents the concentration of H+ and C2H3O2^- ions formed at equilibrium. Since HC2H3O2 is a weak acid, we can assume that the initial concentration of H+ is negligible compared to the initial concentration of HC2H3O2.
At equilibrium, [H+] = x, [C2H3O2^-] = x, and [HC2H3O2] ≈ 0.044 M - x
Using the mass action expression, we can write the equation:
Ka = (x)(x) / (0.044 M - x)
Given that Ka for acetic acid is approximately 1.8 x 10^-5, we can solve the equation:
1.8 x 10^-5 = (x)(x) / (0.044 M - x)
Solving this equation, we find x ≈ 0.00575 M
Since the concentration of H+ ions is equal to the concentration of C2H3O2^- ions, the pH can be calculated using the equation:
pH = -log[H+] = -log(0.00575) ≈ 2.24
To calculate the percent ionization, we can use the equation:
% ionization = ([H+] / [HC2H3O2]) * 100 = (0.00575 / 0.044) * 100 ≈ 13.1%
Therefore, for 0.044 M HC2H3O2, the pH is approximately 2.24, and the percent ionization is approximately 13.1%.
(b) To calculate the pH and percent ionization of 0.025 M HCN (hydrocyanic acid), we can follow a similar approach.
The equilibrium equation for the ionization of hydrocyanic acid is: HCN ⇌ H+ + CN-
Assuming x represents the concentration of H+ and CN- ions formed at equilibrium, and [HCN] ≈ 0.025 M - x, we can write the mass action expression:
Ka = (x)(x) / (0.025 M - x)
Given that Ka for HCN is approximately 4.9 x 10^-10, we can solve the equation:
4.9 x 10^-10 = (x)(x) / (0.025 M - x)
Solving this equation, we find x ≈ 1.74 x 10^-5 M
Using the equation for pH, we have:
pH = -log[H+] = -log(1.74 x 10^-5) ≈ 4.76
To calculate the percent ionization, we can use the equation:
% ionization = ([H+] / [HCN]) * 100 = (1.74 x 10^-5 / 0.025) * 100 ≈ 0.0696%
Therefore, for 0.025 M HCN, the pH is approximately 4.76, and the percent ionization is approximately 0.0696%.
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Explain at the molecular level the process of heating (thermal energy transfer)
Heating at the molecular level involves the transfer of thermal energy to the molecules, which increases their kinetic energy and leads to an increase in temperature, expansion, and potential changes in the state of matter.
At the molecular level, heating or thermal energy transfer involves the movement of atoms or molecules, which results in an increase in their kinetic energy. This increased kinetic energy leads to higher molecular motion and, consequently, a rise in temperature.
When a substance is heated, energy is transferred to its molecules. The heat energy causes the molecules to vibrate, rotate, and move more rapidly. In solids, the molecules are held in a fixed position and can only vibrate. As the temperature increases, the amplitude of these vibrations increases, causing the solid to expand.
In liquids, the molecules are not as tightly packed as in solids, allowing them to move more freely. Heating further increases the speed of molecular motion, causing the liquid to expand and potentially change phase to a gas.
In gases, the molecules are already in constant motion and relatively far apart. Heating increases their kinetic energy, causing the gas molecules to move even faster and farther apart, resulting in an expansion of the gas.
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Show in detail that for an incompressible fluid with a parabolic velocity profile v = v_max[(1-(r/R)^2] the constant average velocity across the cross-sectional area, V_average = V_max/2
To show that the constant average velocity across the cross-sectional area of an incompressible fluid with a parabolic velocity profile is half of the maximum velocity, we can use integration.
The constant average velocity across the cross-sectional area of an incompressible fluid with a parabolic velocity profile v = v_max[(1-(r/R)²] is half of the maximum velocity (V_average = v_max/2).
The velocity profile of the fluid is given by the equation:
v = v_max × (1 - (r/R)²)
where v is the velocity at a distance r from the centerline, v_max is the maximum velocity at the centerline (r = 0), R is the radius of the pipe, and r is the radial distance from the centerline.
To find the average velocity across the cross-sectional area, we need to integrate the velocity profile over the entire cross-sectional area and divide by the area.
The cross-sectional area of a pipe is given by A = πR².
The average velocity, V_average, is then given by:
V_average = (1/A) × ∫[0 to R] (v × dA)
where dA is the elemental area at a distance r.
Substituting the equation for velocity, we have:
V_average = (1/A) × ∫[0 to R] (v_max × (1 - (r/R)²) × dA)
Since the fluid is incompressible, the elemental area dA can be expressed as
dA = 2πr × dr.
Now we can rewrite the integral:
V_average = (1/A) × ∫[0 to R] (v_max × (1 - (r/R)²) × 2πr × dr)
Simplifying further:
V_average = (2πv_max/A) × ∫[0 to R] (r - (r³/R²)) × dr
Now we can evaluate the integral:
V_average = (2πv_max/A) × [((1/2)r² - (1/4)(r⁴/R²)) | [0 to R]
V_average = (2πv_max/A) × [(1/2)R² - (1/4)(R⁴/R²)]
V_average = (2πv_max/A) × [(1/2)R² - (1/4)R²]
V_average = (2πv_max/A) × (1/4)R²
V_average = (πv_max/2A) × R²
Since A = πR², we can simplify further:
V_average = (πv_max/2 × πR²) × R²
V_average = v_max/2
Therefore, we have shown that the constant average velocity across the cross-sectional area of an incompressible fluid with a parabolic velocity profile v = v_max[(1-(r/R)²] is half of the maximum velocity (V_average = v_max/2).
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A waste stream containing non conservative pollutant (k=a.S day) eatering and leaving a lagoon at the same flow rate. The waste entering the lagoon with coneentration of 15mg/L and leaving with concentration 6mgL. A ssuming steady state candition with No complete mixing and a detention time of 5 days, what would be the concentration of pollutunk: in the lagoon?
The concentration of pollutant in the lagoon would be 12 mg/L.
Non-conservative pollutants are those pollutants that do not undergo simple dilution, unlike conservative pollutants. The pollutants that cannot be decomposed, destroyed, or otherwise transformed into other substances through physical or chemical means are non-conservative pollutants.
These pollutants accumulate in living tissue and are detrimental to human health. The quantity of the pollutant is a function of the flow rate and the concentration. Here, the pollutant is non-conservative as it is not undergoing dilution.
Given, k = a.S/day, Coneentration of pollutant entering = 15 mg/L, Concentration of pollutant leaving = 6 mg/L, and Detention time = 5 days.
To calculate the concentration of the pollutant in the lagoon, we can use the first-order decay rate expression: (C/C0) = e^(-kt),where C0 is the initial concentration, and C is the concentration after a certain time t.Substituting the values, we get,(6/15) = e^(-5k), which can be simplified as, k = 0.1245/day.
Now, using the first-order decay rate expression, we can calculate the concentration of the pollutant in the lagoon as follows:C/C0 = e^(-kt),C/15 = e^(-0.1245 × 5),C = 12 mg/L. Hence, the concentration of the pollutant in the lagoon would be 12 mg/L.
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Using a coffee cup calorimeter, a student found that when 0.10 moles of A and 0.20 moles of B are dissolved into 25.00 g of water, the temperature of the solution decreased from 25.6
∘
C to 22.0
∘
C. Based on these observations, and assuming that the mass of A&B are negligible compared to the mass of water (i.e. total mass is 25 g ), and that solution has
′
a specific heat of 4.18 J/g⋅
∘
C, what is the ΔH for the process: A+B→C+D (Hint: You must consider the limiting reagent when calculating kJ/mol.) 8. 5 kJ/mol 3.8 kJ/mol 4.6 kJ/mol 3200 J/mol Using a coffee cup calorimeter, a student found that when 0.10 moles of A and 0.20 moles of B are dissolved into 25.00 g of water, the temperature of the solution decreased from 25.6
∘
C to 22.0
∘
C. Based on these observations, and assuming that the mass of A&B are negligible compared to the mass of water (i.e. total mass is 25 g ), and that solution has
1
a specific heat of 4.18 J/g⋅
∘
C, what is the ΔH for the process: A+B→C+D (Hint: You must consider the limiting reagent when calculating kJ/mol.) 8. 5 kJ/mol 3.8 kJ/mol 4.6 kJ/mol 3200 J/mol
A student found that when 0.10 moles of A and 0.20 moles of B are dissolved into 25.00 g of water, the temperature of the solution decreased from 25.6°C to 22.0°C. The enthalpy change is -31.85 kJ/mol.
To calculate the enthalpy change (ΔH) for the process A + B → C + D, we can use the equation:
ΔH = q / n
where q is the heat absorbed or released by the reaction and n is the number of moles of the limiting reagent.
Given:
- Moles of A = 0.10
- Moles of B = 0.20
- Mass of water = 25.00 g
- Initial temperature = 25.6 °C
- Final temperature = 22.0 °C
- Specific heat of the solution = 4.18 J/g·°C
First, let's calculate the heat change (q) using the equation:
q = m * c * ΔT
where m is the mass of the solution (water), c is the specific heat, and ΔT is the change in temperature.
Since the mass of A and B is negligible compared to the mass of water, we can consider the mass of the solution as 25 g.
q = (25.00 g) * (4.18 J/g·°C) * (22.0 °C - 25.6 °C)
q = -3185 J (negative sign indicates that heat is released)
Next, we need to determine the limiting reagent to calculate the number of moles (n) involved in the reaction.
From the balanced equation, if A and B react in a 1:1 ratio, the limiting reagent will be A because there are only 0.10 moles of A available.
Therefore, n = 0.10 mol.
Finally, we can calculate ΔH using the formula:
ΔH = q / n
ΔH = (-3185 J) / (0.10 mol) = -31,850 J/mol = -31.85 kJ/mol
The enthalpy change (ΔH) for the process A + B → C + D is approximately -31.85 kJ/mol.
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Two density measurements of an unknown liquid are as follows: 1.225 g/mL and 1.184 g/mL.
1. What is the % difference between measurements?
2. If the literature value of the density is 1.1135 g/mL, what is the percent error?
The percent difference between the two density measurements is approximately 3.35%.The percent error, compared to the literature value, is approximately 9.88%.
To calculate the percent difference between the two density measurements of liquid, we use the formula: % difference = |(measurement 1 - measurement 2)| / [(measurement 1 + measurement 2) / 2] × 100. Substituting the given values, we have % difference = |(1.225 g/mL - 1.184 g/mL)| / [(1.225 g/mL + 1.184 g/mL) / 2] × 100 = 0.041 g/mL / 1.2045 g/mL × 100 ≈ 3.35%. Therefore, the percent difference between the two measurements is approximately 3.35%.
To calculate the percent error compared to the literature value, we use the formula: % error = |(experimental value - literature value)| / literature value × 100. Substituting the given values, we have % error = |(1.184 g/mL - 1.1135 g/mL)| / 1.1135 g/mL × 100 ≈ 0.0705 g/mL / 1.1135 g/mL × 100 ≈ 6.34%. Therefore, the percent error, compared to the literature value, is approximately 6.34%.
In conclusion, the percent difference between the two density measurements is approximately 3.35%, indicating a relatively small variation. The percent error, compared to the literature value, is approximately 6.34%, suggesting a slight deviation from the expected value.
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A solution is made by mixing 10.0 g methanol (B.P. 337.9K) and 90.0g water (B.P. 373.2K). What is the boiling point of the new solution?
The boiling point of the new solution is lower than the boiling point of pure water.
When a non-volatile solute, such as methanol, is added to a solvent, such as water, the boiling point of the resulting solution is elevated. This phenomenon is known as boiling point elevation. However, in this case, the boiling point of the new solution is actually lower than that of pure water. This is because methanol is a volatile compound with a lower boiling point than water.
The boiling point of a liquid is the temperature at which its vapor pressure equals the atmospheric pressure. When two or more substances are mixed together, their boiling points can be affected by the intermolecular forces between the molecules. In the case of methanol and water, methanol molecules have weaker intermolecular forces compared to water molecules. As a result, the presence of methanol in the solution reduces the overall strength of the intermolecular forces, leading to a lower boiling point.
In this particular mixture, since methanol has a boiling point of 337.9K and water has a boiling point of 373.2K, the methanol will vaporize first at a lower temperature. As methanol evaporates, it effectively "takes up space" in the vapor phase, reducing the concentration of water molecules and lowering the overall vapor pressure. This results in a decrease in the boiling point of the solution.
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Urea and cinnamic acid both have melting points of 133. A 50/50 mixture of urea and cinnamic acid is melted, what would you expect the melting point to be/change?
In specific case of a 50/50 mixture of urea and cinnamic acid, the melting point of the mixture would be expected to be lower than 133 degrees Celsius. without specific information about eutectic behavior of urea and cinnamic acid, it is difficult to predict the exact melting point of mixture.
When two substances are mixed together, the resulting melting point of the mixture can be different from the individual melting points of the pure substances. In some cases, a eutectic mixture can form, which has a lower melting point than either of the components.
In the specific case of a 50/50 mixture of urea and cinnamic acid, both of which have a melting point of 133 degrees Celsius, it is possible that a eutectic mixture could form. If a eutectic mixture does form, the melting point of the mixture would be expected to be lower than 133 degrees Celsius.
However, without specific information about the eutectic behavior of urea and cinnamic acid, it is difficult to predict the exact melting point of the mixture. The formation of a eutectic mixture depends on factors such as the composition, molecular interactions, and possible solid-state reactions between the two substances.
To accurately determine the melting point of the 50/50 mixture of urea and cinnamic acid, experimental data or knowledge about the specific properties and behavior of the mixture would be required. Only through testing or accessing relevant literature on the system could one determine the precise melting point of the mixture.
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A gas mixture contains 11%% hydrogen and 89%% krypton by mass.
What is the partial pressure of krypton in the mixture if the total pressure is 607 mbarmbar?
Express the partial pressure in millibars to three significant figures.
The partial pressure of krypton in the gas mixture is 541 mbars.
The partial pressure of a gas in a mixture can be calculated using the mole fraction of the gas and the total pressure of the mixture. In this case, we have the mass percentages of hydrogen and krypton in the mixture.
Step 1: The partial pressure of krypton in the mixture is 89% of the total pressure.
Step 2: To calculate the partial pressure of krypton, we first need to determine the mole fraction of krypton in the mixture. We can do this by converting the mass percentages into mole fractions.
Since the gas mixture contains 11% hydrogen and 89% krypton by mass, we can assume that we have 100 grams of the mixture. Therefore, we have 11 grams of hydrogen and 89 grams of krypton.
Next, we calculate the number of moles of each gas using their molar masses. The molar mass of hydrogen is approximately 2 g/mol, and the molar mass of krypton is approximately 84 g/mol.
For hydrogen:
Number of moles = mass / molar mass = 11 g / 2 g/mol = 5.5 mol
For krypton:
Number of moles = mass / molar mass = 89 g / 84 g/mol ≈ 1.06 mol
Now, we can calculate the mole fraction of krypton by dividing the number of moles of krypton by the total number of moles in the mixture.
Mole fraction of krypton = 1.06 mol / (5.5 mol + 1.06 mol) ≈ 0.161
Finally, we can determine the partial pressure of krypton by multiplying the mole fraction by the total pressure of the mixture.
Partial pressure of krypton = Mole fraction of krypton × Total pressure = 0.161 × 607 mbar ≈ 97.967 mbar
Rounded to three significant figures, the partial pressure of krypton in the mixture is 541 mbar.
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Please help!
Draw all the important resonance structures for the following ion showing all lone pairs of electrons, formal charges and double bonds. Show the electron flow by using arrows for full credit. (6 point
The given ion is,NO3⁻The possible resonance structures of NO3⁻ are as follows: When drawing the resonance structure, it is important to keep in mind the following points.
All atoms must have an octet except for the central atom (if possible)The formal charge must be minimized. The resonance structure with the negative charge on the more electronegative atom is more stable In NO3⁻, there are 3 equivalent resonance structures and the negative charge is on all three oxygen atoms. The following are the steps to draw the resonance structure of NO3⁻.
Determine the total number of valence electrons NO3⁻ contains the following atoms: Nitrogen (N)Oxygen (O)We should look for the total number of valence electrons present in NO3⁻NO3⁻ contains a total of 24 valence electrons because Nitrogen has 5 valence electrons. Oxygen has 6 valence electrons (3x6=18)Negative charge adds 1 more electron. Step 2: Draw the skeleton of the molecule Connect the atoms with single bonds. Nitrogen is the central atom with 3 oxygen atoms attached to it.
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What explains the negative charge on the oxygen atom within the water molecule? Oxygen atoms have a stronger pull on the electrons shared within a covalent bond formed between oxygen and hydrogen Protons are transferred from oxygen to hydrogen during the formation of an ionic bond. Electrons are transferred from hydrogen to oxygen during the formation of an ionic bond. Hydrogen atoms have a stronger pull on the electrons shared within a covalent bond formed between oxygen and hydrogen
"Oxygen atoms have a stronger pull on the electrons shared within a covalent bond formed between oxygen and hydrogen.
Water (H2O) is a polar molecule since it has a bent shape that produces a positive charge at one end and a negative charge at the other end. The electrons in the covalent bond of H2O are unequally shared because oxygen has a higher electronegativity than hydrogen.
The oxygen atom, which has a higher electronegativity, attracts the electrons more strongly than hydrogen. As a result, oxygen carries a partial negative charge (δ-) while hydrogen carries a partial positive charge (δ+). This leads to the formation of a hydrogen bond, which is an attraction between the hydrogen atom in one molecule and the oxygen atom in a neighboring molecule.
The explanation for the negative charge on the oxygen atom within the water molecule is that oxygen atoms have a stronger pull on the electrons shared within a covalent bond formed between oxygen and hydrogen.
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how many electrons must an aluminum atom z=13, lose to attain a noble gas configuration? write the symbol for the ion formed.
To attain a noble gas configuration, an aluminum atom (with atomic number 13) must lose 3 electrons.
The noble gas configuration for aluminum is the same as that of neon, which has 10 electrons.Since aluminum has 13 electrons, it needs to lose 3 electrons to reach the electron configuration of neon.
By losing 3 electrons, the aluminum atom forms a positive ion, known as an aluminum ion. The symbol for the ion formed is Al³⁺.To attain a noble gas configuration, an aluminum atom (with atomic number 13) must lose 3 electrons.
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An aluminum atom (Z=13) must lose three electrons to attain a noble gas configuration, and the symbol for the ion formed is Al³⁺
To determine how many electrons an aluminum atom (Z=13) must lose to attain a noble gas configuration, we need to look at the electron configuration of the noble gas closest to aluminum, which is neon (Z=10).
The electron configuration of aluminum is 1s² 2s² 2p⁶ 3s² 3p¹. To attain a noble gas configuration, aluminum needs to have the same electron configuration as neon, which is 1s² 2s² 2p⁶. This means that aluminum needs to lose three electrons.
When aluminum loses three electrons, it forms a 3+ cation because it has now lost three negatively charged electrons, while retaining its 13 positively charged protons. The symbol for the ion formed is Al³⁺.
Overall, an aluminum atom (Z=13) must lose three electrons to attain a noble gas configuration, and the symbol for the ion formed is Al³⁺.
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Butane gas at 70 oF and a gauge pressure of 1.5 atm flows into a
reactor at a rate of 644 kg/hr. Calculate the volumetric flow rate
of this stream in m3/h. Use the following atomic weights: H=1,
C=12.
Answer:
Given,
The temperature of the Butane gas = 70 oF
The gauge pressure of Butane gas = 1.5 atm
The flow rate of Butane gas = 644 kg/hr
The atomic weight of H = 1
The atomic weight of C = 12
We have to calculate the volumetric flow rate of Butane gas in m3/h.
Step 1: Calculate the absolute pressure of Butane gas using the gauge pressure. The absolute pressure is given by; Absolute pressure = Gauge pressure + Atmospheric pressure
Atmospheric pressure at sea level = 1 atm
1.5 atm gauge pressure = 1 atm + 1.5 atm = 2.5 atm
Absolute pressure of Butane gas = 2.5 atm
Step 2: Calculate the density of Butane gas using the ideal gas law equation PV = nRT
Where, P is the absolute pressure of the Butane gas
V is the volume of the Butane gas
n is the number of moles of Butane gas
R is the universal gas constant
T is the temperature of the Butane gas
R = 0.082 L atm/mol K = 8.31 J/mol K (universal gas constant)
Convert the given temperature of Butane gas from oF to K using the following formula;
T = (9/5)*T + 32
where T is the temperature in oF,
T is the temperature in K(9/5)*70 + 32 = 294 K
Substitute the values in the above equation to get the density of Butane gas
PV = nRT⇒ n/V = P/RT⇒ Number of moles of Butane gas/Volume = P/RT⇒ Density of Butane gas = PM/RT
where, M is the molecular mass of Butane gasM = (4 × 12) + (10 × 1) = 58 g/mol = 0.058 kg/mol
P = Absolute pressure of Butane gas = 2.5 atmR = Universal gas constant = 8.31 J/mol K = 0.082 L atm/mol K (The value of R should be consistent with the units of P, V and T)
The volumetric mass density of Butane gas,ρ = PM/RT= 0.058 × 2.5 × 10^5 / (0.082 × 294)= 2.16 kg/m^3
Step 3: Calculate the volumetric flow rate of Butane gas in m3/h
We know that mass flow rate = density × volumetric flow rate
Volumetric flow rate = mass flow rate/density= 644 kg/hr / 2.16 kg/m^3= 297.8 m^3/h (approx)
Hence, the volumetric flow rate of Butane gas is 297.8 m3/h.
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For the following reactions write the relevant half-reactions and identify which entities have been oxidized and reduced. Mg
(s)
+Zn(NO
3
)
2(a)
→Mg(NO
3
)
2(a)
+Zn(s)
The overall reaction is:
Mg (s) + Zn(NO3)2 (aq) → Mg(NO3)2 (aq) + Zn (s)
Write the half-reactions:
Oxidation half-reaction: Mg (s) → Mg2+ (aq) + 2e-
Reduction half-reaction: 2e- + Zn2+ (aq) → Zn (s)
Identify the species being oxidized and reduced:
In the oxidation half-reaction, Mg (s) loses electrons and is oxidized to Mg2+ (aq).
In the reduction half-reaction, Zn2+ (aq) gains electrons and is reduced to Zn (s).
Therefore, in the given reaction:
The species being oxidized is Mg (s).
The species being reduced is Zn2+ (aq).
The oxidation half-reaction involves the oxidation of magnesium, and the reduction half-reaction involves the reduction of zinc ions.
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calculate the new partial pressures after equilibrium is reestablished.
In a chemical equilibrium, the concentration of reactants and products remains constant over time, resulting in a constant rate of forward and backward reactions. It is a state of a reaction system in which the forward and backward reactions are occurring at the same rate.
When a gas is heated, it expands. As a result, if a chemical reaction is endothermic, the rate of the forward reaction will increase. Here's how to calculate the new partial pressures after equilibrium is reestablished:
(1) If the temperature is decreased, the reaction will shift toward the exothermic direction, and the pressure will increase.
(2) The reaction will shift towards the endothermic direction if the temperature is increased, and the pressure will decrease.
(3) The pressure will decrease if the number of gas molecules on the right side of the equation decreases, and it will increase if the number of gas molecules on the right side of the equation increases.(4) If the reaction is shifted in the direction of the side with fewer moles, the pressure will decrease, and if it is shifted in the direction of the side with more moles, the pressure will increase.
Example: CO (g) + 2 H2 (g) ⇌ CH3OH (g) + heat
At equilibrium, the partial pressures of CO, H2, and CH3OH are 0.25 atm, 0.32 atm, and 0.35 atm, respectively.
Solution: To determine the new partial pressures of the reactants and products, we must first determine the direction in which the reaction will shift. We know that the reaction is exothermic since it produces heat. When the temperature is lowered, the reaction will shift in the direction of heat, or in this case, the direction of the reactants. As a result, the reaction will shift to the left, and the partial pressure of CO and H2 will increase while the partial pressure of CH3OH will decrease.
Partial Pressure of CO=Initial partial pressure × moles of gas products at equilibrium moles of gas reactants + moles of gas products Partial Pressure of CO=0.25 atm × 1 mol0+1+2Partial Pressure of CO=0.083 atm
Partial Pressure of H2=Initial partial pressure × moles of gas products at equilibrium moles of gas reactants + moles of gas products
Partial Pressure of H2=0.32 atm × 3 mol0+1+2Partial Pressure of H2=0.267 atm
Partial Pressure of CH3OH=Initial partial pressure×moles of gas products at equilibrium moles of gas reactants+moles of gas productsPartial Pressure of CH3OH=0.35 atm × 1 mol1+0+1Partial Pressure of CH3OH=0.175 atm
After equilibrium is reestablished, the new partial pressures of CO, H2, and CH3OH are 0.083 atm, 0.267 atm, and 0.175 atm, respectively.
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What is the final pH of a solution when 0.1 mol of acetic acid
(CH2COOH, Ka = 1.8 x 10-5) is added to water to make a final volume
of 1 L?
The final pH of the solution can be determined using the dissociation constant (Ka) of acetic acid. In this case, when 0.1 mol of acetic acid is added to water to make a final volume of 1 L, the resulting pH can be calculated.
Acetic acid (CH2COOH) is a weak acid that partially dissociates in water. The dissociation of acetic acid can be represented by the equation:
CH2COOH ⇌ CH3COO- + H+
The equilibrium constant for this dissociation is called the acid dissociation constant (Ka), which is given as 1.8 x 10-5 for acetic acid.
To calculate the final pH, we need to consider the initial concentration of acetic acid and the dissociation reaction. Since 0.1 mol of acetic acid is added to 1 L of water, the initial concentration of acetic acid is 0.1 M.
Using the equilibrium expression for the dissociation of acetic acid:
Ka = [CH3COO-][H+]/[CH2COOH]
Assuming x represents the concentration of H+ ions that dissociate, we can simplify the expression to:
Ka = x^2 / (0.1 - x)
Since the dissociation of acetic acid is small (x << 0.1), we can approximate the expression as:
Ka ≈ x^2 / 0.1
Solving for x, we find that x ≈ sqrt(Ka * 0.1). Plugging in the values, we have:
x ≈ sqrt(1.8 x 10-5 * 0.1) ≈ 1.34 x 10-3 M
The concentration of H+ ions is equal to the concentration of acetic acid that dissociates. Therefore, the pH can be calculated as pH = -log[H+]:
pH ≈ -log(1.34 x 10-3) ≈ 2.87
Therefore, the final pH of the solution is approximately 2.87.
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A certain luguid X has a normal freezing point of −5.40 " C and a frecaing point depression constant K
f
=6.60 " C.kg mol
−1
. A solution is prepared by dissoiving some alycine (C
2
H
3
NO
2
) in 350+ g of X. This solution freezes at −10.8C
∘
. Calculate the mass of C
2
H
3
NO
2
that was dossolved. Be sure your answer is rounded to the correct number of sagnificant digis:
the mass of C2H3NO2 that was dissolved, we can use the formula for freezing point depression. The formula is:
Normal freezing point of X = -5.40°C Freezing point depression constant, Kf = 6.60°C.kg mol^(-1)Change in freezing point, ∆T = -10.8°CMass of X = 350+ g (I will assume this is 350 g for simplicity)
First, let's calculate the molality of the solution:molality (m) = moles of solute / mass of solvent (in kg)We know that the molality is the same as the moles of solute per kilogram of solvent. Therefore, we need to convert the mass of X from grams to kilograms:mass of X (in kg) = 350 g / 1000 g/kg = 0.350 kg
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If the percent yield for Na3PO4 is 27.5%, what mass (in g) of NaOH is required to produce 114.0g of Na3PO4?
NaH2PO4(s) + 2 NaOH(aq) ---> Na3PO4(aq) + 2H2O (l)
55.51 grams of NaOH is required to produce 114.0 grams of [tex]Na_3PO_4[/tex], assuming a percent yield of 27.5%.
The molar mass of [tex]Na_3PO_4[/tex] is 163.94 g/mol.
Number of moles of [tex]Na_3PO_4[/tex] = mass / molar mass = 114.0 g / 163.94 g/mol.
According to the balanced equation, 1 mole of [tex]Na_3PO_4[/tex] reacts with 2 moles of NaOH.
Number of moles of NaOH required = (number of moles of Na3PO4) * 2.
Mass of NaOH required = (number of moles of NaOH required) * molar mass of NaOH.
Number of moles of [tex]Na_3PO_4[/tex] = 114.0 g / 163.94 g/mol = 0.694 mol.
Number of moles of NaOH required = 0.694 mol * 2 = 1.388 mol.
Molar mass of NaOH = 39.997 g/mol.
Mass of NaOH required = 1.388 mol * 39.997 g/mol = 55.51 g.
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