Solutions to select questions can be found online.
5.7
Calculate the reduced mass of HCl molecule given that the mass of H atom is 1.0078 amu and the mass of Cl atom is 34.9688 amu. Note that 1 amu = 1.660565*10-27 kg.
Atomic mass is defined as the number of protons and neutrons in an atom, where each proton and neutron has a mass of approximately 1 amu (1.0073 and 1.0087, respectively). The electrons within an atom are so miniscule compared to protons and neutrons that their mass is negligible. Not to be confused with the mass number, the atomic mass is the mass of an entire atom, including its protons, neutrons and electrons. For example, the atomic mass of oxygen is 16 amu (atomic mass unit). Magnesium reacts with hydrochloric acid according to the following equation: #Mg(s) + 2HCl(aq) - MgCl2(aq) +H2(g)#.
How to find the Molecular Weight for HCl: a step-by-step video tutorial. Using the Periodic Table look up the atomic weight of each of the elements in HCl. The mass of one hydrogen atom is 1gm and the atomic mass of chlorine is 35.5 gm. The atomic mass of HCl will be 35.5+1 =36.5 gm. An element “z” has a mass number of 33 and has 18 neutrons a) What is the atomic number of element Z? (1mk) b) Write an equation to show how atom of “z” forms an ion. (1mk) a) Name 30. Study the flow chart below and answer the questions that follows-: Iron and Sulphur mixture Gas P Heat Compound T HCL Gas U + Fecl 2 HCL.
[mu = dfrac{m_1m_2}{m_1+m_2} nonumber ]
[mu = dfrac{1.0078; amu times 34.9688; amu}{1.0078; amu +34.9688; amu} = 0.9796 ;amu nonumber ]
[mu =0.9796 ;amu times dfrac{1.660565 cdot 10^{-27} ;kg}{1;amu} =1.627 times 10^{-27} kg nonumber ]
5.8
Calculate the reduced mass for the Br2, Cl2, and I2 diatomics.
From the periodic table, the atomic masses for Br, Cl, and I are 79.904, 35.453, and 126.904 respectively.
Covert the atomic mass to kg.
(Br= (79.904 , amu)(1.6606 times 10^{-27} ,amu/kg)=1.327 times 10^{-25};kg) (Cl= (35.453 , amu )(1.6606 times 10^{-27} ,amu/kg )=5.887 times 10^{-26};kg) (I= (126.904 , amu )(1.6606 times 10^{-27} ,amu/kg )=2.107 times 10^{-25};kg)[mu=dfrac{m}{2} nonumber ]
therefore
μBr2=1.327x10-25 kg/2=6.635x10-26 kg μCl2= 5.887x10-26kg/2 = 2.9435x10-26kg μI2=2.107x10-25kg=1.0535x10-25kgThe equation for a reduced mass ((mu)) of a diatomic is
[ mu= dfrac{m_1m_2}{m_1+m_2} nonumber ]
for a diatomic molecule with identical atoms ((m_1=m_2=m)) so
5.14
79Br79Br has a force constant of (240,N cdot m^{-1}). Given this information:
- Calculate the fundamental vibrational frequency and
- Calculate the 79Br79Br zero point energy.
We must first know which formula to use which is
[nu_{obs}=dfrac{1}{2π} sqrt{dfrac{k}{mu}} nonumber ]
calculate the reduced mass
[mu=dfrac{(79; amu)^2}{79 ;amu+ 79 ;amu}=39.5 amu nonumber ]
and convert to Kg:
1.66e-27 kg•amu-1
substitute the given values
[nu=dfrac{1}{2π} sqrt{dfrac{ 240 kg ;m ;s^{-2} ;s }{39.5 amu times 1.66 times 10^{-27} kg ;amu^{-1}}} = 9.63 times 10^{12} s^{-1} nonumber ]
It can also be convert to wavenumber (inverse centimeter (cm^{-1})):
[ nu_{cm^{-1}}=dfrac{1}{lambda}=dfrac{nu}{c}=dfrac{9.63times 10^{12} s^{-1}}{3.0times 10^{10} cm;s^{-1}}=321 cm^{-1} nonumber ]
Zero Point Energy:
[E_{0}=dfrac{1}{2}hnu= dfrac{1}{2}hcnu_{cm^{-1}} nonumber ] (formula to use)
E0=1/2(6.626e-34J•s)(2.998e10cm•s-1)(321cm-1)
E0= 3.19e-21J
5.19
Prove that the second derivative of an even function is even and odd function is odd.
This is an example..not a proof
The following is an even function:
[y(x) = a + bx^2 + cx^4 + dx^6 nonumber ]
so
[dfrac{dy}{dx}= 2bx + 4cx^3 + 6dx^5 nonumber ]
and
[dfrac{d^2y}{dx^2}= 2b + 12cx^2 + 30dx^4 nonumber ]
which is an even function.
The following is an odd function:
[f(x)= ax + bx^3 + cx^5 nonumber ]
so
[dfrac{df}{dx}= a + 3bx^2 + 5cx^4 nonumber ]
and
[ dfrac{d^2f}{dx^2}= 6bx + 10cx^3 nonumber ]
which is an odd function.
5.27
The Harmonic oscillator Hamiltonian obeys the reflective property:
[hat{H}(x) = hat{H}(-x) nonumber ]
What does this say about the nature of the harmonic oscillator wave function?
The harmonic oscillator switches from odd to even due to the fact that the reflective property will alternate.
5.28
If (langle x rangle) is an odd function, what does that say about (p_x)?
Hint: use
[dfrac{d langle p_x rangle }{dt} = leftlangledfrac{-dV}{dx} right rangle nonumber ]
also known as Ehrenfest's Theorem, where (V) is the potential of a one dimensional harmonic oscillator.
Hence, (langle p_x rangle ) does not depend on time.
5.32
Convert (nabla^2) from Cartesian coordinates to cylindrical coordinates.
We have to start with the conversion of Cartesian coordinates ({x, y, z}) to cylindrical coordinates ({r, theta , z})
(x = rcostheta) (y = rsintheta) (z = z)Now putting it all together
(nabla^2 = dfrac{d^2}{dr^2} + dfrac{1}{r}dfrac{d}{dr} + dfrac{1}{r^2}dfrac{d^2}{dtheta^2} + dfrac{d^2}{dz^2})
(r = sqrt{x^2+y^2})
(costheta = dfrac{x}{sqrt{x^2+y^2}})
(sintheta = dfrac{x}{sqrt{x^2+y^2}})
Now by chain rule we get
(dfrac{d}{dx} = dfrac{dr}{dx}dfrac{d}{dr} + dfrac{dtheta}{dx}dfrac{d}{dtheta})
(dfrac{d}{dy} = dfrac{dr}{dy}dfrac{d}{dr} + dfrac{dtheta}{dy}dfrac{d}{dtheta})
(dfrac{dr}{dx} = dfrac{x}{r} = costheta)
(dfrac{dr}{dx} = sintheta)
using implicit differentiation and taking the second derivatives will yield
(dfrac{d^2}{dx^2} = left(costhetadfrac{d}{dr} - dfrac{sintheta}{r} dfrac{d}{dtheta}right)left(costhetadfrac{d}{dr} - dfrac{sintheta}{r} dfrac{d}{dtheta}right))
(dfrac{d^2}{dy^2} = left(sinthetadfrac{d}{dr} + dfrac{costheta}{r} dfrac{d}{dtheta}right)left(sinthetadfrac{d}{dr} + dfrac{costheta}{r} dfrac{d}{dtheta}right))
(dfrac{d^2}{dz^2} = dfrac{d^2}{dz^2})
5.37
Find the magnitude of angular momentum and the (z) component of angular momentum for electrons in a hydrogen-like species with
- quantum numbers (n = 1), (l = 0), ( m = 0); and
- ( n = 2), ( l = 0), ( m = 0).
Compare your answers and explain your results.
The wave function for this problem is given by:
[psi_{100}=R(r)_{10}Y(theta,phi)_{00}=2left(frac{Z}{2a_0}right)^frac{3}{2}e^frac{-Zr}{a_0} nonumber ]
Using that:
[hat{L}^{2}Y_{lm}(theta, phi)=hbar^{2}l(l+1)Y_{lm}(theta,phi), nonumber ]
and
[hat{L_z}=mhbar nonumber ]
Then (hat{L}^{2}=0) and (hat{L_z}=0). Given that the values for (l) and (m) are the same as above, the answers would also be the same.
The reason why both answers are the same is that the operators for angular momentum only act on the angular part of the wave function. Since only the quantum number (n) varied between these two states, the angular momentum eigenvalues did not change.
5.38
Apply the angular momentum operator in the x direction to the following functions ((Y(theta,phi))).
- (dfrac{5pi}{4} + 7exp(pi^2) )
- (3pi sin(theta) )
- (dfrac{3}{2}cos(theta)exp(iphi))
Let us begin by stating the angular momentum operator in terms of (theta) and (phi).
[hat{L_x} = ihbarBig(sin(phi)dfrac{partial}{partial theta} + cot(theta)cos(phi)dfrac{partial}{partial phi}Big) nonumber ]
a) (Y(theta,phi) = dfrac{5pi}{4} + 7exp(pi^2) )
[hat{L_x}(dfrac{5pi}{4} + 7exp(pi^2)) = ihbarBig(sin(phi)dfrac{partial dfrac{5pi}{4} + 7exp(pi^2)}{partial theta} + cot(theta)cos(phi)dfrac{partial dfrac{5pi}{4} + 7exp(pi^2)}{partial phi}Big) nonumber ]
[ = 0 nonumber ]
The function does not depend on (theta) or (phi) so when the angular momentum operator is applied to the function, it equals 0.
b) (Y(theta,phi) = 3pi sin(theta) )
[hat{L_x}(3pi sin(theta)) = ihbarBig(sin(phi)dfrac{partial}{partial theta}3pi sin(theta) + cot(theta)cos(phi)dfrac{partial}{partial phi}3pi sin(theta)Big) nonumber ]
[ = 3ipihbar sin(phi)cos(theta) nonumber ]
c) (Y(theta,phi) = dfrac{3}{2}cos(theta)exp(iphi))
[hat{L_x}(3pi sin(theta)) = ihbarBig(sin(phi)dfrac{partial}{partial theta}dfrac{3}{2}cos(theta)exp(iphi) + cot(theta)cos(phi)dfrac{partial}{partial phi}dfrac{3}{2}cos(theta)exp(iphi)Big) nonumber ]
[ = ihbarBig( dfrac{-3}{2}sin(phi)sin(theta)exp(iphi) + dfrac{3i}{2}cot(theta)cos(phi)cos(theta)exp(iphi)Big) nonumber ]
[ = dfrac{3ihbar exp(iphi)}{2}(icot(theta)cos(phi)cos(theta) - sin(phi)sin(theta)) nonumber ]
5.41
Use the fact that (hat x) and (hat p) are Hermitian in the number operator
[ hat a_- = dfrac{1}{sqrt{2}}(hat x +ihat p) nonumber ]
[hat a_+ = dfrac{1}{sqrt{2}}(hat x -ihat p) nonumber ]
and
[hat H=dfrac{hbar w}{2}(hat a_-hat a_+ + hat a_+hat a_-) nonumber ]
Show that
[int psi^*_v hat{v} psi dx geq 0 nonumber ]
5.43
Determine the unnormalized wave function (psi_circ big(xbig)) given that (hat{a}_- = 2^{-1/2}big(hat{x}+ihat{p}big) ) and that (hat{a_-}psi_circ = 0) Then find the unnormalized wave function for (psi_1big(xbig)) using (hat{a}_+).
It was given that (hat{a}_-psi_circ = 0), so substituting in (hat{a}_-) so we know
[hat{a_-} = 2^{-1/2}big(hat{x}+ihat{p}big) psi_circ = 0 nonumber ]
We can expand and simplify this expression to a first order partial differential equation
[xpsi_circ + dfrac{dpsi_circ}{dx} = 0 nonumber ]
Solve by separating like terms
[dfrac{dpsi_circ}{psi_circ} = -xdx nonumber ]
Solving this equation for (psi_circ big(xbig)) we find that
[psi_circ = e^dfrac{-x^2}{2} nonumber ]
To solve for (psi_1) we understand that (psi_1backsimhat{a_+}psi_circbacksimhat{x}-ihat{p}psi_circ), as well as that
[hat{x}-ihat{p}psi_circ = xpsi_circ - dfrac{dpsi_circ}{dx} = 2xe^dfrac{-x^2}{2}=2xpsi_circ nonumber ]
So then we can say
[boxed{psi_1backsim xe^dfrac{-x^2}{2}} nonumber ]
5.46
Find the reduced mass of an electron in a Tritium atom. Set the mass of the Tritium to be (5.008267 times 10^{-27}, kg). Then find the value of the Rydberg constant for the Tritium atom.
To solve, use the reduced mass equation, and for mass 1 enter the mass of the electron, and for mass 2 enter the mass of the Tritium atom:
[mu = dfrac{m_1m_2}{m_1+m_2} nonumber ]
For which one attains a value of (mu = 9.1077 x 10^{-31} kg)
5.46
The mass of a deuterium atom is (3.343586 times 10^{-27}; kg). First calculate the reduce mass of the deuterium atom. Then using the reduced mass calculated find the Rydberg constant for a deuterium atom.
(mu) = reduced mass
[mu_{deuterium} = dfrac{(9.109390 times 10^{-31} kg) ( 3.343586times 10^{-27}kg)}{(9.109390 times 10^{-31} kg + 3.343586times 10^{-27}kg )} nonumber ]
[mu_{deuterium} = 9.106909 times 10^{-31}kg = 0.9997277 m_e nonumber ]
(R_H) = Rydberg constant
[R_H = (109,737.2 cm^{-1}) (0.9997277 m_e) = 109,707.3 cm^{-1} nonumber ]
5.47
What is the ratio of the frequency of spectral lines of C-14 that has been ionized 5 times and C-12 that has been ionized 5 times?
Carbon that has been ionized 5 times is a hydrogen like ion, so we can use the Bohr model to find the desired ratio.
[E = dfrac{uZ^2e^4n^2}{8ε_0^2h^3c} nonumber ]
gives the placement of spectral lines. The coefficient of n2 is proportional to the frequency of these lines, so the ratio of EC-14/Ec-12 will give the ratio of frequency of the lines. The only difference between these two isotopes is the reduced mass u. So the problem reduces to uC-14/ uC-12. Mass in amu is used below. me = mass of electron = 5.4858*10-4 amu.
[mu_{C-14} = dfrac{m_em_{c-14}}{m_e + m_{c-14}} = dfrac{(14.003)(5.4858 times 10^{-4})}{14.003 + 5.4858 times 10^{-4}} = 5.485585 times 10^{-4} nonumber ]
[mu_{C-12} = dfrac{m_em_{c-12}}{m_e + m_{c-12}}= dfrac{(12)(5.4858 times 10^{-4})}{12 + 5.4858 times 10^{-4}} = 5.485549 times 10^{-4} nonumber ]
[dfrac{mu_{C-14}}{mu_{C-12}} = 1.0000065 nonumber ]
5.47
Calculate the Rydberg constant for a deuterium atom and atomic hydrogen given the reduced mass of a deuterium atom is (9.106909 times 10^{-31} kg) and the reduced mass of hydrogen is (9.104431 times 10^{-31} kg). Compare both of these answers with the experimental result ((109677.6 cm^{-1})). Then determine the ratio of the frequencies of the lines in the spectra of atomic hydrogen and atomic deuterium.
The Rydberg constant is found using
[R_H=dfrac{me^4}{8epsilon_o^2 ch^3} nonumber ]
For a deuterium atom[R_H=dfrac{(9.104431 times 10^{-31}kg)(1.602 times 10^{-19} C)^4}{8(8.854 times 10^{-12} dfrac{F}{m})^2(2.998 times 10^{8} dfrac{m}{s})(6.626 times 10^{-34} J cdot s)^3} nonumber ]
[R_H=109707.3 cm^{-1} nonumber ]
This is different by ( 2.7 times 10^{-2}%).
The ratio of the frequencies of the lines in the spectra of atomic hydrogen and atomic deuterium is equivalent to the ratio of the Rydberg constants we just found.
[dfrac{109707.3; cm^{-1}}{109677.5; cm^{-1}}=1.000272 nonumber ]
[R_H=dfrac{(9.106909 times 10^{-31}kg)(1.602 times 10^{-19} C)^4}{8(8.854 times 10^{-12} dfrac{F}{m})^2(2.998 times 10^{8} dfrac{m}{s})(6.626 times 10^{-34} J cdot s)^3} nonumber ]
[R_H=109677.5 cm^{-1} nonumber ]
This is different by ( 9.1times 10^{-5}%).
For a hydrogen atom5.46
Find the reduced mass of HCl where the mass of hydrogen in 1 amu and the mass of chloride is 35 amu.
[mu = dfrac{m_1 times m_2}{m_1 + m_2} nonumber ]
[mu = dfrac{(1.00)(35.00)}{36.00} 1.603 times 10^{-27} kg = 1.558 times 10^{-27} kg nonumber ]
What is the atomic mass of NaOH ?
2 Answers
Explanation:
For each element in the compound, multiply its subscript times its molar mass, then add the results to get the molar mass of the compound. The molar mass of an element is its atomic weight (mean relative atomic mass) on the periodic table in g/mol.
The subscript for each element in the compound
Relative Atomic Mass Of Hydrochloric Acid
Explanation:
Atomic mass of Na is 22.99 u ( unified atomic mass unit ).
Atomic mass of O is 16 u.
Atomic mass of H is 1 u.
Atomic Mass Of Hydrochloric Acid
Then the molecular mass of NaOH is :22.99+16+1=39.99u
Relative Atomic Mass Of Hydrogen Chloride
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