Answer: Molality of solution is 0.89 mole/kg
Explanation:
Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.
[tex]Molarity=\frac{n}{W_s}[/tex]
where,
n = moles of solute
[tex]W_s[/tex] = weight of solvent in kg
moles of nitrobenzene (solute) =[tex]\frac{\text {given mass}}{\text {Molar Mas}}=\frac{1.1}{123.06g/mol}=0.0089[/tex]
mass of napthalene (solvent )= 10.0 g = 0.0100 kg
Now put all the given values in the formula of molality, we get
[tex]Molality=\frac{0.0089}{0.0100kg}=0.89mole/kg[/tex]
Therefore, the molality of solution is 0.89 mole/kg
Calculate the wavelength in nanometers of the light emitted by a hydrogen atom when it's electron falls from the n= 7 to the n= 4 principal energy level. Recall that the engergy levels of the H atom are given by En = -2.18 x 10 to the negative 18 J ( 1/n to the second power) . ( c= 3.00 x 10 to to the 8 th power m/s ; h=6.63 x 10 to the negative 34 j.s
Answer:
2.165 x 10^3 nm.
Explanation:
Using Rybergs equation,
1/lambda = R * (1/n1^2 - 1/n2^2)
Where,
R = rybergs constant = 109737.32 cm^-1
n1 = 7
n2 = 4
= 109737.32 * (1/7^2 - 1/4^2)
= 4619.05
Lambda = 2.165 x 10^-4 cm
Since 100 cm = 1m, 1 nm = 10^-9 m
= 2.165 x 10^-4 cm * 1 m/100 cm * 1 nm/10^-9 m
= 2.165 x 10^3 nm.
Equal volumes of 0.250 M acetic acid and water are combined; a 50.0 mL portion of this solution is titrated to the endpoint with 0.125 M NaOH. Calculate the volume of NaOH required to reach the endpoint.
Answer: The volume of NaOH required to reach the endpoint is 100 mL
Explanation:
To calculate the volume of NaOH, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]CH_3COOH[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=1\\M_1=0.250M\\V_1=50.0mL\\n_2=1\\M_2=0.125M\\V_2=?mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.250\times 50.00=1\times 0.125\times V_2\\\\V_2=\frac{1\times 0.250\times 50.0}{1\times 0.125}=100mL[/tex]
Hence, the volume of NaOH required to reach the endpoint is 100 mL
Why are methanol and acetone not suitable solvents for extracting organic compounds from aqueous solutions?
Answer: methanol and acetone are not suitable solvents for extracting organic compounds because they are miscible with water virtually in all proportions
Explanation:
Methane and acetone both are polar solvents, which means they are soluble in water. Hence, not suitable for the extraction from aqueous solutions.
What are methanol and acetone?
Methanol is the simplest alcohol which is volatile, colorless, and inflammable.
Acetone is an organic compound whose chemical formula is CH3CH3CO.
Acetone and methanol both contain polar ends and are miscible in water.
Polar solvents are not suitable for the extraction of any compound.
Thus, Methanol and acetone are not suitable for extracting organic compounds from aqueous solutions.
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Which of the following is NOT part of the Kinetic Molecular Theory? A) Gas particles do not repel each other. B) There is a large distance between gas particles as compared to their relative size. C) The size of the actual gas particles is small compared to the volume of the whole gas. D) The average energy of the particles is dependent on the molecular mass of the particle. E) All of the above statements are part of the Kinetic Molecular Theory.
Answer:The average energy of the particles is dependent on the molecular mass of the particle is not part of Kinetic Molecular Theory.
Explanation:
The kinetic theory of gases was postulate by Clausius in 1857. Clausius speculated on how heat energy, temperature, and molecule motion could explain gas behavior. The kinetic theory of gases includes the following:
- A gas consists of molecules in constant random motion.
-Gas molecules influence each other only by collision; they exert no other forces on each other.
-All collisions between gas molecules are perfectly elastic; all kinetic energy is conserved.
- The volume actually occupied by the molecules of a gas is negligibly small; the vast majority of the volume of the gas is empty space through which the gas molecules are moving.
The Kinetic Molecular Theory includes assumptions about gas particles' size, motion, lack of forces of attraction or repulsion, and the dependence of average kinetic energy on temperature. The statement 'Gas particles do not repel each other' is actually part of the theory.
Explanation:The Kinetic Molecular Theory makes several general assumptions about gases. These include :
Gas particles are far apart from each other and their size is considered to be insignificant compared to the volume of the empty space. These particles are always in constant rapid motion in random directions. There are no forces of attraction or repulsion between the gas particles. As such, their motion is completely independent of other particles. The average kinetic energy of the particles is dependent on the temperature of the gas. If the temperature increases, the kinetic energy of particles increases proportionately.Looking at the options provided, the statement which is NOT part of the Kinetic Molecular Theory is option A) Gas particles do not repel each other. This is because the theory does indeed state there are no forces of attraction or repulsion between gas particles.
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Why can’t we overcome the uncertainty predicted by Heisenberg’s principle by building more precise devices to reduce the error in measurements below the h/4π limit?
Answer:
Explanation:
This limit is a consequence of Heisenberg´s uncertainty principle:
Δp x Δx > = h
This state that the product of the uncertainty in momentum ( or velocity since p = mv ) times the uncertainty in position, Δx , must be greater or equal to Planck´s constant ( 6.626 x 10⁻³⁴ J·s ).
Later models refined this equation to:
Δp x Δx > = h/4π
This is the consequence of duality wave matter of the electron and Schrodinger´s equation, in which we can talk of probabilities of finding an electron and not confined to specific distances from the nucleus as in the Bohr atom.
Now think of think of this relation in terms of the uncertainty it describes. If we know the position of the electron with great exactitude, the velocity of the particle will be very high since the mass of hte electron is very small.
This a principle in nature and has nothing to do with the precision of our instruments for particles at the subatomic level.
The reason we do not observe this effect with everyday objects is that the obbects have masses so large compare to subatomic particles that the term mΔv becomes large enough, allowing us to know the position and velocity of macroscopic objects with small uncertainties:
Δp x Δx > = h/4π, Δp very large ( because the mass is very big ) then Δx is very small
The same does not have with small masses of the subatomic levels.
7.55 grams of P4 and 7.55 grams of O2 react according to the following reaction:
P4 + O2--> P4O6
If enough oxygen is available, then the P4O6 reacts further:
P4O6 + O2 --> P4O10
a. Find the limiting reagent in the formation of P4O10.
b. What mass of P4O10 is produced?
c. What mass of excess reactant remains?
Answer:
a) The limiting reactant is O2.
b) 7.57 grams of P4O10 is produced
c) 7.53 grams P4O6 remains
Explanation:
Step 1: Data given
Mass of P4 = 7.55 grams
Mass of O2 = 7.55 grams
Molar mass of P4 = 123.90 g/mol
Molar mass of O2 = 32 g/mol
Step 2: The balanced equations:
P4 + 3O2-→P4O6
P4O6 + 2O2 → P4O10
Step 3: Calculate moles P4
Moles P4 = mass P4 / molar mass P4
Moles P4 = 7.55 grams / 123.90 g/mol
Moles P4 = 0.0609 moles
Step 4: Calculate moles O2
Moles O2 = 7.55 grams / 32.0 g/mol
Moles O2 = 0.236 moles
Step 5: Calculate the limiting reactant
For 1 mol P4 we need 3 moles O2 to produce 1 mol P4O6
P4 is the limiting reactant. It will completely be consumed. (0.0609 moles)
O2 is in excess. There will react 3*0.0609 = 0.1827 moles
There will remain 0.236 - 0.1827 = 0.0533 moles O2
Step 6: Calculate moles P4O6
For 1 mol P4 we need 3 moles O2 to produce 1 mol P4O6
For 0.0609 moles P4 we will have 0.0609 moles P4O6
Step 7: Calculate limting reactant
There remain 0.0533 moles O2 and there are 0.0609 moles P4O6 produced
For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O6
The limiting reactant is O2. It will completely be reacted (0.0533 moles)
There will react 0.0533/2 = 0.02665 moles
There will remain 0.0609 - 0.02665 = 0.03425 moles P4O6
This is 0.03425 moles * 219.88 g/mol = 7.53 grams P4O6
Step 8: Calculate moles P4O10
For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O6
For 0.0533 moles O2, we'll have 0.0533/2 = 0.02665 moles P4O10
Step 9: Calculate mass P4O10
Mass P4O10 = 0.02665 moles * 283.89 g/mol
Mass P4O10 = 7.57 grams
Calculate the pHpH of a 0.10 MM solution of barium hydroxide, Ba(OH)2Ba(OH)2. Express your answer numerically using two decimal places.
Answer:
13.301
Explanation:
To calculate the pH of the solution, we must obtain the pOH of the solution as illustrated below:
The dissociation equation is given below
Ba(OH)2 <==> Ba^2+ + 2OH^-
Since Ba(OH)2 dissociate to produce 2moles of OH^-, the concentration of OH^- = 2x0.1 = 0.2M
pOH = - Log[OH^-]
pOH = - Log 0.2
pOH = 0.699
But
pH + pOH = 14
pH = 14 — pOH
pH = 14 — 0.699
pH = 13.301
Answer:
The pH of this barium hydroxide solution is 13.30
Explanation:
Step 1: Data given
Concentration Ba(OH)2 = 0.10 M
Step 2: Calculate [OH-]
Ba(OH)2 ⇒ Ba^2+ + 2OH-
[OH-] = 2*0.10 M
[OH-] = 0.20 M
Step 3: Calculate pOH
pOH = -log[OH-]
pOH = -log(0.20)
pOH = 0.70
Step 4: Calculate pH
pH + pOH = 14
pH = 14 -pOH
pH = 14 - 0.70
pH = 13.30
The pH of this barium hydroxide solution is 13.30
Consider two aqueous nonvolatile and nonelectrolyte solutions, each with a solute concentration of 1 M. One contains glucose, while the other contains an unidentified covalent solid. Which of the following are sure to be identical in each solution? Select all that apply: a. Their freezing points b. The identity of the solvent c. The identity of the solute d. Their densities
Answer:
a and b are correct
Explanation:
This because both are aqueous solutions,therefore, identity of solvent is same that is water.
And because both solutions are non electrolyte they would not ionize in solution, and for the same concentration, the freezing point of both solution would also be same. Since depression in freezing point is a colligative property this means that it depends on number of solute particles not nature of particles .
Hence answer is that their freezing points and Identity of the solvent shall remain the same.
What concentration of SO 2 − 3 SO32− is in equilibrium with Ag 2 SO 3 ( s ) Ag2SO3(s) and 7.10 × 10 − 3 7.10×10−3 M Ag + Ag+?
The question happens to be in an incorrect order but the correct question can be seen below;
What concentration of [tex]SO^{2-}_3[/tex] is in equilibrium with [tex]Ag_2SO_{3(S)}[/tex] and [tex]7.10*10^{-3}M[/tex] [tex]Ag^+[/tex]? (The [tex]K_{sp}[/tex] of
Answer:
[tex]2.96*10^{-10}M[/tex]
Explanation:
The concentration of [tex]SO^{2-}_3[/tex] can be determined by using the solubility concept.
Given ionic solid is [tex]Ag_2SO_{3(S)}[/tex] ;
The Equilibrium Equation for the ionic compound will be:
[tex]Ag_2SO_{3(S)}[/tex] ⇄[tex]2Ag_{(aq)}[/tex] + [tex]SO^{2-}_3_{(aq)}[/tex]
Now, the solubility product ([tex]K_{sp}[/tex]) of the ionic compound will be;
[tex]K_{sp}[/tex] [tex]= [Ag^+]^{2}[SO^{2-}_3][/tex]
Given that;
the concentration [tex]Ag^+[/tex] is [tex]7.10*10^{-3}M[/tex] ; &
solubility product of the given ionic solid is [tex]1.5*10^{-14}[/tex]
∴
[tex]K_{sp}[/tex] [tex]= [Ag^+]^{2}[SO^{2-}_3][/tex]
[tex]1.5*10^{-14}[/tex] [tex]= (7.10*10^{-3})^2[/tex] [tex][SO^{2-}_3][/tex]
[tex][SO^{2-}_3][/tex] = [tex]\frac{1.5*10^{-14}}{ (7.10*10^{-3})^2}[/tex]
= 2.97560008 × 10⁻¹⁰
≅ [tex]2.96*10^{-10}M[/tex]
Thus, the concentration of [tex][SO^{2-}_3][/tex] is [tex]2.96*10^{-10}M[/tex]
Give all possible ml values for orbitals that have each of the following: (a) l = 3; (b) n = 2; (c) n = 6, l = 1.
Answer : All possible values of 'ml' for the following orbitals are:
(a) At l = 3, [tex]m_l=+3,+2,+1,0,-1,-2,-3[/tex]
(b) l = 0, 1
At l = 0, [tex]m_l=0[/tex]
At l = 1, [tex]m_l=+1,0,-1[/tex]
(c) At l = 1, [tex]m_l=+1,0,-1[/tex]
Explanation:
There are 4 quantum numbers :
Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....
Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...
Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as m_l. The value of this quantum number ranges from [tex](-l\text{ to }+l)[/tex]. When l = 2, the value of [tex]m_l[/tex] will be -2, -1, 0, +1, +2.
Spin Quantum number : It describes the direction of electron spin. This is represented as [tex]m_s[/tex]The value of this is [tex]+\frac{1}{2}[/tex] for upward spin and [tex]-\frac{1}{2}[/tex] for downward spin.
(a) l = 3 then the value of 'ml' is,
At l = 3, [tex]m_l=+3,+2,+1,0,-1,-2,-3[/tex]
(b) n = 2 then the value of 'ml' is,
l = 0, 1
At l = 0, [tex]m_l=0[/tex]
At l = 1, [tex]m_l=+1,0,-1[/tex]
(c) n = 6 and l = 1 then the value of 'ml' is,
n = 6
l = 0, 1, 2, 3, 4, 5
At l = 1, [tex]m_l=+1,0,-1[/tex]
The orbital refers to a region in space where there is a high probability of finding an electron.
What are orbitals?The term orbital refers to a region in space where there is a high probability of finding an electron. Within each energy level, there are orbitals.
Let us consider each of the levels shown;
(a) l = 3
The ml values for this orbital are; -3, -2, -1, 0, 1, 2, 3
(b) n = 2
The ml values for this orbital are; -1, 0, 1
(c) n = 6, l = 1
The ml values for this orbital are; -1, 0, 1
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Write a chemical equation that illustrates the autoionization of water. Express your answer as a chemical equation. Identify all of the phases in your answer.
Answer : The balanced chemical reaction will be:
[tex]H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)[/tex]
Or,
[tex]H_2O(l)+H_2O(l)\rightleftharpoons H_3O^+(aq)+OH^-(aq)[/tex]
Explanation :
Autoionization of water : The autoionization of water means that the reaction water with water means self ionization.
In the autoionization of water, one water molecule loses an hydrogen ion and another one gains it.
The balanced chemical reaction will be:
[tex]H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)[/tex]
Or,
[tex]H_2O(l)+H_2O(l)\rightleftharpoons H_3O^+(aq)+OH^-(aq)[/tex]
The autoionization of water is a chemical reaction where two water molecules react to form hydronium and hydroxide ions. This is represented as H₂O(l) + H₂O(l) = H3O+(aq) + OH¯(aq). The ion-product constant for water (Kw) is a measure of this process.
Explanation:The autoionization of water is a process in which water molecules, acting as both an acid and a base, react with each other to form ions. This can be represented by the chemical reaction: H₂O(l) + H₂O(l) = H3O+(aq) + OH¯(aq). In this equation, the (l) denotes the liquid state of water and (aq) designates aqueous, or water-dissolved, ions.
The ion-product constant for water, Kw, is equal to the product of the concentrations of the hydronium and hydroxide ions: Kw = [H3O+][OH¯]. At 25 °C, the value of Kw is approximately 1.0 × 10-14, indicating very slight ionization. This value increases with temperature, indicating an endothermic reaction.
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Using condensed electron configurations, write reactions showing the formation of the common ions of the following elements:
(a) Ba (Z = 56)
(b) O (Z = 8)
(c) Pb (Z = 82)
Answer:
a) Ba (Z = 56), [Xe].6s² - 2e ⇒ Ba²⁺, [Xe].6s⁰
b) O (Z = 8), [He].2s².2p⁴ + 2e ⇒ O²⁻, [He].2s².2p⁶
c) Pb (Z = 82), [Xe].4f¹⁴.5d¹⁰.6s².6p² - 2e ⇒ Pb²⁺, [Xe].4f¹⁴.5d¹⁰.6s².6p⁰
Explanation:
The condensed electron configurations of given elements are below
a) Ba (Z = 56), [Xe].6s²
b) O (Z = 8), [He].2s².2p⁴
c) Pb (Z = 82), [Xe].4f¹⁴.5d¹⁰.6s².6p²
Since atoms tend to donate/receive more electrons to achieve the saturated or half-saturated orbital. So in our case it happens as below
a) Ba (Z = 56), [Xe].6s² - 2e ⇒ Ba²⁺, [Xe].6s⁰
b) O (Z = 8), [He].2s².2p⁴ + 2e ⇒ O²⁻, [He].2s².2p⁶
c) Pb (Z = 82), [Xe].4f¹⁴.5d¹⁰.6s².6p² - 2e ⇒ Pb²⁺, [Xe].4f¹⁴.5d¹⁰.6s².6p⁰
C2H6O2 is infinitely miscible (soluble) in water. Ethylene glycol is a nonelectrolyte that is used as antifreeze. What is the lowest possible melting point for engine coolant that is 24.9 % (by mass) ethylene glycol?
Answer:
- 7.4 ºC
Explanation:
The change in melting temperature is given by:
ΔTm = Kf m
where kf is the molal freezing point depresion constant, and m is the molality.
The molality of a solution is calculated as
m = mol solute/ kg solvent
Since we have the % composition it is easy to calculate the molality :
In 100 g of solution we have 24.9 ethylene glycol
mol glycol = 24.9 g / 62.07 g/ mol = 0.40 mol
molality = 0.40 mol / 0.1 kg = 4 m
Km for water is 1.86 ºC/m,
therefore,
ΔTm = Kf m = 1.86 ºC/m x 4 = 7.4 ºC
Tm = -7.4 ºC for a solution
A 7.07 7.07 L cylinder contains 1.80 1.80 mol of gas A and 4.86 4.86 mol of gas B, at a temperature of 30.4 30.4 °C. Calculate the partial pressure of each gas in the cylinder. Assume ideal gas behavior.
Answer: The partial pressure of gas A is 6.34 atm and that of gas B is 17.1 atm
Explanation:
To calculate the pressure of the gas, we use the equation given by ideal gas, which follows:
[tex]PV=nRT[/tex] ......(1)
where,
P = pressure of the gas
V = Volume of the gas
T = Temperature of the gas
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles of gas
For Gas A:We are given:
[tex]V=7.07L\\T=30.4^oC=[30.4+273]K=303.4K\\n=1.80mol[/tex]
Putting values in equation 1, we get:
[tex]p_A\times 7.07L=1.80mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 303.4K\\\\p_{A}=\frac{1.80\times 0.0821\times 303.4}{7.07}=6.34atm[/tex]
For Gas B:We are given:
[tex]V=7.07L\\T=30.4^oC=[30.4+273]K=303.4K\\n=4.86mol[/tex]
Putting values in equation 1, we get:
[tex]p_B\times 7.07L=4.86mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 303.4K\\\\p_{B}=\frac{4.86\times 0.0821\times 303.4}{7.07}=17.1atm[/tex]
Hence, the partial pressure of gas A is 6.34 atm and that of gas B is 17.1 atm
Answer:
The partial pressure of gas A is 6.34 atm
The partial pressure of gas B is 17.12 atm
Explanation:
Step 1 :Data given
Volume of cylinder = 7.07 L
Number of moles gas A = 1.80 moles
Number of moles gas B = 4.86 moles
Temperature =30.4 ° C = 303.55 K
Step 2: Calculate pressure of gas A
p*V = n*R*T
p =(n*R*T)/V
⇒ with p = the partial pressure of gas A
⇒ with V = The volume of the cylinder = 7.07 L
⇒ with n = the number of moles gas A = 1.80 moles
⇒ with R = the gas constant = 0.08206 L*atm/K*mol
⇒ with T = the temperature = 303.55 K
p = (1.80 *0.08206 *303.55)/7.07
p = 6.34 atm
Step 3: Calculate pressure of gas B
p*V = n*R*T
p =(n*R*T)/V
⇒ with p = the partial pressure of gasB
⇒ with V = The volume of the cylinder = 7.07 L
⇒ with n = the number of moles gas B = 4.86 moles
⇒ with R = the gas constant = 0.08206 L*atm/K*mol
⇒ with T = the temperature = 303.55 K
p = (4.86 *0.08206 *303.55)/7.07
p = 17.12 atm
The partial pressure of gas A is 6.34 atm
The partial pressure of gas B is 17.12 atm
If a student weighs out 0.744 g Fe ( NO 3 ) 3 ⋅ 9 H 2 O , what is the final concentration of the ∼0.2 M Fe ( NO 3 ) 3 solution that the student makes?
Answer:
Molar concentration of Fe(NO3)3 . 9H2O = 0.12M
Explanation:
Fe(NO3).9H2O --> Fe(NO3)3 + 9H2O
By stoichiometry,
1 mole of Fe(NO3)3 will be absorb water to form 1 mole of Fe(NO3)3 . 9H2O
Therefore, calculating the mass concentration of Fe(NO3)3;
Molar mass of Fe(NO3)3 = 56 + 3*(14 + (16*3))
= 242 g/mol
Mass concentration of Fe(NO3)3 = molar mass * molar concentration
= 242 * 0.2
= 48.4 g/L
Molar mass of Fe(NO3)3 . 9H2O = 56 + 3*(14 + (16*3)) + 9* ((1*2) + 16)
= 242 + 162 g/mol
= 404g/mol
Concentration of Fe(NO3)3 . 9H2O = mass concentration/molar mass
= 48.4 /404
= 0.12 mol/l
Molar concentration of Fe(NO3)3 . 9H2O = 0.12M
A 10.0-cm interference wedge is to be built that has a linear dispersion from 400 to 700 nm. Describe details of its construction. Assume that a dielectric with a refractive index of 1.32 is to be used.
Answer:
∆=2dn/n
Explanation:
Where∆ = lamda
the construction of the 10.20 cm into feet so which which Alina dispersion form 400-700 NM assuming the dielectric has a refractive index of 1.32.
construction of the interference which is described by
∆=2dn/n
the concept of interference which can be described by the calculation of the thickness of the wage at both end to do this report you to the wavelengths of absorption band and the thickness of dielectric constant.
That is
∆=2dn/n
d=∆R/2n
Where ∆ is equal wavelength d = thickness
n = interference order
R = refractive index of dielectric medium
range of linear dispersion is 400nm to 700nm
∆1 = 400nm
∆2 = 700nm
n = 1.32
d = ∆1R/2n
d = 400*10^-9m*1.32/2*1
d = 2.64*10^-7m
For ∆2
d = 700*10^-9m*1.32/2*1
d=4.62*10^-7m
Details of the construction is simply calculation of the thickness of the wedge at both ends
Final answer:
To build a 10.0-cm interference wedge with linear dispersion from 400 to 700 nm using a dielectric with refractive index 1.32 involves precision crafting of a wedge where the thickness incrementally increases, causing a linear change in path length and hence wavelength dispersion.
Explanation:
Constructing a 10.0-cm interference wedge with a linear dispersion from 400 to 700 nm involves creating a transparent object with a slight and uniform increase in thickness from one end to the other, using a dielectric material with a known refractive index, which in this case is 1.32.
The wedge must be carefully designed so that the path length difference between the top and bottom surfaces changes by exactly 300 nm (the range from 400 to 700 nm) over the 10 cm length. This incremental change in path length creates a linear dispersion of wavelengths when light is shone through the wedge.
At the thinnest point, light with a wavelength of 400 nm should experience constructive interference, whereas at the thickest point, light with a wavelength of 700 nm should interfere constructively. The process involves precision cutting and polishing to ensure consistent graduation of thickness across the wedge.
A 226 mL solution containing 22 g of a protein in toluene has an osmotic pressure of 0.053 atm at 27 oC. What is the molar mass (in g/mol) of the protein
Answer:
4.4 × 10⁴ g/mol
Explanation:
The osmotic pressure (π) is a colligative property that can be calculated using the following expression.
π = M × R × T
where,
M: molarity
R: ideal gas constant
T: absolute temperature (27°C + 273.15 = 300 K)
Let's use it to find the molarity of the protein.
M = π / R × T
M = 0.053 atm / (0.082 atm.L/mol.K) × 300 K
M = 2.2 × 10⁻³ M
The molarity of the protein is:
M = mass of the protein / molar mass of the protein × liters of solution
molar mass of the protein = mass of the protein / M × liters of solution
molar mass of the protein = 22 g / 2.2 × 10⁻³ mol/L × 0.226 L
molar mass of the protein = 4.4 × 10⁴ g/mol
The molar mass of the protein is 4.4 * 10⁴ g/mol
Osmotic pressure :It is a colligative property that can be calculated using the following expression.
π = M × R × T
where,
M= molarityR= ideal gas constantT= absolute temperature (27°C + 273.15 = 300 K)Calculation for the molarity of the protein.[tex]M = \frac{\pi}{R * T} \\\\M = \frac{0.053 atm}{(0.082 atm.L/mol.K) * 300 K}\\\\ M = 2.2 * 10^{-3} M[/tex]
The molarity of the protein is:M = mass of the protein / Molar mass of the protein * liters of solution
Molar mass of the protein = mass of the protein / M * liters of solution
Molar mass of the protein = [tex]\frac{22 g}{2.2 * 10^{-3} mol/L * 0.226 L}[/tex]
Molar mass of the protein = [tex]4.4 * 10^4 g/mol[/tex]
Thus, the molar mass of the protein is 4.4 * 10⁴ g/mol.
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A first-order decomposition reaction has a rate constant of 0.00440 yr−1. How long does it take for [reactant] to reach 12.5% of its original value? Be sure to report your answer to the correct number of significant figures.
A first-order decomposition reaction has a rate constant of 0.00440 yr−1. Hence the correct answer is 473.418years. Rounded to the correct number of significant figures, the time it takes for the reactant to reach 12.5% of its original value is approximately 470 years.
A first-order reaction follows the exponential decay equation:
[tex]\[ [A][/tex] = [tex][A]_0 \times e^{-kt}[/tex]
Where:
[tex]\([A]\)[/tex] is the concentration of reactant at time [tex]\(t\)[/tex]
[tex]\([A]_0\)[/tex] is the initial concentration of the reactant.
[tex]\(k\)[/tex] is the rate constant.
[tex]\(t\)[/tex] is time.
It is given that the rate constant[tex]\(k\)[/tex] is 0.00440 y[tex]r^(^-^1^)[/tex] and we want to find out how long it takes for the reactant concentration to reach 12.5% of its original value, which means [tex]\([A][/tex] = [tex]0.125 \times [A]_0\).[/tex]
The equation to solve for time[tex]\(t\)[/tex]:
[tex]\[ t[/tex] = [tex]-\frac{1}{k} \ln\left(\frac{[A]}{[A]_0}\right)[/tex]
Substitute the given values:
[tex]\[ t = -\frac{1}{0.00440 \, \text{yr}^{-1}} \ln\left(\frac{0.125 \times [A]_0}{[A]_0}\right)[/tex]
Simplifying:
[tex]\[ t = -\frac{1}{0.00440 \, \text{yr}^{-1}} \ln(0.125) \][/tex]
Now calculate the value:
[tex]\[ t = -\frac{1}{0.00440 \, \text{yr}^{-1}} \times (-2.07944) \][/tex]
[tex]\[ t \approx 473.418 \, \text{years} \][/tex]
Rounded to the correct number of significant figures, the time it takes for the reactant to reach 12.5% of its original value is approximately 470 years.
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A first-order decomposition reaction is when the rate of the reaction is proportional to the concentration of the reactant. To find the time it takes for the reactant to reach 12.5% of its original value, we can use the equation ln([reactant] / [initial]) = -kt, where t represents time and k is the rate constant. Using the given rate constant, we find that it would take approximately 32.48 years for the reactant to reach 12.5% of its original value.
Explanation:A first-order decomposition reaction is one in which the rate of the reaction is proportional to the concentration of the reactant. The rate equation for a first-order reaction is given by: rate = k[reactant], where k is the rate constant and [reactant] is the concentration of the reactant.
In this case, the rate constant is given as 0.00440 yr-1. To find the time it takes for the reactant to reach 12.5% of its original value, we can use the equation ln([reactant] / [initial]) = -kt, where [initial] is the initial concentration. Rearranging the equation, we have t = ln([reactant] / [initial]) / -k. Plugging in the percentage values, we get t = ln(0.125) / -0.00440 yr-1.
Calculating this value, we find that t ≈ 32.48 years.
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Which statements are true regarding redox reactions? (Note that in redox reactions, the molecule that "causes" another to gain or lose electrons is referred to as the agent.) Select all that apply.
A. Oxidizing agents accept electrons.
B. Reducing agents may accept H+ ions.
C. If a molecule accepts electrons, it has been reduced.
D. Redox reactions may involve the transfer of hydrogen ions (H+).
E. A molecule that has gained H atoms is said to be reduced.
F. Oxidizing agents may accept H+ ions.
Answer:A, C, D, E, F
Explanation:
A. True: Oxidizing agents are electron acceptors. They accept electrons and the get reduced. This means their oxidation number reduces
B. False: Reducing agents do not accept H+ ions. Reducing agents remove oxygen from another substance or give hydrogen to it.
C. True: oxidizing agents oxidizes other molecules but they accept electrons and get reduced themselves. If a molecule accepts electrons it has been reduced.
D True: Redox reactions MAY and may not involve the transfer of hydrogen ions depending on the reactants (H+). But redox in terms of acid and base means the donating and receiving of protons(H+)
E. True: A molecule that has gained H atoms is said to be reduced. Oxidizing agents are always the proton acceptor.
F. True: Oxidizing agents May and may not accept H+. In terms of acid and base oxidizing agents accept protons(H+)
Regarding redox reactions, statements C, D, and E are generally true. Statement B and F can be true in specific scenarios but not universally, while statement A is incorrect because oxidizing agents actually donate electrons.
Explanation:In redox reactions, oxidation and reduction occur simultaneously. We look at the given statements:
A. Oxidizing agents accept electrons. - This statement is incorrect. Oxidizing agents are substances that cause the oxidation of another substance, in the process, they actually get reduced. Therefore, oxidizing agents donate electrons not accept.B. Reducing agents may accept H+ ions - This statement can be true in specific redox reactions where proton (H+) transfer occurs. But always remember that their primary role is involved in the donation or loss of electrons.C. If a molecule accepts electrons, it has been reduced. - This statement is correct. Reduction in terms of redox reactions refers to the gain of electrons.D. Redox reactions may involve the transfer of hydrogen ions (H+). - This statement is correct. In fact, many redox reactions also involve proton (H+) transfer.E. A molecule that has gained H atoms is said to be reduced - This statement is correct. The gain of hydrogen is also considered a reduction.F. Oxidizing agents may accept H+ ions - Same as statement B, this can be true in specific scenarios but the primary role of oxidizing agents is to cause oxidation by accepting electrons.Learn more about Redox reactions here:https://brainly.com/question/13978139
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A certain first-order reaction (A→products) has a rate constant of 9.00×10−3 s−1 at 45 ∘C. How many minutes does it take for the concentration of the reactant, [A], to drop to 6.25% of the original concentration?
Answer:
27.8 minutes
Explanation:
The reaction follows a first order
Rate = k[A] = change in concentration/time
k = 9×10^-3s^-1
Let the original concentration of A be y
Concentration of A at time t = 6.25% × y = 0.0625y
Change in concentration = y - 0.0625y = 0.9375y
0.009 × 0.0625y = 0.9375y/t
t = 0.9375y/0.0005625y = 1666.7sec = 1666.7/60 = 27.8 minutes
It takes approximately 12.8 minutes for the concentration of the reactant [A] to drop to 6.25% of the original concentration.
To determine the time it takes for the concentration of a reactant [A] in a first-order reaction to drop to 6.25% of its original concentration, we can use the first-order kinetics equation:
ln([A]₀/[A]) = kt
where ln is the natural logarithm, [A]₀ is the initial concentration of A, [A] is the concentration of A at time t, k is the rate constant, and t is the time in seconds. To find t when [A] is 6.25% of [A]₀, we can set [A]/[A]₀= 0.0625 (since 6.25% is equivalent to 0.0625 in decimal form).
Using the given rate constant 9.00×10⁻³ s⁻¹, the equation becomes:
ln(1/0.0625) = (9.00×10⁻³) × t
Solving this gives t = 768 seconds. Since there are 60 seconds in a minute, this is equivalent to 12.8 minutes.
Therefore, it takes approximately 12.8 minutes for [A] to drop to 6.25% of its original concentration in a first-order reaction at 45 °C with a rate constant of 9.00×10⁻³ s⁻¹.
A solution is prepared by adding 0.01 M acetic acid and 0 .01 M ethylamine to water and adjusting the pH to 7.4. What is the ratio of acetate to acetic acid? What is the ratio of ethylamine to ethylammonium ion?
The ratio of acetate to acetic acid is approximately 1:1, while the ratio of ethylamine to ethylammonium ion is close to 0:1.
Explanation:
The ratio of acetate to acetic acid can be determined using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Since the pH is 7.4, which is close to the pKa of acetic acid (4.76), the ratio of acetate (A-) to acetic acid (HA) will be close to 1:1. The ratio of ethylamine to ethylammonium ion can be determined using the same equation. In this case, the pKa is the pKa of ethylamine (10.64). If the pH is 7.4, the ethylamine (C2H5NH2) will be mostly in its protonated form, ethylammonium ion (C2H5NH3+), giving a ratio close to 0:1.
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Consider the fermentation reaction of glucose: C6H12O6 → 2C2H5OH + 2CO2 A 1.00-mol sample of C6H12O6 was placed in a vat with 100 g of yeast. If 67.7 g of C2H5OH was obtained, what was the percent yield of C2H5OH?
Answer:
% yield = 73.48 %
Explanation:
The fermentation reaction is:
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
The percent yield of C₂H₅OH is given by:
[tex] \% yield = \frac{m_{E}}{m_{T}} * 100 [/tex]
where [tex]m_{E}[/tex]: is the obtained mass of C₂H₅OH = 67.7g and [tex]m_{T}[/tex]: is the theoretical mass of C₂H₅OH.
The theoretical mass of C₂H₅OH is calculated knowing that 1 mol of C₆H₁₂O₆ produces 2 moles of C₂H₅OH:
[tex] m_{T} = mol * M [/tex]
where M: is the molar mass of C₂H₅OH = 46.068 g/mol
[tex] m_{T} = 2 moles * 46.068 g/mol = 92.136 g [/tex]
Hence, the percent yield of C₂H₅OH is:
[tex] \% yield = \frac{67.7 g}{92.136 g}*100 = 73.48 \% [/tex]
I hope it helps you!
Taking into account definition of percent yield, the percent yield for the reaction is 73.58%.
Reaction stoichiometryIn first place, the balanced reaction is:
C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
C₆H₁₂O₆: 1 mole C₂H₅OH: 2 moles CO₂: 2 molesThe molar mass of the compounds is:
C₆H₁₂O₆: 180 g/moleC₂H₅OH: 46 g/moleCO₂: 44 g/moleThen, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
C₆H₁₂O₆: 1 moles× 180 g/mole= 180 gramsC₂H₅OH: 2 moles× 46 g/mole= 92 gramsCO₂: 2 moles× 44 g/mole= 88 gramsPercent yieldThe percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.
The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:
[tex]percent yield=\frac{actual yield}{theoretical yield} x100[/tex]
where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.
Percent yield for the reaction in this caseIn this case, you know:
actual yield= 67.7 gramstheorical yield= 92 gramsReplacing in the definition of percent yield:
[tex]percent yield=\frac{67.7 g}{92 g} x100[/tex]
Solving:
percent yield= 73.58%
Finally, the percent yield for the reaction is 73.58%.
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Many people claim that science is "just" based on theories, and that since theories can change, science shouldn't be considered stable. How could you prove that science is stable and valid?
Answer:
Explanation:
Science is sustainable and reliable, as it is thoroughly tested and updated, impartial individuals will be observing scientific proof, and science only alters when new research justifies shift.The main concerns of many, if not most are scientific and engineering activities stability and transition. Efforts are made to ensure whatever theory is totally valid after a thorough investigation.
An element has three stable isotopes with masses of 27.977 amu, 28.976 amu, and 29.973 amu. The heavier two isotopes have an abundance of 4.68% and 3.09%, respectively. What is the mass of the element?A) 29.251
B) 27.684
C) 28.085
D) 28.991
E) 30.107
Answer:
The mass of the element is 28.085 amu
Explanation:
Step 1: Data given
Masses of isotopes:
27.977 amu ⇒
28.976 amu ⇒ 4.68%
29.973 amu ⇒ 3.09%
Step 2: Calculate the abundance of the other isotope
100% - 4.68% - 3.09 % = 92.23 %
Step 3: calculate the mass othe element
0.9223 * 27.977 + 0.0468 * 28.976 + 0.0309*29.973 = total mass of the element
Total mass of the element = 28.085 amuj
The mass of the element is 28.085 amu
The weighted average atomic mass of the given element is calculated using the masses and relative abundances of its isotopes. The mass is found to be 28.085 amu. The correct answer is Option C.
Explanation:The average atomic mass of an element is calculated by multiplying each isotope's mass by its relative abundance (as a decimal), and then adding up these products. We can use the given information: the element has three stable isotopes with masses of 27.977 amu, 28.976 amu, and 29.973 amu. The abundance of the first isotope can be determined by subtracting the abundances of the other two isotopes from 100% as only these three isotopes are stable. We have: (27.977 * x) + (28.976 * 0.0468) + (29.973 * 0.0309) = average atomic mass of the element. Solving this equation x(assuming it to be in percentage form)= 1- 0.0468 - 0.0309 = 0.9223 (or 92.23% in percentage form). Substituting the value of x we get: (27.977 * 0.9223) + (28.976 * 0.0468) + (29.973 * 0.0309) = 28.085 amu. Hence, the correct answer is (C) 28.085
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The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 5.00 g of naphthalene (C10H8) in 344 g benzene? (Kf of benzene = 4.90°C/m.)
Answer:
4.94°C, the temperature for freezing the solution
Explanation:
Freezing point depression to solve this.
Formula = T° freezing pure solvent - T° freezing solution = Kf . m
With the data given, let's determine m (molality)
Molality → mol/kg (moles of solute in 1kg of solvent)
We need to convert the 344 g to kg → 344 g . 1kg/1000 g = 0.344 kg
Let's determine the moles of solute (naphtalene)
5 g / 128 g/mol = 0.039 mol
Molality → 0.039 mol / 0.344 kg → 0.113
Let's go back to the formula:
5.5°C - T° freezing of solution = 4.90°C /m. 0.113 m
T° freezing of solution = - ( 4.90°C /m. 0.113 m - 5.5°C)
T° freezing of solution = 4.94 °C
The freezing point of the solution is 4.94 °C.
What is freezing point?
The term freezing point refers to the point in which a liquid is converted to a solid.
We know that;
ΔT = K m i
K = freezing constant
m = molality of the solution
i = Van't Hoff factor
ΔT = 4.90°C/m × (5.00 g /128 g/mol)/0.344 Kg × 1
ΔT =0.56°C
Freezing point of solution = 5.5°C - 0.56°C = 4.94 °C
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Determining reaction order : Rate Laws(Chemistry)
The reaction of nitric oxide with hydrogen at 1280C is as follows:
2NO(g) +2H2 (g) ----> N2(g) + 2H2O(g)
From the following experimental data, determine the rate law and the rate constant.
30 POINTS!
Answer:
The order of the reaction is 4.
The rate law of the reaction will be :
[tex]R=k[NO]3[H_2]^1[/tex]
Rate constant of the reaction: k =[tex] 6\times 10^5 M^{-3}s^{-1}[/tex]
Explanation:
[tex]2NO(g) +2H_2 (g)\rightarrow N_2(g) + 2H_2O(g)[/tex]
Let the stoichiometric coefficient of the NO and [tex]H_2[/tex] in rate law be x and y .
Rate of the reaction is given by :
[tex]R=k[NO]^[H_2]^y[/tex]
1) When , [tex][NO]=0.0100 M, [H_2]=0.0100 M[/tex]
R = 0.00600 M/s
[tex]0.00600 M/s=k[0.0100 M]^x[0.0100 M]^y[/tex]..[1]
2) When , [tex][NO]=0.0200 M, [H_2]=0.0300 M[/tex]
R = 0.144 M/s
[tex]0.144 M/s=k[0.0200 M]^x[0.0300 M]^y[/tex]..[2]
3) When , [tex][NO]=0.0100 M, [H_2]=0.0200 M[/tex]
R = 0.0120 M/s
[tex]0.0120 M/s=k[0.0100 M]^x[0.0200 M]^y[/tex]..[3]
Dividing [1] and [3]
[tex]\frac{0.00600 M/s}{0.0120 M/s}=\frac{k[0.0100 M]^x\times [0.0100 M]^y}{k[0.0100 M]^x\times [0.0200 M]^y}[/tex]
y = 1
Dividing [1] and [2]
[tex]\frac{0.00600 M/s}{0.144 M/s}=\frac{k[0.0100 M]^x\times [0.0100 M]^1}{k[0.0200 M]^x\times [0.0300 M]^1}[/tex]
x = 3
The order of the reaction = x + y = 3 + 1 = 4
The rate law of the reaction will be :
[tex]R=k[NO]3[H_2]^1[/tex]
Rate constant of the reaction: k
When , [tex][NO]=0.0100 M, [H_2]=0.0100 M[/tex]
R = 0.00600 M/s
[tex]0.00600 M/s=k[0.0100 M]^3[0.0100 M]^1[/tex]..[1]
[tex]k=\frac{0.00600 M/s}{[0.0100 M]^3[0.0100 M]^1}=6\times 10^5 M^{-3}s^{-1}[/tex]
Final answer:
The reaction between nitrogen monoxide and hydrogen is second-order with respect to NO, first-order with respect to H2, and third-order overall. The rate law is rate = k[NO]²[H₂], where k is the rate constant specific to the reaction conditions.
Explanation:
The rate law expression for the reaction between hydrogen and nitrogen monoxide to form dinitrogen monoxide and water vapor can be determined using the given rate equation. By analyzing the equation rate = k[NO]²[H₂], we can deduce the orders of reaction with respect to each reactant and the overall order.
The rate law shows that the reaction rate is directly proportional to the square of the concentration of nitric oxide (NO) and linearly proportional to the concentration of hydrogen (H₂). Therefore, the order of reaction with respect to NO is 2 (second order), and with respect to H₂, it is 1 (first order). The overall order of the reaction is the sum of these individual orders, which is 2 + 1 = 3 (third order).
The rate constant (k) would be determined experimentally by measuring the reaction rates at known concentrations of reactants. It is specific to the reaction's conditions, such as temperature and pressure, and would be stated in units that correspond to a third-order reaction.
While performing the assays for the LDH kinetics experiment, you will pipet 25ul of 250 ug/ml LDH into 975 ul of assay buffer (Tris, Lactate, and NAD). What is the final concentration of LDH in the assay?
Answer:
6.25 μg/mL
Explanation:
When a dilution is made, the mass of the solute is conserved (Lavoiser's law), so the mass pipetted will be the mass in the assay. The mass is the concentration (C) multiplied by the volume (V). If the pipet solution is called 1, and the assay 2:
m1 = m2
C1*V1 = C2*V2
C1 = 250 μg/mL
V1 = 25 μL
V2 = 975 μL + 25 μL = 1000 μL (is the final volume of the assay after the addition of LDH)
250*25 = C2*1000
C2 = 6.25 μg/mL
The final concentration of LDH in the assay is 6.25 ug/ml.
Explanation:To determine the final concentration of LDH in the assay, you need to consider the volumes of LDH and assay buffer used. In this case, you pipet 25 ul of 250 ug/ml LDH into 975 ul of assay buffer. To calculate the final concentration, you can use the formula:
Final Concentration = (Volume of LDH x Concentration of LDH) / Total Volume of Assay
Substituting the values, the final concentration of LDH in the assay is:
Final Concentration = (25 ul x 250 ug/ml) / (25 ul + 975 ul) = 6.25 ug/ml
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A diver 50 m deep in 10◦C fresh water exhales a 1.0-cm-diameter bubble. What is the bubbles diameter justas it teaches the surface of the lake, where the water is 20◦C? Assume that the bubble is always in thermalequilibrium with the water.
Answer:
1.82 cm
Explanation:
Utilize the equation [tex]\frac{P_{1}V_{1} }{T_{1} } = \frac{P_{2}V_{2} }{T_{2} }[/tex] to calculate the change in volume and size of an air bubble.
P1 = pressure at 50m = [tex]P_{A}[/tex] + ρ*g*h (where [tex]P_{A}[/tex] = atmospheric pressure, ρ = density of water, g = acceleration due to gravity, h = height/depth)
P1 = 1.01 x 10⁵ Pa + (ρ x g x h)
= 1.01 x 10⁵ Pa + (1000 kg/m³ x 9.8 m/s² x 50 m )
= 1.01 x 10⁵ Pa + 4.9 x 10⁵ Pa
= 5.91 x 10⁵ Pa
V1 = [tex]\frac{4}{3}\pi r_{1} ^{3}[/tex] [tex]r_{1}[/tex] = 10 cm = 1 x 10⁻² m
T1 = 10 °C = 10 + 273 = 283 K
P2 = [tex]P_{A}[/tex] = 1.01 x 10⁵ Pa because at the surface, pressure is equal to atmospheric pressure
V2 = [tex]\frac{4}{3}\pi r_{2} ^{3}[/tex] [tex]r_{2}[/tex] = ??
T2 = 20 °C = 20 + 273 = 293 K
[tex]\frac{P_{1}V_{1} }{T_{1} } = \frac{P_{2}V_{2} }{T_{2} }[/tex]
V₂ = P₁V₁T₂
P₂T₁
[tex]\frac{4}{3}\pi r_{2} ^{3}[/tex] = P₁ x [tex]\frac{4}{3}\pi r_{1} ^{3}[/tex] x T₂
P₂T₁
cancel out common terms
[tex]r_{2}[/tex]³ = 5.91 x 10⁵ Pa x (1 x 10⁻² m)³ x 293 k
1.01 x 10⁵ Pa x 283 k
[tex]r_{2}[/tex]³ = 757.9 x 10⁻⁹
[tex]r_{2}[/tex] = 9.1 x 10⁻³ m
[tex]r_{2}[/tex] = 0.91 cm
Therefore, bubbles diameter = 2r = 1.82 cm
Answer:
1.82 cm
Explanation:
The pressure done by a column of a liquid is called the hydrostatic pressure (Ph) and it can be calculated by:
Ph = Patm + ρgh
Where Patm is the atmospheric pressure under the column (101325 Pa), ρ is the density of the liquid (1000 kg/m³ for water), g is the gravity acceleration (9.8 m/s²), and h is the depth (50 m), so:
Ph = 101325 + 1000*9.8*50
Ph = 591325 Pa
Because the bubble is in equilibrium with the surroundings, its pressure is the same as the surroundings. Supposing a perfect sferic bubble, its volume is:
V = (4/3)*π*r³
Where r is the radius, which is half of the diameter, so r = 0.5 cm.
V = (4/3)*π*(0.5)³
V = 0.52 cm³
According to the ideal gas law, the multiplication of the pressure (P) by the volume (V) divided by the temperature (T) of a gas is constant, so if 1 is the state where the bubble is 50 m depth, and 2 the state at the surface:
P1*V1/T1 = P2*V2/T2
P1 = Ph = 591325 Pa
V1 = 0.52 cm³
T1 = 10°C + 273 = 283 K
P2 = 101325 Pa (atmosferic pressure)
T2 = 20°C + 273 = 293 K
591325*0.52/283 = 101325*V2/293
101325V2 = 318,354.3357
V2 = 3.14 cm³
V2 = (4/3)*π*r³
(4/3)*π*r³ = 3.14
r³ = 0.75
r = ∛0.75
r = 0.91 cm
The diameter is then 2*r = 1.82 cm.
A student was asked to prepare exactly 250 mL of a 0.500 M aqueous potassium hydroxide solution. What mass of potassium hydroxide (molar mass = 56.10 g/mol) must the student dissolve in the 250 mL of solution? 1. 28.1 g 2. 3.0 g 3. None of these 4. 14.0 g 5. 7.01 g 6. 56.1 g
Answer:
We need 7.01 grams of KOH (option 5)
Explanation:
Step 1: Data given
Volume aqueous KOH solution = 250 mL = 0.250 L
Molarity = 0.500 M
Molar mass of KOH = 56.10 g/mol
Step 2: Calculate moles KOH
Moles KOH = molarity * volume
Moles KOH = 0.500 M * 0.250 L
Moles KOH = 0.125 moles
Step 3: Calculate mass of KOH
Mass KOH = moles KOH * molar mass KOH
Mass KOH = 0.125 moles * 56.10 g/mol
Mass KOH = 7.01 grams
We need 7.01 grams of KOH
Write the condensed ground-state electron configurations of these transition metal ions, and state which are paramagnetic:
(a) Mo³⁺ (b) Au⁺ (c) Mn²⁺ (d) Hf²⁺
Final answer:
The condensed ground-state electron configurations of transition metal ions are given, and it is stated which ions are paramagnetic.
Explanation:
The condensed ground-state electron configurations of the transition metal ions are as follows:
(a) Mo³⁺: [Kr]4d³
(b) Au⁺: [Xe]5d¹⁰
(c) Mn²⁺: [Ar]3d⁵
(d) Hf²⁺: [Xe]4f¹⁴5d²×
Out of these, Mo³⁺ and Mn²⁺ are paramagnetic because they have unpaired electrons in their d orbitals, which allows them to be attracted to a magnetic field.