Answer :
[tex]Cl_2[/tex] is reduced species.
[tex]KI[/tex] is oxidized species.
[tex]Cl_2[/tex] is oxidizing agent.
[tex]KI[/tex] is reducing agent.
Explanation :
Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.
Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.
Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.
Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.
Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.
The balanced redox reaction is :
[tex]Cl_2(aq)+2KI(aq)\rightarrow 2KCl(aq)+I_2(aq)[/tex]
The half oxidation-reduction reactions are:
Oxidation reaction : [tex]2I^-\rightarrow I_2+2e^-[/tex]
Reduction reaction : [tex]Cl_2^++2e^-\rightarrow 2Cl^-[/tex]
From this we conclude that the [tex]'KI'[/tex] is the reducing agent that loses an electron to another chemical species in a redox chemical reaction and itself gets oxidized and [tex]'Cl_2'[/tex] is the oxidizing agent that gain an electron to another chemical species in a redox chemical reaction and itself gets reduced.
Thus, [tex]Cl_2[/tex] is reduced species.
[tex]KI[/tex] is oxidized species.
[tex]Cl_2[/tex] is oxidizing agent.
[tex]KI[/tex] is reducing agent.
Iodine from potassium iodide (KI) is oxidized, and chlorine from chlorine gas (Cl2) is reduced; Cl2 is the oxidizing agent and KI is the reducing agent in the reaction.
Explanation:In the reaction Cl2(aq) + 2 KI(aq) → 2 KCl(aq) + I2(aq), the oxidized species is iodine (I) from potassium iodide (KI), and the reduced species is chlorine (Cl) from chlorine gas (Cl2). During the reaction, I- loses electrons and is therefore oxidized to I2; its oxidation number changes from -1 to 0. Conversely, each Cl atom in Cl2 gains one electron to become Cl-, going from an oxidation state of 0 to -1. Thus, the oxidizing agent is Cl2 because it accepts electrons (is reduced), and the reducing agent is KI because it donates electrons (is oxidized).
Melamine, C3N3(NH2)3, is used in adhesives and resins. It is manufactured in a two-step process: CO(NH2)2(l) → HNCO(l) + NH3(g) 6 HNCO(l) → C3N3(NH2)3(l) + 3 CO2(g) What mass of melamine, C3N3(NH2)3, will be obtained from 161.2 kg of urea, CO(NH2)2, if the yield of the overall reaction is 76.5 % ?
Answer:
43.13Kg of melamine
Explanation:
The problem gives you the mass of urea and two balanced equations:[tex]CO(NH_{2})_{2}_{(l)}=HNCO_{(l)}+NH_{3}_{(g)}[/tex]
[tex]6HNO_{l}=C_{3}N_{3}(NH_{2})_{3}_{(l)}+3CO_{2}_{(g)}[/tex]
First we need to calculate the number of moles of urea that are used in the reaction, so:
molar mass of urea = [tex]60.06\frac{g}{mol}*\frac{1kg}{1000g}=0.06006\frac{Kg}{mol}[/tex]
The problem says that you have 161.2Kg of urea, so you take that mass of urea and find the moles of urea:
161.2Kg of urea[tex]*\frac{1molofurea}{0.06006Kgofurea}=[/tex]2684 moles of urea
Now from the stoichiometry you have:
2684 moles of urea[tex]*\frac{1molofHNCO}{1molurea}*\frac{1molofmelamine}{6molesofHNCO}[/tex] = 447 moles of melamine
The molar mass of the melamine is [tex]126.12\frac{g}{mol}[/tex] so we have:
[tex]447molesofmelamine*\frac{126.12g}{1molofmelamine}[/tex] = 5637.64 g of melamine
Converting that mass of melamine to Kg:
5637.64 g of melamine *[tex]\frac{1Kg}{1000g}[/tex] = 56.38 Kg of melamine, that is the theoretical yield of melamine.
Finally we need to calculate the mass of melamine with a yield of 76.5%, so we have:
%yield = 100*(Actual yield of melamine / Theoretical yield of melamine)
Actual yield of melamine = [tex]\frac{76.5}{100}*56.38Kg[/tex] = 43.13Kg of melamine
Using stoichiometry, the actual yield of Melamine from a reaction starting with 161.2 kg of urea with an overall yield of 76.5% can be found to be approximately 43.2 kg.
Explanation:The question asks about the actual yield of Melamine, C3N3(NH2)3, from a reaction starting with 161.2 kg of urea, CO(NH2)2, with an overall yield of 76.5%. To understand this prediction, we must use the concept of stoichiometry, which is a method used in chemistry to calculate the quantities of reactants or products in a chemical reaction.
First, calculate the mole of urea used: it is the mass of the urea divided by the molar mass of urea. The molar mass of urea (CO(NH2)2) is approximately 60 g/mole. So, you have 161.2*1000/60 = approximately 2686.67 moles of urea.
From the balanced chemical equation, we can see that one mole of urea produces one mole of HNCO, and six moles of HNCO produce one mole of Melamine. Therefore, in theory, 2686.67/6 = 447.78 moles of Melamine can be produced.
However, the reaction yield is only 76.5%, so the actual yield of the Melamine will be less. Using the percentage yield, we can calculate the actual yield: 447.78 * 0.765 = approximately 342.65 moles.
Finally, to convert from moles to mass, multiply by the molar mass of melamine, which is approximately 126 g/mole: 342.65*126 = approximately 43173.9g or 43.2 kg.
Learn more about Chemical Reaction Yield here:https://brainly.com/question/30762778
#SPJ3
Consider the reaction of peroxydisulfate ion (S2O2−8) with iodide ion (I−) in aqueous solution: S2O2−8(aq)+3I−(aq)→2SO2−4(aq)+I−3(aq). At a particular temperature the rate of disappearance of S2O2−8 varies with reactant concentrations in the following manner: Experiment S2O2−8(M) I−(M) Initial Rate (M/s) 1 0.018 0.036 2.6×10−6 2 0.027 0.036 3.9×10−6 3 0.036 0.054 7.8×10−6 4 0.050 0.072 1.4×10−5 What is the rate of disappearance of I− when [S2O2−8]= 1.8×10−2 M and [I−]= 5.0×10−2 M ?
The rate law for the reaction is rate = k[S2O8²⁻], and the rate of disappearance of I- can be determined using the stoichiometric coefficients.
Explanation:The rate law for the reaction of peroxydisulfate ion (S2O8²⁻) with iodide ion (I⁻) can be determined by using the initial rates data. In the given data, we can see that the initial rate varies linearly with the concentration of S2O8²⁻ and is independent of the concentration of I⁻. This means that the rate law for this reaction is rate = k[S2O8²⁻].
Using the given rate law, we can determine the rate of disappearance of I⁻ when [S2O8²⁻] = 1.8×10⁻²M and [I⁻] = 5.0×10⁻²M. Since the stoichiometric coefficient of I⁻ in the balanced chemical equation is 3, the rate of disappearance of I⁻ is 3 times the rate of appearance of S2O8²⁻. Therefore, the rate of disappearance of I⁻ is (3)(1.4×10⁻⁵M/s) = 4.2×10⁻⁵M/s.
Learn more about rate law here:https://brainly.com/question/35884538
#SPJ3
A torsion balance has a sensitivity requirement (SR) of 4.5 mg. What is the MWQ of this balance if the maximum error permitted in using it is 3.6%?
Explanation:
MWQ means the minimum weighable quantity.
Mathematically, MWQ = [tex]\frac{Sensitivity}{1 - \text{fraction of accuracy}}[/tex]
or, MWQ = [tex]\frac{sensitivity}{\text{fractional error}}[/tex]
It is given that sensitivity is 4.5 mg and maximum permitted error is 3.6%.
Therefore, fraction error = [tex]\frac{3.6}{100}[/tex] = 0.036
Hence, we will calculate MWQ as follows.
MWQ = [tex]\frac{sensitivity}{\text{fractional error}}[/tex]
= [tex]\frac{4.5 mg}{0.036}[/tex]
= 125 mg
Thus, we can conclude that the MWQ of the given balance is 125 mg.
Define the following: Bronsted-Lowry acid - Lewis acid- Strong acid - (5 points) Problem 6: Consider the following acid base reaction HCI + H20 → H30+ + Cl- a) Is this a strong acid? b) Clearly label the acid, base, conjugate acid and conjugate base. (5 points)
Answer: Yes, [tex]HCl[/tex] is a strong acid.
acid = [tex]HCl[/tex] , conjugate base = [tex]Cl^-[/tex] , base = [tex]H_2O[/tex], conjugate acid = [tex]H_3O^+[/tex]
Explanation:
According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.
Yes [tex]HCl[/tex] is a strong acid as it completely dissociates in water to give [tex]H^+[/tex] ions.
[tex]HCl\rightarrow H^++Cl^-[/tex]
For the given chemical equation:
[tex]HCl+H_2O\rightarrow H_3O^-+Cl^-[/tex]
Here, [tex]HCl[/tex] is loosing a proton, thus it is considered as an acid and after losing a proton, it forms [tex]Cl^-[/tex] which is a conjugate base.
And, [tex]H_2O[/tex] is gaining a proton, thus it is considered as a base and after gaining a proton, it forms [tex]H_3O^+[/tex] which is a conjugate acid.
Thus acid = [tex]HCl[/tex]
conjugate base = [tex]Cl^-[/tex]
base = [tex]H_2O[/tex]
conjugate acid = [tex]H_3O^+[/tex].
How do i estimate the density of a solution of two compounds, when im given wt% and the specific gravity?
Answer:
[tex]\rho_s=\frac{\rho_w}{\frac{wt1}{SpGr1}+\frac{wt2}{SpGr2}}[/tex]
The only other data you need is the density of water ρw.
Explanation:
We can start by the volume balance
[tex]V_s=V_1+V_2[/tex]
We can replace the volumes with V=M/ρ
[tex]\frac{M_s}{\rho_s}=\frac{M_1}{\rho_1}+\frac{M_2}{\rho_2}[/tex]
If we divide every term by Ms
[tex]\frac{M_s/M_s}{\rho_s}=\frac{M_1/M_s}{\rho_1}+\frac{M_2/M_s}{\rho_2}[/tex]
By definition, wt=Mi/Msol, so we can replace that in the expression
[tex]\frac{1}{\rho_s}=\frac{wt1}{\rho_1}+\frac{wt2}{\rho_2}[/tex]
Then we have the expression of the density of the solution
[tex]\rho_s=\frac{1}{\frac{wt1}{\rho_1}+\frac{wt2}{\rho_2}}[/tex]
To replace ρ1 and ρ2, you have to multiply the specific gravity of the components and the density of water.
[tex]\rho_s=\frac{1}{\frac{wt1}{SpGr1\rho_w}+\frac{wt2}{SpGr2\rho_w}}\\\\\rho_s=\frac{\rho_w}{\frac{wt1}{SpGr1}+\frac{wt2}{SpGr2}}[/tex]
Given the densities of the following pure liquids, what volume of each is necessary to make 250 mL of a 350 mM solution? a. Ethanol - 0.789 g/cm? b. Acetone -0.791 g/cm c. Formic acid - 1.220 g/cm3 d. tert-Butylamine - 0.696 g/cm?
Answer :
(a) The volume of ethanol liquid needed are, 2.4 L
(b) The volume of ethanol liquid needed are, 1.9 L
(c) The volume of ethanol liquid needed are, 1.6 L
(d) The volume of ethanol liquid needed are, 1.7 L
Explanation : Given,
Volume of solution = 250 mL = 0.250 L (1 l = 1000 mL)
Concentration of solution = 350 mM = 0.350 M (1 mM = 0.001 M)
First we have to calculate the moles of solution.
[tex]\text{Moles of solution}=\text{Concentration of solution}\times \text{Volume of solution}=0.350M\times 0.250L=0.0875mole[/tex]
(a) For ethanol liquid :
To we have to calculate the mass of ethanol.
[tex]\text{Mass of ethanol}=\frac{\text{Moles}}{\text{Molar mass of ethanol}}=\frac{0.0875mole}{46g/mole}=0.0019g[/tex]
Now we have to calculate the volume of ethanol.
[tex]Volume=\frac{Mass}{Density}=\frac{0.0019g}{0.789g/cm^3}=0.0024cm^3=0.0024mL=2.4L[/tex]
(b) For acetone liquid :
To we have to calculate the mass of acetone.
[tex]\text{Mass of acetone}=\frac{\text{Moles}}{\text{Molar mass of acetone}}=\frac{0.0875mole}{58g/mole}=0.0015g[/tex]
Now we have to calculate the volume of acetone.
[tex]Volume=\frac{Mass}{Density}=\frac{0.0015g}{0.791g/cm^3}=0.0019cm^3=0.0019mL=1.9L[/tex]
(c) For formic acid liquid :
To we have to calculate the mass of formic acid.
[tex]\text{Mass of formic acid}=\frac{\text{Moles}}{\text{Molar mass of formic acid}}=\frac{0.0875mole}{46g/mole}=0.0019g[/tex]
Now we have to calculate the volume of formic acid.
[tex]Volume=\frac{Mass}{Density}=\frac{0.0019g}{1.220g/cm^3}=0.0016cm^3=0.0016mL=1.6L[/tex]
(d) For tert-Butylamine liquid :
To we have to calculate the mass of tert-Butylamine.
[tex]\text{Mass of tert-Butylamine}=\frac{\text{Moles}}{\text{Molar mass of tert-Butylamine}}=\frac{0.0875mole}{73g/mole}=0.0012g[/tex]
Now we have to calculate the volume of tert-Butylamine.
[tex]Volume=\frac{Mass}{Density}=\frac{0.0012g}{0.696g/cm^3}=0.0017cm^3=0.0017mL=1.7L[/tex]
To determine the volume of each pure liquid necessary to make a 350 mM solution with a volume of 250 mL, calculate the required mass of each substance and then use the density to find the volume.
Explanation:To determine the volume of each pure liquid necessary to make a 350 mM solution with a volume of 250 mL, we need to use the formula:
Volume (mL) = (mass (g) / density (g/cm³)) * 1000
For each pure liquid, calculate the mass of the substance needed by multiplying its molarity with the desired volume in moles:
Mass (g) = molarity (mol/L) * volume (L) * molar mass (g/mol)
Then, use the formula for volume to find the necessary volume of each pure liquid:
Volume (mL) = (mass (g) / density (g/cm³)) * 1000
A balloon filled with helium gas at 1.00 atm occupies 12.9 L. What volume would the balloon occupy in the upper atmosphere, at a pressure of 0.19 atm and a constant temperature? Enter your answer in the box provided.
Answer:
67,9 L
Explanation:
Boyle's Law indicates that the pressure of a fixed amount of gas at a constant temperature is inversely proportional to the volume of a gas, for a constant amount of gas we can write:
P1V1=P2V2
For the problem:
P1= 1 atm, V1= 12,9 L
P2=0,19 atm, V2=?
Therefore:
V2=P1V1/P2.................... V2=1 atm*12,9L/0,19 atm = 67,9 L
The balloon would occupy a volume of 67,9 L in the upper atmosphere.
A sailor circumnavigated the earth and covered 4.264,000 meters. Express this number in standard scientific notation A) 4.264 . 10-6 m B ) 4.264 . 10 m C) 4.264 . 106m D) 4.264 . 10-7m
Answer:
[tex]4.264*10^{6}[/tex]
Explanation:
Hello !
The correct answer is option C) [tex]4.264*10^{6}[/tex]
To write in scientific notation, numbers between 0 and 9 are used before the point and the rest of the numbers afterwards.
Then we count how many places we ran the point. If you run to the right, the exponent of 10 is positive and if you run to the left, it is negative.
This way it is [tex]4.264*10^{6}[/tex]
In 2006, a Russian team discovered an interesting molecule they called "sulflower" because of its shape and because it was based on sulfur. It is composed of 57.17% S and 42.83% C and has a molar mass of 448.70 g/mol. Determine the empirical and molecular formulas of "sulflower."
Answer: The empirical and molecular formula for the given organic compound is [tex]C_2S[/tex] and [tex]C_{16}S_8[/tex] respectively.
Explanation:
We are given:
Percentage of C = 42.83 %
Percentage of S = 57.17 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of C = 42.83 g
Mass of S = 57.17 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.83g}{12g/mole}=3.57moles[/tex]
Moles of Sulfur = [tex]\frac{\text{Given mass of Sulfur}}{\text{Molar mass of Sulfur}}=\frac{57.17g}{32g/mole}=1.79moles[/tex]
Step 2: Calculating the mole ratio of the given elements.For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.79 moles.
For Carbon = [tex]\frac{3.57}{1.79}=1.99\approx 2[/tex]
For Sulfur = [tex]\frac{1.79}{1.79}=1[/tex]
Step 3: Taking the mole ratio as their subscripts.The ratio of C : S = 2 : 1
Hence, the empirical formula for the given compound is [tex]C_2S_1=C_2S[/tex]
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is:
[tex]n=\frac{\text{molecular mass}}{\text{empirical mass}}[/tex]
We are given:
Mass of molecular formula = 448.70 g/mol
Mass of empirical formula = 56 g/mol
Putting values in above equation, we get:
[tex]n=\frac{448.70g/mol}{56g/mol}=8[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_{(8\times 2)}S_{(8\times 1)}=C_{16}S_8[/tex]
Thus, the empirical and molecular formula for the given compound is [tex]C_2S[/tex] and [tex]C_{16}S_8[/tex]
Final answer:
The empirical formula of 'sulflower' is CS2, found by using the given percentages to calculate the simplest whole number ratio of moles of carbon to sulfur. Dividing the compound's molar mass by the empirical formula's molar mass indicates that the molecular formula of 'sulflower' is C6S12.
Explanation:
To determine the empirical formula of the molecule known as 'sulflower', we can start by using the given percentages to find the moles of each element in the compound. Assuming we have 100 grams of the substance, we would have 57.17 grams of sulfur (S) and 42.83 grams of carbon (C). Using their molar masses (atomic sulfur is 32 g/mol and carbon is 12 g/mol), we can calculate the moles of each:
Sulfur: 57.17 g S × (1 mol S / 32 g) = 1.786 moles SCarbon: 42.83 g C × (1 mol C / 12 g) = 3.569 moles CTo find the empirical formula, we then find the simplest whole number ratio by dividing by the smallest number of moles, which gives us:
1.786 moles S / 1.786 moles = 1 S3.569 moles C / 1.786 moles = 2 CSo, the empirical formula is CS2. To determine the molecular formula, we use the molar mass of the empirical formula and divide the given molar mass of the compound by this value. The empirical formula mass of CS2 is (12.01 g/mol × 2) + (32.07 g/mol × 1) = 76.09 g/mol. The molecular formula is then found by dividing the compound's molar mass by the empirical formula's molar mass:
448.70 g/mol / 76.09 g/mol = 5.9 ≈ 6
Therefore, the molecular formula is C6S12, as the empirical formula must be multiplied by 6 to get the molecular formula.
If I want to accelerate a mass of 3 kg at 5 m/s2 then how much force should I apply?
Answer: The force that must be applied is 15 N.
Explanation:
Force exerted on the object is defined as the product of mass of the object and the acceleration of the object.
Mathematically,
[tex]F=m\times a[/tex]
where,
F = force exerted = ?
m = mass of the object = 3 kg
a = acceleration of the object = [tex]5m/s^2[/tex]
Putting values in above equation, we get:
[tex]F=3kg\times 5m/s^2=15N[/tex]
Hence, the force that must be applied is 15 N.
A sample of argon (Ar) gas occupies 65.0 mL at 22.0°C and 725 torr. What is the volume of this Ar gas at STP? Enter your answer in the provided box.
Answer: The volume of argon gas at STP is 57.4 mL
Explanation:
STP conditions are:
Pressure of the gas = 1 atm = 760 torr
Temperature of the gas = 273 K
To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law. The equation follows:
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1,V_1\text{ and }T_1[/tex] are the initial pressure, volume and temperature of the gas
[tex]P_2,V_2\text{ and }T_2[/tex] are the final pressure, volume and temperature of the gas
We are given:
[tex]P_1=725torr\\V_1=65.0mL\\T_1=22^oC=(22+273)K=295K\\P_2=760torr\\V_2=?\\T_2=273K[/tex]
Putting values in above equation, we get:
[tex]\frac{725torr\times 65.0mL}{295K}=\frac{760torr\times V_2}{273K}\\\\V_2=57.4mL[/tex]
Hence, the volume of argon gas at STP is 57.4 mL
Explanation:
The given data is as follows.
[tex]V_{1}[/tex] = 65.0 mL = 0.065 L (as 1 ml = 0.001 L),
[tex]T_{1}[/tex] = [tex]22.0^{o}C[/tex] = (22 + 273) K = 295 K,
[tex]P_{1}[/tex] = 725 torr = 0.954 atm (as 1 torr = 0.00131579 atm),
[tex]V_{2}[/tex] = ?, [tex]T_{2}[/tex] = 273 K,
[tex]P_{2}[/tex] = 1 atm
And, according to ideal gas equation,
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
[tex]\frac{0.954 atm \times 0.065 L}{295 K} = \frac{1 atm \times V_{2}}{273 K}[/tex]
[tex]V_{2}[/tex] = 0.0574 L
As, 1 L = 1000 ml. So, 0.0574 L = 57.4 ml.
Thus, we can conclude that the volume of this Ar gas at STP is 57.4 L.
Draw the resonance structures for SO3 2-
SO3 2- has three possible resonance structures, each with one of the three oxygen atoms having a double bond to the central sulfur atom, while the other two have single bonds. The sulfur atom is surrounded by two bonds and one lone pair of electrons in any given structure.
Explanation:The student asked to draw the resonance structures for SO3 2-. Resonance structures are a way of illustrating the delocalization of electrons within certain molecules, often containing double bonds. In the case of the SO3 2- ion, we have sulfur connected to three oxygen atoms.
In each of the resonance structures, sulfur has an oxidation number of +6 and each oxygen atom has an oxidation number of -2. There are three different resonance structures possible for this ion. This is because one of the oxygen atoms can have a double bond to the sulfur atom while the other two oxygen atoms have single bonds, with sulfur carrying one lone pair. The double bond can be with any of the three oxygen atoms, leading to three possible structures, hence the term resonance structures.
The sulfur atom is surrounded by two bonds and one lone pair of electrons in either resonance structure. Therefore, the electron-pair geometry is trigonal planar, and the hybridization of the sulfur atom is sp².
Learn more about Resonance Structures here:https://brainly.com/question/35369195
#SPJ3
why is it necessay to remove the stopper form a
separatoryfunnel when liquid is being drained from it through
thestopcock?
Answer:
It is necessary in order to let air in.
Explanation:
When the stopper is not removed, a vacuum is formed inside the separatory funnel as the liquid drains out. The lower air pressure inside the separatory funnel then causes the liquid to drain improperly.
If you have 0.08157 moles of Al(CIO4) 3... a) How many moles of Al ions do you have? b) How many moles of CIO4 ions do you have? c) How many moles of oxygen atoms do you have? 3+
Answer: a) 0.08157 moles of [tex]Al^{3+}[/tex]
b) 0.24471 moles of [tex]ClO_4^{-}[/tex]
c) 0.97884 moles of oxygen atoms
Explanation:-
The dissociation of the given compound is shown by the balanced equation:
[tex]Al(ClO_4)_3\rightarrow Al^{3+}+3ClO_4^{-}[/tex]
According to stoichiometry:
a) 1 mole of [tex]Al(ClO_4)_3[/tex] produces 1 mole of [tex]Al^{3+}[/tex]
Thus 0.08157 mole of [tex]Al(ClO_4)_3[/tex] produces=[tex]\frac{1}{1}\times 0.08157=0.08157moles[/tex] of [tex]Al^{3+}[/tex]
b) 1 mole of [tex]Al(ClO_4)_3[/tex] produces 3 moles of [tex]ClO_4^{-}[/tex]
Thus 0.08157 mole of [tex]Al(ClO_4)_3[/tex] produces=[tex]\frac{3}{1}\times 0.08157=0.24471moles[/tex] of [tex]ClO_4^{-}[/tex]
c) 1 mole of [tex]Al(ClO_4)_3[/tex] produces 12 moles of oxygen atoms
Thus 0.08157 mole of [tex]Al(ClO_4)_3[/tex] produces=[tex]\frac{12}{1}\times 0.08157=0.97884moles[/tex] of oxygen atoms
A process stream contains 450 mol/s benzene and 375 mol/s toluene. Calculate the mole fraction of benzene in the process stream.
Answer:
0.5455
Explanation:
The moles of benzene in the process stream in 1 sec = 450 moles
The moles of toluene in the process stream in 1 sec = 375 moles
So, according to definition of mole fraction:
[tex]Mole\ fraction\ of\ benzene=\frac {n_{benzene}}{n_{benzene}+n_{toluene}}[/tex]
Applying values as:
[tex]Mole\ fraction\ of\ benzene=\frac {450}{450+375}[/tex]
Mole fraction of benzene in the process stream = 0.5455
The concentration of dye in Solution A is 20.422 M. A serial dilution is performed to make Solutions B and C. In the first dilution, 2 mL of Solution A is diluted with 12 mL water to make Solution B. Then, 2 mL of Solution B is then diluted with 1 mL of water to make Solution C. What is the concentration of dye in Solution C? Provide your response to three significant figures in units of molarity.
Answer:
1.95 M
Explanation:
Considering
Moles of A = Moles of B
Or,
[tex]Molarity_{A}\times Volume_{A}=Molarity_{B}\times Volume_{B}[/tex]
Given that:
[tex]Molarity_{A}=20.422\ M[/tex]
[tex]Volume_{A}=2\ mL[/tex]
[tex]Volume_{B}=2+12=14\ mL[/tex]
[tex]Molarity_{B}=?\ M[/tex]
So,
[tex]20.422\times 2=Molarity_{B}\times 14[/tex]
Molarity of solution B = 2.92 M
Again Considering,
Moles of B = Moles of C
Or,
[tex]Molarity_{B}\times Volume_{B}=Molarity_{C}\times Volume_{C}[/tex]
Given that:
[tex]Molarity_{B}=2.92\ M[/tex]
[tex]Volume_{B}=2\ mL[/tex]
[tex]Volume_{C}=2+1=3\ mL[/tex]
[tex]Molarity_{C}=?\ M[/tex]
So,
[tex]2.92\times 2=Molarity_{C}\times 3[/tex]
Molarity of solution C = 1.95 M
The concentration of dye in Solution C is 1.945 M after performing the serial dilutions.
Explanation:To find the concentration of dye in Solution C after serial dilutions, we can use the dilution equation which states that the concentration times the volume before dilution (C1V1) is equal to the concentration times the volume after dilution (C2V2), or C1V1 = C2V2.
For Solution B, the initial concentration is 20.422 M (C1) and the initial volume is 2 mL (V1). This solution is diluted with 12 mL water, so the final volume (V2) is 2 mL + 12 mL = 14 mL. Using the dilution equation, we calculate the concentration of Solution B (C2).
(20.422 M)(2 mL) = (C2)(14 mL)
C2 = (20.422 M)(2 mL) / (14 mL) = 2.917 M
Next, we perform a second dilution to create Solution C. We take 2 mL of Solution B and add 1 mL of water, giving a final volume of 3 mL. Again, we use the dilution equation to find the concentration of Solution C.
(2.917 M)(2 mL) = (C3)(3 mL)
C3 = (2.917 M)(2 mL) / (3 mL) = 1.945 M
Therefore, the concentration of dye in Solution C is 1.945 M, rounded to three significant figures.
Consider the following reaction: 2 NO(g) + 2H2(g) → N2(g) + 2 H2O(g) The rate law for this reaction is first order in H2 and second order in NO. What would happen to the rate if the initial concentration of NO tripled while all other factors stayed the same? The rate will increase by a factor of 9. The rate will decrease by a factor of 3. The rate will double. The rate will triple. The rate will remain constant.
Answer: The rate will increase by a factor of 9.
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]2NO(g)+2H_2(g)\rightarrow N_2(g)+2H_2O(g)[/tex]
Given: Order with respect to [tex]NO[/tex] = 2
Order with respect to [tex]H_2[/tex] = 1
Thus rate law is:
[tex]Rate=k[NO]^2[H_2]^1[/tex]
k= rate constant
It is given that the initial concentration of NO is tripled while all other factors stayed the same
[tex]Rate'=k[3\times NO]^2[H_2]^1[/tex]
[tex]Rate'=k[3]^2[NO]^2[H_2]^1[/tex]
[tex]Rate'=k\times 9[NO]^2[H_2]^1[/tex]
[tex]Rate'=9\times Rate[/tex]
Thus the rate will increase by a factor of 9.
The correct answer is that the rate will increase by a factor of 9.
To understand why the rate increases by a factor of 9 when the concentration of NO is tripled, we need to consider the rate law for the reaction, which is given as first order in H2 and second order in NO. The rate law can be expressed as:
[tex]\[ \text{Rate} = k[\text{NO}]^2[\text{H2}] \][/tex]
where [tex]\( k \)[/tex] is the rate constant, and [NO] and [H2] are the concentrations of NO and H2, respectively.
Since the reaction is first order with respect to H2, the rate is directly proportional to the concentration of H2. However, the reaction is second order with respect to NO, which means the rate is proportional to the square of the concentration of NO.
If the concentration of NO is tripled (let's denote the initial concentration of NO as [NO]_initial, then the new concentration is [tex]3[NO]_initial)[/tex], the new rate can be calculated as follows:
[tex]\[ \text{New Rate} = k(3[\text{NO}]_{\text{initial}})^2[\text{H2}] \][/tex]
[tex]\[ \text{New Rate} = k \cdot 9 \cdot [\text{NO}]_{\text{initial}}^2 \cdot [\text{H2}] \][/tex]
[tex]\[ \text{New Rate} = 9 \cdot k[\text{NO}]_{\text{initial}}^2[\text{H2}] \][/tex]
Comparing the new rate to the initial rate:
[tex]\[ \text{Initial Rate} = k[\text{NO}]_{\text{initial}}^2[\text{H2}] \][/tex]
we see that the new rate is 9 times the initial rate because the concentration of NO has been tripled, and due to the second-order dependence on NO, the rate is multiplied by the square of the factor by which the concentration of NO has increased [tex](3^2 = 9).[/tex]
Therefore, the rate will increase by a factor of 9 when the initial concentration of NO is tripled, while keeping all other factors constant.
If you feed 100 kg of N2 gas and 100 kg of H2 gas into a
reactor: What is the maximum number of kg of NH3 that can be
produced?
Answer : The mass of ammonia produced can be, 121.429 k
Solution : Given,
Mass of [tex]N_2[/tex] = 100 kg = 100000 g
Mass of [tex]H_2[/tex] = 100 kg = 100000 g
Molar mass of [tex]N_2[/tex] = 28 g/mole
Molar mass of [tex]H_2[/tex] = 2 g/mole
Molar mass of [tex]NH_3[/tex] = 17 g/mole
First we have to calculate the moles of [tex]N_2[/tex] and [tex]H_2[/tex].
[tex]\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles[/tex]
[tex]\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]N_2+3H_2\rightarrow 2NH_3[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]N_2[/tex] react with 3 mole of [tex]H_2[/tex]
So, 3571.43 moles of [tex]N_2[/tex] react with [tex]3571.43\times 3=10714.29[/tex] moles of [tex]H_2[/tex]
From this we conclude that, [tex]H_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]N_2[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]NH_3[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]N_2[/tex] react to give 2 mole of [tex]NH_3[/tex]
So, 3571.43 moles of [tex]N_2[/tex] react to give [tex]3571.43\times 2=7142.86[/tex] moles of [tex]NH_3[/tex]
Now we have to calculate the mass of [tex]NH_3[/tex]
[tex]\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3[/tex]
[tex]\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg[/tex]
Therefore, the mass of ammonia produced can be, 121.429 kg
What is the theoretical yield of bismuth when 198 g of Bi2O3 reacts with excess carbon as shown below? Bi2O3 C(s)>Bi(s) + CO(g) (not balanced)
Answer:
179.4306 g
Explanation:
Given that:
Mass of [tex]Bi_2O_3[/tex] = 198 g
Molar mass of [tex]Bi_2O_3[/tex] = 465.96 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{198\ g}{465.96\ g/mol}[/tex]
[tex]Moles\ of\ Bi_2O_3= 0.4293\ mol[/tex]
From the balanced reaction,
[tex]Bi_2O_3+3C\rightarrow 2Bi+3CO[/tex]
1 mole of [tex]Bi_2O_3[/tex] on reaction produces 2 moles of bismuth
So,
0.4293 mole of [tex]Bi_2O_3[/tex] on reaction produces 2 × 0.4293 moles of bismuth
Moles of bismuth = 0.8586 moles
Molar mass of bismuth = 208.9804 g/mol
So, mass of bismuth = Moles × Molar mass = 0.8586 × 208.9804 g = 179.4306 g
Final answer:
The theoretical yield of bismuth in the given reaction is 0.848 moles or 177.2 grams.
Explanation:
To determine the theoretical yield of bismuth (Bi) in the reaction between 198 g of Bi2O3 and excess carbon, we need to balance the equation first. The balanced equation for the reaction is:
Bi2O3 + 3C → 2Bi + 3CO
From the balanced equation, we can see that for every mole of Bi2O3, we get 2 moles of Bi. To calculate the theoretical yield, we need to convert the given mass of Bi2O3 to moles using its molar mass (465.96 g/mol) and then use the stoichiometry to find the moles of Bi.
Mass of Bi2O3 = 198 g
Molar mass of Bi2O3 = 465.96 g/mol
Moles of Bi2O3 = (198 g) / (465.96 g/mol) = 0.424 mol
Moles of Bi = 2 * (0.424 mol) = 0.848 mol
The theoretical yield of bismuth in this reaction is 0.848 moles or you can convert it to grams using the molar mass of bismuth (208.98 g/mol) to get the theoretical yield in grams.
Question 5 (1 point) You need to prepare 1.0 L of solution containing 1.0 mM MgCl2 and 0.15 M NaCl form stock solutions of 1.0 M MgCl2 and 0.50 M NaCl. You will: use 1.0 mL MgCl2, 0.30 L NaCl, fill up with water to 1 L Ouse 100 mL MgCl2, 300 ml NaCl, fill up with water to 1 L use 1.0 mL MgCl2, 30.0 ml NaCl, fill up with water to 1 L use 0.0010 L MgCl2, 0.030 L NaCl, fill up with water to 1 L use 100 ml MgCl2, 300 ml NaCl, fill up with water to 1L
Answer:
Use 1.0 mL MgCl2, 0.30 L NaCl, fill up with water to 1 L
Explanation:
The dilution equation relates the concentration C₁ and volume V₁ of an undiluted solution to the concentration C₂ and volume V₂ as follows:
C₁V₁ = C₂V₂
We want to solve for V₁, the amount of stock solution that is required for the dilution to 1.0L.
V₁ = (C₂V₂) / C₁
For the MgCl₂ stock solution, the following volume is required:
V₁ = (C₂V₂) / C₁ = (1.0mM)(1.0L) / (1.0M) = 1 mL
For the NaCl stock solution, the following volume is required:
V₁ = (C₂V₂) / C₁ = (0.15M)(1.0L) / (0.50M) = 0.30 L
The volumes V₁ are the diluted until the total volume reaches V₂.
Calculate the number of atoms of bromine in 1.37 g of bromine. Give your answer in scientific notation. Enter your answer in the provided box. x 10 (select)atoms Br
Answer: The number of atoms of bromine present in given number of mass is [tex]1.03\times 10^{22}[/tex]
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of bromine = 1.37 g
Molar mass of bromine = 79.904 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of bromine}=\frac{1.37g}{79.904g/mol}=0.0171mol[/tex]
According to mole concept:
1 mole of an element contains [tex]6.022\times 10^{23}[/tex] number of atoms.
So, 0.0171 moles of bromine will contain = [tex]0.0171\times 6.022\times 10^{23}=1.03\times 10^{22}[/tex] number of bromine atoms.
Hence, the number of atoms of bromine present in given number of mass is [tex]1.03\times 10^{22}[/tex]
Final answer:
To find the number of bromine atoms in 1.37 g of bromine, calculate the atomic mass using isotope information, convert grams to moles, and then use Avogadro's number, yielding approximately 1.03 × 10²² bromine atoms.
Explanation:
To calculate the number of bromine atoms in 1.37 g of bromine, we first need to determine the molar mass of bromine. The atomic mass of bromine is calculated based on the isotopic composition, considering Bromine has two isotopes, 79Br and 81Br with masses of 78.9183 amu and 80.9163 amu respectively, and relative abundances of 50.69% and 49.31%. Using these values, we calculate the average atomic mass of bromine:
Atomic mass = (0.5069 × 78.9183 amu) + (0.4931 × 80.9163 amu) = 79.904 amu.
We then convert grams of bromine to moles by dividing by the molar mass:
Number of moles = 1.37 g ÷ 79.904 g/mol = 0.01715 moles.
Using Avogadro's number, 6.022 × 10²³ atoms/mol, we multiply the number of moles of bromine by this constant to find the number of atoms:
Number of atoms = 0.01715 moles × 6.022 × 10²³ atoms/mol = 1.03 × 10²² atoms.
Thus, there are 1.03 × 10²² atoms of bromine in 1.37 g of bromine. Remember, when giving your answer in scientific notation, it is important to express it in the form of a number between 1 and 10 multiplied by a power of 10.
A gas of potassium chlorate molecules KClO3 all decompose into potassium chloride, KCl, and diatomic oxygen, O2. The products and reactants are in a closed container and can all be treated as ideal gases.A. Fill in the smallest possible integers that allows the stoichiometry of the reaction equation to be correct:__ KClO3 → ___ KCl ___ O2B. If there are N molecules of potassium chlorate in the initial state, how many product molecules are there
Answer: A. [tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]
B. [tex]2.5\times N[/tex] molecules of product.
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
A. The balanced reaction for decomposition of potassium chlorate is:
[tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]
B. According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
According to stoichiometry :
2 moles of reactant give 5 moles of products
Thus [tex]2\times 6.023\times 10^{23}[/tex] molecules of reactant give [tex]5\times 6.023\times 10^{23}[/tex] molecules of product
Thus N molecules of reactant give= [tex]\frac{5\times 6.023\times 10^{23}}{2\times 6.023\times 10^{23}}\times N=\frac{5}{2}\times N[/tex] molecules of product.
Thus [tex]\frac{5}{2}\times N=2.5\times N[/tex] molecules of product are there.
Final answer:
The balanced chemical equation for the decomposition of potassium chlorate is 2 KClO3 → 2 KCl + 3 O2. If there are N molecules of potassium chlorate in the initial state, the number of product molecules of O2 can be calculated using the ratio of 2:3 in the equation.
Explanation:
The balanced chemical equation for the decomposition of potassium chlorate is:
2 KClO3 → 2 KCl + 3 O2
The stoichiometry of the reaction equation can be balanced by filling in the smallest possible integers as follows:
A. 2 KClO3 → 2 KCl + 3 O2
B. If there are N molecules of potassium chlorate in the initial state, the number of product molecules can be calculated as follows:
The ratio of KClO3 to O2 in the balanced equation is 2:3. So, for every 2 molecules of KClO3 that decompose, 3 molecules of O2 are formed. Therefore, the number of O2 molecules produced is given by the equation:
(3 / 2) N = (3 / 2) × N molecules of O2
The vapour pressure of water at 20 C is 2.34 kPa. Given that the heat of vaporisation is 2537.4 kJ/kg, use the Clausius-Clapeyron equation to give the vapour pressure at 40 C.
The Clausius-Clapeyron equation is used to calculate the vapor pressure of water at 40°C given its vapor pressure at 20°C and heat of vaporisation. By rearranging the equation and inserting the known values, the vapor pressure at 40°C can be found, which will be higher than that at 20°C.
Explanation:The Clausius-Clapeyron equation relates the vapor pressure and temperature of a substance to its heat of vaporisation. To calculate the vapor pressure of water at 40°C, given the vapor pressure at 20°C (2.34 kPa) and the heat of vaporisation (2537.4 kJ/kg), we can rearrange the equation to solve for the new vapor pressure:
Clausius-Clapeyron Equation: ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)
Where P1 is the initial vapor pressure, P2 is the final vapor pressure, ΔHvap is the molar heat of vaporisation, R is the universal gas constant in J/(mol·K), T1 is the initial temperature in Kelvin, and T2 is the final temperature in Kelvin.
The calculation involves converting both temperatures from Celsius to Kelvin and plugging the values into the equation to solve for P2. After using this equation, we find that the vapor pressure at 40°C is significantly higher than at 20°C.
The complete combustion of 1.00 kg of coal releases about 3.0 x 10E7 J of energy. The conversion of 1.00 kg of mass into energy is equivalent to the burning of how many metric tonnes of coal?
Answer: 0.001 metric tonnes
Explanation:
According to Einstein equation, Energy and mass are inter convertible.
[tex]E=mc^2[/tex]
E= Energy
m = mass
c= speed of light
Given : Complete combustion of 1.00 kg of coal releases about [tex]3.0\times 10^7 J[/tex] of energy.
Given mass: 1 kg
Converting kg to metric ton using the conversion factor:
1 kg=0.001 metric tonnes
Thus 1 kg of coal would also be equal to 0.001 metric tonnes.
Answer:
qrqwrqewq
Explanation:
You can obtain a rough estimate of the size of a molecule with the following simple experiment: Let a droplet of oil spread out on a fairly large but smooth water surface. The resulting "oil slick" that forms on the surface of the water will be approximately one molecule thick. Given an oil droplet with a mass of 9.00 × 10−7 kg and a density of 918 kg/m3 that spreads out to form a circle with a radius of 41.8 cm on the water surface, what is the approximate diameter of an oil molecule?
Answer:
The diameter of the oil molecule is [tex]4.4674\times 10^{-8} cm[/tex] .
Explanation:
Mass of the oil drop = [tex]m=9.00\times 10^{-7} kg[/tex]
Density of the oil drop = [tex]d=918 kg/m^3[/tex]
Volume of the oil drop: v
[tex]d=\frac{m}{v}[/tex]
[tex]v=\frac{m}{d}=\frac{9.00\times 10^{-7} kg}{918 kg/m^3}[/tex]
Thickness of the oil drop is 1 molecule thick.So, let the thickness of the drop or diameter of the molecule be x.
Radius of the oil drop on the water surface,r = 41.8 cm = 0.418 m
1 cm = 0.01 m
Surface of the sphere is given as: a = [tex]4\pi r^2[/tex]
[tex]a=4\times 3.14\times (0.418 m)^2=2.1945 m^2[/tex]
Volume of the oil drop = v = Area × thickness
[tex]\frac{9.00\times 10^{-7} kg}{918 kg/m^3}=2.1945 m^2\times x[/tex]
[tex]x= 4.4674\times 10^{-10} m= 4.4674\times 10^{-8} cm[/tex]
The thickness of the oil drop is [tex]4.4674\times 10^{-8} cm[/tex] and so is the diameter of the molecule.
The diameter of an oil molecule can be estimated by considering the oil slick as a cylinder and using the formula for the volume of a cylinder. From the given, we calculate the volume of the oil droplet. Dividing this volume by the area of the slick gives us the height of the cylinder, which represents the diameter of the oil molecule.
Explanation:The oil slick forms a cylindrical shape with a height equal to the diameter of an oil molecule or two times the radius. We know the volume of the oil droplet, oil slick, and the diameter of the oil slick.
The volume of oil droplet (V) = Mass (m) / Density (ρ)
= 9 x 10⁻⁷ kg / 918 kg/m³
= 9.81 x 10⁻¹⁰ m³
The volume of the cylindrical oil slick (V) = πr²h, where r is the radius and h is the height of the cylinder.
So, h (height of the oil molecule) = V / (πr^2)
= 9.81 x 10⁻¹⁰ m³ / (3.14 x (0.418 m)²)
= 1.74 x 10⁻¹⁰ m or 174 pm (picometers)
Since the height of the cylinder is approximately equal to the diameter of an oil molecule, the approximate diameter of an oil molecule is 174 pm.
Learn more about Diameter of Oil Molecule here:https://brainly.com/question/14553021
#SPJ3
Write 23,665,700 in Scientific Notation with 4 significant figures.
Answer: The given number in scientific notation is [tex]2.367\times 10^{7}[/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 = 23,665,700
Converting this into scientific notation, we get:
As, the decimal is shifting to left side, the power of 10 will be positive.
[tex]\Rightarrow 23,665,700=2.367\times 10^{7}[/tex]
Hence, the given number in scientific notation is [tex]2.367\times 10^{7}[/tex]
When measuring a solid metal block at constant temperature, which measurement will change in numerical value depending on the location where it is taken? A) weight B) length C) volume D) mass
Answer:
A) weight
Explanation:
The weight is the force that gravity makes in a body, and it is given by the equation:
W = mxg
Where W is the weight, m is the mass, and g is the gravity in the local. So, the length, the volume, and the mass are constant and don't depend on the local that is measured, but the weight depends on the local gravity.
Describe how you would make 250 ml of a 3 M solution of sodium acetate (NaOAc = 82.03 g/mol). First figure out how much sodium acetate you would need, then describe how you would make the solution if you were given a bottle of solid sodium acetate, a volumetric flask, and DI water.
Answer:
You need to do the following conversion to pass from 3M in 250 mL to g of sodium acetate
[tex]3 M (mol/L)*(1L/1000 mL)*(250 mL)*(82.03 g/1 mol)=61.52 g[/tex]
Explanation:
First, you need to dissolve 61.52 g of solid sodium acetate (MW 82.03 g/mol) in 200 ml of DI water. Then, using a volumetric flask add water to bring the total volume of the solution to 250 mL.
Water enters a 4.00-m3 tank at a rate of 6.33 kg/s and is withdrawn at a rate of 3.25 kg/s. The tank is initially half full.
How long does it take for the tank to overflow?
Explanation:
The given data is as follows.
Volume of tank = 4 [tex]m^{3}[/tex]
Density of water = 1000 [tex]kg/m^{3}[/tex]
Since, the tank is initially half-filled. Hence, the volume of water in the tank is calculated as follows.
[tex]\frac{1}{2} \times 4 = 2 m^{3}[/tex]
Also, density of a substance is equal to its mass divided by its volume. Therefore, initially mass of water in the tank is as follows.
Mass = [tex]Density \times initial volume[/tex]
= [tex]1000 \times 2[/tex]
= 2000 kg
Whereas mass of water in tank when it is full is as follows.
Mass = [tex]Density \times final volume[/tex]
= [tex]1000 \times 4[/tex]
= 4000 kg
So, net mass of the fluid to be filled is as follows.
Net mass to be filled = Final mass - initial mass
= 4000 kg - 2000 kg
= 2000 kg
Mass flow rate [tex](m_{in})[/tex] = 6.33 kg/s
Mass flow rate [tex](m_{out})[/tex] = 3.25 kg/s
Time needed to fill tank = [tex]\frac{\text{net mass to be filled}}{\text{net difference of flow rates}}[/tex]
= [tex]\frac{2000 kg}{m_{in} - m_{out}}[/tex]
= [tex]\frac{2000 kg}{6.33 kg/s - 3.25 kg/s}[/tex]
= 649.35 sec
Thus, we can conclude that 649.35 sec is taken by the tank to overflow.
Which equation or rule can be used to estimate the pseudo-critical properties of mixtures based on the pure-component critical constants? Select one: a. Newton's correlations b. Soave-Redlich-Kwong equation c. Kay's rule d. van der Waals equation of state e. Peng-Robinson
Answer:
c. Kay's rule
Explanation:
Kay's rule -
The rule is used to determine the pseudo reduced critical parameters of mixture , with the help of using the critical properties of the components of a given mixture .
The equation for Kay's rule is as follows ,
PV = Z RT
Where Z = The compressibility factor of the mixture .
Hence from the given options , the correct answer is Kay's rule .
To estimate the pseudo-critical properties of mixtures based on pure-component critical constants, Kay's rule is used. It estimates the properties based on mole fractions of components, offering a simpler alternative to complex equations like Van der Waals or Peng-Robinson equations.
Explanation:The equation that can be used to estimate the pseudo-critical properties of mixtures based on the pure-component critical constants is Kay's rule. This rule is a simple and common method for calculating the critical properties of mixtures, which assumes that the critical properties are proportional to the mole fractions of the components in the mixture. It is not as complex as the Van der Waals equation of state or the Peng-Robinson equation, which are more accurate but also more complicated.
For ideal solutions, Raoult's Law is often used to determine the vapor pressures and hence the equilibrium states of the components in a solution, given by the equation PA = XAPA, where PA is the partial pressure of component A, XA is the mole fraction of component A, and PAo is the vapor pressure of the pure component A.