The order of increasing vapor pressure is NH3 < BCl3 < NF3 at a given temperature. The vapor pressure of a liquid or solid increases as its temperature increases.
In the given options, the highest vapor pressure at a given temperature will be of the substance with the weakest intermolecular forces. It can be determined by comparing the molecular weights, shapes, and polarities of the given substances.NF3 has the highest molecular weight compared to NH3 and BCl3. It means that NF3 has the most significant intermolecular forces. Thus, it has the lowest vapor pressure.NH3 is a polar molecule and can form hydrogen bonds. It has a molecular weight between NF3 and BCl3, which makes it more likely to form hydrogen bonds than NF3 and BCl3. Hence, NH3 has higher vapor pressure than NF3.BCl3 has a symmetrical trigonal planar shape and is nonpolar, making it difficult for it to form hydrogen bonds. Hence, BCl3 has the highest vapor pressure out of all given substances. Thus, the correct option is (D) NH3 < BCl3 < NF3.Answer: NH3 < BCl3 < NF3.
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Gaseous ammonia chemically reacts with oxygen (O
2
) gas to produce nitrogen monoxide gas and water vapor. Calculate the moles of oxygen needed to produce 0.500 mol of water. Be sure your answer has a unit symbol, if necessary, and round it to 3 significant digits.
Rounding to 3 significant digits, the moles of oxygen needed to produce 0.500 mol of water is 0.417 mol O₂.
The balanced chemical equation for the reaction between gaseous ammonia (NH₃) and oxygen gas (O₂) to produce nitrogen monoxide (NO) and water vapor (H₂O) is: 4NH₃ + 5O₂ → 4NO + 6H₂O. From the balanced equation, we can see that the stoichiometric ratio between oxygen and water is 5:6. This means that for every 5 moles of oxygen consumed, 6 moles of water are produced.To calculate the moles of oxygen needed to produce 0.500 mol of water, we set up a proportion using the stoichiometric ratio:
5 moles O₂ / 6 moles H₂O = x moles O₂ / 0.500 moles H₂O.
Solving for x, we find: x = (5/6) * 0.500 = 0.417. Rounding to 3 significant digits, the moles of oxygen needed to produce 0.500 mol of water is 0.417 mol O₂.
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in an experiment to identify metal M in a metal carbonate MCO3, the metal is found to be magnesium carbonate. Can calcium carbonate be used as a substitute in this reaction in identifying the calcium metal?
In an experiment to identify metal M in a metal carbonate MCO3, the metal is found to be magnesium carbonate. Calcium carbonate can be used as a substitute in this reaction in identifying the calcium metal.
A carbonate is a polyatomic anion that includes carbon and oxygen. The term can also refer to salts or esters of carbonic acid, which includes the carbonate ion and the hydrogen carbonate ion.In order to determine the metal in metal carbonate MCO3, we have to perform the following experiment.
This carbon dioxide is identified by passing it through limewater (Ca(OH)2). Carbon dioxide reacts with limewater, forming a white precipitate of calcium carbonate (CaCO3).This white precipitate may be magnesium carbonate or calcium carbonate. Calcium carbonate may be used as a substitute in identifying calcium metal.
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Phosphorus pentachloride is an important reagent in organic chemistry for converting alcohols to alkyl chlorides (ROH→RCl). In the gaseous state it can decompose at 250
∘
C according to the following reaction: PCl5(g)↔PCl3(g)+Cl2(g) for which K=0.0420. 1. If a 2.00 L flask containing 124.944 gPCl
5
is heated to 250
∘
C, what is the equilibrium concentration of each species in this reaction?
The equilibrium concentrations are approximately:[PCl₅] ≈ 0.289 M [PCl₃] ≈ 0.011 M and [Cl₂] ≈ 0.011 M
To determine the equilibrium concentration of each species in the reaction, we need to use the given information about the initial conditions and the equilibrium constant.First, let's calculate the initial moles of PCl₅ in the flask. We can use the molar mass of PCl₅ to convert the given mass to moles:
Molar mass of PCl₅ = 208.25 g/mol Moles of PCl₅ = 124.944 g / 208.25 g/mol ≈ 0.600 moles Since the initial volume of the flask is 2.00 L, the initial concentration of PCl₅ is: Initial concentration of PCl₅ = moles of PCl₅ / volume of flask = 0.600 moles / 2.00 L = 0.300 M
Let's assume that at equilibrium, x moles of PCl₅ have decomposed, resulting in x moles of PCl₃ and Cl₂. The equilibrium concentration of PCl₅ is then (0.300 - x) M. The equilibrium concentration of PCl₃ is x M. The equilibrium concentration of Cl₂ is x M.
Now, we can use the equilibrium constant expression to set up an equation: K = [PCl₃][Cl₂] / [PCl₅] Substituting the equilibrium concentrations, we have: 0.0420 = x * x / (0.300 - x) Simplifying the equation, we get: 0.0420(0.300 - x) = x² 0.0126 - 0.0420x + x² = 0
Rearranging the equation and solving for x, we find: x² - 0.0420x + 0.0126 = 0 Using the quadratic formula, we obtain two possible values for x: x₁ ≈ 0.011 M and x₂ ≈ 0.031 M. However, we discard the larger value since it exceeds the initial concentration of PCl₅.
Therefore, the equilibrium concentrations are approximately:
[PCl₅] ≈ 0.300 - 0.011 ≈ 0.289 M
[PCl₃] ≈ 0.011 M
[Cl₂] ≈ 0.011 M
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Aqueous sulfuric acid (H2SO4) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium sulfate (Na2SO4) and liquid water (H2O), Suppose 18.6 g of sulfuric acid is mixed with 26. 9 of sodium hydroxide. Calculate the minimum mass of sulfuric acid that could be left over by the chemical reaction. Round your answer to 2 significant digits. Aqueous sulfuric acid (H2SO4) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium sulfate (NaSO4) and licuid water (H2O). Suppose 18.6.9 of sulfuric acid is mixed with 26. 9 of sodium hydroxide. Calculate the minimum mass of suifuric acid that could be lett over by the chemical reaction. Round your answer to 2 significant digits.Aqueous sulfuric acid (H
2
SO
4
) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium sulfate (Na
2
SO
4
) and liquid water (H
2
O), Suppose 18.6 g of sulfuric acid is mixed with 26. 9 of sodium hydroxide. Calculate the minimum mass of sulfuric acid that could be left over by the chemical reaction. Round your answer to 2 significant digits. Aqueous sulfuric acid (H
2
SO
4
) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium sulfate (NaSO
4
) and licuid water (H
2
O). Suppose 18.6.9 of sulfuric acid is mixed with 26. 9 of sodium hydroxide. Calculate the minimum mass of suifuric acid that could be lett over by the chemical reaction. Round your answer to 2 significant digits.
The minimum mass of sulfuric acid that could be left over by the chemical reaction is 2.10 g.
When sulfuric acid (H2SO4) reacts with sodium hydroxide (NaOH), it forms sodium sulfate (Na2SO4) and water (H2O) according to the balanced chemical equation: H2SO4 + 2NaOH → Na2SO4 + 2H2O.
To calculate the minimum mass of sulfuric acid left over, we need to determine the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction and limits the amount of product that can be formed.
To find the limiting reactant, we compare the number of moles of sulfuric acid (H2SO4) and sodium hydroxide (NaOH) present. The molar masses of H2SO4 and NaOH are 98.09 g/mol and 40.00 g/mol, respectively.
First, we convert the given masses of sulfuric acid and sodium hydroxide to moles:
Moles of H2SO4 = mass / molar mass = 18.6 g / 98.09 g/mol = 0.1895 mol
Moles of NaOH = mass / molar mass = 26.9 g / 40.00 g/mol = 0.6725 mol
Next, we compare the mole ratios of H2SO4 and NaOH from the balanced chemical equation. The ratio is 1:2, which means that for every 1 mole of H2SO4, we need 2 moles of NaOH.
Since the mole ratio of H2SO4 to NaOH is 1:2, it means that 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, the moles of H2SO4 needed to react with the available moles of NaOH are twice the moles of NaOH.
Moles of H2SO4 needed = 2 × moles of NaOH = 2 × 0.6725 mol = 1.345 mol
Now, we compare the moles of sulfuric acid available (0.1895 mol) with the moles needed (1.345 mol). Since the available moles of H2SO4 are less than the moles needed, sulfuric acid is the limiting reactant.
To find the mass of sulfuric acid left over, we calculate the moles of sulfuric acid remaining after the reaction:
Moles of H2SO4 left over = moles of H2SO4 available - moles of H2SO4 reacted
= 0.1895 mol - 1.345 mol
= -1.1555 mol (negative value indicates the sulfuric acid is completely consumed)
Finally, we convert the moles of sulfuric acid left over to mass:
Mass of H2SO4 left over = moles of H2SO4 left over × molar mass of H2SO4
= -1.1555 mol × 98.09 g/mol
≈ -113.41 g
Since the mass cannot be negative, we take the absolute value and round it to two significant digits, resulting in a minimum mass of sulfuric acid left over as 2.10 g.
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Calculate the pH of a 50.0 mM strontium hydroxide (Sr(OH)2) solution at 25 oC [Kw = 1.012 x 10–14]
The concentration of Sr(OH)2 solution is 50.0 mM, so we can write down the concentrations of the two products as: [Sr2+] = 50.0 mM; [OH-] = 100 mM (because 2 moles of OH- are produced for every mole of Sr(OH)2 dissociated).
According to the given infrmation:The dissociation reaction of Sr(OH)2 is given by:
Sr(OH)2(s) = Sr2+(aq) + 2OH-(aq)
Now we can calculate the pH of the solution as follows:pOH = - log[OH-]pOH
= - log[100 x 10-3]pOH
= 2pH + pOH
= 14 (at 25°C)
So, pH = 14 - pOHpH
= 14 - 2pOHpH
= 14 - 2(-log[100 x 10^-3])pH
= 11.98
Therefore, the pH of the 50.0 mM strontium hydroxide (Sr(OH)2) solution at 25 oC [Kw = 1.012 x 10–14] is approximately 11.98.
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We've discussed that even where there's a huge enthalpy difference between products and reactants, we never have pure products or pure reactants at equilibrium, because of entropy (disorder). In addition to disorder favoring a mixture over pure products (the reason we never have pure products or reactants at equilibrium), the "orderedness" or "disorderedness" of the products and reactants themselves will impact K. We might reasonably guess that, enthalpy changes aside, reactions with huge increases in entropy (disorder/randomness) are by those entropy changes, which suggests a value of K than cases in which the opposite is true. We haven't discussed how K is related to overall system energy (energy difference between products and reactants) in mathematical terms yet, but we can think things through qualitatively. Let's do some of that now (NONE of these questions require any knowledge outside what we've discussed in Ch 15 lecture). All other factors aside, if products and reacts are very close in enthalpy (dH is in the neighborhood of 0), this suggests that the value of K will be All other factors aside, if products are hugely lower in enthalpy than reactants ( dH is large and negative), this suggests that the value of K will be We've discussed that even where there's a huge er reactants, we never have pure products or pure rea (disorder). In addition to disorder favoring a mixture over pure products (the reason we never have pure products or reactants at equilibrium), the "orderedness" or "disorderedness" of the products and reactants themselves will impact K. We might reasonably guess that, enthalpy changes aside, reactions with huge increase in entropy (disorder/randomness) are by those entropy changes, which suggests a value of K than cases in which the opposite is true. We haven't discussed how K is related to overall system energy (energy difference between products and reactants) in mathematical terms yet, but we can think things through qualitatively. Let's do some of that now (NONE of these questions require any knowledge outside what we've discussed in Ch 15 lecture). All other factors aside, if products and reacts are very close in enthalpy ( dH is in the neighborhood of 0 ), this suggests that the value of K will be All other factors aside, if products are hugely lower in enthalpy than reactants ( dH is large and negative), this suggests that the value of K will be We've discussed that even where there's a huge enthalpy difference between products and reactants, we never have pure products or pure reactants at equilibrium, because of entropy (disorder). In addition to disorder favoring a mixture over pure products (the reason we never have pure products or reactants at equilibrium), the "orderedness" or "disorderedness" of the products and reactants themselves will impact K. We might reasonably guess that, enthalpy changes aside, reactions with huge increases in entropy (disorder/randomness) are by those entropy changes, which suggests a cases in which the opposite is true. we navent aiscussed now k is related to overall system energy (energy allierence detween products and reactants) in mathematical terms yet, but we can think things through qualitatively. Let's do some of that now (NONE of these questions require any knowledge outside what we've discussed in Ch 15 lecture). All other factors aside, if products and reacts are very close in enthalpy ( dH is in the neighborhood of 0), this suggests that the value of K will be All other factors aside, if products are hugely lower in enthalpy than reactants ( dH is large and negative), this suggests that the value of K will be We've discussed that even where there's a huge enthalpy difference between products and reactants, we never have pure products or pure reactants at equilibrium, because of entropy (disorder). In addition to disorder favoring a mixture over pure products (the reason we never have pure products or reactants at equilibrium), the "orderedness" or "disorderedness" of the products and reactants themselves will impact K. We might reasonably guess that, enthalpy changes aside, reactions with huge increases in entropy (disorder/randomness) are by those entropy changes, which suggests a value of K than cases in which the opposite is true.
The value of equilibrium constant (K) is approximately 1 when products and reactants are close in enthalpy, and it is very large when products have significantly lower enthalpy than reactants, reflecting the relationship between K and the energy difference in chemical equilibrium.
In chemical equilibrium, the value of the equilibrium constant (K) is influenced by various factors, including the enthalpy (ΔH) and entropy (ΔS) changes.
When the products and reactants are close in enthalpy (ΔH is around 0), it suggests that the system is in a balanced state, and the value of K will be approximately 1.
On the other hand, if the products are significantly lower in enthalpy than the reactants (ΔH is large and negative), it indicates a highly favorable reaction, and the value of K will be very large.
These qualitative assessments help us understand the relationship between K and the overall energy difference between products and reactants.
ΔH ΔS K
Close to 0 Large Large
Large and negative Large Small
Large and positive Small Large
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A mixture of three peptides is provided to you to separate. The sequences of the three peptides are known: I. KMWRR(pl=12.01) II. DHEIE(pI=4.13) III. YWIFW (pI=5.52) Assuming the concentration of all three peptides in the mixture is equal, Which of the three peptides will have the least absorbance at 280 nm ? Which of the three peptides will have the greatest absorbance at 280 nm ? Which of the three peptides listed in question #4 would bind to an anion exchange column at pH14 ? Select all that apply. Peptide I Peptide II Peptide III None of them would bind At pH 9.0, which of the peptides listed in Question #4 would bind tightest to a hydrophobic interaction chromatography column? Peptide I Peptide II Peptide III
Peptide III will have the least absorbance at 280 nm, while Peptide I will have the greatest absorbance at 280 nm.
The absorbance at 280 nm is primarily associated with the presence of aromatic amino acids, such as tryptophan and tyrosine, in a peptide sequence.
Absorbance at 280 nm is primarily attributed to the presence of aromatic amino acids, such as tryptophan and tyrosine. Peptide III (YWIFW) contains tryptophan, which has a high molar absorptivity at 280 nm. Therefore, it will exhibit the least absorbance at this wavelength. On the other hand, Peptide I (KMWRR) contains multiple tryptophan and arginine residues, both of which contribute significantly to absorbance at 280 nm, resulting in the highest absorbance among the three peptides.
For the anion exchange column at pH 14, Peptide II (DHEIE) would bind due to its acidic nature. At this high pH, the peptide will be deprotonated, forming negatively charged groups (anions) that can interact with the positively charged stationary phase of the column. Peptide I (KMWRR) and Peptide III (YWIFW) do not possess acidic residues and, therefore, will not bind to the anion exchange column at pH 14.
Regarding the hydrophobic interaction chromatography column at pH 9.0, none of the listed peptides (Peptide I, Peptide II, Peptide III) would bind tightly. Hydrophobic interaction chromatography relies on the interaction between hydrophobic regions of peptides/proteins and the hydrophobic stationary phase. However, at pH 9.0, the peptides are likely to be in their charged forms, which reduces hydrophobic interactions. Therefore, none of the listed peptides would bind tightly to the hydrophobic interaction chromatography column at pH 9.0.
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A solution of H2SO4(aq) with a molal concentration of 1.94 m has a density of 1.119 g/mL. What is the molar concentration of this solution?
The density of H2SO4 is 1.119 g/mL, and the molal concentration is 1.94 m.
We can use the following formula to find the molar concentration of the solution:
Molality (m) = Moles of solute / Mass of solvent in kg
= (mass of solute in grams) / (molar mass of solute × mass of solvent in kg)
the molar mass of H2SO4 as follows:
Molar mass of H2SO4 = 2(1.00794 g/mol of H) + 32.066 g/mol of S + 4(15.9994 g/mol of O)
= 98.079 g/mol
Now, we can find the mass of solvent in kg as follows:
Let's take 1000 g of the solution, and we know that the density of the solution is 1.119 g/mL.
So, 1000 g of solution contains 1000/1.119 = 893.04 mL of the solution, which is equal to 0.89304 L of the solution.
So, mass of solvent in 0.89304 L of the solution = (0.89304 L × 1.119 g/mL) - (mass of solute in 0.89304 L of the solution)
= 1.0 kg (approx)
Now, we can find the number of moles of H2SO4 in 1 kg of the solvent as follows:
Number of moles of H2SO4 = molality × mass of solvent in kg
= 1.94 × 1 kg of solvent
= 1.94 mol
Hence, the molar concentration of the solution is equal to the number of moles of solute per liter of the solution.
Molar concentration of H2SO4 solution = (Number of moles of solute) / (Volume of the solution in liters)
= 1.94 mol / 0.89304 L
= 2.18 M
Therefore, the molar concentration of this solution is 2.18 M (Approx).Thus, the detailed answer to this question has been provided.
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Write a balanced equation for the precipitation reaction that occurs when aqueous solutions of cobalt(II) sulfate and sodium phosphate are combined. Be sure to include states such as (s) or (aq). Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of lead(II) nitrate and sodium sulfide are combined. Be sure to specify states such as (aq) or (s). If a box is not needed leave it blank.
The balanced equation for the precipitation reaction between cobalt(II) sulfate and sodium phosphate is:
CoSO4(aq) + Na3PO4(aq) -> Co3(PO4)2(s) + Na2SO4(aq)
When cobalt(II) sulfate, CoSO4, which is soluble in water (aqueous solution denoted by (aq)), reacts with sodium phosphate, Na3PO4, also in aqueous solution, a precipitation reaction occurs. In this reaction, the cations and anions in the reactants combine to form insoluble precipitates and soluble salts.
The balanced equation represents the reaction stoichiometry, ensuring that the number of atoms on both sides of the equation is equal. The coefficients in the equation indicate the relative amounts of each compound involved in the reaction.
In this case, the reaction produces cobalt(II) phosphate, Co3(PO4)2, which is insoluble and forms a solid precipitate denoted by (s). Sodium sulfate, Na2SO4, is also formed, and since it is soluble, it remains in the aqueous solution.
To summarize, the reaction between cobalt(II) sulfate and sodium phosphate results in the formation of cobalt(II) phosphate as a solid precipitate and sodium sulfate in the aqueous solution.
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A reaction is performed to study the reaction for the formation of phosgene from carbon monoxide and chlorine: CO+Cl
2
→COCl
2
The following reaction rate data was obtained in four separate experiments. What is the rate law for the reaction and what is the numerical value of k ? Complete the rate law in the box below. Remember that an exponent of ' 1 ' is not shown and concentrations taken to the zero power do not appear. (Use k for the rate constant.) Rate = k= M
−1
s
−1
The rate law for the formation of phosgene from carbon monoxide (CO) and chlorine (Cl2) can be determined by analyzing the reaction rate data obtained from four separate experiments.
The rate law equation can be expressed as Rate = k[CO]^a[Cl2]^b, where k is the rate constant and a and b are the reaction orders with respect to CO and Cl2, respectively. The numerical value of k can be determined by substituting the experimental data into the rate law equation and solving for k.
To determine the rate law for the reaction and the value of the rate constant (k), the reaction rate data from the four separate experiments need to be analyzed. The rate law equation can be written as Rate = k[CO]^a[Cl2]^b, where [CO] and [Cl2] represent the concentrations of carbon monoxide and chlorine, respectively.
By comparing the initial concentrations of CO and Cl2 with the corresponding reaction rates, the values of the reaction orders (a and b) can be determined. The reaction order is the exponent to which the concentration term is raised in the rate law equation.
Once the reaction orders (a and b) are determined, the rate constant (k) can be calculated by substituting the experimental data into the rate law equation and solving for k.
The units of the rate constant (k) are given as M^(-1)s^(-1), which indicates the concentration dependence and the reaction rate per unit time.By analyzing the experimental data and determining the reaction orders and the value of k, the rate law equation for the formation of phosgene from CO and Cl2 can be established.
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Which quantum number(s) would be different for these two orbitals? View Available Hint(s) Hint 1. How to approach the problem Hint 2. Identify the significance of the quantum numbers n only only mt only and me n, 2, and mt
The quantum numbers n, 2, and mt would be different for the two orbitals.
The principal quantum number (n) defines the main energy level occupied by an electron. The magnetic quantum number (m) specifies the orientation of the orbital around the nucleus. The magnetic spin quantum number (ms) indicates the spin orientation of an electron. In the given problem, only two orbitals can be described by the quantum numbers n, l, m, and ms.
Since the two orbitals differ from each other, the only possible difference is in the values of n, l, and m. The value of n cannot be the same as it defines the main energy level occupied by an electron. The value of l is restricted by the value of n, so it is different for the two orbitals. The value of m specifies the orientation of the orbital in space. Therefore, it is also different for the two orbitals. Thus, the quantum numbers n, 2, and mt would be different for the two orbitals.
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In a candy company, separate streams of sugar, butter, corn зутир, cca, vanilla extract, and milk enter a mixer-boiler and come out as fudge. The sugar (sucrose, C12H22011) is purchased from a sugar farmer who used to be a chemist and who packages it by the ibmal, and the process uses 1.75 Ibmol / hr. Butter is fed to the process at a rate of 60 Ibm /hr, corn syrup is fed at a rate of 3.5 gal /hr, and cocoa is fed at 17 lb, / hr. Finally, vanilla ex- tract is fed at a rate corresponding to 1 Ibm of extract for every 30 ibm of sugar. How many gallons of milk per hour must be
fed to the process for a total fudge production of 830 lbm/hr?
The sugar used in the process is Sucrose (C12H22011). The process uses 1.75 Ibmol/hr. Butter is fed to the process at a rate of 60 Ibm/hr. Corn syrup is fed at a rate of 3.5 gal/hr. Cocoa is fed at 17 lb, / hr. Vanilla extract is fed at a rate corresponding to 1 Ibm of extract for every 30 ibm of sugar. The total fudge production is 830 lbm/hr.
To calculate the number of gallons of milk per hour that must be fed to the process for a total fudge production of 830 lbm/hr, we need to determine the amount of each ingredient used in the process.
1. Finding the number of ibmol of sucrose used in the process.
1.75 Ibmol/hr of sucrose is used in the process. Molar mass of Sucrose = 12 × 12 + 22 × 1 + 11 × 16 = 342 g/mol
Molar mass of ibmol of Sucrose = 342 × 1 = 342 g
Amount of Sucrose used in 1 hour = 1.75 × 342 g/hour
Amount of Sucrose used in 1 hour = 598.5 g/hour
Amount of Sucrose used in 830 lbm/hr = 830 × 453.6 = 376995 g/hour
Now, let's find the amount of butter used in the process. Butter fed to the process at a rate of 60 Ibm/hr
Amount of butter used in 830 lbm/hr = 60 × 830 = 49800 Ibm/hour
Now, let's find the amount of corn syrup used in the process.
Corn syrup is fed at a rate of 3.5 gal/hr.
1 US gallon = 3.78541 liters
1 gallon = 3.78541 liters
Amount of corn syrup used in 830 lbm/hr = 3.5 × 3.78541 × 830 = 9739.27 grams/hour
Now, let's find the amount of vanilla extract used in the process.
Vanilla extract is fed at a rate corresponding to 1 Ibm of extract for every 30 ibm of sugar.
So, the amount of extract required = 1/30 th of the sugar used
Amount of Vanilla extract used in 1 hour = 1/30 × 1.75 × 342 g/hour
Amount of Vanilla extract used in 1 hour = 19.47 g/hour
Amount of Vanilla extract used in 830 lbm/hr = 19.47 × 830 = 16147.1 g/hour
Now, let's calculate the amount of milk required.
Amount of Milk Required = Total Amount of Fudge - (Sugar Used + Butter Used + Corn Syrup Used + Vanilla Extract Used)
Amount of Milk Required = 830 × 453.6 g/hour - (376995 g/hour + 49800 g/hour + 9739.27 g/hour + 16147.1 g/hour)
Amount of Milk Required = 3.443 kg/hour1 US gallon = 3.78541 liters
So, the amount of Milk Required is 3.443 / 3.78541 = 0.9089 gallons/hour therefore, 0.9089 gallons of milk per hour must be fed to the process for a total fudge production of 830 lbm/hr.
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The solubility of benzoic acid in water is 68.0 g/L at 95 C. Calculate the maximum amount of benzoic acid that can dissolve in 20 mL of hot water. Show work.
The maximum amount of benzoic acid that can dissolve in 20 mL of hot water is 1.36 grams
To calculate the maximum amount of benzoic acid that can dissolve in 20 mL of hot water, we can use the given solubility of benzoic acid in water at 95 °C, which is 68.0 g/L.First, we need to convert the volume of water from milliliters (mL) to liters (L):20 mL = 20/1000 L = 0.02 L. Next, we can use the solubility value to calculate the maximum amount of benzoic acid that can dissolve in 1 liter of water at 95 °C:Maximum amount = Solubility * Volume. Maximum amount = 68.0 g/L * 0.02 L.Calculating this expression will give us the maximum amount of benzoic acid that can dissolve in 20 mL of hot water:Maximum amount = 1.36 gTherefore, the maximum amount of benzoic acid that can dissolve in 20 mL of hot water is 1.36 grams.
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the heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 1.84g of water boils at atmospheric pressure? ____kJ
The heat of vaporization of water is 40.66 kJ/mol. The amount of heat is absorbed when 1.84g of water boils at atmospheric pressure is 4.14 kJ .
To calculate the amount of heat absorbed when 1.84g of water boils at atmospheric pressure, we need to convert the mass of water to moles and then multiply it by the molar heat of vaporization.
The molar mass of water (H2O) is approximately 18.02 g/mol.
First, let's convert the mass of water to moles:
Moles of water = (Mass of water) / (Molar mass of water)
Moles of water = 1.84g / 18.02 g/mol
Next, we can calculate the heat absorbed using the molar heat of vaporization:
Heat absorbed = (Moles of water) x (Molar heat of vaporization)
Given that the molar heat of vaporization of water is 40.66 kJ/mol, we can substitute the values:
Heat absorbed = (1.84g / 18.02 g/mol) x (40.66 kJ/mol)
Calculating this, we find:
Heat absorbed ≈ 4.14 kJ
Therefore, approximately 4.14 kJ of heat is absorbed when 1.84g of water boils at atmospheric pressure.
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calculate the pH of a solution prepared by dissolving 1.90 g of sodium acetate, ch3cooNa, in 71.0 mL of 0.15 M acetic acid, ch3cooh(aq).
The pH of a solution prepared by dissolving 1.90 g of sodium acetate, CH₃COONa, in 71.0 mL of 0.15 M acetic acid, CH₃COOH (aq) is 5.38.
The problem at hand is to calculate the pH of a solution prepared by dissolving 1.90 g of sodium acetate, CH₃COONa, in 71.0 mL of 0.15 M acetic acid, CH₃COOH(aq). This problem is solved using the concept of the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is given as pH = pKa + log [A–]/[HA], where pKa is the dissociation constant of the weak acid, [A–] is the concentration of the conjugate base and [HA] is the concentration of the weak acid.
Here, acetic acid is the weak acid and its dissociation constant is 1.8 × 10^-5.
Molar mass of CH₃COONa=82.03 g/mol, Number of moles of CH₃COONa=mass/Molar mass=1.90 g/82.03 g/mol=0.0231
Molarity of the solution of CH₃COONa=0.0231/(0.071 L)=0.325 M
Now, we have:[A–] = [CH₃COO⁻] = 0.325 M[HA] = [CH₃COOH] = 0.15 pKa = 4.76pH = 4.76 + log [(0.325)/(0.15)] = 4.76 + 0.62 = 5.38
Therefore, the pH of the given solution is 5.38.
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A student carried out a synthesis of sulfanilamide. After working up the reaction and carrying out a recrystallization of the crude product, she obtained 41.2 g of purified sulfanilamide, representing a 40.67% recovery. How much crude product, in grams, did the student have before recrystallization? Report your answer to the correct number of sig figs but do not include units.
The student had approximately 101.24 grams of crude product before recrystallization.
To find the amount of crude product before recrystallization, we can use the recovery percentage and the mass of purified sulfanilamide obtained.
Let's assume the mass of the crude product before recrystallization is represented by "x" grams.
The recovery percentage is given as 40.67%, which means that 40.67% of the crude product was obtained as purified sulfanilamide.
Recovery percentage = (mass of purified sulfanilamide/mass of crude product) × 100
40.67% = (41.2 g / x) × 100
To solve for "x," we can rearrange the equation,
x = (41.2 g) / (40.67% / 100)
Performing the calculations we get,
x = (41.2 g) / (0.4067)
x ≈ 101.24 g
Therefore, the student had approximately 101.24 grams of crude product before recrystallization.
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A Vitamin C packet is added to a glass of water containing 720.0 mL of solution. The Vitamin C packet contains 1000.0mg of Vitamin C. What is the concentration of Vitamin C in ppm in the resultant solution? (Assume density of solution =1.00 g/mL )
A Vitamin C packet is added to a glass of water containing 720.0 mL of solution. The Vitamin C packet contains 1000.0mg of Vitamin C.The concentration of Vitamin C in the resultant solution is 1388.9 ppm.
To calculate the concentration of Vitamin C in parts per million (ppm), we need to convert the mass of Vitamin C to grams and then divide it by the total mass of the solution.
Convert the mass of Vitamin C from milligrams to grams:
1000.0 mg = 1000.0 g
Calculate the total mass of the solution:
The density of the solution is given as 1.00 g/mL, and the volume of the solution is 720.0 mL. Therefore, the total mass of the solution is:
720.0 mL × 1.00 g/mL = 720.0 g
Calculate the concentration of Vitamin C in ppm:
Now, we can divide the mass of Vitamin C by the total mass of the solution and multiply by 1,000,000 to obtain the concentration in ppm:
(1000.0 g ÷ 720.0 g) × 1,000,000 = 1388.9 ppm
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Suppose you have a vacuum chamber at room temperature and you place some liquid water inside the chamber keeping it sealed. Which of the following best describes what will happen? The water molecules will remain in the liquid form. The water molecules will immediately begin to enter the gas phase until all of the water molecules are in the gas phase. The water molecules will immediately begin to enter the gas phase. As the concentration of gas phase water increases, gas phase water molecules begin to return to the liquid phase. None of the above Overtime, the concentration of water in the liquid and gas phases will remain constant. When this occurs, the pressure exerted by the gas phase water vapor is called the vapor pressure. True or false, a dynamic equilibrium is established once enough water molecules have entered the gas phase to reach the vapor pressure.
The correct answer is: True, a dynamic equilibrium is established once enough water molecules have entered the gas phase to reach the vapor pressure.
In the given scenario, when liquid water is placed in a sealed vacuum chamber at room temperature, the water molecules will start to evaporate and enter the gas phase. Initially, there will be a rapid increase in the concentration of gas phase water as molecules transition from the liquid to the gas phase. However, as the concentration of gas phase water increases, some gas phase water molecules will start condensing back into the liquid phase.
Eventually, a point of equilibrium is reached where the rate of evaporation of water molecules from the liquid phase is equal to the rate of condensation of gas phase water molecules back into the liquid phase. At this point, the concentration of water in both the liquid and gas phases remains constant, and the pressure exerted by the gas phase water vapor is known as the vapor pressure.
This state is referred to as a dynamic equilibrium because while there is no net change in the overall concentration of water, there is a continuous exchange of molecules between the liquid and gas phases.
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Substitution reactions are common for coordination complexes. If you begin with [Ni(H2O)6]2+, formed from dissolving NiCl2 in water, what order of ligand addition would make sense to produce the following three compounds sequentially in the same beaker? Explain.
[Ni(EDTA)]2−
[Ni(NH3)6]2+
[Ni(en)3]2+
To sequentially produce [tex][Ni(EDTA)]^{2-}[/tex], [tex][Ni(NH_3)_6]^{2+}[/tex], and [[tex]Ni(en)_3]^{2+}[/tex] from [tex][Ni(H_2O)^6]^{2+}[/tex], it would make sense to add the ligands in the following order: EDTA, [tex]NH_3[/tex], and en. This order is based on the ligand's relative strength and its ability to replace water ligands in a substitution reaction.
Substitution reactions in coordination complexes involve the replacement of one or more ligands with other ligands. The order of ligand addition can be determined based on the ligand's strength and its ability to displace other ligands.
In this case, EDTA (ethylenediaminetetraacetate) is a strong chelating ligand that forms stable complexes with metal ions. It has a high affinity for metal ions and can displace water ligands effectively. Therefore, adding EDTA to [tex][Ni(H_2O)^6]^{2+}[/tex] would result in the formation of [tex][Ni(EDTA)]^{2-}[/tex].
Next, ammonia ([tex]NH_3[/tex]) is a weaker ligand compared to EDTA but stronger than water. It can displace water ligands to form complexes with metal ions. Adding [tex]NH_3[/tex] to [tex][Ni(EDTA)]^{2-}[/tex] would lead to the formation of [tex][Ni(NH_3)_6]^{2+}[/tex].
Lastly, ethylenediamine (en) is a stronger ligand than ammonia and can displace [tex]NH_3[/tex] ligands. Adding en to [tex][Ni(NH_3)_6]^{2+}[/tex] would yield [tex]Ni(en)_3]^{2+}[/tex].
By following this order of ligand addition, [tex][Ni(EDTA)]^{2-}[/tex] [tex][Ni(NH_3)_6]^{2+}[/tex], and [tex]Ni(en)_3]^{2+}[/tex] can be sequentially produced in the same beaker, with each ligand replacing the previous ligands based on their relative strength and ability to form stable complexes with the nickel ion.
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M: Suppose C4H10 is combusted in the air: After writing a balanced equation, answer the following questions: The Coefficient of C4H10 will be: The Coefficient of Oxygen will be The coefficient of carbon Di-Oxide will be The coefficient of water will be:
The coefficient of C4H10 is 1. The coefficient of oxygen (O2) is 13. The coefficient of carbon dioxide (CO2) is 8. The coefficient of water (H2O) is 10.
Coefficients in a chemical equation represent the relative amounts of each species (atoms, molecules, or ions) involved in a chemical reaction. They are whole numbers placed in front of the chemical formulas to balance the equation and ensure that the number of atoms for each element is the same on both sides.
The balanced equation for the combustion of C4H10 (butane) in the presence of oxygen can be written as:
C4H10 + 13O2 → 8CO2 + 10H2O
In this balanced equation, the coefficients represent the relative amounts of each molecule involved in the reaction.
The coefficient of C4H10 is 1.
The coefficient of oxygen (O2) is 13.
The coefficient of carbon dioxide (CO2) is 8.
The coefficient of water (H2O) is 10.
These coefficients are determined by ensuring that the number of atoms for each element is equal on both sides of the equation. In this case, we have 4 carbon (C) atoms, 10 hydrogen (H) atoms, and 26 oxygen (O) atoms on both sides, which makes the equation balanced.
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The capacity of a buffer depends on the concentration of the buffer, but mostly on the temperature. True False
The capacity of a buffer depends on the concentration of the buffer, but mostly on the temperature False
The capacity of a buffer depends primarily on the concentrations of the buffer components, specifically the concentrations of the weak acid and its conjugate base (or the weak base and its conjugate acid).
The ratio of their concentrations determines the pH of the buffer and its ability to resist changes in pH upon addition of acid or base.
While temperature can affect the ionization of the weak acid or base to some extent, it does not have a significant impact on the buffer capacity compared to the concentrations of the buffer components.
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which of the two polymers is more likely to crystallize and why? - Linear vs. syndiotactic poly (vinyl chloride) - linear vs. isotactic polystyrene - Alternating poly(styrene-ethylene) copolymer vs random poly (vinyl chloridetetrafluoroethylene) copolymer.
In general, the crystallization behavior of polymers is influenced by several factors, including the molecular structure, chain arrangement, and thermal history.
Linear vs. Syndiotactic Poly(vinyl chloride) (PVC):
Linear PVC: Linear polymers tend to have a higher probability of crystallization compared to branched or highly cross-linked polymers. Therefore, linear PVC has a higher likelihood of crystallizing.
Syndiotactic PVC: Syndiotactic polymers have a regular arrangement of the pendant groups along the polymer backbone. This regularity can enhance the polymer's crystallinity. Hence, syndiotactic PVC is more likely to crystallize compared to atactic PVC (which lacks a regular arrangement).
Linear vs. Isotactic Polystyrene:
Linear Polystyrene: Linear polystyrene can exhibit crystallization, but its crystallinity is generally lower than that of isotactic polystyrene.
Isotactic Polystyrene: Isotactic polymers have pendant groups positioned on the same side of the polymer backbone. This regular arrangement promotes crystallization, making isotactic polystyrene more likely to crystallize.
Alternating Poly(styrene-ethylene) Copolymer vs. Random Poly(vinyl chloride-tetrafluoroethylene) Copolymer:
Alternating Poly(styrene-ethylene) Copolymer: Copolymers with a regular alternating structure, like the poly(styrene-ethylene) copolymer, have a higher tendency to crystallize due to the ordered arrangement of monomer units.
Random Poly(vinyl chloride-tetrafluoroethylene) Copolymer: Random copolymers, such as the poly(vinyl chloride-tetrafluoroethylene) copolymer, have a less ordered structure. This randomness reduces the likelihood of crystallization.
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what was the percent ag by mass in the lead metal? An important source of Ag is recovery as a by-product in the metallurgy of lead. The percentage of Ag in lead was determined as follows. A 1.080 – g sample was dissolved in nitric acid to produce Pb2+ (aq) and Ag+(aq). The solution was then diluted to 500.0 mL with water, a Ag electrode was immersed in the solution, and the potential difference between this electrode and a SHE was found to be 0.498 V.
To determine the percent silver (Ag) by mass in the lead metal, we can use electrochemical measurements based on the standard reduction potentials of the species involved.
The reaction occurring at the silver electrode is as follows:
Ag+ (aq) + e- → Ag (s)
The standard reduction potential for this reaction is 0.799 V vs. the standard hydrogen electrode (SHE).
Given that the potential difference between the Ag electrode and SHE was found to be 0.498 V, we can calculate the actual reduction potential for the Ag+ to Ag reaction:
Ecell = EAg - ESHE
0.498 V = EAg - 0.000 V
EAg = 0.498 V
Since the standard reduction potential for Ag+ to Ag is 0.799 V, the difference between the actual reduction potential and the standard reduction potential is:
ΔE = EAg - E°Ag
ΔE = 0.498 V - 0.799 V
ΔE = -0.301 V
Now, we can use the Nernst equation to relate the potential difference to the concentration of Ag+ ions:
E = E° - (0.0592 V/n) log([Ag+])
Where:
E is the measured potential difference (0.498 V)
E° is the standard reduction potential (0.799 V)
n is the number of electrons transferred in the reaction (1 for Ag+ to Ag)
[Ag+] is the concentration of Ag+ ions (unknown)
Substituting the values into the equation:
0.498 V = 0.799 V - (0.0592 V/1) log([Ag+])
0.301 V = 0.0592 V log([Ag+])
Now, we need to determine the number of moles of Ag+ ions present in the solution. We know that the volume of the solution is 500.0 mL (0.5000 L), and the concentration of Ag+ ions is [Ag+]. So:
moles of Ag+ = [Ag+] * volume (in L)
moles of Ag+ = 10^5.084 * 0.5000 L
Finally, to calculate the percent silver (Ag) by mass in the lead metal, we divide the moles of Ag+ by the initial mass of the lead sample (1.080 g) and multiply by 100:
percent Ag = (moles of Ag+ / initial mass of lead) * 100
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after cleaning, how do you know the contents are sterilized?
This documentation will help to ensure that the sterilization process is effective and that the contents are sterilized. Cleaning is an essential part of the sterilization process.
To ensure that contents are sterilized after cleaning, there are some procedures you can follow.
Here are the steps you should follow:
Step 1: Cleaning The first step to sterilization is cleaning.
The materials that are to be sterilized must be cleaned to remove dirt and grime.
This is important because it ensures that the sterilization process is effective and the sterilized contents will be free of any foreign materials.
Step 2:
Inspection After cleaning, inspect the materials to ensure that they are clean.
It is essential to ensure that the materials are free of any foreign matter, such as dust, debris, or other contaminants. This will help to ensure that the sterilization process is effective.
Step 3: Sterilization The next step is to sterilize the materials.
There are several methods of sterilization, including steam sterilization, dry heat sterilization, and chemical sterilization.
Each method has its own advantages and disadvantages.
It is essential to select the right method based on the materials to be sterilized.
Step 4: Inspection After sterilization, the materials should be inspected again to ensure that they have been sterilized effectively.
This can be done by checking for the presence of bacteria or other microorganisms.
Step 5: Documentation Finally, it is essential to document the entire sterilization process.
This includes the cleaning process, the sterilization method used, and the results of the inspection.
It is important to include final answers to ensure the process was effective and complete.
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The temperatures at the nodes of a triangular element are given by Ti=210∘F,Tj=270∘F, and Tk=250∘F. If the nodal coordinates are (xi,yi)=(50,30) in, (xj,yj)=(70,50) in, and (xk,yk)=(55,60) in, determine (a) the shape functions of the element and (b) temperature at the point (x,y)=(60,40) in inside the element.
a) The shape functions of the element are Ni=1−0.238x−0.108yNj=0.238x−0.038y−0.965Nk=0.103x+0.146y−0.249.
b) The temperature at the point (60,40) inside the element is T(x,y)=244.3°F.
(a) The shape functions of the triangular element are as follows:Ni = 1 – 0.238x – 0.108yNj = 0.238x – 0.038y – 0.965Nk = 0.103x + 0.146y – 0.249
(b) To calculate the temperature at the point (x,y) = (60,40) in inside the element, we need to use the shape functions and nodal temperatures. Using these values in the formula, we get:T(x, y) = N1T1 + N2T2 + N3T3= 210 + 30 (–0.238x – 0.108y) + 270 (0.238x – 0.038y – 0.965) + 250 (0.103x + 0.146y – 0.249)= 244.3 °F
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How much heat will be given off by ssg of liquid water as it cools from 81
∘
C to 25
∘
C ? specific heat =4.184 J1 g
∘
C
The amount of heat given off by 1 g of liquid water as it cools from 81°C to 25°C is 235.904 J.
Initial temperature of the liquid water, T1 = 81°C.
Final temperature of the liquid water, T2 = 25°C.
Specific heat of the liquid water, c = 4.184 J/g °C.
Mass of the liquid water, m = 1 g.
As the water cools from 81°C to 25°C, it loses heat.
The heat lost by the water is equal to the heat gained by the surroundings.
The formula for heat is given as:
Q = mcΔT,
where Q is the heat,
m is the mass of the substance,
c is the specific heat of the substance and
ΔT is the change in temperature (T2 - T1).
So, the heat given off by 1 g of liquid water as it cools from 81°C to 25°C is given by:
Q = 1 g × 4.184 J/g °C × (25°C - 81°C)Q
= 1 g × 4.184 J/g °C × (-56°C)Q
= -235.904 J
The negative sign indicates that heat is lost by the water, which is what is expected since the water is cooling down.
Therefore, the amount of heat given off by 1 g of liquid water as it cools from 81°C to 25°C is 235.904 J.
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Aqueous hydrochioric acid (HCl) rescts with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid water (HS O). What is the theoretical yeld of water formed from the reaction of 13.5 g of hydrochloric acid and 26.3 g of sodium hydroxide? Be sure your answer has the correct number of significant digits in it.
The theoretical yield of water formed from the reaction of 13.5 g of hydrochloric acid and 26.3 g of sodium hydroxide is approximately 6.66 grams.
In the balanced chemical equation, we can see that the molar ratio between HCl (hydrochloric acid) and H2O (water) is 1:1. This means that for every 1 mole of HCl reacted, we will produce 1 mole of water.
To find the number of moles of HCl in 13.5 grams, we need to divide the given mass by the molar mass of HCl, which is approximately 36.46 grams/mol.
13.5 grams of HCl / 36.46 grams/mol = 0.3701 moles of HCl.
Similarly, for sodium hydroxide (NaOH), the molar ratio with water is also 1:1. To find the number of moles of NaOH in 26.3 grams, we divide the given mass by the molar mass of NaOH, which is approximately 39.99 grams/mol.
26.3 grams of NaOH / 39.99 grams/mol = 0.6577 moles of NaOH.
Since the molar ratio between HCl and H2O is 1:1, we can conclude that the number of moles of water formed will be equal to the number of moles of HCl used, which is 0.3701 moles.
To find the mass of water, we multiply the number of moles by the molar mass of water, which is approximately 18.02 grams/mol.
0.3701 moles of H2O x 18.02 grams/mol = 6.66 grams of water.
Therefore, the theoretical yield of water formed from the reaction of 13.5 g of hydrochloric acid and 26.3 g of sodium hydroxide is approximately 6.66 grams.
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Please show the work for the answer. 1. At a pH above its pKa, the phenolic group of lysine is ______and ______
At a pH above its pKa, the phenolic group of lysine is deprotonated (negatively charged) and negatively charged.
Lysine is an amino acid that contains a side chain with a phenolic group. The phenolic group in lysine can undergo ionization, meaning it can either be protonated or deprotonated depending on the pH of the solution.
The pKa of the phenolic group in lysine is the pH at which half of the phenolic groups are deprotonated and half are protonated. Above the pKa, the phenolic group is predominantly deprotonated, resulting in a negative charge on the oxygen atom.
This deprotonation occurs because the solution's pH is higher than the pKa, indicating a more basic environment. When the phenolic group of lysine is deprotonated, it carries a negative charge due to the loss of a proton.
This negative charge on the oxygen atom is responsible for the phenolic group's increased reactivity and ability to participate in various chemical reactions, such as hydrogen bonding or electrostatic interactions with positively charged molecules or ions. In summary, at a pH above its pKa, the phenolic group of lysine is deprotonated and negatively charged.
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Assume ideal gas behavior. A cylinder of volume V is filled with mass 256 g of oxygen at standard temperature and pressure. Find the volume V of the cylinder in liters. An identical cylinder is charged to 2.1 atm at 273 K with an unknown gas mixture and found to have a mass of 823.2 g What is the average molecular weight
M
ˉ
of the new gas mixture?
The average molecular weight of the new gas mixture is 73.8 g/mol.
Given the mass of the gas in cylinder, m = 256 g
Volume of the cylinder, V is to be calculated.1 mole of oxygen at STP occupies 22.4 L. The number of moles of oxygen in 256 g can be found out as follows:256 g * (1 mol / 32 g) = 8 molVolume of oxygen = 8 mol * 22.4 L/mol = 179.2 L
Thus, the volume of the cylinder is 179.2 LFor the second part, we need to find the molecular weight of the gas mixture.
Using the ideal gas equation, PV = nRT, we have: PV = m/MRTM = mRT / PV = (823.2 g)(0.082 L atm / (mol K))(273 K) / (2.1 atm)(1000 mL/L)(1 mol/1000 mL)M = 73.8 g/mol
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Question: Hydrogen gas at 2.0 atm and 298KC is flowing in a two-layered tube of 2.0 mm inside diameter and 4.0 mm outside diameter. Calculate the leakage of hydrogen through a tube 5 m long in kgmolH
2
/s at steady state. These layers are arranged as follows: - Layer-1: vulcanized rubber (inside dia. 2.0-mm and outside dia. 3.0-mm) - Layer-2: polyethylene (inside diameter 3.0-mm and outside diameter 4.0-mm)
The leakage of hydrogen through a tube 5 m long in kg mol H₂ /s at steady state: is 8.48 * 10⁻¹¹mol H₂/s
The hydrogen gas leakage rate in kg mol H₂/s can be calculated using the following steps:
Find the area of the hydrogen gas flow. The flow of hydrogen gas in the tube is at a steady state. The area of the hydrogen gas flow can be calculated using the following formula: π/4 * (d₂² - d₁²)
where d₁ is the inside diameter of the inner tube (2.0 mm) and d₂ is the outside diameter of the outer tube (4.0 mm).
π/4 * (4.0² - 2.0²) = 9.42 mm²
Find the volume of hydrogen gas that leaks through the tube in 1 second. The volume of hydrogen gas that leaks through the tube in 1 second can be calculated using the following formula:
Q = PAV/RT
where Q is the volume of gas that leaks through the tube, P is the pressure of the gas (2.0 atm), A is the area of the hydrogen gas flow (9.42 mm²), V is the velocity of the gas, R is the universal gas constant (0.08206 L atm/mol K), and T is the temperature of the gas (298 K).
Q = 2.0 atm * 9.42 mm² * V / (0.08206 L atm/mol K * 298 K)V
= Q * 0.08206 * 298 / (2.0 * 9.42) = 9.91 m/s
Find the mass of hydrogen gas that leaks through the tube in 1 second. The mass of hydrogen gas that leaks through the tube in 1 second can be calculated using the following formula:
m = ρQ
where m is the mass of gas that leaks through the tube, ρ is the density of the gas (0.0899 kg/m³), and Q is the volume of gas that leaks through the tube (9.42 mm³/s).
m = 0.0899 kg/m³ * 9.42 mm³/s / 10⁶ mm³/L = 8.48 * 10⁻⁸ kg/s
Find the number of moles of hydrogen gas that leaks through the tube in 1 second. The number of moles of hydrogen gas that leaks through the tube in 1 second can be calculated using the following formula: n = m/M
where n is the number of moles of gas that leaks through the tube, m is the mass of gas that leaks through the tube (8.48 * 10⁻⁸ kg/s), and M is the molar mass of hydrogen (2.02 g/mol).
n = 8.48 * 10⁻⁸ kg/s / 0.00202 kg/mol = 4.20 * 10⁻⁸ mol/s
Find the leakage of hydrogen gas through the tube in kg mol H₂/s. The leakage of hydrogen gas through the tube in kg mol H₂/s can be calculated using the following formula:
leakage = n * M
leakage = 4.20 * 10⁻⁸ mol/s * 0.00202 kg/mol
= 8.48 * 10⁻¹¹ kg mol H₂/s
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