Using the Nernst equation with given concentrations and standard cell potential, the [tex]\(H^+\)[/tex] concentration in the cathode compartment is approximately 7916 M.
To solve this problem, we can use the Nernst equation, which relates the cell potential[tex](\(E_{\text{cell}}\))[/tex] to the standard cell potential[tex](\(E^\circ\)),[/tex] the reaction quotient ((Q)), the temperature ((T)), and the gas constant ((R)).
The Nernst equation is given by:
[tex]\[E_{\text{cell}} = E^\circ - \frac{0.0592}{n} \log(Q)\][/tex]
Where:
[tex]- \(E_{\text{cell}}\)[/tex]= cell potential under non-standard conditions
[tex]- \(E^\circ\)[/tex]= standard cell potential
- (n) = number of moles of electrons transferred in the balanced redox reaction
- (Q) = reaction quotient
- (R) = gas constant[tex](\(8.314 \, \text{J/mol} \cdot \text{K}\))[/tex]
- (T) = temperature in Kelvin
Given:
[tex]- \(E^\circ = 0.76 \, \text{V}\)[/tex]
[tex]- \(E_{\text{cell}} = 0.53 \, \text{V}\)[/tex]
[tex]- \(P_{\text{H}_2} = 1.0 \, \text{atm}\)[/tex]
[tex]- \([\text{Zn}^{2+}] = 1.0 \, \text{M}\)[/tex]
- We know that the number of moles of electrons transferred (\(n\)) is 2 because of the balanced equation.
First, let's find the reaction quotient (\(Q\)) using the given concentrations:
[tex]\[Q = \frac{[\text{Zn}^{2+}][\text{H}_2]}{[\text{H}^+]^2}\][/tex]
Given [tex]\([\text{Zn}^{2+}] = 1.0 \, \text{M}\), \([\text{H}_2] = 1.0 \, \text{atm}\)[/tex], and[tex]\([\text{H}^+]\)[/tex]as the unknown concentration in the cathode compartment, we can substitute these values into the equation.
[tex]\[Q = \frac{(1.0 \, \text{M})(1.0 \, \text{atm})}{[\text{H}^+]^2}\][/tex]
Now, we can use the given cell potentials and the Nernst equation to solve for [tex]\([\text{H}^+]\).[/tex]
[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - \frac{0.0592}{2} \log\left(\frac{(1.0 \, \text{M})(1.0 \, \text{atm})}{[\text{H}^+]^2}\right)\][/tex]
Let's solve this equation step by step:
[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - \frac{0.0592}{2} \log\left(\frac{(1.0 \, \text{M})(1.0 \, \text{atm})}{[\text{H}^+]^2}\right)\][/tex]
[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - \frac{0.0296}{2} \log\left(\frac{(1.0 \, \text{M})(1.0 \, \text{atm})}{[\text{H}^+]^2}\right)\][/tex]
[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - 0.0148 \log\left(\frac{1.0}{[\text{H}^+]^2}\right)\][/tex]
[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - 0.0148 \left(\log(1.0) - \log([\text{H}^+]^2)\right)\][/tex]
[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - 0.0296 \log([\text{H}^+]^2)\][/tex]
Now, let's simplify and solve for [tex]\([\text{H}^+]\):[/tex]
[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - 0.0296 \log([\text{H}^+]^2)\][/tex]
[tex]\[0.53 \, \text{V} = 0.76 \, \text{V} - 0.0592 \log([\text{H}^+])\][/tex]
Now, let's isolate \(\log([\text{H}^+])\):
[tex]\[0.53 \, \text{V} - 0.76 \, \text{V} = - 0.0592 \log([\text{H}^+])\][/tex]
[tex]\[-0.23 \, \text{V} = - 0.0592 \log([\text{H}^+])\][/tex]
Now, divide by (-0.0592):
[tex]\[\frac{-0.23 \, \text{V}}{-0.0592} = \log([\text{H}^+])\][/tex]
[tex]\[3.89189 \approx \log([\text{H}^+])\][/tex]
Now, we can find[tex]\([\text{H}^+]\)[/tex] by taking the antilog of (3.89189):
[tex]\([\text{H}^+] = 10^{3.89189}\)[/tex]
[tex]\([\text{H}^+] \approx 7915.82 \, \text{M}\)[/tex]
So, the concentration of [tex]\(H^+\)[/tex] in the cathode compartment is approximately[tex]\(7915.82 \, \text{M}\).[/tex]
According to Raoult's law, which statement is FALSE? a) The vapor pressure of a solvent over a solution decreases as its mole fraction increases. b) The vapor pressure of a solvent over a solution is less than that of pure solvent. c) The greater the pressure of a gas over a solution the greater its solubility. d) The solubility of a gas increases as the temperature decreases. e) Ionic solutes dissociate in solution causing an enhancement of all colligative properties.
Answer:
a) The vapor pressure of a solvent over a solution decreases as its mole fraction increases.
According to Raoult's law, the statement which is FALSE is:
A) The vapor pressure of a solvent over a solution decreases as its mole fraction increases.Raoult's law states that if the pressure of the content of a liquid remains constant to the temperature, then it is proportional to the mole fraction of the mixture.
As a result of this, the vapor pressure of a solvent over a solution is less than the pure solvent and the greater the pressure of a gas over a solution, the greater the solubility.
Therefore, the correct answer is option A because it is false.
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Which of the following is true according to the kinetic theory of gases? A) Molecules move randomly. B) Molecules have elastic collisions. C) Molecules occupy negligible volume. D) all of the above E) none of the above
Answer: Option (D) is the correct answer.
Explanation:
According to Kinetic theory of gases, molecules of a gas move in rapid and random motion. That is, particles are constantly in linear motion.
When these molecules colloid with each other then no energy is gained or lost by them. Also, these molecules occupy negligible amount of volume as compared to the volume of the container in which they are placed.
Moreover, as there is no energy loss taking place so, these molecules of gas undergo perfect elastic motion.
Therefore, we can conclude that all of the above given options are true according to the kinetic theory of gases.
A grist mill of the 1800s employed a water wheel that was 8 m high; 490 liters per minute of water flowed onto the wheel near the top. How much power, in kW, could this water wheel have produced? Take the density of water to be 1,000 kg/m kW
Answer:
0.64 kW
Explanation:
The potential energy of a mass (M) at some height (h) is computed from ...
PE = Mgh
At 1 kg/liter, the available power is the rate at which that energy is available ...
(490 kg/min)×(1 min/(60 s))×(9.8 m/s²)(8 m) ≈ 640.3 kg·m²/s³
= 640.3 W
In kilowatts, that is 0.64 kW.
Final answer:
The water wheel could have produced approximately 0.641 kW of power by converting the gravitational potential energy of water. This is calculated using the water's mass flow rate and the height of the wheel, considering the density of water is 1000 kg/m³.
Explanation:
The power that the water wheel could have produced can be calculated using the principles of mechanical energy and gravitational potential energy (GPE). The formula for GPE is given by mgh, where m is the mass of the water, g is the acceleration due to gravity (9.8 m/s²), and h is the height.
The flow rate of water is 490 liters per minute, which we convert to cubic meters per second (m³/s) for our calculations, as 490 liters per minute is equal to 0.49 m³/min or 0.00817 m³/s, given that 1 cubic meter equals 1000 liters. Using the density of water, which is 1000 kg/m³, the mass flow rate of water is calculated as the product of the flow rate by the density (0.00817 m³/s * 1000 kg/m³ = 8.17 kg/s).
Thus, the power (P) in watts (W) is P = mgh which translates into P = 8.17 kg/s * 9.8 m/s² * 8 m. Upon calculation, this gives a power of approximately 640.88 watts, which we convert to kilowatts (kW) by dividing by 1000, resulting in approximately 0.641 kW.
True or False Materials that have predominantly ionic or covalent bonds have many free electrons.
Answer:
False
Explanation:
Atomic bonding can be of two types:
Ionic bonds: The bond formed between two opposite charged ions. The strong electrostatic force acting between the ions make them compounds of high stability . Example: NaCl
Covalent bonds: The bond formed by sharing of the electrons between the atoms. The balance between the attractive force (nucleus, electron) and repulsive force (electron, electron or nucleus, nucleus) makes the compound stable. Example: [tex]CH_4[/tex]
The bonds are either formed by transferring (losing and gaining) or sharing of electrons. Thus, compound forming these bonds share or transfer electrons to attain noble gas configuration. Thus, they lack free electrons.
Enter the balanced complete ionic equation for HCl(aq)+K2CO3(aq)→H2O(l)+CO2(g)+KCl(aq). Express your answer as a chemical equation. Identify all of the phases in your answer.
The complete net ionic equation for the reaction between HCl(aq) and K₂CO₃(aq) is:
2H⁺(aq) + CO₃²¯(aq) —> H₂O(l) + CO₂(g)We'll begin by writing the dissociation equation for HCl and K₂CO₃. This is illustrated below:
HCl(aq) —> H⁺(aq) + Cl¯(aq)
K₂CO₃(aq) —> 2K⁺(aq) + CO₃²¯(aq)
In solution, the reaction will proceed as follow:
HCl(aq) + K₂CO₃(aq) —>
2H⁺(aq) + 2Cl¯(aq) + 2K⁺(aq) + CO₃²¯(aq) —> H₂O(l) + CO₂(g) + 2Cl¯(aq) + 2K⁺(aq)
Cancel out the spectator ions (i.e Cl¯ and K⁺) to obtain the net ionic equation.
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Using the Antoine equation, prepare two plots of Psat versus T for Benzene over the range of temperature for which the parameters are valid. One plot should present Psat on a linear scale and the other should present Psat on a log scale. Make these plots using appropriate software of your choice (Excel, Matlab, etc.), not by hand. In both cases, T should be on the horizontal axis (independent variable) and Psat nwww ww www. should be on the vertical axis. Please show all work and use Excel or Matlab. The parameters for Benzene are given below. A 13.7819 B 2726.81 C 217.572 Temp Range( C) 6-104 AHn (kJ/mol) 30.72 Latent heat of Vaporization at normal boiling point a0000nd Normal boiling point Tr (oC) 80.0
Answer:
Here's what I get.
Explanation:
The Antoine equation is
[tex]\log p = A - \dfrac{B }{C+T}[/tex]
A = 13.7819
B = 2726.81
C = 217.572
I did the calculations and the plots in Excel.
Figure 1 shows the calculations, Figure 2 is the linear plot, and Figure 3 is the log plot.
A solution has a pOH of 7.1 at 10∘C. What is the pH of the solution given that Kw=2.93×10−15 at this temperature? Remember to report your answer with the correct number of significant figures
Answer : The pH of the solution is, 7.4
Explanation : Given,
pOH = 7.1
[tex]K_w=2.93\times 10^{-15}[/tex]
First we have to calculate the value of [tex]pK_w[/tex].
The expression used for the calculation of [tex]pK_w[/tex] is,
[tex]pK_w=-\log [K_w][/tex]
Now put the value of [tex]K_w[/tex] in this expression, we get:
[tex]pK_w=-\log (2.93\times 1-^{-15})[/tex]
[tex]pK_w=15-\log (2.93)[/tex]
[tex]pK_w=14.5[/tex]
Now we have to calculate the pH of the solution.
As we know that,
[tex]pH+pOH=pK_w[/tex]
Now put all the given values in this formula, we get:
[tex]pH+7.1=14.5[/tex]
[tex]pH=7.4[/tex]
Therefore, the pH of the solution is, 7.4
A solution has a pOH of 7.1 and a pH of 7.4 at 10 °C, being Kw = 2.93 × 10⁻¹⁵ at that temperature.
What is pH?pH is a figure expressing the acidity or alkalinity of a solution on a logarithmic scale on which 7 is neutral, lower values are more acid and higher values are more alkaline.
A solution has a pOH of 7.1 at 10 °C. We can calculate the pH at this temperature using the following expression.
pH + pOH = pKw
pH = pKw - pOH
pH = -log (2.93 × 10⁻¹⁵) - 7.1 = 7.4
where,
pH = -log [H⁺].pOH = -log [OH⁻].Kw is the ionic product of water.A solution has a pOH of 7.1 and a pH of 7.4 at 10 °C, being Kw = 2.93 × 10⁻¹⁵ at that temperature.
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Urea is an organic compound widely used as a fertilizer. Its solubility in water allows it to be made into aqueous fertilizer solutions and applied to crops in a spray. What is the maximum theoretical number of water molecules that one urea molecule can hydrogen bond with? Ignore shape for the purposes of this answer.
Answer:
8 water molecules
Explanation:
The hydrogen bond may be H-O~H-N or H-N~H-O; in the first one, the hydrogen bond is between an oxygen atom and a hydrogen which is covalently bonded to a nitrogen atom. The second one is the hydrogen bond of a nitrogen atom with a hydrogen covalently bonded to a oxygen one. The first case would be the hydrogen bonds that water may form with the hydrogen of the urea; the second ones would be the hydrogen bonds that urea may form with water molecules. So, for each nitrogen in urea there would be a hydrogen bond, and for each hydrogen too. Finally, the oxygen in the urea molecule may form hydrogen bonds with water as well, but it has two lone pairs to donate, so the oxygen atom may form hydrogen bond with 2 water molecules:
N=(2 because of the oxygen atom of the urea)+(4 because of the hydrogen bonded to nitrogen)+2(because of the nitrogens).
N=8.
One urea molecule can theoretically form a maximum of four hydrogen bonds with water molecules, two from its NH2 groups and two from the lone pairs of electrons on its oxygen atom.
This capacity to bond with water makes urea an effective compound in the formulation of agricultural fertilizers.
Explanation:Urea is an organic compound that has the formula (NH2)2CO. It can form hydrogen bonds with water due to the presence of hydrogen atoms in NH2 groups and a lone pair of electrons on the oxygen atom. Through these groups, each urea molecule can form four hydrogen bonds with water molecules: two through the NH2 groups (each nitrogen can form a bond with water) and two through the oxygen atom (each lone pair can form a bond).
NH2 groups in urea can form a bond with water because nitrogen being a more electronegative element compared to hydrogen, can draw electrons towards itself and cause partial positive charge on hydrogen auxiliaries which can then attract the oxygen part of water molecules. Similarly, oxygen in urea can attract hydrogen parts from water molecules due to its lone pair of electrons on it.
Thus, understanding the interaction between urea and water molecules through hydrogen bonding is not only essential in chemistry but also has practical applications, such as in the formulation of fertilizers for agricultural use. It's through this principle that urea can deliver the necessary nutrients for plant growth when mixed with water and applied to soils.
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The temperature versus time graph of a solid substance absorbing heat is shown. A graph is shown with Temperature followed by degree Celsius in parentheses labeled on the y axis and Time followed by minutes in parentheses labeled on the x axis. An upwards arrow is shown between y axis and the label and a right pointing arrow is shown between the x axis and the label. A slanting graph line starting at a point A near the intersection of the two axis is shown. The slanting graph after a point B starts running parallel to the x axis till point C. The line after point C slopes upwards till point D and then runs parallel to the x axis till point E after which it again slopes upwards to finally terminate at point F. What best describes the change taking place in section CD of the graph? The intermolecular bonds of the solid state are being broken as particles vibrate faster. The intermolecular bonds of the liquid state are being broken as particles flow faster. The particles of the solid vibrate faster as the kinetic energy of the particles increases. The particles of the liquid slide around faster as the kinetic energy of the particles increases.
Answer:
The particles of the liquid slide around faster as the kinetic energy of the particles increases.
Explanation:
After all the bonds in the solid state are broken in part CD, the more free particles in the liquid state gain more kinetic energy with increase in energy supplied.
The increase in kinetic energy is indicated by the temperature increase thus the positive gradient of the part CD.
Kinetic energy means more vibrations thus the particles slide more and more against each other.
The following equation IS balanced: HNO3+ NaHSo3-NaNO3+ H2o (T/F)
Answer : The given equation are not balanced equation.
Explanation :
Balanced chemical equation : It is defined as the number of atoms of individual elements present on the reactant side must be equal to the number of atoms of individual elements present on the product side.
The given chemical reaction is,
[tex]HNO_3+NaHSO_3\rightarrow NaNO_3+H_2O[/tex]
This chemical reaction is an unbalanced reaction because in this reaction, the number of atoms of oxygen are not balanced and the molecule [tex]SO_2[/tex] are missing on the product side.
So, in order to balanced the chemical reaction, the molecule [tex]SO_2[/tex] are added on the product side.
Thus, the balanced chemical reaction will be,
[tex]HNO_3+NaHSO_3\rightarrow NaNO_3+H_2O_SO_2[/tex]
You decided to prepare a phosphate buffer from solid sodium dihydrogen phosphate (NaH2PO4) and disodium hydrogen phosphate (Na2HPO4)and you need 1L of the buffer at pH 7.00 with a total phosphate concentration(NaH2PO4+ Na2HPO4) of 0.100 M. Hint:Phosphoric acid (H3PO4), a triprotic acid, has 3 pKa values: 2.14, 6.86, and12.4. Determine the weight in grams of sodium dihydrogen phosphate (NaH2PO4;138g/mol) and disodium hydrogen phosphate (Na2HPO4;142g/mol) needed to prepare this buffer.
Answer:
For disodium hydrogen phosphate:
5.32g Na2HPO4
For sodium dihydrogen phosphate:
7.65g Na2HPO4
Explanation:
First, you have to put all the data from the problem that you going to use:
-NaH2PO4 (weak acid)
-Na2HPO4 (a weak base)
-Volume = 1L
-Buffer pH = 7.00
-Concentration of [NaH2PO4 + Na2HPO4] = 0.100 M
What we need to find the pKa of the weak acid, in this case NaH2PO4, for that you need to find the Ka (acid constant) of NaH2PO4, and for this we use the pKa of the phosphoric acid as follow:
H3PO4 = H2PO4 + H+ pKa1 = 2.14
H2PO4 = HPO4 + H+ pKa2 = 6.86
HPO4 = PO4 + H+ pKa3 = 12.4
So, for the preparation of buffer, you need to use the pKa that is near to the value of the pH that you want, so the choice will be:
pKa2= 6.86
Now we going to use the Henderson Hasselbalch equation for the pH of a buffer solution:
pH = pKa2 + log [(NaH2PO4)/(Na2HPO4)]
The solution of the problem is attached to this answer.
To prepare 1L of a phosphate buffer at pH 7 with a total concentration of 0.1 M, one would need to mix equal molar amounts of NaH2PO4 and Na2HPO4, resulting in 6.9 g of NaH2PO4 and 7.1 g of Na2HPO4.
The relevant pKa value for the pH target of 7.00 is 6.86, which is close to the second pKa of phosphoric acid. The Henderson-Hasselbalch equation is as follows:
pH = pKa + log([A-]/[HA]), where:
At pH 7.00, the ratio [A-]/[HA] is 1:1 because pH = pKa. Thus, we need equal molar amounts of NaH2PO4 and Na2HPO4. Since the total molarity is 0.1 M, this means we need 0.05 M of each component.
Therefore, you would need to weigh 6.9 g of NaH2PO4 and 7.1 g of Na2HPO4 to prepare your buffer.
Which statement describes the action of a buffer composed of acetic acid (CH3COOH) and sodium acetate (NaCH3COO) ? Acetic acid neutralizes added base, and sodium acetate neutralizes added acid. Both components, acetic acid and sodium acetate, neutralize added acid. Sodium acetate neutralizes added base, and acetic acid neutralizes added acid. Both components, acetic acid and sodium acetate, neutralize added base.
Answer:
Acetic acid neutralizes added base, and sodium acetate neutralizes added acid.
Explanation:
Acetic acid is the acid that will dissociate to release H⁺ ion which will react and neutralize the added base.
CH₃COOH → H⁺ + CH₃COO⁻
H⁺ + OH⁻ → H₂O
Sodium acetate will dissociate to release the acetate ion (CH₃COO⁻) which will react and neutralize the added acid.
CH₃COONa → Na⁺ + CH₃COO⁻
H⁺ + CH₃COO⁻ → CH₃COOH
Based on the information given, the correct option will be A. Acetic acid neutralizes added base, and sodium acetate neutralizes added acid.
An acetic acid simply means the acid that will dissociate to release H⁺ ion which will react and neutralize the added base. Acetic acid is used for manufacturing acetic anhydride, cellulose acetate, acetic esters, plastics, dyes, etc.In conclusion, the correct option is A.
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Commercial grade HCl solutions are typically 39.0% (by mass) HCl in water. Determine the molality of the HCl, if the solution has a density of 1.20 g/mL
To determine the molality of the commercial grade HCl solution, we calculate the mass of HCl and water in the solution, and the moles of HCl. We plug these values into the molality formula, which gives us a molality of approximately 17.5 m (mol/kg).
Explanation:To determine the molality of the HCl solution, we first need to understand that molality (m) is defined as the number of moles of solute per kg of solvent. Therefore we need to know the amount of HCl in moles and the weight of water in kg.
Commercial grade HCl solutions are 39.0% HCl by mass. This means that in 1000g of solution there is 390g of HCl. Since the molar mass of HCl is roughly 36.5 g/mol, this means we have about 10.68 moles of HCl in 1000g solution.
The HCl solution has a density of 1.20 g/mL. This means that 1000g (1kg) solution occupy 1000/1.20 = 833.3 mL. Since the solution is made up of HCl and water, we can say that the mass of water = total mass of solution - mass of HCl = 1000g - 390g = 610g (or 0.61 kg).
Then we find the molality of HCl solution by using the formula: Molality (m) = moles of HCl/kg of water = 10.68 moles/0.61 kg = 17.5 m (mol/kg).
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Which of the following four statements is/are accurate with respect to glycolysis?
A. Glycolysis involves the conversion of monosaccharides into glucose.
B. Glycolysis involves the breakdown of glucose into pyruvate.
C. Glycolysis involves the conversion of pyruvate into glucose to glyceraldehyde-3-phosphate.
D. Glycolysis involves the conversion of pyruvate into acetyl CoA.
Answer:
B. Glycolysis involves the breakdown of glucose into pyruvate.
Explanation:
Glycolysis is the biological process whereby one glucose molecule is broken down to two pyruvate molecules. There are a lot of enzymatic processes that are involved in this. It is one of the most important reactions in the world because it allows living cells to harness chemical energy from organic molecules
A sample of potassium phosphate octahydrate (K3PO4•8H2O) is heated until 7.93 grams of water are released. How many grams did the original hydrate weigh?
The original sample of potassium phosphate octahydrate had a mass of 19.6 grams. When it was heated, it released 7.93 grams of water.
Further Explanation:
For every mole of the compound potassium phosphate octahydrate, there are 8 moles of water of hydration which can be removed from the crystal by heating without altering the chemical composition of the substance.
To determine how much original sample was used, the amount of water released upon heating may be used as well as the mole ratio of the water of hydration with the compound itself following the steps below:
Convert mass of water released to moles.Use the mole ratio of water of hydration to the compound (8 mol water for every mol of potassium phosphate octahydrate) to get the moles of original sample.Convert the moles of original sample to grams.STEP 1: Convert 7.93 g water to moles.
[tex]moles \ of\ H_{2}O \ = 7.93 \ g \ H_{2}O \ (\frac{1 \ mol \ H_{2}O}{18.00 \ g \ H_{2}O})\\\boxed {moles \ of \ H_{2}O \ = 0.4406 \ mol}[/tex]
STEP 2: Calculate the moles of original sample using the mole ratio: 1 mol K3PO4 8H2O : 8 mol H2O.
[tex]moles \ of \ K_{3}PO_{4}\ 8H_{2}O \ = 0.4406 \ mol \ H_{2}O \ (\frac{1 \ mol \ K_{3}PO_{4}\ 8H_{2}O \ }{8 \ mol \ H_{2}O})\\\\\boxed {moles \ of \ K_{3}PO_{4}\ 8H_{2}O \ = 0.0551 \ mol}[/tex]
STEP 3: Convert the moles of original sample to mass.
[tex]mass \ of \ K_{3}PO_{4}\ 8H_{2}O = 0.0551 \ mol \ K_{3}PO_{4}\ 8H_{2}O \ (\frac{356.3885 \ g}{1 \ mol\ K_{3}PO_{4}\ 8H_{2}O})\\ mass \ of \ K_{3}PO_{4}\ 8H_{2}O \ = 19.637 \ g[/tex]
Following the significant figures of the given, the final answer should be:
[tex]\boxed {mass \ of \ K_{3}PO_{4}\ 8H_{2}O = 19.6 \ g}[/tex]
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Answer: The mass of original hydrate is 19.63 grams.
Explanation:
We are given:
Mass of water released = 7.93 grams
We are given a chemical compound known as potassium phosphate octahydrate having chemical formula of [tex]K_3PO_4.8H_2O[/tex]
Mass of [tex]K_3PO_4.8H_2O[/tex] = 356.4 grams
Mass of 8 water of crystallization = (8 × 18) = 144 grams
By applying unitary method, we get:
144 grams of water is released when 356.4 grams of salt is heated.
So, 7.93 grams of water will be released when = [tex]\frac{356.4g}{144g}\times 7.93g=19.63g[/tex] of salt is heated.
Hence, the mass of original hydrate is 19.63 grams.
Copy of A buffer consists of 0.33 M Na2HPO4 and 0.28 M Na3PO4. Given that the K values for H3PO4 are, Ka1 = 7.2 x 10-3, Ka2 = 6.3 x 10-8, and Ka3 = 4.2 x 10-13, calculate the pH for this buffer
Answer:
Use Ka3
henderson-hasselbach equation: pH = pKa + log [base]/[acid]
pKa3 = - log Ka3 = - log 4.2 x 10^-13 = 12.38
therefore: pH = 12.38 + log (0.28/0.33) = 12.30
the pH is 12.30
Explanation:
phosphoric acid is a polyprotic acid meaning it donates more than one proton
weak Acid ↔ conjugate Base
H3PO4 ↔ H2PO4^- corresponding to Ka1
H2PO4^- ↔ HPO4^2- corresponding to Ka2
HPO4^2- ↔ PO4^3- corresponding to Ka3
A buffer consist of a weak acid and its conjugate base, the given buffer has the combination of HPO4^2- and PO4^3- thud we used Ka3
knowing that we used henderson-hasselbach equation to get the pH which is 12.30
Determine the amount of heat (in kJ) given off when 1.26 × 104 g of ammonia are produced according to the equation N2(g) + 3H2(g) ⟶ 2NH3(g) ΔH°rxn = −92.6 kJ/mol Assume that the reaction takes place under standardstate conditions at 25°C.
Answer: The enthalpy of the reaction for given amount of ammonia will be -3431.3 kJ.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
For ammonia:
Given mass of ammonia = [tex]1.26\times 10^4g=1260g[/tex]
Molar mass of ammonia = 17 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of ammonia}=\frac{1260g}{17g/mol}=74.11mol[/tex]
We are given:
Moles of ammonia = 74.11 moles
For the given chemical reaction:
[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g);\Delta H^o_{rxn}=-92.6kJ[/tex]
By Stoichiometry of the reaction:
If 2 moles of ammonia produces -92.6 kJ of energy.
Then, 74.11 moles of ammonia will produce = [tex]\frac{-92.6kJ}{2mol}\times 74.11mol=-3431.3kJ[/tex] of energy.
Thus, the enthalpy of the reaction for given amount of ammonia will be -3431.3 kJ.
Answer:
-34317.56 Kj
Explanation:
Moles of ammonia = mass/molar mass=1.26 x 10^4/17
= 741.2 moles
If 2 moles of ammonia gives - 92.6 Kj/mol
What about 741.2 moles
741.2/2 x - 92.6
= - 34317.56 KJ
For a stationary fluid, the pressure will vary in the x, y and z directions.T/F
Answer:
True
Explanation:
For a stationary fluid, the pressure will vary in the x, y and z directions.
This statement is true.
A 50% antifreeze solution is to be mixed with a 90% antifreeze solution to get 200 liters of a 80% solution. How many liters of the 50% solution and how many liters of the 90% solution will be used?
Answer:
50 ltr 150 ltr
Explanation:
this problem can be solved by the mixture and allegation concept which can be clearly understand from bellow figure in which the concentration of solution 1 is 50% and concentration of solution 2 is 90% before mixing after mixing with help bellow concept the ratio of concentration become 10:30
ratio of solution 1 and solution 2 =10:30
=1:3
total mixture is 200 liters
part of solution 1=[tex]\frac{1}{4}[/tex] ×200
=50 liters
part of solution 2=[tex]\frac{3}{4}[/tex] ×200
=150 liters
If the fugacity coefficient of the components of a binary mixture are = 0.784, and =0.638 and mole fraction of component 1 is 0.4, write down and expression for the fugacity coefficient Inp, for the mixture.
Answer : The expression for the fugacity coefficient [tex]\ln \phi[/tex], for the mixture is, -0.3669.
Explanation : Given,
Fugacity coefficient of component 1 = 0.784
Fugacity coefficient of component 2 = 0.638
Mole fraction of component 1 = 0.4
First we have to calculate the mole fraction of component 2.
As we know that,
[tex]\text{Mole fraction of component 1}+\text{Mole fraction of component 2}=1[/tex]
[tex]\text{Mole fraction of component 2}=1-0.4=0.6[/tex]
Now we have to calculate the expression for the fugacity coefficient [tex]\ln \phi[/tex].
Expression used :
[tex]\ln \phi=X_1\ln \phi_1+X_2\ln \phi_2[/tex]
where,
[tex]\phi[/tex] = fugacity coefficient
[tex]\phi_1[/tex] = fugacity coefficient of component 1
[tex]\phi_2[/tex] = fugacity coefficient of component 2
[tex]X_1[/tex] = mole fraction of of component 1
[tex]X_2[/tex] = mole fraction of of component 2
Now put all the give values in the above expression, we get:
[tex]\ln \phi=0.4\times \ln(0.784)+0.6\times \ln(0.638)[/tex]
[tex]\ln \phi=-0.3669[/tex]
Therefore, the expression for the fugacity coefficient [tex]\ln \phi[/tex], for the mixture is, -0.3669.
Determine the calcium carbonate equivalent (CCE) of the following compounds: (amount that has the same neutralizing value as 100 g pure CaCO3) (a) MgO (b) Mg(OH)2 (c) and CaMg(CO3)2.
Answer:
a) 40 g
b) 58 g
c) 184 g
Explanation:
The calcium carbonate equivalent of any compound is calculated by calculating it molar mass.
The molar mass of the compound is the calcium carbonate equivalent as it corresponds to 1 mole of the compound and equals to one mole of calcium carbonate then.
a) MgO
Atomic mass of Mg = 24
Atomic mass of O = 16
Molar mass = CCE = 24+16 = 40 g
b) Mg(OH)₂
Atomic mass of Mg = 24
Atomic mass of O = 16
Atomic mass of H = 1
Molar mass of Mg(OH)₂= CCE = 24 + (2X16)+2(X1) = 58g
c) CaMg(CO₃)2
Atomic mass of Mg = 24
Atomic mass of O = 16
Atomic mass of C = 12
Atomic mass of Ca = 40
Molar mass = CCE = 40 + 24 + (2X12) + (6X16) = 184 g.
The half-life of krypton-91 (91Kr) is 10 s. At time t = 0 a heavy canister contains 7 g of this radioactive gas. (a) Find a function m(t) = m02−t/h that models the amount of 91Kr remaining in the canister after t seconds
Answer:
misteri Cell ini quest ia half-life of beauty of misteri best, of Cell can't answer =
Explanation:
[tex] \sqrt[ \geqslant { { | \geqslant | \geqslant \sqrt[ \gamma \% log_{ \tan( \sqrt[ < \pi \sqrt[ | \geqslant \sqrt[ < \leqslant |x| ]{y} | \times \frac{?}{?} ]{?} ]{?} ) }(?) ]{?} | | }^{2} }^{?} ]{ \sqrt[ < \gamma log_{ \frac{ | \geqslant y \sqrt[ |x \sqrt{ |?| } | ]{?} | }{?} }(?) ]{?} } [/tex]
A function m(t) = m02−t/h that models the amount of 91Kr remaining in the canister after t seconds is m (t) = 9 x 2⁻t/¹⁰.
What is half life?Half life is defined as the amount of time it takes for a radioactive substance (or half its atoms) to break down or change. The time it takes for roughly half of the radioactive atoms in a sample to transform into a more stable form is known as the half-life. The half-life of each radioactive element varies. Half-life is the length of time it takes for a radioactive element to decay to half of its initial value. This suggests that a source's activity has a half-life when it takes time for it to decrease to half its initial value.
The half-life of krypton-91 = 10 s
At time t = 0 a heavy canister contains 7 g of this radioactive gas.
h = 10
m_0 = 7
m(t) = m₀ x 2⁻t/h
m(t) = 7 x 2⁻t/10
Thus, a function m(t) = m02−t/h that models the amount of 91Kr remaining in the canister after t seconds is m (t) = 9 x 2⁻t/¹⁰.
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Calculate the values of ΔG, ΔF and ΔSuniv for:
a) vaporization of 39 g of benzene at its boiling temperature (80.1 °C, 1 bar);
b) adiabatic expansion of 0.100 mol of an ideal gas, in vacuo (Joule's experiment), with an initial temperature of 300 K, if the initial volume is 2.00 dm3 and the final volume is 6.00 dm3.
Note: process a) is reversible, isothermal and isobaric.
Data:
R = 8.314 J mol-1 K-1
1bar = 105 Pa
M (C6H6) = 78.66 g mol-1
Answer:
a.
ΔF = w = 2.94 kJ
ΔSuniv = 0
ΔG = 0
b.
ΔS = 0.1000×8.315×1.099 = 0.913 J/mol K
ΔQ =0
ΔH = 0
Explanation:
a. You have to find the ΔG, ΔF, who are two forms of free energy
G: Gibbs free energy
F: Helmholtz free energy
-G: Gibbs free energy:
For solve these, you have the following equation:
ΔG = ΔH – T ΔS with T constant (Eq 1)
Where:
ΔG = change in Gibbs free energy
ΔH= change in enthalpy
T = temperature
ΔS = change in entropy
This process is irreversible and isothermic, it last means that temperature doesn’t change.
For that reason:
ΔS = q/T with p constant. (Eq 2)
Where:
q = heat
And, with p constant, it just making P-V work, so:
ΔH = qp =q (Eq 3)
where:
qp = heat at constant pressure.
-F: Helmholtz free energy
To find ΔF, you have to use the following equation:
ΔF = ΔU – T ΔS With T constant, (Eq 4)
Where:
ΔF = change in Helmholtz free energy
ΔU= change in internal energy
T = temperature
ΔS = change in entropy
And also, you have to use the equation for internal energy:
ΔU = q + w (Eq 5)
The complete exercise is on the document attached.
b. For this problem we have to establish two states, A and B, based on the data given from the problem:
State A:
V1 = 2 dm³
T1 = 300K
State B:
V2 = 6 dm³
T2 = ?
Due to the adiabatic properties of the process, this expansion makes that change on heat “q” equals to 0:
Δq= 0
SO we have to ask ourselves what is the value of the change in entropy. But we don’t know if the process is reversible or not. Also, we don’t know if the process is static or not, and the volume could be hard to define.
The complete exercise is on the document attached.Consider the reaction: HCN (aq) + H2O (l) ⇄ CN- (aq) + H3O + (aq) Which of the following statements will decrease the amount of work the system could perform? (a) Boil off water from the container (increasing the concentration of all species) (b) Add solid NaOH to the reaction (assume no volume change) (c) Selectively remove CNfrom the solution (d) Add water to the reaction vessel (e) Increase the concentration of the HCN
Answer: Option (b) is the correct answer.
Explanation:
The given chemical reaction shows that hydrogen cyanide acid has been added to water which results in the formation of hydronium ion and cyanide ion.
Also, when we add a base like sodium hydroxide (NaOH) to HCN then it will help in accepting a proton ([tex]H^{+}[/tex]) from hydrogen cyanide. As a result, formation of [tex]CN^{-}[/tex] anion will be rapid and easy.
This will make the system not to do any extra work. So, amount of work done by system will decrease.
Thus, we can conclude that out of the given options, add solid NaOH to the reaction (assume no volume change) will decrease the amount of work the system could perform.
Boiling off water from the container will increase the concentration of all species and shift the equilibrium toward the reactants, thus decreasing the potential work the system could perform, aligning with Le Chatelier's Principle. Option A is correct.
Which, action would decrease the amount of work a system, involving the equilibrium reaction of HCN (aq) with water to produce CN⁻ (aq) and H₃O⁺ (aq), could perform. Various actions can shift the equilibrium of this reaction, affecting the system's potential work. According to Le Chatelier's Principle, the system will respond to minimize the effect of any change.
Boiling off water ((a)) from the container will increase the concentration of all species, shifting the equilibrium to the left, thus diminishing the potential work by favoring reactants.Adding solid NaOH ((b)) will increase the hydroxide ion concentration, which consumes H₃O⁺ ions, driving the reaction to the right and potentially increasing work, contradicting the asked condition.Removing CN- ((c)) selectively from the solution will drive the reaction to the right, to replenish CN⁻ ions, potentially increasing the system's capacity to do work, which is against the question's condition.Adding water ((d)) to the reaction vessel will dilute all reactive species, shifting the equilibrium towards CN⁻ and H₃O⁺ formation, which doesn't directly answer the question of decreasing work.Increasing the concentration of HCN ((e)) will push the equilibrium to the right, thus increasing the amount of work the system could perform by forming more products.The action that will decrease the amount of work the system could perform is (a) boiling off water from the container, as it favors the formation of reactants over products.
Hence, A. is the correct option.
According to the following balanced reaction, how many moles of NO are formed from 12.66 moles of NO2 if there is plenty of water present? 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g
Answer:
[tex]\boxed{\text{4.220 mol}}[/tex]
Explanation:
3NO₂ + H₂O → 2HNO₃ + NO
n/mol: 12.66
You get 1 mol of NO from 3 mol of NO₂
[tex]\text{Moles of NO} = \text{12.66 mol NO}_{2}\times \dfrac{\text{1 mol NO}}{\text{3 mol NO}_{2}} = \textbf{4.220 mol NO}\\\\\text{The reaction forms } \boxed{\textbf{4.220 mol}} \text{ of NO}[/tex]
Answer: The moles of NO produces are 4.22 moles
Explanation:
We are given:
Moles of nitrogen dioxide = 12.66 moles
The given chemical equation follows:
[tex]3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)[/tex]
By Stoichiometry of the reaction:
3 moles of nitrogen dioxide produces 1 mole of NO
So, 12.66 moles of nitrogen dioxide will produce = [tex]\frac{1}{3}\times 12.66=4.22mol[/tex] of NO
Hence, the moles of NO produces are 4.22 moles
Calcium hydroxide is a strong base but is not very soluble ( Ksp = 5.02 X 10-6 ). What is the pH of a saturated solution of Ca(OH)2 ?
Answer : The pH of a saturated solution is, 12.33
Explanation : Given,
[tex]K_{sp}[/tex] = [tex]5.02\times 10^{-6}[/tex]
First we have to calculate the solubility of [tex]OH^-[/tex] ion.
The balanced equilibrium reaction will be:
[tex]Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-[/tex]
Let the solubility will be, 's'.
The concentration of [tex]Ca^{2+}[/tex] ion = s
The concentration of [tex]OH^-[/tex] ion = 2s
The expression for solubility constant for this reaction will be,
[tex]K_{sp}=[Ca^{2+}][OH^-]^2[/tex]
Let the solubility will be, 's'
[tex]K_{sp}=(s)\times (2s)^2[/tex]
[tex]K_{sp}=(4s)^3[/tex]
Now put the value of [tex]K_[sp}[/tex] in this expression, we get the solubility.
[tex]5.02\times 10^{-6}=(4s)^3[/tex]
[tex]s=1.079\times 10^{-2}M[/tex]
The concentration of [tex]Ca^{2+}[/tex] ion = s = [tex]1.079\times 10^{-2}M[/tex]
The concentration of [tex]OH^-[/tex] ion = 2s = [tex]2\times (1.079\times 10^{-2}M)=2.158\times 10^{-2}M[/tex]
First we have to calculate the pOH.
[tex]pOH=-\log [OH^-][/tex]
[tex]pOH=-\log (2.158\times 10^{-2})[/tex]
[tex]pOH=1.67[/tex]
Now we have to calculate the pH.
[tex]pH+pOH=14\\\\pH=14-pOH\\\\pH=14-1.67=12.33[/tex]
Therefore, the pH of a saturated solution is, 12.33
Cavendish prepared hydrogen in 1766 by the novel method of passing steam through a red-hot gun barrel: 4H2 O(g) + 3Fe(s) ⟶ Fe3 O4 (s) + 4H2 (g) (a) Outline the steps necessary to answer the following question: What volume of H2 at a pressure of 745 torr and a temperature of 20 °C can be prepared from the reaction of 15.O g of H2O? (b) Answer the question.
Answer : The volume of [tex]H_2[/tex] will be, 0.2690 L
Solution :
(a) Steps involved for this problem are :
First we have to calculate the moles of [tex]H_2O[/tex].
Now we have to calculate the volume of hydrogen gas by using the ideal gas equation.
(b) First we have to calculate the moles of [tex]H_2O[/tex].
[tex]\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{15g}{18g/mole}=0.833moles[/tex]
The balanced chemical reaction is,
[tex]4H_2O(g)+3Fe(s)\rightarrow Fe_3O_4(s)+4H_2(g)[/tex]
From the balanced chemical reaction, we conclude that the moles of hydrogen is equal to the moles of water.
Thus, the moles of hydrogen gas = 0.833 mole
Now we have to calculate the volume of hydrogen gas.
Using ideal gas equation,
[tex]PV=nRT[/tex]
where,
n = number of moles of gas = 0.833 mole
P = pressure of the gas = [tex]745torr=\frac{745}{760}=0.98atm[/tex]
conversion used : (1 atm = 760 torr)
T = temperature of the gas = [tex]20^oC=273+20=293K[/tex]
R = gas constant = 0.0821 Latm/moleK
V = volume of gas = ?
Now put all the given values in the above equation, we get :
[tex](0.98atm)\times V=(0.833mole)\times (0.0821Latm/moleK)\times (293K)[/tex]
[tex]V=0.2690L[/tex]
Therefore, the volume of [tex]H_2[/tex] will be, 0.2690 L
Answer:
V = 20.4 L
Explanation:
Step 1: Write the balanced equation
4 H₂O(g) + 3 Fe(s) ⟶ Fe₃O₄(s) + 4 H₂(g)
Step 2: Find the moles of H₂
We can establish the following relations.
The molar mass of H₂O is 18.0 g/molThe molar ratio of H₂O to H₂ is 4:4.The moles of H₂ obtained from 15.0 g of H₂O is:
[tex]15.0gH_{2}O.\frac{1molH_{2}O}{18.0gH_{2}O} .\frac{4molH_{2}}{4molH_{2}O} =0.833molH_{2}[/tex]
Step 3: Find the volume of H₂
We will use the ideal gas equation.
P = 745 torr × (1 atm/760 torr) = 0.980 atm
V = ?
n = 0.833 mol
R = 0.08206 atm.L/mol.K
T = 20°C + 273 = 293 K
P × V = n × R × T
0.980 atm × V = 0.833 mol × (0.08206 atm.L/mol.K) × 293 K
V = 20.4 L
You are holding a container of 28.7L of dinitrogen tetroxide. How many grams of gas are inside? Step by step.
Answer:
=71.76 grams
Explanation:
At Room temperature and pressure, 1 mole of an ideal gas occupies a volume of 24 liters.
Therefore, the number of moles occupied by 28.7 L is:
(28.7×1)/24=1.196 moles.
Mass=Number of moles× RMM
RMM of N₂O₂ is 60
mass=1.196 moles× 60 grams/mol
=71.76 grams
Hemoglobin, a protein in red blood cells, carries O2 from the lungs to the body's cells. Iron (as ferrous ion, Fe2+) makes up 0.33 mass % of hemoglobin. If the molar mass of hemoglobin is 6.8 × 104 g/mol, how many Fe2+ ions are in one molecule?
Answer: The number of [tex]Fe^{2+}[/tex] ions in one molecule of hemoglobin are 4.
Explanation:
According to mole concept:
1 mole of an element contains [tex]6.022\times 10^{23}[/tex] number of atoms.
We are given:
Mass of 1 mole of hemoglobin = [tex]6.8\times 10^4g[/tex]
Using above equation:[tex]6.022\times 10^{23}[/tex] number of molecules have a mass of [tex]6.8\times 10^4g[/tex]
So, 1 molecule of hemoglobin will have a mass of [tex]\frac{6.8\times 10^4g}{6.022\times 10^{23}}\times 1=1.129\times 10^{-19}g[/tex]
It is also given that 0.33 mass % of hemoglobin has [tex]Fe^{2+}[/tex] ions
So, mass of [tex]Fe^{2+}[/tex] ions will be = [tex]\frac{0.33}{100}\times 1.129\times 10^{-19}g=3.7257\times 10^{-22}g[/tex]
To calculate the number of moles, we use the equation:[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of iron ion = [tex]3.7257\times 10^{-22}g[/tex]
Molar mass of iron ion = 55.85 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of }Fe^{2+}\text{ ion}=\frac{3.7257\times 10^{-22}g}{55.85g/mol}=6.67\times 10^{-24}mol[/tex]
Using mole concept:1 mole of an element contains [tex]6.022\times 10^{23}[/tex] number of atoms.
So, [tex]6.67\times 10^{-24}[/tex] moles of hemoglobin will contain = [tex]6.022\times 10^{23}\times 6.67\times 10^{-24}=4[/tex]
Hence, the number of [tex]Fe^{2+}[/tex] ions in one molecule of hemoglobin are 4.
There are [tex]4[/tex] Fe2+ ions in one molecule of hemoglobin. Each hemoglobin molecule contains 4 iron atoms, and all these iron atoms are in the ferrous ion (Fe2+) form
To determine how many Fe2+ ions are present in one molecule of hemoglobin, we can proceed with the following steps:
1. Calculate the mass of Fe2+ in one molecule of hemoglobin:
Given:
- Iron (Fe) makes up 0.33 mass % of hemoglobin.
- Molar mass of hemoglobin (Hb) = 6.8 × 10^4 g/mol.
Mass of Fe in one mole of hemoglobin:
[tex]\[ \text{Mass of Fe in 1 mol Hb} = 0.33\% \times \text{Molar mass of Hb} \] \[ \text{Mass of Fe in 1 mol Hb} = 0.33 \times 10^{-2} \times 6.8 \times 10^4 \text{ g} \] \[ \text{Mass of Fe in 1 mol Hb} = 2244 \text{ g} \][/tex]
2. Calculate the number of moles of Fe in one mole of hemoglobin:
Now, determine the number of moles of Fe:
[tex]\[ \text{Moles of Fe} = \frac{\text{Mass of Fe}}{\text{Molar mass of Fe}} \][/tex]
The molar mass of Fe (Fe2+) is approximately 55.845 g/mol (atomic mass of Fe).
[tex]\[ \text{Moles of Fe} = \frac{2244 \text{ g}}{55.845 \text{ g/mol}} \] \[ \text{Moles of Fe} \approx 40.17 \text{ mol} \][/tex]
3. Calculate the number of Fe2+ ions in one molecule of hemoglobin:
Hemoglobin (Hb) contains 4 iron atoms per molecule, and each iron atom in Hb is present as Fe2+.
Therefore, the number of Fe2+ ions in one molecule of hemoglobin:
[tex]\[ \text{Number of Fe}^{2+} \text{ ions per molecule of Hb} = 4 \][/tex]
Estimate the vapor pressure of methanol at 25°C. The heat of vaporization for methanol is 37,400 J/mol and the boiling point is 65°C
The vapor pressure of methanol at 25°C can be estimated using the Clausius-Clapeyron equation. By plugging in the values for methanol's boiling point, enthalpy of vaporization, and the desired temperature, we can solve for the vapor pressure. The estimated vapor pressure of methanol at 25°C is approximately 0.147 atm.
Explanation:In order to estimate the vapor pressure of methanol at 25°C, we can use the Clausius-Clapeyron equation. This equation relates the vapor pressure, temperature, and enthalpy of vaporization. The equation is:
ln(P2/P1) = -ΔHvap/R × (1/T2 - 1/T1)
Where P1 is the known vapor pressure at a known temperature (T1), P2 is the vapor pressure at the desired temperature (T2), ΔHvap is the enthalpy of vaporization, and R is the ideal gas constant. Plugging in the values for methanol:
P1 = vapor pressure at boiling point = 1 atm, T1 = boiling point = 65°C = 338 K, T2 = desired temperature = 25°C = 298 K, ΔHvap = 37,400 J/mol, R = 8.314 J/(mol·K)
the equation becomes:
ln(P2/1) = -37,400 J/mol / (8.314 J/(mol·K)) × (1/298 K - 1/338 K)
Solving for P2:
P2 = 1 × e ^ (-37,400 J/mol / (8.314 J/(mol·K)) × (1/298 K - 1/338 K))
P2 ≈ 0.147 atm
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