Answer: The number of moles of disulfur decafluoride is [tex]6.62\times 10^{-20}[/tex]
Explanation:
We are given:
Number of disulfur decafluoride molecules = [tex]3.99\times 10^4[/tex]
According to mole concept:
[tex]6.022\times 10^{23}[/tex] number of molecules are contained in 1 mole of a compound.
So, [tex]3.99\times 10^4[/tex] number of molecules will be contained in = [tex]\frac{1mol}{6.022\times 10^{23}}\times 3.99\times 10^{4}=6.62\times 10^{-20}mol[/tex] of disulfur decafluoride.
Hence, the number of moles of disulfur decafluoride is [tex]6.62\times 10^{-20}[/tex]
Find the theoretical oxygen demand for the
followingsolutions?
a. 200mg/L of acetic acid, CH3COOH.
b. 30mg/L of ethanol, C2H5OH.
C. 50mg/L of sucrose C2H12O6.
Answer:
a) 213.3 mg/L
b) 62.61 mg/L
c) 0.0225 mg/L
Explanation:
Theoretical oxygen demand (ThOD)is essentially the amount of oxygen required for the complete degradation of a given compound into the final oxidized products
a) Given:
Concentration of acetic acid,[tex][CH3COOH][/tex] = 200 mg/L
[tex]CH3COOH + 2O2 \rightarrow 2CO2 + 2H2O[/tex]
[tex]ThOD = \frac{mass\ O2}{mass\ CH3COOH} * conc. CH3COOH[/tex]
Based on the reaction stoichiometry:
mass of [tex]CH3COOH[/tex] = 60 g
mass of [tex]O2[/tex]= 2(32) = 64 g
[tex]ThOD = \frac{64 g}{60 g} * 200mg/L = 213.3 mg/L[/tex]
b) Given:
Concentration of ethanol, [tex][C2H5OH][/tex] = 30 mg/L
[tex]C2H5OH + 3O2 \rightarrow 2CO2 + 3H2O[/tex]
[tex]ThOD = \frac{mass\ O2}{mass\ C2H5OH} * conc. C2H5OH[/tex]
Based on the reaction stoichiometry:
mass of [tex]C2H5OH[/tex] = 46 g
mass of [tex]O2[/tex]= 3(32) = 96 g
[tex]ThOD = \frac{96 g}{46 g} * 30mg/L = 62.61 mg/L[/tex]
c) Given:
Concentration of sucrose, [tex][C12H22O11][/tex] = 50 mg/L
[tex]C12H22O11 + 12O2 \rightarrow 12CO2 + 11H2O[/tex]
[tex]ThOD = \frac{mass\ O2}{mass\ C22H22O11} * conc. C12H22O11[/tex]
Based on the reaction stoichiometry:
mass of [tex]C12H22O11[/tex] = 342 g
mass of [tex]O2[/tex]= 12(32) = 384 g
[tex]ThOD = \frac{384 g}{342 g} * 50mg/L = 0.0225 mg/L[/tex]
A population of rabbits triples every 2 months If there are 2 rabbits initially, how long will it take for the population to increase to 500 rabbits? Round your answer to the nearest whole number (5 marks)
Answer:
The rabbit population will reach 500 after 10 months.
Explanation:
According to the given data:
The initial number of rabbit's equals 2.
Number of rabbit's after 2 months =2x3= 6
Number of rabbit's after 4 months = 6x3=18
Number of rabbit's after 6 months = 18x3=54
Number of rabbit's after 8 months = 54x3=162
Thus we can see that the number of rabbit's form a Geometric series with common ratio =3 and initial term = 2
Now the general term of a geometric series with first term 'a' and common ratio 'r' is given by
[tex]T_{n+1}=ar^{n}[/tex]
Thus we need to find when the term becomes 500 thus using the given data we have
[tex]500=2\cdot 3^{n}\\\\3^{n}=250\\\\(n)log_3(3)=log_3(250)\\\\(n)=5.025\\\\[/tex]
Thus the fifth term (excluding the start term) will have a rabbit count of 500 now since each term has a time difference of 2 months thus sixth term will occur after [tex]5\times 2=10months[/tex]
You have a stock solution of epinephrine at a concentration of 1 mg/mL. Knowing that the pipette you will use delivers 20 drops/mL, a. calculate the number of drops of the stock solution that must be added to a smooth muscle bath containing 25 mL of Locke’s solution so that the final concentration of epinephrine in the muscle bath will be 100 µg/mL, and
If you want to achieve a final concentration of 100 µg/mL of epinephrine in a 25 mL solution, when using a stock solution of 1 mg/mL and a pipette that delivers 20 drops/mL, you need to add 50 drops of your stock solution.
Explanation:Since we are asked to find the number of drops of stock solution required to achieve a final concentration of 100 µg/mL in a 25 mL solution, the first step is to convert the concentration of the stock solution to the same units, µg/mL. Hence, 1 mg/mL is equal to 1000 µg/mL. Further, we know that 1 mL of the stock solution contains 1000 µg of epinephrine, and our pipette delivers 20 drops/mL, so 1 drop of stock solution contains 1000 µg / 20 drops = 50 µg. Thus, if we need a 100 µg/mL concentration in 25 mL, we need a total of 100 µg/mL * 25 mL = 2500 µg of epinephrine. Therefore, to achieve this, we must add 2500 µg / 50 µg/drop = 50 drops of our stock solution. Hence,
50 drops
of the stock solution should be added to achieve the desired concentration.
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To achieve a final concentration of 100 µg/mL of epinephrine in 25 mL of Locke's solution, you need to add 50 drops of the 1 mg/mL stock solution. The dilution factor for one drop of the stock solution in the muscle bath is 500
a) To determine the required number of drops to achieve a specific concentration in Locke's solution, we follow these steps:
Calculate the total amount of epinephrine needed: With a final concentration of 100 µg/mL in 25 mL, the total amount of epinephrine necessary is:b) Dilution Factor
To calculate the dilution factor, we need to consider how much one drop dilutes in the 25 mL smooth muscle bath:Volume of one drop:
1 mL / 20 drops = 0.05 mL/drop
The dilution factor, therefore, is the ratio of the final volume to the volume of one drop:25 mL / 0.05 mL = 500
The correct dilution factor is 500.
Therefore, you must add 50 drops of the stock epinephrine solution to the muscle bath in Locke's solution to achieve the desired concentration of 100 µg/mL and dilution factor of 500.
Complete question.
Assume that you have a stock solution of epinephrine at a concentration of 1mg/ml. Further assume that there are 20 drops/ml.
a. Calculate the number of drops of the stock solution that must be added to a smooth muscle bath containing 25 ml of Locke's solution so that the final concentration of epinephrine in the muscle bath will be 100 µg/mL.
b. Compute the dilution factor which describes the extent to which a single drop of stock drug solution is diluted when added to the smooth muscle bath.
HNO3 is a strong acid. Calculate the pH of an aqueous 0.60 M solution.
Answer:
pH = 0.22
Explanation:
The pH of the a solution is related to the concentration of the hydronium ion as follows:
pH = -log([H₃O⁺])
HNO₃ is a strong acid that reacts to completion with water as follows:
HNO₃ + H₂O ⇒ H₃O⁺ + NO₃⁻
The molar ratio between nitric acid and the hydronium ion is 1:1, so a 0.60 M nitric acid solution has a hydronium ion concentration of 0.60 M.
The pH is calculated:
pH = -log([H₃O⁺]) = -log(0.60) = 0.22
Predict the sign of ΔS°for the following reaction.
Cu2O(s) + C(s) → 2Cu(s) + CO(g)
ΔS° ≈ 0
Δ S°< 0
Δ S°> 0
More information is needed to make a reasonable prediction.
Answer:
Δ S° > 0
Explanation:
Entropy -
In a system, the randomness is measured by the term entropy .
Randomness basically refers as a form of energy that can not be used for any work.
The change in entropy is given by the change in heat per change in temperature.
When solid is converted to gas entropy increases,As the molecules in solid state are tightly packed and has more force of attraction between the molecules, but as it is converted to gas, the force of attraction between the molecule decreases and hence entropy increases.
So,
The particles of the substance , if are tightly held by strong force of attraction will decrease the entropy ,
And
If the particles are loosely held , the entropy will increase .
From the reaction given in the question ,
Cu₂O (s) + C (s) → 2 Cu (s) + CO (g)
Considering the states of the reactant and the product ..
The product formed i.e. CO is in gaseous species , hence , the entropy increases .
Calculate the specific volume of Helium using the compressibility factor. 500 kPa, 60 degrees Celsius
Explanation:
Formula for compressibility factor is as follows.
z = [tex]\frac{P \times V_{m}}{R \times T}[/tex]
where, z = compressibility factor for helium = 1.0005
P = pressure
[tex]V_{m}[/tex] = molar volume
R = gas constant = 8.31 J/mol.K
T = temperature
So, calculate the molar volume as follows.
[tex]V_{m} = \frac{z \times R \times T}{P}[/tex]
= [tex]\frac{1.0005 \times 8.314 \times 10^{-3} m^{3}.kPa/mol K \times (60 + 273)K}{500 kPa}[/tex]
= 0.0056 [tex]m^{3}/mol[/tex]
As molar mass of helium is 4 g/mol. Hence, calculate specific volume of helium as follows.
[tex]V_{sp} = \frac{V_{m}}{M_{w}}[/tex]
= [tex]\frac{0.0056 m^{3}/mol}{4 g/mol}[/tex]
= 0.00139 [tex]m^{3}/g[/tex]
= 0.00139 [tex]m^{3}/g \times \frac{1 g}{10^{-3}kg}[/tex]
= 1.39 [tex]m^{3}/kg[/tex]
Thus, we can conclude that the specific volume of Helium in given conditions is 1.39 [tex]m^{3}/kg[/tex].
Describe how you could separate and purify compound A from a mixture of two neutral compounds (A and B) when A comprises 95% of the total and B the other 5% of the total. Assume that A and B have similar polarities.
Answer:
We could use fractional distillation, solvent extraction, partial crystallization, chromatography, and another.
Explanation:
The only information given is that the compounds have similar polarities, which means they form a homogeneous mixture, so, to separate them, we should use some of the processes of separation of a homogeneous mixture.
Without knowing any of the physical properties of A and B (physical state, density, boiling point, for example) it's impossible to determinate the better process of separation. However, we can describe some of the processes that are used to separate homogeneous mixtures: fractional distillation, solvent extraction, partial crystallization, and chromatography, for example.
What is the appropriate chemical formula for Iron(III) Sulfate? Sulfate is SO_4^2 Fe3SO4 O a . Fe2(SO4)3 O b. O c. Fe2SO4 O d. Fe2(SO4)2 O e. Fe3(SO4)2
Answer: The chemical formula of Iron (III) sulfate is [tex]Fe_2(SO_4)_3[/tex]
Explanation:
Iron is the 26th element of periodic table having electronic configuration of [tex][Ar]3d^64s^2[/tex].
To form [tex]Fe^{3+}[/tex] ion, this element will loose 3 electrons.
Sulfate ion is a polyatomic ion having chemical formula of [tex]SO_4^{2-}[/tex]
By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.
So, the chemical formula for Iron (III) sulfate is [tex]Fe_2(SO_4)_3[/tex]
The osmotic pressure of an aqueoussolution of urea at 300 K is
120 kPa. Calculate the freezing pointof the same solution.
The freezing point of the aqueous solution of urea at 300 K with an osmotic pressure of 120 kPa is -0.09°C
To calculate the freezing point of the solution, we can use the following equation:
ΔTf = Kf * m
where:
ΔTf is the freezing point depression in C
Kf is the molal freezing point constant of water in C·kg/mol (1.86 C·kg/mol)
m is the molality of the solution in mol/kg
The molality of the solution can be calculated from the osmotic pressure using the following equation:
Π = MRT
where:
Π is the osmotic pressure in Pa
M is the molarity of the solution in mol/m³
R is the ideal gas constant in J/mol·K (8.314 J/mol·K)
T is the temperature in K
Substituting the given values into the equation, we get:
M = Π / (RT) = 120 kPa / (8.314 J/mol·K * 300 K) = 0.4918 mol/m³
The molality of the solution is equal to the molarity divided by the density of the solvent. The density of water is 1 kg/L, so the molality of the solution is:
m = M / d = 0.4918 mol/m³ / 1 kg/L = 0.4918 mol/kg
Now that we know the molality of the solution, we can calculate the freezing point depression using the first equation:
ΔTf = Kf * m = 1.86 C·kg/mol * 0.4918 mol/kg = 0.09 °C
Finally, we can calculate the freezing point of the solution by subtracting the freezing point depression from the freezing point of pure water:
Freezing point of solution = freezing point of pure water - freezing point depression
Freezing point of solution = 0 °C- 0.09 °C = -0.09°C
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The freezing point of an aqueous solution of urea at 300 K with an osmotic pressure of 120 kPa is -0.08928 °C.
1. To find the freezing point depression (ΔTf) of the solution, we can use the formula: ΔTf = i x k x m, where:
i is the van 't Hoff factor (for urea, i = 1 because it does not dissociate in solution). Kf is the freezing point depression constant for water (1.86 °C kg/mol).m is the molality of the solution.2. We need to calculate the molality (m) using osmotic pressure (π), which we can find from the formula: π / (i x R x T), given:
Osmotic pressure (π) = 120 kPa (convert to atm: 120 kPa / 101.325 = 1.184 atm).R is the universal gas constant = 0.0821 L atm/mol K.T is the temperature = 300 K.M is the molarity.Rearranging for M (molarity): M = π / (i x R x T)Substitute values: M = 1.184 atm / (1 x 0.0821 L atm/mol K x 300 K) = 0.048 mol/L3. Since the density of water is approximately 1kg/L, the molality (m) ≈ molarity (M) in this case: m ≈ 0.048 mol/kg
Now, we can find the freezing point depression (ΔTf):ΔTf = 1 x 1.86 °C kg/mol x 0.048 mol/kg = 0.08928 °C
Hydrogen gas and oxygen gas react to form water. The balanced equation for the reaction is the following. 2 H2(g) + O2(g) → 2 H2O(g) Write equations to describe the rate of disappearance of H2(g) and the rate of disappearance of O2(g). Then write a third equation to show how the rates are related. Rate of disappearance of H2(g): rate of disappearance of H2 = − Δ[H2] Δt rate of disappearance of H2 = Δ[H2] Δt rate of disappearance of H2 = Δt Δ[H2] rate of disappearance of H2 = − Δt Δ[H2] Rate of disappearance of O2(g): rate of disappearance of O2 = Δ[O2] Δt rate of disappearance of O2 = Δt Δ[O2] rate of disappearance of O2 = − Δ[O2] Δt rate of disappearance of O2 = − Δt Δ[O2] Relationship between the rates: rate of disappearance of H2 = − 2 mol H2 1 mol O2 × rate of disappearance of O2 rate of disappearance of H2 = 2 mol H2 1 mol O2 × rate of disappearance of O2 rate of disappearance of H2 = 1 mol H2 2 mol O2 × rate of disappearance of O2 rate of disappearance of H2 = − 1 mol H2 2 mol O2 × rate of disappearance of O2
Answer: Thus rate of disappearance of [tex]H_2=2\times {\text {rate of disappearance of}O_2}[/tex]
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
[tex]2H_2(g)+O_2(g)\rightarrow 2H_2O(g)[/tex]
Given: Order with respect to [tex]H_2[/tex] = 2
Order with respect to [tex]O_2[/tex] = 1
Thus rate law is:
[tex]Rate=k[H_2]^2[O_2]^1[/tex]
k= rate constant
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
[tex]Rate=-\frac{1d[H_2]}{2dt}=-\frac{1d[O_2]}{dt}[/tex]
[tex]Rate=-\frac{1d[H_2]}{dt}=2\times -\frac{1d[O_2]}{dt}[/tex]
Thus rate of disappearance of [tex]H_2=2\times {\text {rate of disappearance of}O_2}[/tex]
How many moles of solute particles are present in 7.94 mL of 0.887 M NaNO3? When you have your answer, take the LOG (base 10) of your answer and enter it with 2 decimal places.
To calculate the number of moles of solute particles in a solution, multiply the molarity by the volume of the solution. In this case, the result is 0.0070 mol. Taking the LOG (base 10) of the result gives -2.15.
Explanation:To determine the number of moles of solute particles in a solution, you need to consider the molarity and volume of the solution. In this case, the molarity is 0.887 M and the volume is 7.94 mL. To calculate the number of moles, you can use the formula:
Moles = Molarity x Volume
Converting mL to L:
7.94 mL = 7.94/1000 L = 0.00794 L
Now, substitute the values into the formula:
Moles = 0.887 M x 0.00794 L = 0.0070 mol
Taking the LOG (base 10) of 0.0070, we get -2.15 (rounded to 2 decimal places).
The molar mass of monopotassium phosphate is 136.09 g/mol. How many moles of monopotassium phosphate are needed to make 250.0 ml of 0.15 M monopotassium phosphate?
Answer:
0.0375 moles
Explanation:
Given that:
Molar mass of monopotassium phosphate = 136.09 g/mol
Given that volume = 250.0 mL
Also,
[tex]1\ mL=10^{-3}\ L[/tex]
So, Volume = 250 / 1000 L = 0.25 L
Molarity = 0.15 M
Considering:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
[tex]0.15=\frac{Moles\ of\ solute}{0.25}[/tex]
Thus, moles of monopotassium phosphate needed = 0.0375 moles
7. Aspirin has a pKa of 3.4. What is the ratio of A to HA in: (a) the blood (pH 7.4) (b) the stomach (pH 1.4) =
Answer : a) [tex]\frac{[A^-]}{[HA]}=10^{4}[/tex]
b) [tex]\frac{[A^-]}{[HA]}=0.01[/tex]
Explanation : Given,
a) pH = 7.4
[tex]pK_a[/tex] for aspirin = 3.4
Using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]
Concentration of salt [tex](A^-)[/tex] = ?
Concentration of acid [tex](HA)[/tex] = ?
[tex]pH=pK_a+\log \frac{[A^-]}{[HA]}[/tex]
Now put all the given values in this expression, we get:
[tex]7.4=3.4+\log \frac{[A^-]}{[HA]}[/tex]
[tex]\frac{[A^-]}{[HA]}=10^{4}[/tex]
b) pH = 1.4
Using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[A^-]}{[HA]}[/tex]
Now put all the given values in this expression, we get:
[tex]1.4=3.4+\log \frac{[A^-]}{[HA]}[/tex]
[tex]\frac{[A^-]}{[HA]}=0.01[/tex]
The pKa is the acid dissociation constant. The ratio of A to HA in the blood is, 10000, and in the stomach is 0.01.
What is the acid dissociation constant?The acid dissociation constant or pKa has been the dissociation of the compound into ions at specified pH.
The relation between pKa and pH can be given as:
[tex]\rm pH=pKa\;+\;log\;\dfrac{acid}{salt}[/tex]
The pKa of aspirin is 3.4.
The ratio of acid (A) to salt (HA) in blood is given as:[tex]\rm 7.4=3.4\;+\;log\;\dfrac{A}{HA} \\\dfrac{A}{HA}=10000[/tex]
The ratio of acid (A) to salt (HA) in the stomach is given as:[tex]\rm 7.4=1.4\;+\;log\;\dfrac{A}{HA} \\\dfrac{A}{HA}=0.01[/tex]
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Liquids are agitated for a number of purposes, depending on the objectives of the process. Name at least four
Answer:
To dissolve a solid, for example, adding sugar to water agitating until it completely dissolves.To promote heat transference.To blend two miscible liquids such as water and alcohol.To disperse a solid in a liquid to form a suspension.I hope you find these examples useful! good luck!
Define "Mutarotation"
Answer:
Mutarotation refers to the change in the optical rotation or optical activity of a solution due to the change in the equilibrium of the two anomers. It depends upon the optical activity and ratio of the anomeric forms in the solution.
To measure the optical rotation of a given solution, a polarimeter can be used and thus the ratio of the anomeric forms can be calculated.
Which of the following statements is not true regarding molecular orbital theory? O Bonding molecular orbitals have lower energy than the atomic orbitals from which they are formed. O Molecular orbitals are formed by the combination of atomic orbitals. Two atomic orbitals combine to form one molecular orbital. O Molecular orbitals are regions of a molecule where the electrons are most likely to be found.
Answer:
Option c, Two atomic orbitals combine to form one molecular orbital
Explanation:
Molecular orbitals are formed by linear combination of atomic orbitals.
Some of the important facts of molecular orbital theories are as follows:
No. of the molecular orbitals formed are equal to the no. of atomic orbitals participated.Half of the molecular orbitals are bonding molecular orbitals and half of the molecular orbitals are anti bonding molecular orbitals.Anti bonding molecular orbitals have energy higher than participating atomic orbitals. Bonding molecular orbitals have energy lower than participating atomic orbitals.Molecular orbitals are that region in the molecule where electrons are most likely to found.So, among given, option c which is 'atomic orbitals combine to form one molecular orbital' is incorrect.
All the statements provided are true with regards to Molecular Orbital Theory, except for 'Two atomic orbitals combine to form one molecular orbital.' In fact, two atomic orbitals combine to form two molecular orbitals, one being a lower-energy bonding molecular orbital and the other one being a higher-energy antibonding orbital.
Explanation:In the context of Molecular Orbital Theory, three out of the four statements made in the question are correct. Namely, molecular orbitals are indeed formed by the combination of atomic orbitals, and they represent regions of a molecule where electrons are most likely to be found. Additionally, bonding molecular orbitals do have lower energy than the atomic orbitals from which they are formed.
However, the statement suggesting that 'Two atomic orbitals combine to form one molecular orbital' is not accurate. In reality, when two atomic orbitals combine, they form two molecular orbitals: one lower-energy (bonding) molecular orbital and one higher-energy (antibonding) orbital.
This nuanced concept is a cornerstone of MO theory and helps explain a range of phenomena, from the paramagnetism of the oxygen molecule to violations of the octet rule. Understanding this allows us to further appreciate the differences between bonding and antibonding orbitals and their roles in the structure and stability of molecules.
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How does the energy of the activated complex compare with the energies of reactants an products? a. It is lower than the energy of both reactants and products. b. It is lower than the energy of reactants but higher than the energy of products. c. It is higher than the energy of reactants but lower than the energy of products. d. It is higher than the energy of both reactants and products.
Answer:
Option d, It is higher than the energy of both reactants and products.
Explanation:
When reactant molecules collide with each other during the course of the reaction, an intermediate state is reached before the formation of product. This intermediate state is also called activated complex.
Activated complex consist energy higher than that of both the reactants and products. Because of possessing higher energy, it is unstable and temporary.
Hence, among given, option d is correct.
Answer:
d. It is higher than the energy of both reactants and products.
Explanation:
The activated complex is a transient state, or intermediate phase, between reagents (weak or not), in which the final product has not yet been formed. This effective shock, like any chemical bond, needs energy. To form this transient state requires an activation energy, a minimum energy for this intermediate phase to occur.
Within this phenomenon, the activation energy must occur by some criteria, which characterizes a chemical bond in an activated complex:
Binding reagent molecules need to be involved; This collision in the activation energy must be as close as possible to the geometric formation of the complex in order to activate it; The collision cannot have a lower energy than the activation energy. Must be equal to or greater than propagated during reaction.Within these three rules, it is clearer to define this complex as the moment when a molecule of a given atom collides with another molecule of any other atom and breaks the bond established between them.
For it to happen, the energy needs to be high. When this potential energy is present at a high level during the reaction, it is also necessary to have a high energy charge to complete the complex and also the collision of the reactant molecules to form the binding products.
What is the effect on the outlet temperatures in both the hot and cold streams, when the surface area (length and diameter) is reduced?
a- The outlet temperature of the cold stream decreases and the outlet temperature of the hot stream increases.
b- Both the outlet temperature increases
c- The outlet temperature of cold fluid increases and the outlet temperature of hot fluid decreases
Answer:
The right answer is option A: The outlet temperature of the cold stream decreases and the outlet temperature of the hot stream increases.
Explanation:
In a heat exchanger, when the exchange area is reduced, and at the same condition for the rest of the variables, the amount of heat exchanged by the two fluids is reduced.
In the case of the hot fluid, it will rise at a higher temperature, compared to the temperature at which it would come out with a larger exchange area.
In the case of the hot fluid, it will come out at a lower temperature, compared to the temperature at which it would come out with a larger exchange area.
The correct answer is option A.
Answer:
a- The outlet temperature of the cold stream decreases and the outlet temperature of the hot stream increases.
Explanation:
Hello,
In heat transfer processes, specific units such as heat exchangers are useful to either heat up a cold stream with a hot one or vise-versa. However, the heat transfer efficiency increases as surface area increases, it means that the higher the surface area the higher the heat exchanger's efficiency, therefore, if the surface area is reduced, the efficiency is reduced as well, due to the lack of more space for the heat transfer.
In such a way, the outer temperature of the cold fluid will be lower (less heating up) and the outer one of the hot fluid will be higher (less cooling down), thus, the outlet temperature of the cold stream decreases and the outlet temperature of the hot stream increases (a option).
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The solubility constant of AgBr is 5.0 x 10-13. The formation constant of the silver(I) ammonia complex, Ag(NH3)2, is 1.6 x 107 Calculate the solubility, in moles/L, of AgBr in 1.00 Mammonia solution.
Answer:
S AgBr = 2.82 E-3 mol/L
Explanation:
AgBr ↔ Ag+ + Br- .....(1)∴ Ksp = [ Ag+ ] * [ Br- ] = 5.0 E-13
Ag+ + 2NH3 ↔ [ Ag(NH3)2 ]+ ..........(2)∴ Kf = 1.6 E7 = α[Ag(NH3)2]+ / ( αAg+ )*( αNH3 )²
∴ C NH3(sln) = 1.00 M
from (1) + (2):
AgBr(s) + 2NH3(aq) ↔ Ag(NH3)2+(aq) + Br-(aq)1 M 0 0
1 - 2x x x
∴ K = Ksp*Kf = ( 5.0 E-13 )*( 1.6 E7 ) = 8.0 E-6
⇒ K = ( [ Br- ] * [ Ag(NH3)2+] ) / [ NH3 ]² = 8.0 E-6
⇒ K = (( x )*( x )) / ( 1 - 2x ) = 8.0 E-6
⇒ x² = 8.0 E-6*( 1- 2x )
⇒ x² + 1.6 E-5x - 8.0 E-6 = 0
⇒ x = 2.82 E-3 M
⇒ [ Br- ] = [ AgBr ] = 2.82 E-3 M
To calculate the solubility of AgBr in a 1.00 M ammonia solution, we can use the equation for the formation constant of the silver(I) ammonia complex.
Explanation:The solubility constant of AgBr is 5.0 x 10-13. The formation constant of the silver(I) ammonia complex, Ag(NH₃)₂, is 1.6 x 107. To calculate the solubility of AgBr in a 1.00 M ammonia solution, we can use the equation:
AgBr(s) → Ag+(aq) + Br-(aq)
Let x be the concentration of AgBr(s). Then, the equilibrium concentrations of Ag+ and Br- would be x and x, respectively.
Using the equation for the formation constant:
Kf = [Ag(NH₃)₂]+ / [Ag+][NH₃]₂
we can substitute the given values and solve for [Ag+] and [NH₃].
For light with a frequency of 5.371 x 10^14 s^-1, what is the energy (in J)?
Answer:
The energy is 3.559 * 10⁻¹⁹ J
Explanation:
For this problem we need to use Planck's equation, this equation describes the relationship between energy radiated by light (a photon, to be precise) and its frequency. The equation is:
E = h * v
Where E is energy (in Joules), h is Planck's constant (6.626 * 10⁻³⁴ J·s), and v is the frequency (in s⁻¹).
Now we solve the equation, using the data given in the problem:
E = 6.626 * 10⁻³⁴ J·s * 5.371 * 10¹⁴ s⁻¹ = 3.559 * 10⁻⁴⁷ J
Thus the energy is 3.559 * 10⁻¹⁹ J
Determine the Darcy friction factor for a fluid flowing in a pipe with diameter 0.01 metres, flowing at 0.03 m/s. The fluid density is 36.4 kg/m and viscosity is 0.651 Pa*s. The pipe absolute roughness is 0.00062459 m. Give your answer rounded to the nearest whole number.
Explanation:
The given data is as follows.
Diameter of pipe, D = 0.01 m
Velocity V = 0.03 m/s
Fluid density = 36.4 [tex]kg/m^{3}[/tex]
Viscosity = 0.651 Pa-s
Formula to calculate Reynold number is as follows.
Re = [tex]diameter \times velocity \times \frac{density}{viscosity}[/tex]
= [tex]0.01 \times 0.03 \times \frac{36.4}{0.651}[/tex]
= 0.01677
Since, Re < 2100. This means that the flow is laminar.
So, for laminar flow in pipes,
Darcy friction factor f = [tex]\frac{64}{Re}[/tex]
= [tex]\frac{64}{0.01677}[/tex]
= 3816
Thus, we can conclude that the Darcy friction factor value is 3816.
explain why sodium (metal) is soft and can be bend, whereas NaCl is hard and brittle.
Answer:
Sodium chloride has a crystalline face-centered cubic structure whereas metallic sodium body-centered cubic structure.
Explanation:
Hello, atomic arrangements provide the molecules different features and behaviors, since the sodium metal has a body-centered cubic structure (https://en.wikipedia.org/wiki/Cubic_crystal_system#/media/File:Cubic-body-centered.svg) the lack of inner atoms, allows the material to be soft and bendable. On the other hand the compacted sodium chloride's face-centered cubic structure (https://en.wikipedia.org/wiki/Cubic_crystal_system#/media/File:Cubic-face-centered.svg), provides it a crystalline structure which is hard and brittle since the atoms are closer.
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Give the ΔH value for the combustion of butane as shown in the reaction 2C4H10(g)+13O2(g)→8CO2(g)+10H2O(g)+5315 kJ.
Express your answer using four significant figures. If the value is positive, do not include the + sign in your answer.
The ΔH value for the combustion of butane is given as 5315 kJ for the reaction as written, which is for 2 moles of butane. To find the enthalpy of combustion per mole, divide this number by 2, yielding -2658 kJ/mol.
Explanation:The student asked to give the ΔH value for the combustion of butane as shown in the reaction 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g) + 5315 kJ. The ΔH value for the reaction is given as +5315 kJ, indicating the amount of energy released during the combustion of butane.
Now, the enthalpy of combustion is typically expressed on a per mole basis, referring to the amount of heat released when one mole of a substance is burned. Since the reaction above shows the combustion for 2 moles of butane, we can calculate the enthalpy of combustion per mole by dividing the total energy by 2. So, the enthalpy of combustion for one mole of butane would be 5315 kJ / 2 = 2657.5 kJ/mol, which is typically reported in kilojoules per mole (kJ/mol).
Expressed with four significant figures, the enthalpy of combustion per mole of butane would be -2658 kJ/mol (the negative sign indicates that energy is released).
Our system is the liquid water contained in a bath tub. The drain is open at the bottom while water is being poured into it at an equivalent rate so that the water level within the bathtub is not rising. Is this process steady-state?
Answer:
The process described in the problem is steady-state.
Explanation:
The steady-state can be described by the conservation of mass. In a bathtub it will be the same as In=Out+ Accumulation. Since the water has already reached the adequate level and the drain is open, the accumulation is equal to zero, and therefore it is at steady state. This also means that In = Out which is also described in the problem: "the water is poured into the bathtub at an equivalent rate of the water being drained".
A gas mixture of 20 moles nitrogen and 70 moles hydrogen is fed to a reactor in which ammonia is produced. Assuming that the reaction proceeds to completion, determine the composition of the exit gas stream both in mole and mass %. If 75% of nitrogen gets converted, what will be the corresponding fractions?
Answer:
Complete reaction
mole %: 0% N₂; 20% H₂; 80% NH₃
Gram %: 0% N₂; 2,9% H₂; 97,1% NH₃
75% reaction:
mole%: 8,3% N₂; 41,7% H₂; 50% NH₃
Gram: 20% N₂; 7,2% H₂; 72,8% NH₃
Explanation:
The global reaction in the reactor is:
N₂(g) + 3 H₂ (g) → 2 NH₃ (g)
For a complete reaction of 20 moles of N₂(g) you need:
20 moles N₂ ₓ [tex]\frac{3 mol H_2}{1 mol N_2}[/tex] = 60 moles of H₂(g)
So, 10 moles of H₂(g) will stay in the end.
The moles produced of NH₃ are:
20 moles N₂ ₓ [tex]\frac{2 mol NH_3}{1 mol N_2}[/tex] = 40 moles of NH₃(g)
Thus, the final composition in moles is:
0 moles N₂; 10 moles H₂; 40 moles of NH₃
The mole % is:
0% N₂; ¹⁰/₅₀ ₓ 100 = 20% H₂; ⁴⁰/₅₀ ₓ100 = 80% NH₃
In grams you have:
10 moles H₂ [tex]\frac{2,02 g}{1 mol}[/tex] = 20,2 g of H₂
40 moles NH₃ [tex]\frac{17,03 g}{1 mol}[/tex] = 681,2 g of NH₃
0% N₂; [tex]\frac{20,2 g}{701,4}[/tex] ₓ 100 = 2,9% H₂; [tex]\frac{681,2 g}{701,4}[/tex] ₓ100 = 97,1% NH₃
With a 75% of conversion you have that the moles produced of NH₃ are:
40 moles of NH₃(g) × 75% = 30 moles of NH₃
The necessary moles to produce these moles of NH₃ are:
30 moles NH₃ ₓ [tex]\frac{3 mol H_2}{2 mol NH_3}[/tex] = 45 moles of H₂(g)
So, 25 moles of H₂(g) will stay in the end.
30 moles NH₃ ₓ [tex]\frac{1 mol N_2}{2 mol NH_3}[/tex] = 15 moles of N₂(g)
So, 5 moles of H₂(g) will stay in the end.
Thus, the final composition in moles is:
5 moles N₂; 25 moles H₂; 30 moles of NH₃
The mole % is:
⁵/₆₀ ₓ 100 =8,3% N₂; ²⁵/₆₀ ₓ 100 = 41,7% H₂; ³⁰/₆₀ ₓ100 = 50% NH₃
In grams you have:
5 moles N₂ [tex]\frac{28 g}{1 mol}[/tex] = 140 g of N₂
25 moles H₂ [tex]\frac{2,02 g}{1 mol}[/tex] = 50,5 g of H₂
30 moles NH₃ [tex]\frac{17,03 g}{1 mol}[/tex] = 510,9 g of NH₃
[tex]\frac{140 g}{701,4}[/tex] ₓ 100 =20% N₂; [tex]\frac{ 50,5g}{701,4}[/tex] ₓ 100 = 7,2% H₂; [tex]\frac{510,9 g}{701,4}[/tex] ₓ100 = 72,8% NH₃
I hope it helps!
What is the value for the potential energyfor a n = 6 Bohr orbit electron in Joules?
Answer:
Potential energy for n = 6 Bohr orbit electron is -1.21*10⁻¹⁹J
Explanation:
As per the Bohr model, the potential energy of electron in an nth orbit is given as:
[tex]PE_{n} = -\frac{kZe^{2}}{r_{n}}[/tex]
here:
k = Coulomb's constant = 9*10⁹ Nm2/C2
Z = nuclear charge
e = electron charge = 1.6*10⁻¹⁹ C
r(n) = radius of the nth orbit = n²(5.29*10⁻¹¹m)
Substituting for k, Z(= 1), e and r(n) in the above equation gives:
[tex]PE_{6} = -\frac{9*10^{9}Nm2/C2*1*(1.6*10^{-19}C)^{2}}{(6)^{2}*5.29*10^{-11}m}=-1.21*10^{-19}J[/tex]
A chemist prepares a solution of nickel(II) chloride (NiCly) by measuring out 61.3 umol of nickel(II) chloride into a 200 ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in umol/L of the chemist's nickel(II) chloride solution. Round your answer to 3 significant digits. x 5 ? Explanation Check 2009 Mew Econ Aires Reserved. Terms of Use
The concentration of the nickel(II) chloride solution is determined by dividing the amount of substance (in umol) by the volume of the solution (in liters). This results in a concentration of 306.5 umol/L.
Explanation:The concentration of a solution is determined by the formula c = n/V, where 'c' is the concentration, 'n' is the amount of substance (in moles), and 'V' is the volume of the solution (in liters).
In this case, the chemist has used 61.3 umol of nickel(II) chloride (NiCl2), and the total volume of the solution is 200 ml. We need to convert this volume to liters, giving us 0.2 L. Now, using the formula, we calculate: c = 61.3 umol / 0.2 L, which equals 306.5 umol/L.
Make sure to always convert the units appropriately before you substitute the values into the formula.
Learn more about Solution Concentration here:https://brainly.com/question/10720472
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A heparin drip (IV) has been ordered that contains 10,000 units of heparin in 1000ml of D5W (5% dextrose in water). The drip is set to run at 50 ml/hr. How many units of heparin does the patient receive in 24 hrs? (Dimensional analysis)
Answer:
12,000 units of heparin are received by patient in 24 hours.
Explanation:
10,000 units of heparin in 1000ml of D5W.
Units of heparin in 1 mL = [tex]\frac{10,000}{1000}=10 units[/tex]
Rate at heparin drip set at = 50 ml/hour
Units of heparin in 50 ml of solution = [tex]50\teimes 10 units =500 units[/tex]
500 units of heparin are received by patient in an every hour.So, in 24 hours patient will receive:
[tex]500units\times 24 = 12,000 units[/tex] of heparin
12,000 units of heparin are received by patient in 24 hours.
A solution is prepared by dissolving 20 g NaOH (FW = 400 g/mol) to 255 ml of solution. If the density of the solution is 1.15 gim, what is the mass percent NaOH in the solution? 0.68% 0.78% 0.90% 6.8% 78% nt Navigator
Answer:
6.82% is the mass percent NaOH in the solution.
Explanation:
Mass of the solute that is NaOH = m = 20 g
Mass of the solution = M
Volume of the solution = V = 255 mL
Density of the solution = d = 1.15 g/mL
[tex]Density=\frac{Mass}{Volume}[/tex]
[tex]1.15 g/mL=\frac{M}{255 mL}[/tex]
M = 293.25 g
Mass percentage of solute :(w/w)%
[tex]\frac{m}{M}\times 100[/tex]
[tex](w/w)\%=\frac{20 g}{293.25 g}\times 100=6.82\%[/tex]
6.82% is the mass percent NaOH in the solution.
Write 0.000000093425 in Scientific Notation with 4 significant figures.
Answer: The given number in scientific notation is [tex]9.342\times 10^{-8}[/tex]
Explanation:
Scientific notation is the notation where a number is expressed in the decimal form. This means that the number is always written in the power of 10 form. The numerical digit lies between 0.1.... to 9.9.....
If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.
We are given:
A number having value = 0.000000093425
Converting this into scientific notation, we get:
[tex]\Rightarrow 0.000000093425=9.342\times 10^{-8}[/tex]
Hence, the given number in scientific notation is [tex]9.342\times 10^{-8}[/tex]
Answer:
[tex]9.343\times 10^{-8}[/tex]
Explanation:
Scientific notation is the way of writing numbers which are either large or small. The number is written in the scientific notation when the number is between 1 and 10 and then multiplied by the power of 10.
The given number:
0.000000093425 can be written as [tex]9.3425\times 10^{-8}[/tex]
Answer upto 4 significant digits = [tex]9.343\times 10^{-8}[/tex]